ECE 370 SIGNALS AND SYSTEMS - FALL 2012

Text: These lecture notes and board notes.

Optional Texts: H.P. Hsu, Signals & Systems. Shaum's Outline: McGraw{Hill, 1995.

This is essentially a good source of worked examples, all in the standard EE notation that

we use in ECE 370. Reasonably priced!

C.L. Phillips, J.M. Parr, and E.A. Riskin, Signals, Systems, & Transforms. 4th Edition,

Prentice-Hall, 2008. Only get this if you are inclined to read additional books!

Prerequisites: ECE 225 Electric Circuits, MA 238 Di®erential Equations.

Course Goals: To provide electrical engineers with physical understanding and compe-

tence in mathematical signal analysis and systems theory.

Instructor: Robert W. Scharstein, Houser 209

phone 205-348-1761, e-mail rscharst@bama.ua.edu

O±ce hours: to be announced and by appointment.

Lecture: 12:00 noon { 12:50 pm, Mon, Wed, Fri; SERC 1013.

Problem Session: Wednesday 7:00-8:00 pm, Houser 301, optional. Bring your questions!

Daily Homework: Homework problems and textbook reading will be assigned for each

class.

FFF !!!THESE ARE YOUR PRIMARY RESPONSIBILITY!!! FFFTests will consist of homework problems, modi¯ed homework problems, new problems,

and concepts and derivations from the text and lecture. Details and discussion of the

MATLAB homework problems will appear on tests.

Tentative Grades: All quizzes and tests are closed-book.

Eleven Friday quizzes : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1/3

Test 1, Friday September 21 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :1/6

Test 2, Friday November 2 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1/6

Final Exam, Monday December 10 11:30 am { 2:00 pm : : : : : : : : : : : : : : : : : : : : : : : : : : 1/3

Instructor may determine ¯nal course grades using any combination of the above. However,

the ¯nal exam will not be weighted less than 1/3.

Friday quiz dates: 1: Aug 24, 2: Aug 31, 3: Sept 7, 4: Sept 14, 5: Sept 28, 6: Oct 12

7: Oct 19, 8: Oct 26, 9: Nov 9, 10: Nov 16 11: Nov 30

\Our goal is not to develop all the applications, but to prepare for them

- and that preparation can only come by understanding the theory."

Gilbert Strang Linear Algebra and Its Applications, Academic Press, 1976 page x

Typeset by AMS-TEX

0

ECE 370 SIGNALS AND SYSTEMS - ROUGH LECTURE NOTES

Robert W. Scharstein

23 July 2012

CONTENTS

BACKGROUND AND TIME-DOMAIN SYSTEMS

Complex Numbers : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1.

A Couple of Singularity Functions : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 7.

Symmetry : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :10.

A Note on Mathematical Functions and Units/Physical Dimensions : : : : : : : : : : : : : : : : : 13.

Linear System : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 15.

Time-Invariant System : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 16.

Input/Output Relationship for a LTI System : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :17.

Convolution : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 18.

Another Convolution Example: The Convolution of Two Gaussian Pulses : : : : : : : : : : : 21.

Yet Another Convolution Example : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :22.

Causal System : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 28.

Bounded Input/Bounded Output Stability : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 28.

Singular Functions and Integrals : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 29.

Homework Problems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :35.

Relationship Between Impulse Response and Step Response : : : : : : : : : : : : : : : : : : : : : : : : : 37.

Time-Domain RC Circuit Analysis : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :38.

Step Response of an RC Circuit : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 41.

Step Response of Another RC Circuit : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :43.

Homework Problems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :45.

Step and Impulse Response of Series RLC Circuit : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 46.

Homework : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :53.

FOURIER INTEGRAL

Evaluation of an Important Integral via the Laplace Transform : : : : : : : : : : : : : : : : : : : : : :54.

Spectral Form of the Dirac-Delta Function : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 57.

Fourier Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 58.

Fourier Transform Exercises : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 59.

Tables of Integral Transforms : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 60.

Fourier Transform of the Gaussian : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 63.

Fourier Transform of the Heaviside Unit Step Function : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :65.

Duality Property of the Fourier Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 68.

Half-Power Bandwidth or Pulse width : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 70.

Homework : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :73.

Uncertainty Principle for the Fourier Integral : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 74.

i

*Riemann-Lebesgue Lemma (two serious treatments) : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 76.

Riemann-Lebesgue Lemma (one reasonable treatment) : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :78.

Convergence of the Fourier Integral Representation : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :79.

System Transfer Function : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :81.

Superposition Interpretation of the System Transfer Function : : : : : : : : : : : : : : : : : : : : : : : 83.

Transfer Function for Linear Electric Circuits : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 84.

Parseval's Theorem for the Fourier Integral : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 87.

Other Conventions for the Fourier Transform Pair : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :88.

Cascade of Two LTI Systems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 90.

Linear Dispersionless Filter : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 90.

Rectangular Pulse Response of Ideal Low Pass Filter : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :91.

Ideal Band-Pass Filter : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :95.

Complex Phasors for Time-Harmonic Circuit Analysis : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :96.

Homework Problem on Uncertainty Principle : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 98.

*Approximate Analysis of Dispersion : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 99.

Fourier Integral Solution of Potential Problem : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 103.

FOURIER SERIES

Fourier Transform of the comb : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :106.

Fourier Transform of a Periodic Signal : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 107.

Periodic Signals and Fourier Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :110.

Orthogonality : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 110.

Kronecker delta : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 111.

Orthogonality of the Fourier basis : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 111.

Derivation of Fourier Coe±cients by Minimizing the Mean Square Error : : : : : : : : : : : :112.

Fourier Coe±cients by a Direct Inner Product : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 114.

Trigonometric Form of the Fourier Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 114.

Two Standard Notations for the Trig Fourier Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 116.

Riemann-Lebesgue Lemma: Fourier Coe±cients : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :117.

Gibbs' Phenomenon: Example Square Wave : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 118.

Pointwise Convergence of the Fourier Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :119.

On the Convergence and Summation of Some Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :126.

Kummer Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 129.

Homework Problem: Convergence of Fourier Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :131.

Fourier Series of Full-Wave Recti¯ed Sine Wave : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :132.

Parseval's Theorem for the Fourier Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 134.

Power Supply Performance : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 135.

Periodic Excitation of a Simple Parallel GLC Circuit : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :139.

Homework Problems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 140.

Simple Low-Pass and Hi-Pass Filter Response to a Periodic Input : : : : : : : : : : : : : : : : : : 141.

The Vibrating String : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 147.

Homework Problems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 150.

Poisson Sum Formula : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 151.

ii

TRANSITION TO DISCRETE-TIME SYSTEMS

Sampling and Reconstruction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :153.

Discrete-Time Signals, Systems, and Transforms : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 157.

Exercises : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 159.

Discrete-Time Systems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 159.

Linear Time-Invariant Discrete-Time Systems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :160.

Input/Output Relationship : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 160.

Example of a LTI Discrete-Time System : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 161.

The Z-Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 163.

Fibonacci Sequence - Di®erence Equation and Z-Transform : : : : : : : : : : : : : : : : : : : : : : : : 165.

Convolution Theorem of the Z-Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 166.

Z-Transform of a Delayed Sequence : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 166.

System Transfer Function : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 166.

Discrete-Time Fourier Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 167.

Discrete Fourier Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 167.

Exercises : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 169.

DFT Approximation of the Continuous-Time Fourier Transform : : : : : : : : : : : : : : : : : : : :170.

Numerical Example of Using the FFT to Evaluate the Fourier Integral : : : : : : : : : : : : : 171.

Example: DFT Approximation of a Fourier Sine Integral : : : : : : : : : : : : : : : : : : : : : : : : : : :174.

Discrete-Time Approximation for the Step Response of Series RLC Circuit : : : : : : : : : 177.

* Sections marked by the asterisk are A+ level, so put these last in your prioritized list of

study topics.

\... mathematics is learned by doing it, not by watching other people do it ... "

M. Reed and B. Simon, Functional Analysis, Academic Press, 1980, page ix.

iii

Complex Numbers

j = +p¡1

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x

y

²

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Á

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..............................r

z

complex z = x+ jy plane

If z 2 C and x; y; r; Á 2 R, then we write the complex number z in rectangular and polarform as

z = x+ jy = rejÁ:

Refzg = x and Imfzg = yThe modulus or magnitude of z is

jzj = rand the argument or angle of z is

argfzg = Á:The complex conjugate of z = x+ jy = rejÁ is

z¤ = x¡ jy = re¡jÁ:

z + z¤ = 2x and z ¡ z¤ = 2jyzz¤ = jzj2 = r2

Euler's identity:

ejÁ = cosÁ+ j sinÁ

Taylor series for the exponential function, written as a function of the complex variable z

is

ez =

1Xn=0

zn

n!:

1

Let z = jy and separate the real and imaginary parts via the even and odd powers in

ejy =

1Xn=0

(jy)n

n!=

1Xm=0

(jy)2m

(2m)!+

1Xm=0

(jy)2m+1

(2m+ 1)!

Note that j2m = (¡1)m and j2m+1 = j(¡1)m so that

ejy =

1Xm=0

(¡1)my2m(2m)!

+ j

1Xm=0

(¡1)my2m+1(2m+ 1)!

= cos y + j sin y

There was nothing special about the real number y, and we also write

ejÁ = cosÁ+ j sinÁ:

The sum of ejÁ and its conjugate gives

ejÁ + e¡jÁ = 2cosÁ

and the di®erence gives

ejÁ ¡ e¡jÁ = 2j sinÁwhich are the important representations

cosÁ =ejÁ + e¡jÁ

2

sinÁ =ejÁ ¡ e¡jÁ

2j

Many (all?) of the trig identities are easily seen in complex form: Consider

1 = jejÁj2 =³ejÁ´³e¡jÁ

´=³cosÁ+ j sinÁ

´³cosÁ¡ j sinÁ

´= cos2 Á+ sin2 Á:

Now observe

z = x+ jy = rejÁ = r£cosÁ+ j sinÁ

¤Equating real and imaginary parts separately gives

x = r cosÁ and y = r sinÁ:

jzj2 = zz¤ = r2 = (x+ jy)(x¡ jy) = x2 + y2r = +

px2 + y2

Á = tan¡1³yx

´Note that the two-argument arctangent function (that maintains full information about

quadrant) is required. For example, observe that

z1 = ¡1 + j =p2ej3¼=4 where tan¡1

μ1

¡1¶=3¼

4

is certainly di®erent from

z2 = 1¡ j =p2e¡j¼=4 where tan¡1

μ¡11

¶= ¡¼

4

Your calculator has the two-argument arctangent function as part of its rectangular-to-

polar conversion. In MATLAB, it's called atan2(y,x) just like Fortran and probably

C§.

2

Consider z1 = x1 + jy1 and z2 = x2 + jy2. Then our four basic operations are:

addition z1 + z2 = x1 + jy1 + x2 + jy2 = (x1 + x2) + j(y1 + y2)

multiplication by a real constant c: cz1 = c(x1 + jy1) = cx1 + jcy1

multiplication z1z2 = (x1 + jy1)(x2 + jy2) = (x1x2 ¡ y1y2) + j(x1y2 + x2y1)and z1z2 = (r1e

jÁ1)(r2ejÁ2) = r1r2e

j(Á1+Á2)

divisionz1

z2=x1 + jy1

x2 + jy2=(x1 + jy1)(x2 ¡ jy2)(x2 + jy2)(x2 ¡ jy2) =

(x1x2 + y1y2) + j(x2y1 ¡ x1y2)x22 + y

22

andz1

z2=r1e

jÁ1

r2ejÁ2=r1

r2ej(Á1¡Á2)

Generally, since the real number line is a subset of the complex plane, all of our ordinary

functions f(x) of a real variable x have extensions (or continuations) to a function f(z) of

a complex variable z. One notable di®erence between the real number line and the more

general complex plane is the lack of ordering on z: That is, there is no inequality on the

complex plane. Inequality is only de¯ned on the reals, i.e.

x1 > x2; y1 · y2; 7 < 13; etc

but something like z1 < z2 is meaningless.

Triangle inequalities.

jz1 + z2j · jz1j+ jz2jjz1 ¡ z2j ¸

¯̄jz1j ¡ jz2j¯̄

3

Proof of the ¯rst triangle inequality:

jzj2 = x2 + y2

jzj =px2 + y2 ¸ jxj

jzj ¸ jRe(z)jjzj ¸ Re(z)

jz1z¤2 j ¸ Re(z1z¤2) (1)

jz1 + z2j2 = (z1 + z2)(z¤1 + z¤2) = z1z¤1 + z2z¤2 + z1z¤2 + z¤1z2= jz1j2 + jz2j2 + 2Re(z1z¤2) (2)

(jz1j+ jz2j)2 = jz1j2 + jz2j2 + 2jz1jjz2j = jz1j2 + jz2j2 + 2jz1z¤2 j (3)

compare (2) and (3) in view of (1)

jz1 + z2j2 · (jz1j+ jz2j)2jz1 + z2j · jz1j+ jz2j

..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

z1

z2

z1 + z2

Extensions ¯̄̄̄¯NXn=0

fn(z)

¯̄̄̄¯ ·

NXn=0

jfn(z)j

¯̄̄̄¯̄bZa

f(t) dt

¯̄̄̄¯̄ ·

bZa

jf(t)j dt

(Sometimes called the monotonicity property of the integral.)

F Prove the second triangle inequality.

Hint: let z1 = ³1 ¡ ³2 and z2 = ³2 in the ¯rst triangle inequality.

4

Multivalued functions.

The nth roots of unity

1 = ej2k¼ (k = 0;§1;§2; : : : )11=n = ej2k¼=n (k = 0; 1; 2; : : : ; n¡ 1) distinct roots

Examples.

11=2 =

½ej0 = 1 (k = 0)

ej¼ = ¡1 (k = 1)...................................................................................

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.......

²²

11=3 =

8>>>><>>>>:ej0 = 1 (k = 0)

ej2¼=3 =¡1 + jp3

2(k = 1)

ej4¼=3 =¡1¡ jp3

2(k = 2)

.......

.......

.......................................................................

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²

²

²

logarithm

ln(z) = ln£rejÁ

¤= ln

£rej(Á+2k¼)

¤= ln(r) + j(Á+ 2k¼) (k = 0;§1;§2; : : : )

The value for k = 0 is called the principal branch of the logarithm.

5

powers: Let z = x + jy and w = u + jv be two complex numbers, that is z;w 2 C andx; y; u; v 2 R.

zw = exp£ln(zw)

¤= exp

£w ln z

¤and we know how to interpret/evaluate

ez = ex+jy = ex£cos y + j sin y

¤

Example.

ln(j) = j(¼=2 + 2k¼)

jj = ej¢j(¼=2+2k¼) = e¡(¼=2+2k¼)

principal value is jj = exp(¡¼=2) = 0:2079 ¢ ¢ ¢

Problem: Show that

μHint: use the ¯nite geometric series

n¡1Pk=0

ak on page 164

¶n¡1Xk=0

ej2¼k=n = 0

Fundamental theorem of algebra. The polynomial

Pn(z) = anzn + an¡1zn¡1 + ¢ ¢ ¢+ a2z2 + a1z + a0

with an 6= 0 is said to be of degree n. It has exactly n roots (zeros) counting multiplicity.

Proposition: If the coe±cients ak (k = 0; 1; 2; : : : ; n) of the polynomial Pn(z) are all real,

then any complex roots must occur in complex conjugate pairs. That is, if Pn(z) = 0 then

Pn(z¤) = 0.

proof:

anzn + an¡1zn¡1 + ¢ ¢ ¢+ a2z2 + a1z + a0 = 0

take the complex conjugate

an(z¤)n + an¡1(z¤)n¡1 + ¢ ¢ ¢+ a2(z¤)2 + a1z¤ + a0 = 0

6

A Couple of Singularity Functions

(or \Distributions" or \Generalized Functions")

..................................................................................................................................................................................................................................................................................................................................................................

0 t

u(t) 1

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....N

0 t

±(t) (1)

Heaviside unit-step function

u(t) =

½0; t < 0

1; t > 0=) u(t¡ T ) =

½0; t < T

1; t > T:

Dirac delta-function (de¯nition)

±(t) = 0 if t6= 0 AND

1Z¡1

±(t) dt = 1

From this strange de¯nition, the sampling or sifting property follows:

bZa

f(t)±(t¡ t0) dt =½f(t0); a < t0 < b

0; o.w.

if f(t) is continuous at t = t0. This is e®ectively an operational de¯nition of the Dirac-delta.

It is under the operation of integration that the Dirac delta-function enjoys meaning, and

thus it is sometimes said that \every Dirac-delta function is begging to be integrated."

Relationship:tZ

¡1±(¿) d¿ = u(t) () ±(t) =

du(t)

dt

Dirac-Delta Sequences: There are lots (1!) of ordinary functions that satisfy, in somelimit, the two-statement de¯nition of the Dirac-delta function. Here are several:

lim²!0

1

²¦

μt

²

¶= ±(t) where ¦

μt

T

¶, u(t+ T=2)¡ u(t¡ T=2) =

½1 jtj < T=20 jtj > T=2

lim²!0

²=¼

t2 + ²2= ±(t) Cauchy or Lorenz distribution

lim®!1

r®

¼e¡®t

2

= ±(t) Gaussian distribution

lim−!1

sin−t

¼t= lim−!1

1

2¼

−Z¡−

e§j!td! = ±(t) Dirichlet kernel

7

Study the time behavior of the above \Dirac-delta sequences" to see how they approach

the ideal delta function. Also examine the Heaviside unit-step sequence

lim®!1

12 [1 + tanh®t] = u(t)

and its time-derivative.

Note: we don't ever try to evaluate ±(0). You might want to, and some books might say

that \±(0) =1 ", but I would prefer we avoid this point. We don't ever have to attach a

value to the Dirac delta function when its argument is zero. But, in the words of Lennon

& McCartney, it's \nothin' to get hung about" from Strawberry Fields Forever.

Another note: (along the same line) If a Dirac-delta function appears in the integrand

of some integral, the integral is evaluated from the sampling property.

Warning: the integration limits cannot \split" a Dirac-delta function \down the middle,"

for example1Z0

±(t) dt

should be either 1Z0¡

±(t) dt = 1 or

1Z0+

±(t) dt = 0:

You either capture all of the delta-function or you don't get any of it. It cannot be divided.

Or you will have to agree on some scheme to cut it up while minimizing bloodshed.

Proof of the SAMPLING PROPERTY: No harm is done in taking t0 = 0, so that

we want to showbZa

f(t)±(t) dt =

½f(0); a < 0 < b

0; o.w.

for f(t) continuous in a neighborhood of t = 0. For de¯niteness, use the limit of the

rectangular pulse to represent

±(t) = lim²!0

1

²¦

μt

²

¶and represent the continuous f(t) by its Taylor series about t = 0:

f(t) = f(0) + tf 0(0) +t2

2!f 00(0) +

t3

3!f 000(0) +O(t4):

Then the integral of interest is, for a < 0 < b,

bZa

f(t)±(t) dt = lim²!0

1

²

²=2Z¡²=2

·f(0) + tf 0(0) +

t2

2!f 00(0) + : : :

¸dt:

8

Note²=2Z

¡²=2

dt = ²;

²=2Z¡²=2

t dt = 0;

²=2Z¡²=2

t2 dt =²3

12

and sobZa

f(t)±(t) dt = lim²!0

1

²

·²f(0) +

²3

24f 00(0) +O(²5)

¸= f(0):

If, for a < b, either 0 < a or b < 0, then the integral is zero since ±(t) = 0 for t6= 0. ¥

F Provide an alternate proof of the sampling property by appealing to the mean value

theorem of the integral.

\Heavy formalism is not required to get across fundamental ideas. If it can't be said simply,

it's not worth being said at all."

Bernard H. Lavenda, Statistical Physics, 1991, p. viii.

\The pursuit of excellence is gratifying and healthy. The pursuit of perfection is frustrating,

neurotic, and a terrible waste of time."

Edwin Bliss

9

Symmetry1

symmetric or even fe(¡x) = fe(x)

..................................................................................................................................................................................................................................................................................................................................................................................................................................................

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x

cos(x)

...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................

......................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...........

asymmetric or odd fo(¡x) = ¡fo(x)

..................................................................................................................................................................................................................................................................................................................................................................................................................................................

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x

sin(x)

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x

x2 ¡ 25.............................................................................................................................................................................................................................................................................................................................................................................................

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x

x

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x

18(35x

4 ¡ 30x2 + 3)....................................................................................................................................................................................................................

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x

12 (5x

3 ¡ 3x)

...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

............................................................................................................................................................................................

1\Symmetry is immunity to a possible change." J. Rosen, Symmetry Rules, Springer, 2008, p. 4

10

If a function f(x) is continuous to all orders (the function f(x) and all of its nth derivatives

f (1)(x); f (2)(x); : : : are continuous) then f(x) can be represented by a Taylor series about

the origin

f(x) =

1Xn=0

f (n)(0)

n!xn:

The Taylor series of an even function will have only even powers of x

fe(x) =

1Xn=0

f (2n)(0)

(2n)!x2n

and therefore all of its odd derivatives (at the origin) vanish

f (2n+1)e (0) =d2n+1f(x)

dx2n+1

¯̄̄̄x=0

= 0:

In particular, note that f 0e(0) = 0 (if fe(x) is continuous and di®erentiable at the origin).

Similarly, the Taylor series of an odd function will have only odd powers of x

fo(x) =

1Xn=0

f (2n+1)(0)

(2n+ 1)!x2n+1

and therefore all of its even derivatives (at the origin) vanish

f (2n)o (0) =d2nf(x)

dx2n

¯̄̄̄x=0

= 0:

In particular, note that fo(0) = 0 (if fo(x) is continuous at the origin).

Even-Order Derivatives Preserve Symmetry and Odd-Order Derivatives Re-

verse Symmetry

f 0(t) = lim¢t!0

f(t+¢t=2)¡ f(t¡¢t=2)¢t

f 0(¡t) = lim¢t!0

f(¡t+¢t=2)¡ f(¡t¡¢t=2)¢t

Even function fe(t) = fe(¡t)

f 0e(t) = lim¢t!0

fe(t+¢t=2)¡ fe(t¡¢t=2)¢t

f 0e(¡t) = lim¢t!0

fe(¡t+¢t=2)¡ fe(¡t¡¢t=2)¢t

= lim¢t!0

fe(t¡¢t=2)¡ fe(t+¢t=2)¢t

= ¡ lim¢t!0

fe(t+¢t=2)¡ fe(t¡¢t=2)¢t

= ¡f 0e(t) =)d

dtfe(t) is odd.

Similarly,d

dtfo(t) is even where fo(¡t) = ¡fo(t) is an odd function.

This is also seen by appealing to the Taylor series for even and odd functions.

11

Consider the integral of f(t) between limits that are symmetrically disposed about the

origin, that is look atTZ

¡Tf(t) dt

where T is some ¯xed, positive constant.

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t

fo(t)

...........................................................................................................................................

.............................................................................................................................................................................................................................................................................................................................................................................................................................

.............................................................................................................................

.........................................................................................

¡T.........................................................................................

T

TZ¡T

fo(t) dt = 0

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......

t

fe(t)

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....

¡T.......................................

T

TZ¡T

fe(t) dt = 2

TZ0

fe(t) dt

Any function f(t) (neither symmetric nor antisymmetric) can always be decomposed into

a linear combination of a symmetric plus an antisymmetric part since

f(t) =f(t) + f(¡t)

2+f(t)¡ f(¡t)

2= fe(t) + fo(t):

Example.

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...

t0 T

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...

¡T T

.......

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..

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..

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.. ............................................

¡TT

= +

t

t

F Homework. Decompose the Heaviside unit step function u(t) into the sum of a

symmetric plus antisymmetric signal.

12

A Note on Mathematical Functions and Units/Physical Dimensions

Consider almost any \ordinary" or \regular" function2 of an independent variable x, such

as

f(x) = cos(x) or sin2(x) or 3x2 ¡ 7x or ex:

The argument or independent variable x is always dimensionless. To see this, examine the

Taylor series expansion, such as

f(x) = cos(x) = 1¡ x2

2!+x4

4!¡ : : : :

If x has units such as meters (m), then the sum would be nonsense since the ¯rst term is

dimensionless, the second term has units (m)2, the third term has units (m)4, etc. Also

note that the function f(x) itself is dimensionless, too.

This may seem strange and somewhat contrary to your previous experience. For example,

you have dealt with a time-domain voltage signal such as

v(t) = 12 cos(t+ 45±) or 12 cos(t+ ¼=4):

If the time variable t is measured in (s), then there really is a transparent (or hidden)

frequency ! = 1 (s¡1) multiplying the t so that

v(t) = V0 cos(!t+ Á):

Now the argument !t+ Á of the cosine is clearly dimensionless. The constant V0 = 12 (V)

carries the dimensions of the voltage signal v(t). Also note that we often say ! has units

of (rad/s), but a radian is not dimensioned: Its use in (rad/s) is simply a reminder that

we are using the angular frequency ! and not the common experimentalist's frequency

f = !=2¼ (cycle/s=Hz).

The operators di®erentiation and integration impart units: Consider the time-

domain current signal

i(t) = I0 cos(!t):

In our standard mksC (SI) units, the current amplitude I0 is in (A), the frequency ! is in

(s¡1), and the time t is in (s). Di®erentiation with respect to the dimensioned variable timparts the units of 1=t or (s¡1) since

di(t)

dt= ¡!I0 sin(!t):

We can also see that the di®erential operator has to have units of inverse time by appealing

to the de¯nition of the derivative

df(t)

dt= lim

¢t!0

f(t+¢t)¡ f(t)¢t

:

The (s¡1) comes from the denominator ¢t that is in (s).

2Not ±(x) or u(x) or jxj, to name a few exceptions.

13

Integration with respect to the dimensioned variable t imparts the units of t or (s) since

tZt0

i(¿) d¿ =

tZt0

I0 cos(!¿) d¿ =I0

!

£sin(!t)¡ sin(!t0)

¤:

Note the good practice of using a dummy time variable for the integration, to avoid con-

fusion with the ¯nal time that is the upper integration limit. We can also see that the

integral operator has to have units of time by appealing to the de¯nition of the integral in

terms of a limit on the Riemann sum

bZa

f(t) dt = limN!1

NXn=1

f(tn)¢tn:

The prime notation for derivatives. When dealing with a generic function such as

f(t) or g(x), we commonly denote derivatives using the prime

df(t)

dt= f 0(t) and

dg(x)

dx= g0(x):

The prime denotes di®erentiation to the entire argument of the function, as in

f 0(») =df(»)

d»or f 0(~) = df(~)

d~ :

Note thatd

dtf(») =

df(»)

d»

d»

dt= f 0(»)

d»

dt:

14

Linear System

x(t) y(t)T

............................................................................................................................................................................................................................................................................................................................................................................

........................................................... ............. ........................................................... .............

Assume the zero-state response y(t) of a system to the input x(t) is described by an

operator T , such that

y(t) = Tfx(t)g:Another name for T is system transformation rule. Examples might be

T1fx(t)g = Ax(t); T2fx(t)g = x(t¡ t0); T3fx(t)g = d

dtx(t)

T4fx(t)g =μc2d2

dt2+ c1

d

dt+ c0

¶x(t); T5fx(t)g =

tZ¡1

x(¿) d¿

T6fx(t)g = x2(t); T7fx(t)g = Ax(t) +B:

Linear System: If the input/output relationship or system operator or system transfor-

mation rule satis¯es the superposition principle

Tfc1x1(t) + c2x2(t)g = c1Tfx1(t)g+ c2Tfx2(t)g;

then the system is said to be linear.

F HW 1 Test the examples T1; T2; : : : ; T7 above for linearity.

15

Time-Invariant System:x(t) y(t)

T

............................................................................................................................................................................................................................................................................................................................................................................

........................................................... ............. ........................................................... .............

x(t¡ t0) y(t¡ t0)T

............................................................................................................................................................................................................................................................................................................................................................................

........................................................... ............. ........................................................... .............

Let the response of a system to the input signal x(t) be the output signal y(t). If the

response to the delayed input x(t¡ t0) is the delayed response y(t¡ t0), the characteristicsof the system did not change in time, and the system is said to be time-invariant.

Note: In our linear RLC circuit analysis, the equation-of-motion is typically a linear

integrodi®erential equation, with constant coe±cients, such as

.......

.......

.......

.......

.......

.......

.......

.......

....................................................................................................................................................................................................................................................................................

............................................................................................................................................................

...............................................................................................................................................................................................................................................................................................................................................................................................

..........................................................................................................................................................

+¡vg(t)

R L..........................................................................

..........................................................................

.................................

.................................

C.....................................................

...................................

..........................................................................................................................................................................................................................................................................................................................................................................................................................

.............................................................................................

......................... ................

i(t) ........................................................................................................................................................................................................................................................................................................

......................................................... ................ ......................................................... ................vg(t) i(t)

Ldi(t)

dt+Ri(t) +

1

C

tZ¡1

i(¿) d¿ = vg(t)

which is easily converted to an ordinary di®erential equation by di®erentiating w.r.t. t so

that

Ld2i(t)

dt2+R

di(t)

dt+1

Ci(t) = v0g(t):

Again, note the constant coe±cients L, R, and 1=C. If the equation-of-motion of a system

is a di®erential equation with non-constant (i.e. time varying) coe±cients, then the system

is a time-variant system.

F HW 2 Give an example of a linear, time-varying system.

16

Input/Output Relationship for a LTI System

(zero-state or driven response)

x(t) y(t) = Tfx(t)gT

............................................................................................................................................................................................................................................................................................................................................................................

........................................................... ............. ........................................................... .............

This fundamental concept is easy to see and derive, starting with a little \trick" by way

of the notation. First, invoke the sampling property of the Dirac-delta function to write

x(t) =

1Z¡1

x(¿)±(t¡ ¿) d¿:

Now exploit the linearity of the operator T and note that Tf¢g operates only on functionsof time t. In particular, T treats x(¿) as a constant, and acts only on the time-domain

function ±(t¡ ¿)

y(t) = Tfx(t)g = T8<:

1Z¡1

x(¿)±(t¡ ¿) d¿9=; =

1Z¡1

x(¿)Tf±(t¡ ¿)g d¿:

The response of our system to the Dirac-delta function or impulse function is called the

system impulse response and it is denoted by the signal h(t). It is a fundamental

characterization of any linear, time-invariant system.

±(t) h(t)T

............................................................................................................................................................................................................................................................................................................................................................................

........................................................... ............. ........................................................... .............

±(t¡ ¿) h(t¡ ¿)T

............................................................................................................................................................................................................................................................................................................................................................................

........................................................... ............. ........................................................... .............

Since the system is also time-invariant,

Tf±(t¡ ¿)g = h(t¡ ¿)and the result above is

y(t) =

1Z¡1

x(¿)h(t¡ ¿) d¿:

This integral operator that takes two functions of time x(t) and h(t), and produces a third

function of time y(t), is called the convolution integral; we write

y(t) = x(t)~ h(t):

17

Convolution

The convolution operation is not restricted to our above LTI system application. It can

operate on any two (reasonably) arbitrary signals, say f(t) and g(t):

f(t)~ g(t) =1Z

¡1f(¿)g(t¡ ¿) d¿:

Example. Evaluate s(t) = f(t) ~ g(t) where f and g are causal, decaying exponentialsignals

f(t) = e¡atu(t) and g(t) = e¡btu(t) with 0 < a < b:

case: t < 0

................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

¿0

f(¿) = e¡a¿u(¿)

...................................................................................................................................................................................................................................................................................................................................................................................................................

................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

¿0

g(t¡ ¿) = e¡b(t¡¿)u(t¡ ¿)

..............................................................................................

..............................................................................................................................................................................................................................................................

t

................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

¿0

f(¿)g(t¡ ¿)

...........

t

s(t) =

1Z¡1

f(¿)g(t¡ ¿) d¿ =1Z

¡10 d¿ = 0

18

case: t > 0

................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

¿0

f(¿) = e¡a¿u(¿)

...................................................................................................................................................................................................................................................................................................................................................................................................................

................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

¿0

g(t¡ ¿) = e¡b(t¡¿)u(t¡ ¿)

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.......................................................................................................................................................................................................................................................................

t

................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

¿0

f(¿)g(t¡ ¿)

.............................................................................................

................................................................................................................................................................................

t

s(t) =

1Z¡1

f(¿)g(t¡ ¿) d¿

=

0Z¡1

0 d¿ +

tZ0

e¡a¿e¡b(t¡¿)d¿ +

1Zt

0 d¿

= e¡bttZ0

e(b¡a)¿d¿ =e¡at ¡ e¡btb¡ a

19

Now combine both expressions (¯rst one for t < 0 and second one for t > 0) into a single

equation valid for all t

s(t) =e¡at ¡ e¡btb¡ a u(t):

.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

t0

s(t)

............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

tm

Let's look for the maximum of s(t) analytically: The derivative

s0(t) =be¡bt ¡ ae¡at

b¡ a u(t) +e¡at ¡ e¡btb¡ a ±(t)

=be¡bt ¡ ae¡at

b¡ a u(t)

sincee¡at ¡ e¡btb¡ a

¯̄̄̄t=0

= 0:

Recall that Ã(t)±(t) = Ã(0)±(t). The derivative vanishes at time t = tm, so write s0(tm) = 0

so that

ae¡atm = be¡btm

e(b¡a)tm =b

a

tm =ln(b=a)

b¡ a

20

Another Convolution Example: The Convolution of Two Gaussian Pulses

Evaluate z(t) = f(t)~ g(t) with f(t) = exp(¡at2) and g(t) = exp(¡bt2).

Both f(t) and g(t) are \on" for all time, so we don't have to worry about sketching them

to see when they turn on or o®.

z(t) =

1Z¡1

f(¿)g(t¡ ¿) d¿ =1Z

¡1e¡a¿

2

e¡b(t¡¿)2

d¿

= e¡bt2

1Z¡1

e¡[(a+b)¿2¡2bt¿ ]d¿

complete the square

(a+ b)¿2 ¡ 2bt¿ =μp

a+ b¿ ¡ btpa+ b

¶2¡ (bt)2

a+ b

z(t) = exp

·μb2

a+ b¡ b¶t2¸ 1Z¡1

exp

"¡μp

a+ b¿ ¡ btpa+ b

¶2#d¿

let

x =pa+ b¿ ¡ btp

a+ b

dx =pa+ b d¿

¿ = §1 ¡! x = §1recall or use (\and it is a trick!"see page 63)

1Z¡1

e¡x2

dx =p¼

z(t) = exp

·μb2

a+ b¡ b¶t2¸ 1Z¡1

e¡x2 dxpa+ b

=

r¼

a+ bexp

·¡μb¡ b2

a+ b

¶t2¸

=

r¼

a+ bexp

·¡ ab

a+ bt2¸

The convolution of two Gaussians is a third Gaussian.

21

Yet Another Convolution Example

Evaluate s(t) = f(t)~ g(t) where f(t) and g(t) are the given rectangular pulses:

f(t) = u(t)¡ u(t¡ a) =½1; 0 < t < a

0; t < 0 or t > a

g(t) = u(t)¡ u(t¡ b) =½1; 0 < t < b

0; t < 0 or t > b

Here the pulse width a of f(t) has been arbitrarily selected to be greater than the pulse

width b of g(t).

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............................................................................................................................................................................................................................................................................................................................................................... t0 a

1f(t)

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..................................................................................................................................................................................................................................................................... t0 b

1g(t)

s(t) = f(t)~ g(t) =1Z

¡1f(¿)g(t¡ ¿) d¿

Since the arguments of our individual signals are ¿ (and a shifted, rotated version of ¿)

in the convolution integral, let's redraw our functions f and g as functions of the dummy

variable ¿ .

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............................................................................................................................................................................................................................................................................................................................................................... ¿0 a

1f(¿)

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..................................................................................................................................................................................................................................................................... ¿0 b

1g(¿)

22

Let's leave f(¿) alone (as in the version of the convolution integral above), and shift and

rotate g(¿) (as it appears in the version of the convolution integral above).

..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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..................................................................................................................................................................................................................................................................... ¿0 t t+b

g(¿ ¡ t)

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..................................................................................................................................................................................................................................................................... ¿0 tt¡b

g(t¡ ¿)

Now let's vary the parameter t (it survives the ¿ integration), compute (graph!) the

product of f(¿) and g(t ¡ ¿), and evaluate the in¯nite range integral from ¿ = ¡1 to

¿ = +1. That is, we must calculate the total area under the curve of f(¿)g(t¡ ¿).

case: t < 0

..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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............................................................................................................................................................................................................................................................................................................................................................... ¿0 a

......

.....

b

f(¿)

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..................................................................................................................................................................................................................................................................... ¿0t¡ b t

g(t¡ ¿)

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¿0

f(¿)g(t¡ ¿)

s(t) = f(t)~ g(t) =1Z

¡1f(¿)g(t¡ ¿) d¿ =

1Z¡1

0 d¿ = 0

23

case: 0 < t < b

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............................................................................................................................................................................................................................................................................................................................................................... ¿0 a

......

.....

b

f(¿)

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..................................................................................................................................................................................................................................................................... ¿0t¡ b t

g(t¡ ¿)

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............................................................................................................................................................................................................... ¿0 a

......

.....

b...........

t

f(¿)g(t¡ ¿)

s(t) = f(t)~ g(t) =1Z

¡1f(¿)g(t¡ ¿) d¿ =

tZ0

d¿ = t

case: b < t < a

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............................................................................................................................................................................................................................................................................................................................................................... ¿0 a

......

.....

b

f(¿)

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..................................................................................................................................................................................................................................................................... ¿0 t¡ b t

g(t¡ ¿)

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..................................................................................................................................................................................................................................................................... ¿0 a

......

.....

b...........

tt¡ b

f(¿)g(t¡ ¿)

s(t) = f(t)~ g(t) =1Z

¡1f(¿)g(t¡ ¿) d¿ =

tZt¡b

d¿ = b

24

case: a < t < a+ b

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............................................................................................................................................................................................................................................................................................................................................................... ¿0 a

......

.....

b...........

a+ b

f(¿)

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..................................................................................................................................................................................................................................................................... ¿0 a

......

................

a+ btt¡b

g(t¡ ¿)

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........................................................................................................................................................................... ¿0 a

......

................

a+ bt...........

t¡b

f(¿)g(t¡ ¿)

s(t) = f(t)~ g(t) =1Z

¡1f(¿)g(t¡ ¿) d¿ =

aZt¡b

d¿ = a+ b¡ t

case: t > a+ b

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............................................................................................................................................................................................................................................................................................................................................................... ¿0 a

......

.....

b...........

a+ b

f(¿)

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..................................................................................................................................................................................................................................................................... ¿0 a

......

................

a+ b tt¡b

g(t¡ ¿)

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¿0

f(¿)g(t¡ ¿)

s(t) = f(t)~ g(t) =1Z

¡1f(¿)g(t¡ ¿) d¿ =

1Z¡1

0 d¿ = 0

25

All together now.

s(t) = f(t)~ g(t) =

8>>>>><>>>>>:

0; t < 0

t; 0 < t < b

b; b < t < a

a+ b¡ t; a < t < a+ b

0; t > a+ b

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....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... t

0 a...........

......

.....

b...........

a+ b

s(t)

...........b

......................................................................................................................................................................................................................................

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26

Homework Problems

1. In the above example on page 18, ¯nd r(t) = f(t)~ f(t), that is let b = a. Show thatyour result can also be obtained from the limit of the above as b! a

r(t) = limb!a

s(t):

Use l'Hopital's rule to disarm the 0=0 form.

2. Show that f(t)~ ±(t¡ t0) = f(t¡ t0).

3. Show that

f(t)~ u(t) =tZ

¡1f(¿) d¿:

4. Evaluate the convolution of the standard rectangular pulse with itself ¦(t=T )~¦(t=T ).Express the result in terms of the standard triangular pulse ¤(t=T ).

..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

¡T=2 T=20

1

¦(t=T )

t.............................................................................................................

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......................................................................................................................................................................................................................................................................................................................................................................................................................................

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......................................................................................................................................................................................................................................................................................................................................................................................................................................

......................................................................................................................................................................................................................................................................................................................................................................................................................................

.........................................................................................................................................................................................................................................................................................................................................................................................................................

.......................... 1

¡T 0 Tt

¤(t=T )

¦(t=T ) =

½1; (jtj < T=2)0; (jtj > T=2)

¤(t=T ) =

½1¡ jtj=T; (jtj < T )0; (jtj > T )

5. Find an expression for, and graph, each of the derivatives

d

dt¤(t=T ) and

d

dt¦(t=T ):

27

Causal system: If the input x(t) = 0 for t < t0, then the zero-state response y(t) = 0

for t < t0. The impulse response of a causal system is therefore a causal signal.

If f(t) and g(t) are both causal signals, then so is their convolution f(t)~ g(t).

Bounded input/bounded output stability: If the input signal is bounded, such that

jx(t)j · A <1 for all time t, then the zero-state output y(t) is also bounded if the impulse

response is absolutely integrable:

jy(t)j =¯̄̄̄¯̄1Z

¡1h(¿)x(t¡ ¿) d¿

¯̄̄̄¯̄ ·

1Z¡1

¯̄h(¿)x(t¡ ¿)¯̄ d¿ = 1Z

¡1

¯̄h(¿)

¯̄¯̄x(t¡ ¿)¯̄ d¿

· A1Z

¡1

¯̄h(¿)

¯̄d¿ <1

if 1Z¡1

¯̄h(t)

¯̄dt <1:

28

Singular Functions and Integrals

Consider a real-valued function f of a real variable x.

With A some positive and ¯nite constant, if jf(x)j · A < 1 for all x, then we say the

function f(x) is bounded.

If jf(x0)j = 1 for some particular x = x0, then we say f is singular at x0, and x0 is a

singular point of f .

Examples.

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x

1=x

1

xis singular at x = 0

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..

x

sin(x)=x

sin(x)

xis bounded at x = 0.....................................

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..

x

cos(x)=x

cos(x)

xis singular at x = 0

................................................................................................................................................................................................................................................................................................................................

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29

......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

......

.......

......

...

0 7

x

1=(x¡ 7)2

1

(x¡ 7)2 is singular at x = 7..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... .........................................................................................................................................................................................................................................................................................................................................................

..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

0

x

1=px

1pxis singular at x = 0

.....................................................................................................................................................................................................................................................................................................................................................................................................

Consider now the de¯nite integral of f(x) between the limits x = a and x = b. These

limits are arbitrary, but with the restriction that a · b.

If

¯̄̄̄¯bZa

f(x) dx

¯̄̄̄¯ · A; then the integral converges, and it exists.

If

¯̄̄̄¯bZa

f(x) dx

¯̄̄̄¯ =1; then the integral diverges, and it does not exist.

From the triangle inequality

¯̄̄̄¯bZa

f(x) dx

¯̄̄̄¯ ·

bZa

jf(x)j dx:

30

If f(x) = jf(x)j (f is non-negative, that is f(x) ¸ 0), and if b > a, thenbZa

f(x) dx ¸ 0:

Therefore, for simplicity, let's temporarily restrict our attention to non-negative f(x).

Basically, there are two ways to get a divergent integral:

(1) A ¯nite range integral (¡1 < a < b <1) having an integrand with a non-integrable(too big!) singularity.

(2) An in¯nite range integral (a = ¡1 and/or b = 1) having a bounded integrandthat doesn't decay fast enough at 1.

Examples.

1Z0

dx

x2diverges because the integrand has a non-integrable singularity at the origin.

With ² > 0, if we avoid the second-order pole at the origin, the integral evaluates to

1Z²

dx

x2=1

x

¯̄̄̄¯²

x=1

=1

²¡ 1:

Clearly this integral is unbounded (singular) if we let ²! 0.

1Z0

dx

x= ln(x)

¯̄̄1x=0

= 0¡ ln(0) =1 is divergent.

1Z0

dxpx= 2x1=2

¯̄̄1x=0

= 2 is convergent.

1Z1

dx

x2=1

x

¯̄̄̄¯1

x=1= 1¡ 0 = 1 is convergent.

31

1Z1

dx

x= ln(x)

¯̄̄̄¯1

x=1

=1¡ 0 =1 is divergent.

Cauchy Principal Value

Sometimes we relax our concept of the integral to obtain a ¯nite result and \dance around"

singularities in the integrand that are, strictly speaking, non-integrable. Here is one exam-

ple that demonstrates what is meant by the Cauchy principal value which is one possible

interpretation of an otherwise divergent integral.

If f(x) is bounded everywhere and if a < c < b, then we can always break the integration

domain from a to b at the point c and write

bZa

f(x) dx =

cZa

f(x) dx+

bZc

f(x) dx:

We have already seen that the function f(x) = 1=x has a non-integrable singularity at the

origin, since1Z0

dx

xis divergent.

Also observe that0Z

¡1

dx

xis divergent.

If we want to integrate 1=x from x = ¡1 to x = 1, we should not break it as1Z

¡1

dx

x6=

0Z¡1

dx

x+

1Z0

dx

x:

But under the Cauchy principal value interpretation, if we exploit the odd symmetry of

the integrand, ignoring the singularity at the origin, then we can say

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.......

......

x

1=x

PV

1Z¡1

dx

x=

1Z¡1|

dx

x= lim²!0

24 ¡²Z¡1

dx

x+

1Z²

dx

x

35 = 0:..............................................................................................................................................................................................................................................

.......

.......

.......

.......

........

........................................................................................

.................................................................

...................................................................

..................

¡11

32

SHOW THAT

1Z0

¯̄̄̄sinx

x

¯̄̄̄dx IS DIVERGENT

x

k¼ (k + 1=2)¼ (k + 1)¼

.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

....................................................................................................................................................

(k+1)¼Zk¼

¯̄̄̄sinx

x

¯̄̄̄dx >

¼=2

(k + 1=2)¼=

1

2k + 1

1Z0

¯̄̄̄sinx

x

¯̄̄̄dx =

¼Z0

¯̄̄̄sinx

x

¯̄̄̄dx+

1Xk=1

(k+1)¼Zk¼

¯̄̄̄sinx

x

¯̄̄̄dx >

¼Z0

¯̄̄̄sinx

x

¯̄̄̄dx

| {z }¯nite

+

1Xk=1

1

2k + 1| {z }divergent

=1

\... better to say something new and stimulating to the interested students than to show

routine steps to the uninterested ones."

E.G. Peter Rowe Geometrical Physics in Minkowski Spacetime, Springer, 2001, p. v

33

SHOW THAT

1Z0

sinx

xdx IS CONVERGENT

.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

................................................................................................................................................................................................................................................................................................................................................................................................................................................

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.....................................

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....................................

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..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... x

.......

.......

......

..

.......

.......

......

..

.......

.......

......

..

2k¼ (2k + 1)¼

(2k + 2)¼

(2k+2)¼Z2k¼

sinx

xdx =

(2k+1)¼Z2k¼

sinx

xdx +

(2k+2)¼Z(2k+1)¼

sinx

xdx

| {z }let t=x¡¼

=

(2k+1)¼Z2k¼

sinx

xdx ¡

(2k+1)¼Z2k¼

sin t

t+ ¼dt

=

(2k+1)¼Z2k¼

sinx

μ1

x¡ 1

x+ ¼

¶dx

= ¼

(2k+1)¼Z2k¼

sinx

x2 + ¼xdx < ¼

(2k+1)¼Z2k¼

dx

x2=

1

4k2 + 2k

1Z0

sinx

xdx =

2N¼Z0

sinx

xdx+

1Xk=N

(2k+2)¼Z2k¼

sinx

xdx <

2N¼Z0

sinx

xdx

| {z }¯nite

+

1Xk=N

1

4k2 + 2k| {z }¯nite

34

Homework Problems

1. Express each of these four functions (signals) in terms of shifted, modulated, unit steps.

..................................................................................................................................................................................................................................................................................................

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¡TT

+A

¡A

f(t)

0

................... 1

..........................................................................................................................................................................................................................................................................................................................................................................................................................................

t¡T T

¤

μt

T

¶

0

.........................................................................................................................................................................................................................................................................

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¡TT

B

¡B

g(t)

0

................................................................................................................................................................................................................................................................

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................................................................................................................................................................................................................................................................

...................

...................

........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

t

T1 T2

.......

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.....

x(t)

0

...................A

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2. Discuss the existence and evaluate (if appropriate!) the following integrals. First sketch

the integrand (except in an extreme case!).

1Z¡1

u(t) dt;

1Z¡1

u(t)e¡¾t dt (¾ > 0);

1Z¡1

u(t)e¡¾te¡j!t dt

1Z¡1

sin(!0¿)±(¿) d¿;

1Z¡1

cos(!0¿)±(¿) d¿;

¼Z0

sin(x2)

ln(x=¼)±(x¡ 2¼) dx

1Z0

cosx

xdx;

1Z¡1

cosx

xdx;

1Z1

dx

x2;

1Z0

dx

x2;

1Z0

dxpx;

1Z0

dxpx

2¼Z¡p¼

·exp[j sin2(!0t)] ±(t+ ¼)

t3 tan(7t) + ln(sin t)¡ ¼2 sin(3t=2) ±(t¡ ¼)t2 + 2t¡ 4 sin¡1(t=¼)

¸dt

35

3. Prove that convolution is commutative:

f(t)~ g(t) = g(t)~ f(t):

4. Evaluate u(t)~ u(t).

5. Show that ±(t) = ±(¡t) is an even function.

6. If f(t) is continuous at t = t0, show that f(t)±(t¡ t0) = f(t0)±(t¡ t0).

36

Relationship Between Impulse Response and Step Response

±(t) h(t)

............................................................................................................................................................................................................................................................................................................................................................................

........................................................... ............. ........................................................... .............

u(t) s(t)

............................................................................................................................................................................................................................................................................................................................................................................

........................................................... ............. ........................................................... .............

±(t) =d

dtu(t) h(t) =

d

dts(t)

u(t) =

tZ¡1

±(¿) d¿ s(t) =

tZ¡1

h(¿) d¿

To see this, consider the general linear system input/output relationship.

x(t) y(t)h(t)

............................................................................................................................................................................................................................................................................................................................................................................

........................................................... ............. ........................................................... .............

y(t) = x(t)~ h(t) =1Z

¡1x(¿)h(t¡ ¿) d¿ =

1Z¡1

h(¿)x(t¡ ¿) d¿

d

dty(t) =

1Z¡1

x(¿)d

dth(t¡ ¿) d¿ =

1Z¡1

h(¿)d

dtx(t¡ ¿) d¿

= x(t)~ h0(t) = x0(t)~ h(t)

The output of these two systems is identical, since the order of di®erentiation and passing

through the linear system having a given impulse response h(t) does not matter. More

concisely, the operators h(t)~ and d=dt commute.

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ex(t) ey(t)h(t)d

dt........................................................................ ..........

.... ....................................................................................... .............. ........................................................................ ..........

....

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ex(t) ey(t)h(t)d

dt........................................................................ ..........

.... ....................................................................................... .............. ........................................................................ ..........

....

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37

Time-Domain RC Circuit Analysis

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.......................................................................................................................................

t = 0

R

C

+

¡

v(t)+

¡V0 cos(!t)u(t) = vg(t)

The capacitor is initially charged, such that v(t) = Vc0 prior to the closing of the switch

at time t = 0. Find an expression for the signal v(t) that is valid for all time t ¸ 0.The constants R, C, !, V0, and Vc0 are given and thus known.

The initial condition on the voltage v(t) that applies to our circuit for time t ¸ 0 is v(0+)and v(0+) = v(0¡) = Vc0 since the voltage across a capacitor must be a continuous functionof time. That is, if the capacitor current is bounded, because the voltage/current terminal

relationship of a capacitor is

iC(t) = Cd

dtvC(t) (1)

where iC(t) and vC(t) are related by the passive sign convention. Inverting this, that is,

writing the voltage as a function of the current, we have

vC(t) =1

C

tZ¡1

iC(¿) d¿ =1

C

0Z¡1

iC(¿) d¿ +1

C

tZ0

iC(¿) d¿ = vC(0) +1

C

tZ0

iC(¿) d¿: (2)

The Kirchho® current law applied to the (single unknown) node of potential v(t), for time

t ¸ 0 isCdv(t)

dt+v(t)¡ vg(t)

R= 0 (3)

which is rearranged asdv(t)

dt+

1

RCv(t) =

V0

RCcos(!t)u(t): (4)

Here we are already working in the time zone t ¸ 0, so with the stipulation that t ¸ 0, theHeaviside unit step is probably unnecessary so we can write

dv(t)

dt+

1

RCv(t) =

V0

RCcos(!t): (5)

The homogeneous solution is the solution to

dvh(t)

dt+

1

RCvh(t) = 0 (6)

38

and it is

vh(t) = Ke¡t=RC (7)

where the constant K cannot be determined until we construct the complete solution. The

particular solution is some speci¯c or particular function vp(t) that yields the right-hand

side (forcing term) when operated upon by the di®erential operatorμd

dt+

1

RC

¶vp(t) =

V0

RCcos(!t): (8)

There are no unknown constants (unknown by the di®erential operator) in the particular

solution.3 Since the forcing function is cos(!t) and since cos(!t) is not a homogeneous

solution, then we see that the particular solution must be a linear combination of cos(!t)

and sin(!t), as in

vp(t) = A cos(!t) +B sin(!t): (9)

The constants A and B come directly from the di®erential equation (8) , which readily

surrenders them (to us): All we have to do is insert the form (9) into the di®erential

equation (8). To do this, ¯rst write the required derivative

dvp(t)

dt= !B cos(!t)¡ !A sin(!t): (10)

Insertion of (9) and (10) into (8) yieldsμA

RC+ !B

¶cos(!t) +

μB

RC¡ !A

¶sin(!t) =

V0

RCcos(!t): (11)

The only way this can be true for all t ¸ 0 is if the coe±cients of the cos(!t) and the

sin(!t) terms are separately equal on both sides of the equation, so that

A

RC+ !B =

V0

RC(12)

B

RC¡ !A = 0: (13)

The solution to this two-by-two system of simultaneous, linear equations is

A =V0

1 + (!RC)2and B =

!RCV0

1 + (!RC)2; (14)

and so our particular solution (9) is

vp(t) =V0

1 + (!RC)2

£cos(!t) + !RC sin(!t)

¤: (15)

3In other words, the particular solution comes only from the di®erential equation, that is, the particular

solution is completely independent and ignorant of any initial conditions.

39

Again, note that vp(t) comes completely from the di®erential equation alone. We now can

assemble the complete solution v(t) = vh(t) + vp(t) or

v(t) = Ke¡t=RC +V0

1 + (!RC)2

£cos(!t) + !RC sin(!t)

¤: (16)

Application of the one initial condition v(0) = v(0+) = Vc0 (note that it is the plus side of

zero that applies to our di®erential equation, because the di®erential equation is valid for

time t ¸ 0) to (16) gives

Vc0 = K +V0

1 + (!RC)2or K = Vc0 ¡ V0

1 + (!RC)2: (17)

Let's write out the complete solution in two forms. Firstly, observe that for t ¸ 0

v(t) =

·Vc0 ¡ V0

1 + (!RC)2

¸e¡t=RC| {z }

homogeneous solution

+V0

1 + (!RC)2

£cos(!t) + !RC sin(!t)

¤| {z }

particular solution

(18)

is exactly in the form in which we constructed the solution, that is, the sum of the homo-

geneous plus particular solutions. If we instead group terms proportional to Vc0 and V0,

we have for t ¸ 0

v(t) = Vc0e¡t=RC| {z }

zero-input response

+V0

1 + (!RC)2

hcos(!t) + !RC sin(!t)¡ e¡t=RC

i| {z }

zero-state response

: (19)

If there is no input signal (or generator voltage vg(t) or forcing term), then V0 = 0 and

the output signal v(t) is caused solely by the initial energy (or non-zero initial state of

the system) stored in the capacitor, and is proportional to the non-zero initial condition

voltage Vc0. On the other hand, if the initial state of the system is zero, so that Vc0 = 0,

then the response due to the input vg(t) that is proportional to V0 is called the zero-state

response.

40

Step Response of an RC Circuit

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..........................................................

R

C

+

¡

v(t)+

¡V0u(t) = vg(t)

The step response is, by de¯nition, a zero-state response so the capacitor C is initially

uncharged prior to the turning on of the input step function at time t = 0. Therefore, by

the continuity of capacitor voltage we have the initial condition

v(0+) = v(0¡) = 0: (20)

Ultimately, the step response s(t) = y(t) should be the response due to a non dimensioned

input signal u(t) = x(t), but in doing circuit analysis I ¯nd it better to explicitly express

all physical signals such as voltages and currents in appropriate mksC (SI) units. At the

end we can normalize by V0, for example. In systems engineering (\black-box") notation,

we have the general form of a zero-state response y(t) to some arbitrary input x(t) like

this:

................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.......................................................................... ................ .......................................................................... ................x(t) y(t)

For the step response, we have:

................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.......................................................................... ................ .......................................................................... ................u(t) s(t)

Our ensuing circuit analysis regards the input as x(t) = vg(t) = V0u(t) and our output

signal is y(t) = v(t). The Kirchho® current law applied to the node of potential v(t) in

the circuit is for time t ¸ 0

Cdv(t)

dt+v(t)¡ vg(t)

R= 0 (21)

which is rearranged as

dv(t)

dt+

1

RCv(t) =

V0

RCu(t): (22)

41

Here we are already working in the time zone t ¸ 0, so with the stipulation that t ¸ 0, theHeaviside unit step is probably unnecessary so we can write

dv(t)

dt+

1

RCv(t) =

V0

RC: (23)

The homogeneous solution is (still) of the form

vh(t) = Ke¡t=RC (24)

where K is some (new) unknown constant. But now the particular solution, due to a

constant forcing term, is itself a constant, call it

vp(t) = D: (25)

This constant D is, again, completely determined (that is known by) the di®erential equa-

tion (24). Insertion of (25) into (23) gives the rather trivial

D

RC=V0

RCor D = V0: (26)

The complete solution is now v(t) = vh(t) + vp(t) or

v(t) = Ke¡t=RC + V0; (27)

whereupon the initial condition (20) gives K = ¡V0 so thatv(t) = V0

£1¡ e¡t=RC¤ (t ¸ 0): (28)

A better (systems engineering style!) way to write this is

v(t) = V0£1¡ e¡t=RC¤u(t): (29)

The unit step not only eliminates the need for the disclaimer t ¸ 0, but its inclusion is

invaluable in emphasizing the causal nature of the response, and it is critically important

if (when!) we di®erentiate the response.

Now we can normalize by V0, as advertised, and obtain the true, non dimensional step

response

s(t) =£1¡ e¡t=RC¤u(t): (30)

The impulse response is the derivative of the step response

h(t) =d

dts(t) =

1

RCe¡t=RCu(t) +

£1¡ e¡t=RC¤±(t)

=1

RCe¡t=RCu(t): (31)

Note that the units of h(t) are [(−)(F)]¡1 = (s)¡1. In this particular instance, the chain-rule di®erentiation that also di®erentiated the u(t) to get a ±(t) might appear super°uous,

since the contribution£1¡ e¡t=RC¤±(t) = £

1¡ e¡t=RC¤¯̄̄t=0

±(t) = 0±(t) = 0 (32)

is zero. But that will not happen in general.

42

Step Response of Another RC Circuit

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..

R

C

+

¡

v(t)+

¡V0u(t) = vg(t)

+ ¡vC(t)

In this circuit, we take our output signal to be the voltage v(t) across the resistor, but the

capacitor voltage vC(t) must also be speci¯cally addressed because of the importance of

continuity of capacitor voltage for determining initial conditions. The zero-state response

demands that there is no initial energy storage in the capacitor so vC(0¡) = 0 prior to the

unit step input turning on at time t = 0. By the continuity of capacitor voltage

vC(0+) = vC(0

¡) = 0: (33)

The application of the Kirchho® current law to the node of potential v(t) gives

v(t)

R+ C

d

dt

£v(t)¡ vg(t)

¤= 0 (34)

ordv(t)

dt+

1

RCv(t) =

dvg(t)

dt: (35)

The forcing term is now the derivative of the voltage generator, which puts us in the

unfamiliar realm of solving a di®erential equation with a wild Dirac-delta forcing term

dv(t)

dt+

1

RCv(t) = V0±(t): (36)

If we stay away from t = 0, then the delta function is zero and we have (note the time

restriction)dv(t)

dt+

1

RCv(t) = 0 (t > 0): (37)

It looks as though we have lost our excitation: The di®erential equation is now homoge-

neous. Information about the excitation, embodied by the source strength V0, is included

in the initial condition. Observe in the circuit that

vg(t) = vC(t) + v(t) (38)

43

and at time t = 0+ this is

vg(0+) = vC(0

+) + v(0+): (39)

The initial condition (33) on the capacitor voltage together with vg(0+) = V0 reveals that

the non-zero initial condition on our output signal v(t) is v(0+) = V0. No particular

solution is required for the complete solution to our homogeneous di®erential equation

(37), which is readily seen to be

v(t) = V0e¡t=RC (t > 0): (40)

Inclusion of the Heaviside unit step function gives

v(t) = V0e¡t=RCu(t) (41)

so that the time zone restriction t > 0 can be removed. The non dimensional step response

is therefore

s(t) = e¡t=RCu(t) (42)

and the corresponding impulse response is

h(t) =d

dts(t) = e¡t=RC±(t)¡ 1

RCe¡t=RCu(t)

= ±(t)¡ 1

RCe¡t=RCu(t) (43)

since

e¡t=RC¯̄̄̄¯t=0

= 1: (44)

Observe that the proper use of chain-rule di®erentiation and explicitly including the u(t)

in the step response s(t) is necessary to get the true impulse response h(t), which most

de¯nitely contains a ±(t) term.

44

Relationship Between the Two RC Circuits

............................

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..................................................................................................................................................................................................................................................................................................

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.......

.......

.......

.......

.......

.......

.......

.......

..

.......

.......

.......

.......

.......

.......

.......

.......

..

............................................

............................................

²

²R

C

+

¡

±(t)

+

¡

ha(t)

............................

............................

............................

............................

..................................................................................................................................................................................................................................................................................................

......................................................................................................

.......

.......

.......

.......

.......

.......

..

.......

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.......

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..

...............................................................................................................................................................

.......

......

.......

.......

........................................................................

........................................................................

..........................................................

......

.......

......

.......

..

²

²C

R

+

¡

±(t)

+

¡

hb(t)

In the lab, simply connect R and C in series and hang the series leg across the output

terminals of a (ideal?) voltage source. Then we see that the output of circuit (a) is the

voltage across C and the output of circuit (b) is the voltage across R. Kirchho®'s voltage

law clearly shows the simple relationship that is displayed by comparing (30) and (42) and

by comparing (31) and (43).

.......

...............................................................................................................................................................................

.........................................................................................................................................................................................................................................................................................................................................................................................

............................................................................................................................................................................................................................

........................................................................

..........................................................

........

......

.......

.......................................................................................................................................................................................................................................................................

+

¡vg(t)

R

+

¡vb(t)

C

+

¡va(t)

vg(t) = va(t) + vb(t)

±(t) = ha(t) + hb(t)

u(t) = sa(t) + sb(t)

Homework Problems

1. Observe that the circuit on page 41 is a special case of the circuit on page 38, if Vc0 = 0

and ! = 0. Therefore, show for yourself that the response (29) is the corresponding special

case of the response (19).

2. Use convolution to ¯nd the zero-state (or driven) response of the RC circuit on page 41

to a rectangular input pulse vg(t) = V0[u(t)¡u(t¡T )]. Graph this response using MATLABfor several numerical values of the important (non dimensional) parameter T=RC.

3. Go backwards, and obtain the two step responses (30) and (42) by integrating the

corresponding impulse responses (31) and (43), since

h(t) =d

dts(t) () s(t) =

tZ¡1

h(¿) d¿:

45

Step and Impulse Response of Series RLC Circuit

...................................................................................................................................................................

..............................................

..............................................

................................................................................................................

.......

.......

.......

.......

.......

.......

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..

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..

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.......

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.......

.......

.......................................................................

........................................................................

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.......

.......

.......

.......

.......

.......

.......

.......

.......

...

²

²

.......

.....................................

........................................................................................................................................................................................

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

................................................................

.............................................................................................................................................

+

¡vg(t)

+ vC(t) ¡L

C

R vR(t)

+

¡

.......................................................................................................................................................................................................................................................................

......................................................................................................................................................................................................

.............................

..............................................................

..........................................................................................................................................................................................................

.................. ................

i(t)

................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

.....

Fig. 1. Circuit layout.

..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

............................................................................................................................. ................ ............................................................................................................................. ................vg(t) = x(t) y(t) = Ri(t)

Fig. 2. System block diagram.

The system step response is a \zero-state" response: No energy is initially stored in the

system prior to the onset of the exciting Heaviside step function that turns on at t = 0.

For de¯niteness and to keep track of the physical units, let

vg(t) = V0u(t) (1)

46

with the (real) constant V0 being the amplitude of the exciting step function. Since the

instantaneous energy stored in the inductor and capacitor at time t are

wL(t) =12Li

2(t) and wC(t) =12Cv

2C(t); (2)

respectively, zero initial (at time t = 0¡) energy storage requires

i(0¡) = 0 and vC(0¡) = 0: (3)

If the inductor voltage

vL(t) = Ldi(t)

dt

is to be bounded, then its current must be a continuous function of time, and hence

i(0+) = i(0¡): (4)

Similarly, if the capacitor current

iC(t) = CdvC(t)

dt

is to be bounded, then its voltage must be a continuous function as in

vC(0+) = vC(0

¡): (5)

The equation-of-motion for the series circuit of Fig. 1 is Kirchho®'s voltage law

Ldi(t)

dt+Ri(t) +

1

C

tZ¡1

i(¿) d¿ = vg(t); (6)

an integrodi®erential equation in the unknown series current i(t). With the forcing function

vg(t) known, the unique solution of equation (6), whose highest derivative term is of degree

one, requires the speci¯cation of one initial condition. The physics of the circuit dictates

i(0+) = 0; (7)

from (3) and (4). If we di®erentiate (6) once to eliminate the integral operator, the resulting

second-order di®erential equation

Ld2i(t)

dt2+R

di(t)

dt+1

Ci(t) = v0g(t) (8)

now requires two initial conditions for a complete (unique) solution. This second initial

condition was already contained in (6), before we di®erentiated. Evaluation of (6) at time

t = 0+ is

Li0(0+) +Ri(0+) + vC(0+) = vg(0+): (9)

The capacitor voltage, from (3) and (5), starts o® at zero

vC(0+) = 0; (10)

so that together with (1) and (7), statement (9) gives

i0(0+) =V0

L: (11)

Note that the dimensions are consistent in that (A/s)= (V/H). X

47

The proper formulation of the above initial value problem, especially stating the correct

number (two) and type of initial conditions (on the current and its derivative), required

our detailed understanding of the electrophysics embodied in the circuit. In one view

the ensuing construction of the solution to the di®erential equation may seem like a pure

exercise in mathematics, but that is not the view we will take. Instead, we want to let each

step in our solution be guided by the physics that the mathematics represents, so that our

mathematics will be e±cient and keep us in touch with the important physics that is our

quest. If we do this right, then our symbols and formulas and solution will be in a form

that not only allows, but even begs for physical interpretation.

Let's restate our initial value problem:

Ld2i(t)

dt2+R

di(t)

dt+1

Ci(t) = v0g(t) (12)

i(0+) = 0 (13)

i0(0+) =V0

L: (14)

Since our voltage generator is vg(t) = V0u(t), the forcing function on the right-hand side

of (12) is a Dirac-delta function, i.e.

Ld2i(t)

dt2+R

di(t)

dt+1

Ci(t) = V0±(t): (15)

That looks troublesome, I believe, but if we stay away from t = 0, then the delta function

is zero and our equation is homogeneous

Ld2i(t)

dt2+R

di(t)

dt+1

Ci(t) = 0 (t > 0): (16)

It looks as though the homogeneous di®erential equation (16) has lost the source, but note

that initial condition (14) has information about the original source in the circuit. If there

is no initial energy storage to drive the \dead" circuit, the excitation of a nonzero response

has to come from an explicit input. And our input (1) is proportional to the constant V0that explicitly appears in the nonzero (\nonhomogeneous") initial condition (14).

Homogeneous constant-coe±cient linear ordinary di®erential equations (LODE's) are par-

ticularly easy to solve, since the solutions are exponentials, say of the form i(t) = exp(rt)

where two such linearly independent forms (i.e. two di®erent r values), are required in our

case of a second-order equation. Before proceeding, it is expedient to condense the notation

and in fact, combine the element values R, L, and C by in a manner that anticipates the

solution. Of course, no one could truly do this in advance if they had not already solved

the problem (or read someone else's solution). But that's not the point: We choose the

best path because it draws attention to what is important and because it is e±cient and

48

because it works. Firstly, normalize (16) so that the coe±cient of the highest derivative is

unityd2i(t)

dt2+R

L

di(t)

dt+

1

LCi(t) = 0

and then introduce constants

® =R

2Land !20 =

1

LC(17)

to writed2i(t)

dt2+ 2®

di(t)

dt+ !20i(t) = 0: (18)

Insertion of i(t) = exp(rt) into (18) gives£r2 + 2®r + !20

¤ert = 0;

and since the exponential cannot equal zero, we arrive at the characteristic polynomial for

the two roots

r2 + 2®r + !20 = 0: (19)

The quadratic formula provides the two roots immediately

r1;2 = ¡®§q®2 ¡ !20 : (20)

Two distinct possibilities exist: The case of an over-damped circuit having ® > !0 and an

under-damped circuit when ® < !0. The case when ® = !0 is called critically-damped, and

can be obtained from either of the other two cases in the limit as ® ! !0. Or, it can be

constructed directly.

Under-damped circuit. If ® < !0, introduce yet another parameter (constant)

¯ = +

q!20 ¡ ®2 (21)

so that the two roots (20) are a complex-conjugate pair

r1;2 = ¡®§ j¯: (22)

The solution to the homogeneous di®erential equation (18) is a linear combination

i(t) = K1er1t +K2e

r2t = K1e¡(®¡j¯)t +K2e

¡(®+j¯)t = e¡®t£K1e

¡j¯t +K2e+j¯t

¤= e¡®t [A cos(¯t) +B sin(¯t)] : (23)

It is critically important to recognize that a linear combination of exp(§jx) is equivalentto a (di®erent) linear combination of cosx and sinx. Although either form is ¯ne, the

49

trig form is better here because it facilitates the application of initial conditions. Initial

condition (13) gives

i(0) = A = 0 (24)

immediately! The derivative of the remaining form

i(t) = Be¡®t sin(¯t) (25)

is

i0(t) = Be¡®t£¯ cos(¯t)¡ ® sin(¯t)¤ (26)

which has initial value

i0(0) = ¯B: (27)

Initial condition (14) is now

¯B =V0

L(28)

and the complete solution to the initial value problem is

i(t) =V0

¯Le¡®t sin(¯t) (t > 0)

or, with attention drawn to its \turned-on" nature,

i(t) =V0

¯Le¡®t sin(¯t)u(t): (29)

Note that the dimensions of ® and ¯ are both (s¡1) and that the dimensions of the product¯L are (H/s= −), so that the current is in (A). The signal that is regarded as the output

in the linear system diagram of Fig. 2 is the voltage vR(t) = Ri(t), so that when the input

vg(t) is regarded as

x(t) = V0u(t); (30)

the output is

y(t) =RV0

¯Le¡®t sin(¯t)u(t): (31)

If we normalize both signals by the amplitude V0, then the nondimensional input/output

signals are

x(t) = u(t) and y(t) = s(t) =R

¯Le¡®t sin(¯t)u(t): (32)

The response to the Heaviside unit step function is called the step response, and is denoted

here as s(t).

The derivative of the convolution

z(t) = f(t)~ g(t) =1Z

¡1f(¿)g(t¡ ¿) d¿ =

1Z¡1

g(¿)f(t¡ ¿) d¿ (33)

50

is

z0(t) = f 0(t)~ g(t) = f(t)~ g0(t): (34)

Therefore, the system impulse response is the derivative of its step response

h(t) = s0(t): (35)

The impulse response of the RLC circuit of Fig. 1 is therefore

h(t) =R

Le¡®t

·cos(¯t)¡ ®

¯sin(¯t)

¸: (36)

Note that the units of h(t) are (s¡1), as are the units of ±(t).

0 2 4 6 8 10 12 14 16 18 20¡0:75

¡0:50

¡0:25

0.00

0.25

0.50

0.75

1.00

¯t

¯L

Rs(t)

.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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..........................................................................................................................................................................................................................................................................................................................................

Fig 2. Normalized step response of underdamped RLC circuit.

Sold curve: ®=¯ = 0:1; Dashed curve: ®=¯ = 0:5

51

Over-damped circuit. If ® > !0, then both roots of the characteristic quadratic poly-

nomial (19) are real, so with

r1 = ¡®1 with ®1 = ®¡q®2 ¡ !20 (37)

and

r2 = ¡®2 with ®2 = ®+

q®2 ¡ !20; (38)

we see that 0 < ®1 < ®2 and the solution to the homogeneous di®erential equation (16)

can be written as the linear combination

i(t) = k1e¡®1t + k2e¡®2t (t > 0): (39)

Initial condition (13) is i(0+) = 0 so that k2 = ¡k1 and therefore

i(t) = k1£e¡®1t ¡ e¡®2t¤: (40)

The required derivative is

i0(t) = k1£®2e

¡®2t ¡ ®1e¡®1t¤

(41)

and at time t = 0+ this must be

i0(0+) = k1£®2 ¡ ®1

¤=V0

L(42)

and therefore

k1 =V0

(®2 ¡ ®1)L: (43)

The solution for the current is now

i(t) =V0

(®2 ¡ ®1)L£e¡®1t ¡ e¡®2t¤u(t) (44)

and the nondimensional step response is

s(t) =R

(®2 ¡ ®1)L£e¡®1t ¡ e¡®2t¤u(t): (45)

52

Homework. Given that a (scaled) rectangular pulse

x(t) =1

T¦

μt

T

¶behaves as the Dirac-delta function in the limit as T ! 0, how small does T have to be so

that the response y(t) of a system to the pulse x(t) is a good approximation to the true

system impulse response h(t) ? Consider the underdamped RLC circuit with parameters ®

and ¯. (Note that specifying ® and ¯ is equivalent to, and in fact more illuminating than,

specifying the circuit element values R, L, and C.) Use MATLAB to compare graphs of

h(t) with y(t) for several values of the pulse-width T . The important parameter(s) of the

problem should be nondimensional.

Homework Exercises. Find the impulse response of these four circuits.

............................

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....................................................................................................................................................................................................

..................................................................

.............................................................................................................................................

.......

......

.......

......

............................................................................................................................................................................

²

²R

L

+

¡

±(t)

+

¡

hc(t)

............................

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.......

.......

.......

.......

..

.......

.......

.......

.......

.......

.......

.......

.......

..

............................................

............................................

²

²R

C

+

¡

±(t)

+

¡

ha(t)

............................

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.......

......

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........................................................................

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.......

.......

......

.......

...

²

²L

R

+

¡

±(t)

+

¡

hd(t)

............................

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.......

.......

.......

.......

.......

.......

..

.......

.......

.......

.......

.......

.......

..

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.......

.......

......

.......

........................................................................

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.......

.......

.......

.......

²

²C

R

+

¡

±(t)

+

¡

hb(t)

What is the relationship between ha(t) and hb(t) ? Between ha(t) and hd(t) ?

Homework. Find the step and impulse responses of the series RLC circuit that is critically

damped: when ® = !0 and the characteristic quadratic polynomaial has a single, repeated

root.

53

Evaluation of an Important Integral via the Laplace Transform

¡0:4¡0:20.0

0.2

0.4

0.6

0.8

1.0

sin(x)

x

¡6¼ ¡4¼ ¡2¼ 0 6¼4¼2¼

x

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...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..................................................................................................................................................................................................................................................................................................................................................................................

..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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...............................................................................................................

sinc(x) , sin(x)

x=1

x

1Xn=0

(¡1)n x2n+1

(2n+ 1)!=1

x

·x¡ x

3

3!+x5

5!¡ : : :

¸=

·1¡ x

2

3!+x4

5!¡ : : :

¸(1)

sinc(0) = limx!0

sin(x)

x= 1 (2)

sinc(¡x) = sinc(x) is even (3)

We want to show that 1Z¡1

sin(x)

xdx = ¼: (4)

Since sinc(x) is even, observe that

1Z¡1

sin(x)

xdx = 2

1Z0

sin(x)

xdx: (5)

54

Our approach is to ¯nd the Laplace transform of the function

Si(t) ,tZ0

sin(x)

xdx; (6)

called the \sine integral." Our desired integral (half of it!) is the ¯nal value of Si(t), that

is Si(1). Perhaps it is helpful to write Si(t) using ¿ as the dummy variable of integration

Si(t) ,tZ0

sin(¿)

¿d¿: (7)

We will use the following Laplace transform properties. If L[f(t)] = F (s), then

L·f(t)

t

¸=

1Zs

F (¸) d¸ (8)

L24 tZ0

f(¿) d¿

35 = F (s)

s(9)

limt!1 f(t) = lim

s!0sF (s) ¯nal value theorem (10)

The Laplace transform of

g(t) = sin(t)u(t) (11)

is

G(s) =1

s2 + 1: (12)

The Laplace transform of

h(t) =g(t)

t= sinc(t)u(t) (13)

is then

H(s) =

1Zs

d¸

¸2 + 1: (14)

This integral can be found in a table (say in your calculus book, or you can use a symbolic

mathematical tool such as MAPLE, or you can use a web-based integrator4). Although

not important for our purposes, let's do it ourselves just for fun.

Consider y = tan(x) so that

dy

dx=d

dx

sin(x)

cos(x)= 1 + tan2(x)

4If you know one, tell the class. If you don't know one, ¯nd one!

55

dy

1 + y2= dxZ

dy

1 + y2=

Zdx = x = tan¡1(y)

1Zs

dy

1 + y2= tan¡1(1)¡ tan¡1(s) = ¼

2¡ tan¡1(s) = tan¡1(1=s)

..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... .......

.......

....................

1

s

μ

¼=2¡ μ

tan¡1(s) + tan¡1(1=s) =¼

2

Therefore the Laplace transform of h(t) = sinc(t)u(t) is

H(s) = tan¡1(1=s): (15)

The Laplace transform of

f(t) = Si(t) =

tZ0

sinc(¿) d¿ (16)

is now

F (s) =tan¡1(1=s)

s(17)

and the ¯nal value theorem gives

Si(1) = f(1) = lims!0

sF (s) = lims!0

tan¡1(1=s) =¼

2: ¥ (18)

56

Proposition (Spectral form of the Dirac-delta). This Fourier integral satis¯es the

de¯nition of the Dirac-delta (distribution or generalized function):

1

2¼

1Z¡1

ej!t d! = ±(t): (1)

Given the integral1Z

¡1

sinx

xdx = ¼; (2)

Dirichlet's discontinuous integral falls out

1Z¡1

sin®x

xdx = ¼ sgn(®) = §¼ (® ? 0): (3)

Integrate the left hand side of (1) from a to b:

bZa

1

2¼

1Z¡1

ej!t d! dt =1

2¼

1Z¡1

bZa

ej!t dt d! =1

2¼

1Z¡1

ejb! ¡ eja!j!

d!

=1

2¼

1Z¡1

·sin b!

!¡ sin a!

!

¸d! =

1

2¼[¼ sgn(b)¡ ¼ sgn(a)] =

½1; a < 0 < b

0; o.w.(4)

This is the sampling property of ±(t) ¤:

57

Fourier Transform

F (!) = F [f(t)] =1Z

¡1f(t)e¡j!t dt F() f(t) = F¡1[F (!)] = 1

2¼

1Z¡1

F (!)ej!t d!

¯̄̄̄¯̄1Z

¡1f(t)e¡j!t dt

¯̄̄̄¯̄ ·

1Z¡1

¯̄f(t)e¡j!t

¯̄dt =

1Z¡1

¯̄f(t)

¯̄dt

Therefore a su±cient (not necessary) condition for the existence of F (!) is the absolute

integrability of f(t)1Z

¡1

¯̄f(t)

¯̄dt <1:

Note that the zero frequency value of the Fourier transform (or spectrum)

F (0) =

1Z¡1

f(t) dt

is the DC or \average value" of the signal f(t).

We can easily demonstrate the legitimacy of the Fourier transform pair by taking the

inverse transform of the forward transform

F¡1nF£f(t)¤o = 1

2¼

1Z¡1

24 1Z¡1

f(¿)e¡j!¿d¿

35| {z }

F (!)

ej!td!

=

1Z¡1

f(¿)1

2¼

1Z¡1

ej!(t¡¿)d!

| {z }±(¿¡t)

d¿

=

1Z¡1

f(¿)±(¿ ¡ t) d¿ = f(t):

58

Fourier Transform Exercises

Find an expression for the Fourier transform of the following time-domain signals, and

sketch BOTH f(t) and F (!):

0. f(t) = e¡atu(t) causal, decaying exponential (a > 0)

1. f(t) = A¦(t=T ) rectangular pulse

2. f(t) = B¤(t=T ) triangular pulse

3. f(t) = C constant

4. f(t) = ±(t¡ t0) delayed Dirac impulse5. f(t) = Ae¡®t

2

Gaussian signal \and it is a trick"

6. f(t) = e¡®jtj

7. f(t) =

½cos

sin

¾(!0t)

8. f(t) = u(t) Heaviside unit-step function (why is this one so hard?)

hint: write u(t) = lima!0

e¡atu(t) and note lima!0

a

a2 + !2= ¼±(!)

9. f(t) = sgn(t) signum (or signature or sign) function

10. f(t) = m(t) cos(!0t) AM modulated signal

i.e. !0=2¼ = 1250 (kHz) and m(t) = \Your Cheatin' Heart"

11. f(t) = sinc(®t)

12. f(t) = ¦(t=T )~ sgn(t)13. f(t) = t¦(t=T )

14. f(t) = D combT (t) = D1P

n=¡1±(t¡ nT ) Woodward's comb function

15. f(t) =1P

k=¡1g(t¡ kT ) periodic replication where F [g(t)] = G(!)

16. f(t) = A1P

k=¡1¦[(t¡ kT )=¿ ] rectangular pulse train

59

Fourier Series

f(t) = a0 +

1Xn=1

[an cosn!0t+ bn sinn!0t] =

1Xn=¡1

cnejn!0t (!0 = 2¼=T )

a0 =1

T

t0+TZt0

f(t) dt

½anbn

¾=2

T

t0+TZt0

f(t)

½cos

sin

¾(n!0t) dt

cn =1

T

t0+TZt0

f(t)e¡jn!0t dt

Fourier Transform

F (!) = F [f(t)] =1Z

¡1f(t)e¡j!t dt F() f(t) = F¡1[F (!)] = 1

2¼

1Z¡1

F (!)ej!t d!

Properties Signal Transform

1. superposition c1f1(t) + c2f2(t) c1F1(!) + c2F2(!)

2. time delay f(t¡ t0) e¡j!t0F (!)

3. time scale f(®t)1

j®jF³!®

´4. duality F (t) 2¼f(¡!)5. frequency translation f(t)ej!0t F (! ¡ !0)6. convolution f(t)~ g(t) F (!)G(!)

7. multiplication f(t)g(t)1

2¼F (!)~G(!)

8. time di®erentiationdf(t)

dtj!F (!)

9. frequency di®erentiation ¡jtf(t) d

d!F (!)

10. integrationtR

¡1f(¿) d¿ ¼F (0)±(!) +

1

j!F (!)

60

1Z¡1

jf(t)j2 dt = 1

2¼

1Z¡1

jF (!)j2 d!

u(t)F() 1

j!+ ¼±(!)

¦(t=¿)F() ¿

sin(!¿=2)

!¿=2= ¿ sinc(!¿=2)

sinc(¯t)F() ¼

¯¦(!=2¯)

combT (t)F() 2¼

Tcomb2¼=T (!)

Z-Transform

Zfx[n]g = X(z) =1X

n=¡1x[n]z¡n

Properties Signal Transform

1. superposition c1x1[n] + c2x2[n] c1X1(z) + c2X2(z)

2. time delay x[n¡m] z¡mX(z)

3. modulation x[n]³n X(z=³)

4. convolution x[n]~ y[n] X(z)Y (z)

5. frequency di®erentiation nx[n] ¡ z ddzX(z)

6. accumulationnP

k=¡1x[k] X(z)

1

1¡ z¡1

61

Lff(t)g = F (s) =1Z0

f(t)e¡st dt is the transform pair f(t)L, F (s)

property t-domain s-domain

linearity ®f(t) + ¯g(t) ®F (s) + ¯G(s)

time delay f(t¡ t0)u(t¡ t0) e¡st0F (s) (t0 ¸ 0)time scaling f(®t) 1

®F (s=®) (® > 0)

exponential modulation e¡®t f(t) F (s+ ®)

time di®erentiation df(t)=dt sF (s)¡ f(0)reprise d2f(t)=dt2 s2F (s)¡ sf(0)¡ f 0(0)time integration

R t0f(¿) d¿ F (s)=s

time integrationR t¡1 f(¿) d¿ F (s)=s+ 1

s

R 0¡1 f(t) dt

s-domain di®erentiation tf(t) ¡dF (s)=dsdivision by t f(t)=t

R1sF (¸) d¸

t-domain convolution f(t)~ g(t) F (s)G(s)

Initial value theorem limt!0

f(t) = lims!1 sF (s)

Final value theorem limt!1 f(t) = lim

s!0sF (s)

Short Table of Laplace Transforms (s = ¾ + j!)

±(t) 1 (8s)u(t)

1

s(¾ > 0)

e¡atu(t)1

s+ a(¾ > ¡Re[a])

te¡atu(t)1

(s+ a)2(¾ > ¡Re[a])

tnu(t)n!

sn+1(¾ > 0)

cos(®t)u(t)s

s2 + ®2(¾ > 0)

sin(®t)u(t)®

s2 + ®2(¾ > 0)

62

Fourier Transform of the Gaussian

If f(t) = exp[¡®t2] then F (!) =

r¼

®exp[¡!2=4®].

This one is a bit involved to show, especially without complex variable theory. But it's

so pervasive in signal analysis and communications that we use it almost daily, so here is

a classical derivation that uses nothing more than integration-by-parts and a simple ¯rst

order di®erential equation.

A preliminary integral

I =

1Z¡1

e¡x2

dx =p¼

is required. The trick5 to derive this is to consider

I2 =

1Z¡1

e¡x2

dx

1Z¡1

e¡y2

dy =

1Z¡1

1Z¡1

e¡(x2+y2)dx dy:

Conversion to polar coordinates via

x = r cosÁ and y = r sinÁ

gives

I2 =

2¼Z0

1Z0

e¡r2

r dr dÁ = (2¼)

μ1

2

¶= ¼ and so I =

p¼:

Our desired Fourier integral is

F (!) =

1Z¡1

e¡®t2

e¡j!tdt = 2Z 1

0

e¡®t2

cos!t dt

since f(¡t) = f(t) is even. Di®erentiation w.r.t. ! gives

dF (!)

d!= ¡2

1Z0

te¡®t2

sin!t dt:

Integration by parts Zudv = uv ¡

Zv du

5\And it is a trick!"

63

with

u = sin!t du = ! cos!t dt

dv = ¡2te¡®t2 v =e¡®t

2

®

givesdF (!)

d!=1

®e¡®t

2

sin!t

¯̄̄̄1t=0

¡ !®

Z 1

0

e¡®t2

cos!t dt

ordF (!)

d!= ¡ !

2®F (!)

dF (!)

d!F (!)

= ¡ !

2®

d

d!lnF (!) = ¡ !

2®

lnF (!) = ¡!2

4®+ C

F (!) = eCe¡!2=4®

eC = F (0) =

1Z¡1

e¡®t2

dt =

r¼

®

F (!) =

r¼

®e¡!

2=4®

solid curves: ® = 1, dashed curves: ® = 2

¡3 ¡2 ¡1 0 1 2 3

t

0.0

0.2

0.4

0.6

0.8

1.0

f(t)

........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

¡8 ¡6 ¡4 ¡2 0 2 4 6 8

!

0.0

0.5

1.0

1.5

2.0

F (!)

.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

......................................................................................................................................................................................................................................................................................................................................

64

Fourier Transform of the Heaviside Unit Step Function

F (!) = F [f(t)] =1Z

¡1f(t)e¡j!t dt F() f(t) = F¡1[F (!)] = 1

2¼

1Z¡1

F (!)ej!t d!

Recall that a su±cient (but not necessary) condition for the Fourier transform F (!) of a

signal f(t) to exist is that the signal be absolutely integrable, that is

1Z¡1

jf(t)j dt <1:

The Heaviside unit step function is not absolutely integrable

1Z¡1

ju(t)j dt =1Z0

dt ¥1

and therefore we might anticipate that its Fourier transform will be both di±cult to eval-

uate and exhibit singular behavior. Recall or note that the periodic signal cos(!0t) is not

absolutely integrable, and its spectrum consists of Dirac-deltas

F [cos(!0t)] =1Z

¡1

ej!0t + e¡j!0t

2e¡j!tdt =

1

2

24 1Z¡1

e¡j(!¡!0)tdt+

1Z¡1

e¡j(!+!0)tdt

35= ¼

£±(! ¡ !0) + ±(! + !0)

¤:

Two preliminary formulae or ideas:

First: lim®!0

®=¼

®2 + !2= ±(!) (1)

This is the Cauchy or Lorentz Dirac-delta sequence (that we already studied back on page

7). Clearly

lim®!0

®=¼

®2 + !2= 0 if !6= 0.

The integral

®

¼

1Z¡1

d!

®2 + !2=1

¼

1Z¡1

dx

1 + x2=1

¼tan¡1(x)

¯̄̄̄1x=¡1

=¼=2¡ (¡¼=2)

¼= 1

gives unity independently of ®, so the truth of (1) is veri¯ed.

65

Second: lim®!0

¡j!®2 + !2

=1

j!(2)

Rewrite it as

lim®!0

1

j![1 + (®=!)2]=1

j!

which is clearly true if ! 6= 0. The only other possibility is that our limit also has a

Dirac-delta at the origin of strength or weight K

lim®!0

1

j![1 + (®=!)2]=1

j!+K±(!):

Integrate both sides of this from ! = ¡² to ! = +²

1

j

²Z¡²

d!

![1 + (®=!)2]=

1

j

²Z¡²

d!

!+ K

²Z¡²±(!) d!

and observe that the integral on the left side and the ¯rst integral on the right side vanish

because of the odd symmetry of their integrands. Therefore the only possibility is K = 0

and the truth of (2) is established.

One of our ¯rst examples (or exercises) of Fourier integral evaluations was the causal,

decaying exponential

F£e¡®tu(t)¤ = 1

®+ j!(® > 0):

The condition that ® > 0 is necessary to ensure convergence of the integral, by making

the time-domain function decay su±ciently fast as t!1. Although the restriction ® > 0enabled us to evaluate the Fourier transform of the decaying exponential, let's try to get

our desired Fourier transform of the plain Heaviside unit-step by taking the limit of the

above result as ®! 0. That is

F [u(t)] = lim®!0

F£e¡®tu(t)¤ = lim®!0

1

®+ j!:

Rewrite1

®+ j!=

®¡ j!(®+ j!)(®¡ j!) =

®

®2 + !2+

¡j!®2 + !2

so that

F [u(t)] = lim®!0

®

®2 + !2+ lim®!0

¡j!®2 + !2

= ¼±(!) +1

j!(F)

from (1) and (2) above.

66

Since

2u(t) = 1 + sgn(t)

we now have the Fourier transform of the signum function as

F [sgn(t)] = 2F [u(t)]¡F [1] = 2·¼±(!) +

1

j!

¸¡ 2¼±(!) = 2

j!:

With

F [u(t)] = ¼±(!) + 1

j!

then by the time-di®erentiation property of the Fourier transform

F·df(t)

dt

¸= j!F [f(t)]

where

±(t) =du(t)

dt;

we have the correct transform

F [±(t)] = j!·¼±(!) +

1

j!

¸= 1:

Using the convolution property of the Fourier transform

F [f(t)~ g(t)] = F (!)G(!)

and the integral identitytZ

¡1f(¿) d¿ = f(t)~ u(t);

we have the Fourier transform of the integral as

F24 tZ¡1

f(¿) d¿

35 = F (!) ·¼±(!) + 1

j!

¸= ¼F (0)±(!) +

F (!)

j!:

67

Duality Property of the Fourier Transform

Recall our notation for the Fourier transform

F (!) = F [f(t)] =1Z

¡1f(t)e¡j!t dt F() f(t) = F¡1[F (!)] = 1

2¼

1Z¡1

F (!)ej!t d!:

For our duality property, we want the Fourier transform of some signal F (t):

F [F (t)] =1Z

¡1F (t)e¡j!t dt: (1)

The fact that our time domain signal is a \big" F of t means that we interpret our big F

as related to some \little" f via a direct Fourier transform

F (x) =

1Z¡1

f(»)e¡jx»d» (2)

and therefore the particular little f that must have spawned our big F must be

f(») =1

2¼

1Z¡1

F (x)ejx»dx: (3)

According to (2), we can also write

F (t) =

1Z¡1

f(»)e¡jt»d» (4)

and insertion of this into (1) yields

F [F (t)] =1Z

¡1

8<:1Z

¡1f(»)e¡jt»d»

9=; e¡j!tdt: (5)

Reverse the orders of the » and t integrations

F [F (t)] =1Z

¡1f(»)

24 1Z¡1

e¡j(»+!)tdt

35| {z }recognize as 2¼±(»+!)

d» (6)

68

and so

F [F (t)] = 2¼1Z

¡1f(»)±(» + !) d» = 2¼f(¡!): (7)

Our desired duality property of the Fourier transform is here F [F (t)] = 2¼f(¡!) wheref(t)

F, F (!) is the notation for the original Fourier transform pair F [f(t)] = F (!).

Example. Since we already know the Fourier transform pair

¦

μt

T

¶F, T sinc(!T=2); (8)

it follows from duality that

F [T sinc(tT=2)] = 2¼¦³¡!T

´(9)

or

F [sinc(T2 t)] =2¼

T¦³!T

´(10)

by the linearity of the Fourier transform operator, and invoking the even symmetry of the

pulse function ¦. Now T is just a remnant parameter from the original pair (8), so let's

introduce a new (better!) constant ® = T=2 such that the above is

F [sinc(®t)] = ¼

®¦³ !2®

´: (11)

Add this entry in your table of Fourier transforms!

Another View. Consider a function ª(x) related to a given function Á(u) through the

direct Fourier transform operator

ª(x) =

1Z¡1

Á(u)e¡juxdu: (12)

Consider a second function Ã(x) related to our given function Á(u) through the inverse

Fourier transform operator

Ã(x) =1

2¼

1Z¡1

Á(u)ejuxdu: (13)

Observe that ª(x) = 2¼Ã(¡x). If we know the forward transform, then we also know thereverse transform of any function, by inspection.

69

Half-Power Bandwidth or Pulse width

..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

x

f1(x)

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x

f2(x)

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f3(x)

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Consider the three functions (or signals) graphed above. The function f1(x) has a clear

maximum and a \function width" or \pulse width" about the maximum where most of its

energy is concentrated. The same observation applies to the signal f2(x), but its \pulse

width" is perhaps not as obvious. The oscillatory and somewhat wild signal f3(x) possesses

no obvious maximum that we see in its graph and it is unclear how to de¯ne any kind of

a pulse width for it.

Often our time-domain signals are one of the \linear" signals in an electric circuit, such as

a voltage v(t) or a current i(t). In a mechanical system, the displacements x(t) or velocities

v(t) are examples of one of the so-called \linear" signals that we might be interested in.

The instantaneous power associated with one of the linear signals f(t) is proportional to

jf(t)j2. If the signal f(t) does possess a clear maximum and most of its energy or power6

6This is an example of hazy (perhaps lazy, too) language. Power p(t) (in Watts) is not equal to energyE(t) (in Joules) but we are approximately thinking that p(t) / E(t) / jf(t)j2.

70

is concentrated around the maximum (such as the functions f1 and f2 above), then we

de¯ne the half-power pulse width as

T1=2 = t2 ¡ t1where

jf(t1;2)j2jfmaxj2 =

1

2or

jf(t1;2)jjfmaxj =

1p2:

When the signal magnitude is given or graphed in units of decibels (dB), the half-power

points t1;2 are also called the \3 (db)" points since

10 log10(1=2) = 20 log10(1=p2) ¼ ¡3:

The half-power pulse width is also then called the \3 (dB) pulse width." This measure of

\signal duration" applies equally well in the frequency domain, where the term bandwidth

is commonly used instead of pulse width. The symbol for bandwidth is typically BW in

communications work, but a simple B1=2 seems ¯ne.

Example: Gaussian pulse. Note in this case that jf(t)j = f(t) and jF (!)j = F (!).

This will not be true in general.

......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

t

f(t) = exp(¡®t2)

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......

......

......

t1 t2

jf jmaxjf jmaxp

2

T1=2............................................... .............. .............................................................

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!

F (!) =

r¼

®exp(¡!2=4®)

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!1 !2

jF jmaxjF jmaxp

2

B1=2.......................................................................... ............................................................ ..............

........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

exp(¡®t22)1

=1p2

exp(¡!22=4®)1

=1p2

t2 =

rln 2

2®!2 =

p2® ln 2

T1=2 = t2 ¡ t1 = 2t2 =r2 ln 2

®B1=2 = !2 ¡ !1 = 2!2 =

p8® ln 2

The \BT product" B1=2T1=2 = 4 ln 2 is independent of ®.

71

Example: The sinc function. Examine the two graphs of j sinc(x)j = j sin(x)=xj, theone on the top using a linear ordinate scale and the one on the bottom using a logarithmic

(dB) ordinate scale. The abscissa is x, which can serve as a scaled frequency ! for the

spectrum of a time-domain rectangular pulse.

¡10 ¡8 ¡6 ¡4 ¡2 0 2 4 6 8 100.0

0.2

0.4

0.6

0.8

1.0

x

jsinc(x)j

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..

1=p2

x1

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¡10 ¡8 ¡6 ¡4 ¡2 0 2 4 6 8 10¡30

¡25

¡20

¡15

¡10

¡5

0

x

jsinc(x)j(dB)

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¡3

x1

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sin(x1)

x1=

1p2

=) x1 =p2 sin(x1) =) x1 = 1:391557378251510

If f(t) = ¦(t=T ) is the standard rectangular pulse of pulse width T , then since the spectrum

is F (!) = T sinc(!T=2), we see that the half-power bandwidth is

B1=2 = 2!1 ¼ 2£ 2£ 1:39T

¼ 5:56

T:

72

Homework

1. Find the pulse width, bandwidth, and BT product of these two signals:

f(t) = exp(¡®jtj) and g(t) = exp(¡®t)u(t):Graph jf(t)j and its amplitude spectrum jF (!)j, and also jg(t)j and jG(!)j. Use at leasttwo values for the constant ®.

2. Compare the half-power pulse width T1=2 described above with the (statistical class)

of pulse width Dt that is used in the mathematical analysis of the Uncertainty Principle

that appears on page 74. If the signal is normalized to have unit energy

1Z¡1

jf(t)j2 dt = 1

then the (square of) pulse width is

D2t =

1Z¡1

t2jf(t)j2 dt:

In other words, simply take Dt as de¯ned here, to be yet another measure of a signal's

pulse width. For a speci¯c signal, use the Gaussian pulse.

73

Uncertainty Principle for the Fourier Integral

To simplify our notation, assume that the signals under consideration are normalized to

have unit energy:1Z

¡1jf(t)j2 dt = 1

2¼

1Z¡1

jF (!)j2 d! = 1:

If we further assume that both f(t) and its Fourier transform F (!) are centered about

t = 0 and ! = 0, respectively, then it is reasonable to de¯ne the t and !-domain durations

by

D2t =

1Z¡1

t2jf(t)j2 dt and D2! =

1Z¡1

!2jF (!)j2 d!:

Our so-called \uncertainty principle" states:

If limt!§1

pjtjf(t) = 0; Then DtD! ¸

r¼

2:

prf: Equality holds in Schwarz's inequality¯̄̄̄¯̄bZa

g1g2 dt

¯̄̄̄¯̄2

·bZa

jg1j2 dtbZa

jg2j2 dt

only if g2(t) = kg1(t). Let g1(t) = tf(t) and g2(t) = f0(t):¯̄̄̄

¯̄1Z

¡1tf(t)

df(t)

dtdt

¯̄̄̄¯̄2

·1Z

¡1jtf(t)j2 dt

1Z¡1

¯̄̄̄df(t)

dt

¯̄̄̄2dt:

Integrate by parts

1Z¡1

tfdf

dtdt =

tf2(t)

2

¯̄̄̄1¡1| {z }

=0 by above

¡12

1Z¡1

jf j2 dt| {z }=1 by above

= ¡12:

From the Fourier transform pair f 0(t)() j!F (!) and Parseval's theorem,

1Z¡1

¯̄̄̄df

dt

¯̄̄̄2dt =

1

2¼

1Z¡1

j!F (!)j2 d!

and so Schwarz's inequality gives

1

4·

1Z¡1

jtf(t)j2 dt 12¼

1Z¡1

j!F (!)j2 d!

74

or¼

2· D2

tD2!: ¥

Equality holds ifdf(t)

dt= ktf(t):

The solution of this di®erential equation is

f(t) = Cekt2=2:

Let k = ¡2®; this is the inescapable Gaussian function

f(t) =

μ2®

¼

¶1=4e¡®t

2 () F (!) =

μ2®

¼

¶1=4r¼

®e¡!

2=4® =

μ2¼

®

¶1=4e¡!

2=4®:

75

Theorem (Riemann-Lebesgue Lemma). Let f(x) 2 L1(¡1;1). That is, f(x) isabsolutely integrable

1Z¡1

jf(x)j dx <1:

Then the integrals1Z

¡1f(x) cos¸xdx;

1Z¡1

f(x) sin¸xdx;

tend to zero as ¸!1.

Proof. Consider the cosine integral. Let ² be a given positive number. Then we can choose

X so large that1ZX

jf(x)j dx < ²;¡XZ¡1

jf(x)j dx < ²:

Hence ¯̄̄̄¯̄1ZX

f(x) cos¸xdx

¯̄̄̄¯̄ < ²;

¯̄̄̄¯̄¡XZ¡1

f(x) cos¸xdx

¯̄̄̄¯̄ < ²

for all values of ¸. Next, we can de¯ne a function Á(x), absolutely continuous in the

interval (¡X;X), such thatXZ

¡Xjf(x)¡ Á(x)j dx < ²:

Then ¯̄̄̄¯̄XZ

¡Xff(x)¡ Á(x)g cos¸xdx

¯̄̄̄¯̄ < ²

for all values of ¸. Finally

XZ¡X

Á(x) cos¸xdx =Á(X) sin¸X

¸+Á(¡X) sin¸X

¸¡ 1

¸

XZ¡X

Á0(x) sin¸xdx;

and (for a ¯xed X) we can choose ¸0 so large that the modulus of this is less than ² for

¸ > ¸0. Then ¯̄̄̄¯̄1Z

¡1f(x) cos¸xdx

¯̄̄̄¯̄ < 4² (¸ > ¸0):

This proves the theorem for the cosine integral; a similar proof applies to the sine integral.

¥

76

The above is the working-class proof of the Riemann-Lebesgue lemma, as given by E.C.

Titchmarsh in Introduction to the Theory of Fourier Integrals, 1937. Perhaps you will

prefer this one, from R.R. Goldberg, Fourier Transforms, Cambridge University Press,

1962.

Definition (the Class Lp). Suppose 1 · p < 1. The function f on (¡1;1) is saidto be of class Lp (written f 2 Lp) if

1R¡1

jf(x)jp dx <1. If f 2 Lp then kfkp is de¯ned tobe 0@ 1Z

¡1jf(x)jp dx

1A1=p

:

The symbol kfkp is read as the Lp norm of f .

Theorem (Riemann-Lebesgue). If f 2 L1 then

lim!!§1F (!) = lim

!!§1

1Z¡1

e¡j!tf(t) dt = 0:

Proof. Since

F (!) =

1Z¡1

e¡j!tf(t) dt; (1)

then

¡F (!) =1Z

¡1e¡j![t+(¼=!)]f(t) dt =

1Z¡1

e¡j!tf(t¡ ¼=!) dt: (2)

Subtracting (2) from (1) we obtain

2F (!) =

1Z¡1

e¡j!t [f(t)¡ f(t¡ ¼=!)] dt:

Hence

2jF (!)j ·1Z

¡1jf(t)¡ f(t¡ ¼=!)j dt: (3)

But since f 2 L1,

lim!!§1

1Z¡1

jf(t)¡ f(t¡ ¼=!)j dt = 0 (4)

by the continuity-in-the-mean theorem. The theorem follows from (3) and (4). ¥

77

Riemann-Lebesgue Lemma

If f 0(t) is absolutely integrable, then F (!)! 0 as ! ! §1.

PROOF: Integrate by parts

F (!) =

1Z¡1

f(t)e¡j!t dt =f(t)e¡j!t

¡j!¯̄̄̄1t=¡1

+1

j!

1Z¡1

f 0(t)e¡j!tdt

If f(§1) is ¯nite, then the boundary terms go to zero as ! ! §1. The integral is theFourier transform of the derivative of f(t)

1Z¡1

f 0(t)e¡j!tdt = F [f 0(t)]

and this Fourier integral exists (has ¯nite magnitude jF [f 0(t)] j < 1) if the derivative isabsolutely integrable

1Z¡1

jf 0(t)j dt · A <1:

If jF [f 0(t)] j <1, then the integral term

F [f 0(t)]j!

¡¡¡¡¡!!!§1

0:

Exercise. Examine the high frequency limiting behavior (! ! §1) of the spectra ofthese signal pairs and interpret in terms of the Riemann-Lebesgue Lemma:

u(t)F() 1

j!+ ¼±(!)

¦(t=T )F() T sinc(!T=2)

sinc(¯t)F() ¼

¯¦(!=2¯)

e¡®tu(t) F() 1

®+ j!

±(t)F() 1

1

t

F() ¼

jsgn(!)

78

Convergence of the Fourier Integral Representation

F (!) = F [f(t)] =1Z

¡1f(t)e¡j!t dt F() f(t) = F¡1[F (!)] = 1

2¼

1Z¡1

F (!)ej!t d!

Assume f 0(t) is absolutely integrable so that the Riemann-Lebesgue lemma guarantees

lim!!1F (!) = 0:

Denote the inverse Fourier transform of the Fourier transform as

ef(t) = lim−!1

1

2¼

−Z¡−

1Z¡1

f(¿)e¡j!¿d¿ ej!t d!

= lim−!1

1Z¡1

f(¿)1

2¼

−Z¡−

ej!(t¡¿)d! d¿

= lim−!1

1Z¡1

f(¿)sin[−(¿ ¡ t)]¼(¿ ¡ t) d¿

= lim−!1

1Z¡1

f(x+ t)sin−x

¼xdx:

If f is continuous at t, then clearly ef(t) = f(t) if we accept the Dirac-delta representationlim−!1

sin−x

¼x= ±(x):

If f su®ers a jump discontinuity at t, in that

limx!0¡

f(t+ x) = f(t¡)6= f(t+) = limx!0+

f(t+ x);

then it is appropriate to break the integral above at x = 0 so that

ef(t) = lim−!1

0Z¡1

f(x+ t)sin−x

¼xdx+ lim

−!1

1Z0

f(x+ t)sin−x

¼xdx

= lim−!1

0Z¡1

f(x+ t)¡ f(t¡)¼x

sin−xdx+ lim−!1

f(t¡)

0Z¡1

sin−x

¼xdx

+ lim−!1

1Z0

f(x+ t)¡ f(t+)¼x

sin−xdx+ lim−!1

f(t+)

1Z0

sin−x

¼xdx

79

The functionf(x+ t)¡ f(t¡)

¼x

is continuous at x = 0¡, and is well-behaved everywhere else by our initial assumption.(We are assuming that f is continuous everywhere except at the particular t of interest. If

f actually has a ¯nite number of such jump discontinuities, then each one can be handled

in a similar fashion.) By the Riemann-Lebesgue lemma, the ¯rst term on the right-hand

side of the above integral vanishes as −!1. The same argument applies to the function

f(x+ t)¡ f(t+)¼x

and its Fourier integral. From the famous integral

1Z¡1

sinu

udu = ¼

and its variation 1Z¡1

sin¯u

udu = ¼ sgn¯;

then for − > 0 we have0Z

¡1

sin−x

¼xdx =

1Z0

sin−x

¼xdx = 1

2 :

Therefore, we have shown

F¡1 fF [f(t)]g = ef(t) = 12

£f(t¡) + f(t+)

¤:

80

System Transfer Function

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................................................... ............. ...................................................................................... .............x(t) y(t)h(t)

The zero-state response (in the time domain) of our linear, time-invariant (LTI) system is

the convolution of the input signal with the system impulse response

y(t) = x(t)~ h(t): (1)

The direct Fourier transform of this equation is

Y (!) = X(!)H(!) (2)

which exhibits the important role of the system transfer function

Ffh(t)g = H(!): (3)

By transforming to the frequency domain, the nontrivial convolution operator has been

replaced by a simple multiplication! Besides algebraic simplicity, working in the frequency

domain also gives us a whole another conceptual view of the input/output dynamics of our

systems. Electrical engineers work so much in the frequency domain that we sometimes

loose touch with the original time domain!

For an arbitrary input signal x(t), the time-domain response y(t) now requires an inverse

Fourier transform

y(t) = F¡1fX(!)H(!)g: (4)

However, the transfer function H(!) immediately gives the response of a LTI system to a

time-harmonic signal. The simplest time-harmonic input signal is the complex exponential

form

x(t) = ej!0t: (5)

(The only other choice would be a real-valued cosine or sine, but as we have seen repeatedly

this semester, the mathematics of signal analysis usually (always?) is much simpler in terms

of the complex exponential form. And in deriving important concepts and principles, we

always take the simpler path.) Note the use of !0 to denote some particular frequency, to

avoid confusion with the Fourier transform frequency variable plain !. The response of

our system to the excitation (5) is

y(t) = h(t)~ x(t) =1Z

¡1h(¿)x(t¡ ¿) d¿ =

1Z¡1

h(¿)ej!0(t¡¿) d¿

= ej!0t1Z

¡1h(¿)e¡j!0¿ d¿ = ej!0tH(!0): (6)

81

Observe that the response (steady-state since the excitation has been on forever) of a

LTI system to a time-harmonic signal is a time-harmonic signal of the same frequency as

the excitation. Only the amplitude and phase of the response have been modi¯ed by the

amplitude and phase of the system transfer function, evaluated (of course) at the excitation

frequency. The amplitude and phase are apparent in the polar representation

H(!) = jH(!)jej£(!): (7)

Note that if the system impulse response h(t) is a real-valued function of time, then we

can recover the separate responses to the real and imaginary parts of a complex input by

inspection:

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................................................... ............. ...................................................................................... .............x(t) = xr(t) + jxi(t) y(t) = xr(t)~ h(t) + jxi(t)~ h(t)h(t)

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................................................... ............. ...................................................................................... .............Refx(t)g = xr(t) yr(t) = Refy(t)gh(t) real

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................................................... ............. ...................................................................................... .............Imfx(t)g = xi(t) yi(t) = Imfy(t)gh(t) real

F Exercise 1. Show that the time-harmonic response of a real LTI system to a cosine

excitation x(t) = A cos(!0t+ Á) of arbitrary amplitude A and phase Á is

y(t) = AjH(!0)j cos[!0t+ Á+£(!0)]where the amplitude and phase of the transfer function are de¯ned in (7).

Hint: write x(t) = RefAejÁej!0tg. In AC circuit analysis, we call AejÁ the complex phasor.

F Exercise 2. Specialize the above result to the case where x(t) = sin(!0t):

F Exercise 3. Now consider the periodic excitation

x(t) =

1Xn=¡1

cnejn!0t

applied to a LTI system having arbitrary (not necessarily real) impulse response h(t). The

transfer function is the generic H(!). What is the form of the response? Think about the

linearity of the system.

82

Superposition Interpretation of System Transfer Function

.......

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...........................................................................................................................................................................................................................

............................................................................................................................................................... ..............

......................................................................................................................................................................................................................................................................................................................................................................................................

............................................................................................................................................................................................................. ..............

§

H(!)

H(!)

H(!)

H(!)

H(!)

x(t) =Pn

X(!n)ej!nt ....................................................................................... ..........

....Pn

H(!n)X(!n)ej!nt = y(t)

X(!2)ej!2t H(!2)X(!2)e

j!2t

X(!1)ej!1t H(!1)X(!1)e

j!1t

X(!n)ej!nt H(!n)X(!n)e

j!nt

........................................................................................................................................................................................................................................................................................................................................

......................................................................................................... .............. ......................................................................................................... .........

.....H(!)x(t) = 12¼

1Z¡1

X(!)ej!td! 12¼

1Z¡1

H(!)X(!)ej!td! = y(t)

83

Transfer Function for Linear Electric Circuits

Example.

............................

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....................................................................................................................................................................................................

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.............................................................................................................................................

.......

......

.......

......

............................................................................................................................................................................

²

²R

L

+

¡

x(t)

+

¡

y(t)

Recall (look back in your notes; no one memorizes such speci¯cs) that the impulse response

of the given RL circuit is

h(t) = ±(t)¡ RLexp

μ¡RLt

¶u(t): (8)

Most of us would agree that the correct derivation of this time-domain result is a nontrivial

exercise. We ¯rst found the step response s(t) and then di®erentiated to get h(t) = s0(t).Fortunately, the Fourier transform allows us to circumvent the time-domain di®erential

equations and work with purely algebraic equations. Temporarily set the input x(t) = vg(t)

and the output y(t) = vL(t) for obvious dimensional consistency and to use signal labels

for voltages that look like voltages. At least this helps my circuit analysis.

............................

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....................................................................................................................................................................................................

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......

.......

......

............................................................................................................................................................................

²

²R

L

+

¡

vL(t)

..................................................................................................................................................................................................................................................................................................

.......

.......

.......

.......

.......

.......

..................................................................

+

¡vg(t)

................................................ ...................

i(t)

Kirchho®'s voltage law is

vg(t) = Ri(t) + Ldi(t)

dt(9)

and the direct Fourier transform of this is

Vg(!) = [R+ j!L]I(!): (10)

Note the natural occurrence of the familiar complex impedance. The ratio of output to

input voltage spectraVL(!)

Vg(!)=

j!L

R+ j!L=

ZL

ZR + ZL(11)

could have been written by inspection as a simple voltage divider. Our transfer function is

H(!) =Y (!)

X(!)=

j!L

R+ j!L=

j!L=R

1 + j!L=R=

j!=!b

1 + j!=!b(12)

where the constant !b = R=L (s¡1) is called the break frequency or half-power frequency

or maybe even cut-o® frequency of this simple high-pass ¯lter.

F Exercise 4. Verify that the Fourier transform of impulse response (8) gives transfer

function (12).

84

¡40

¡35

¡30

¡25

¡20

¡15

¡10

¡5

0

10¡2 10¡1 100 101 102

!=!b

jHj (dB)

.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.........................................................................................................................................................................................................................................................................................................................................

(a) Magnitude response.

10¡2 10¡1 100 101 102

!=!b

argH

0

¼=8

¼=4

3¼=8

¼=2 ............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

(b) Phase response.

Fig. 1. Frequency behavior of single-pole high-pass ¯lter.

85

The magnitude and phase response of the transfer function (12)

H(!) =j!=!b

1 + j!=!b(13)

are graphed versus a logarithmic and normalized frequency abscissa in Fig. 1. We also use

a logarithmic (unnatural or base 10) scale to express the magnitude in decibels according

to

jHj (dB) = 20 log10 jHj: (14)

Note that the low and high frequency approximations

H(!) =j!=!b

1 + j!=!b=

8>><>>:j!=!b; ! ¿ !b

j

1 + j; ! = !b

1; ! À !b

(15)

clearly show that this ¯lter passes the high frequencies unscathed. In the low frequency

regime (! ¿ !b) where this ¯lter has the greatest e®ect between the input and output,

the frequency-domain relationship is

Y (!) ¼ j!X(!)

!b: (16)

Inverse Fourier transformation of (16) to the time domain gives

y(t) ¼ 1

!b

dx(t)

dt(17)

so we see that a high-pass ¯lter acts as a di®erentiator when it is having its biggest e®ect

on the input.

F Exercise 5. Consider the simple circuit having the positions of R and L reversed

relative to the circuit above. Find its transfer function and show that this low-pass ¯lter

acts as an integrator when it is having its greatest e®ect on the input signal.

Homework Problem: Rectangular Pulse Response of Ideal Low-Pass Filter.

Consider an ideal low-pass ¯lter of bandwidth B and subsequent transfer function

H(!) = ¦³ !2B

´:

Find the transient response y(t) of this ¯lter to a rectangular pulse of pulse-width T

x(t) = ¦(t=T ):

Graphical results simplify with normalized time t=T as the independent variable, for various

values of the dimensionless parameter BT . This problem is most readily solved in terms

of a special function, the sine integral7

Si(u) =

uZ0

sin »

»d» Si(¡u) = ¡Si(u) Si(1) = ¼=2:

7M. Abramowitz and I.A. Stegun, Handbook of Mathematical Functions. National Bureau of Standards,1970, pp. 231{232, Table 5.1. (\Big Red"or \Citizen's Handbook"). MATLAB has it built-in, too!

86

Parseval's Theorem for the Fourier Integral

The energy associated with the signal f(t) is

E =1Z

¡1jf(t)j2 dt =

1Z¡1

f(t)f¤(t) dt

=

1Z¡1

24 12¼

1Z¡1

F (!)ej!td!

3524 12¼

1Z¡1

F (−)ej−td−

35¤ dt=

1Z¡1

24 12¼

1Z¡1

F (!)ej!td!

3524 12¼

1Z¡1

F ¤(−)e¡j−td−

35 dt=1

2¼

1Z¡1

F (!)

1Z¡1

F ¤(−)

8<: 1

2¼

1Z¡1

ej(!¡−)tdt

9=; d− d!

=1

2¼

1Z¡1

F (!)

1Z¡1

F ¤(−)±(−¡ !) d− d! = 1

2¼

1Z¡1

F (!)F ¤(!) d!

=1

2¼

1Z¡1

jF (!)j2 d!

F Exercise 6. Derive the generalized Parseval theorem

1Z¡1

f(t)g¤(t) dt =

F Exercise 7. Find a simple evaluation for the integrals

1Z¡1

sinx

xdx

1Z¡1

·sinx

x

¸2dx:

We met the ¯rst one back on page 54. The similarity (more than similar! ) in the numerical

values strikes me as a strange coincidence. What do you think? (You'll need the answers,

¯rst!)

87

Other Conventions for the Fourier Transform Pair

Note that our forward Fourier transform could be scaled by any constant K as long as we

divide by K in performing the inverse:

F (!) = K

1Z¡1

f(t)e¡j!tdt F() f(t) =1

2¼K

1Z¡1

F (!)ej!td!:

The common choice K = 1=p2¼ gives a pleasingly symmetric pair

F (!) =1p2¼

1Z¡1

f(t)e¡j!tdt F() f(t) =1p2¼

1Z¡1

F (!)ej!td!:

One-sided (unilateral) trigonometric forms

Fc(!) =

1Z0

f(t) cos(!t) dtF() f(t) =

2

¼

1Z0

Fc(!) cos(!t) d!

Fs(!) =

1Z0

f(t) sin(!t) dtF() f(t) =

2

¼

1Z0

Fs(!) sin(!t) d!

Note that the sine transform is a natural choice if f(0) = 0, while the cosine transform ¯ts

the case when f 0(0) = 0.

The use of frequency f (Hz=cycle/s) instead of ! (1/s)

Introduce the frequency f (Hz) such that ! = 2¼f , and our usual Fourier transform pair

X(!) =

1Z¡1

x(t)e¡j!tdt F() x(t) =1

2¼

1Z¡1

X(!)ej!td!

becomes

X(f) =

1Z¡1

x(t)e¡j2¼ftdt F() x(t) =

1Z¡1

X(f)ej2¼ftdf:

This form has the advantage that both variables t and f appear symmetrically without

any multiplying factors outside the integrals. Its disadvantage (to me) is the appearance

of a 2¼ in every kernel function exp(§j2¼ft). Its supporters are often experimentalists,because of their natural preference to measure frequency f in (Hz), as opposed to the

natural frequency !. The use of f can also pave the way for a slightly smoother transition

to the discrete-time world. Either way you look at it, the Fourier transform pair has

to have 2¼'s somewhere: The question is, do you place them as harmless scaling factors

outside the integrals, or do you turn them loose as pesky irritants inside the arguments

of the complex exponentials, so that every di®erential and integral operation results in a

factor of 2¼ or (2¼)¡1 ? Seriously, the world is divided and you will inevitably need to

translate between the two forms, but not in our course, where, at least for now, ! rules!

88

ECE 370 QUIZ 7 9 October 1940 Name IgÄor Reiruμof, III

Take-Home: due next class, Wednesday October 31 12:00 noon

Consider a real-valued function f(t) that possesses a de¯nite Fourier transform F (!).

Assume that f(t) is absolutely integrable so that we are assured of the ¯niteness of F (!)

for all frequencies.

1.) If f(t) = f(¡t) is even, what can you deduce about the symmetry (even/odd parity)of F (!)? Is F (!) pure real, pure imaginary, or arbitrarily complex?

2.) If f(t) = ¡f(¡t) is odd, what can you deduce about the symmetry (even/odd parity)of F (!)? Is F (!) pure real, pure imaginary, or arbitrarily complex?

3.) What is the physical (geometric) signi¯cance of F (0)?

F (!) =

1Z¡1

f(t)e¡j!tdt

************************** SOLUTION **************************

F (!) =

1Z¡1

f(t) cos(!t) dt¡ j1Z

¡1f(t) sin(!t) dt = Fr(!)¡ jFi(!)

If f(t) = f(¡t) is even, then Fi(!) = 0 and Fr(!) = 21Z0

f(t) cos(!t) dt = Fr(¡!):

That is, the spectrum of an even, real signal is pure real and is an even function of !.

If f(t) = ¡f(¡t) is odd, then Fr(!) = 0 and Fi(!) = 21Z0

f(t) sin(!t) dt = ¡Fi(¡!):

That is, the spectrum of an odd, real signal is pure imaginary and is an odd function of !.

F (0) =

1Z¡1

f(t) dt is the integral of f(t); the DC value, total area under the curve.

89

Cascade of Two LTI Systems

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................................................... ............. ...................................................................................... .............x(t) y(t)h1(t) h2(t)

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................................................... .............

h(t) = h1(t)~ h2(t)

H(!) = H1(!)H2(!)

Linear Dispersionless Filter

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................................................... ............. ...................................................................................... .............x(t) y(t) = x(t¡ t0)h(t) = ±(t¡ t0)

H(!) = e¡j!t0 \linear phase"

90

Homework Problem: Rectangular Pulse Response of Ideal Low-Pass Filter.

Consider an ideal low-pass ¯lter of bandwidth B and subsequent transfer function

H(!) = ¦³ !2B

´: (1)

Find the transient response y(t) of this ¯lter to a rectangular pulse of pulse-width T

x(t) = ¦(t=T ): (2)

Graphical results simplify with normalized time t=T as the independent variable, for various

values of the dimensionless parameter BT . This problem is most readily solved in terms

of a special function, the sine integral8

Si(z) =

zZ0

sin »

»d» Si(¡z) = ¡Si(z) Si(1) = ¼=2: (3)

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................................................... ............. ...................................................................................... .............x(t) y(t)h(t)

The ¯lter impulse response is

h(t) = F¡1[H(!)] = B

¼sinc(Bt) (4)

which can be found from your table of Fourier transforms or by direct evaluation of the

inverse transform integral. Similarly, the spectrum of the input signal is

X(!) = F [x(t)] = T sinc(!T=2): (5)

At least three di®erent, but related, approaches to ¯nding the output y(t) are worth

considering.

Time-Domain Convolution.

y(t) = x(t)~ h(t) =1Z

¡1x(¿)h(t¡ ¿) d¿ =

T=2Z¡T=2

h(t¡ ¿) d¿ (6)

=B

¼

T=2Z¡T=2

sin[B(t¡ ¿)]B(t¡ ¿) d¿ =

B

¼

T=2Z¡T=2

sin[B(¿ ¡ t)]B(¿ ¡ t) d¿ =

1

¼

B(T=2¡t)ZB(¡T=2¡t)

sin »

»d»

(7)

8M. Abramowitz and I.A. Stegun, Handbook of Mathematical Functions. National Bureau of Standards,1970, pp. 231{232, Table 5.1. (\Big Red"or \Citizen's Handbook"). MATLAB has it built-in, too!

91

y(t) =1

¼

©Si[B(T=2¡ t)]¡ Si[¡B(T=2 + t)]ª = 1

¼

©Si[B(t+ T=2)]¡ Si[B(t¡ T=2)]ª

(8)

=1

¼

½Si

·BT

μt

T+1

2

¶¸¡ Si

·BT

μt

T¡ 12

¶¸¾(9)

Note that y(t ! 1) = 0. Also note that time appears normalized to the incident pulse

width T as in t=T , and that the single (dimensionless!) parameter that characterizes the

¯lter response to the rectangular pulse is the product BT .

Frequency-Domain Multiplication.

y(t) = F¡1[X(!)H(!)] = 1

2¼

1Z¡1

X(!)H(!)ej!td! =1

2¼

BZ¡B

T sinc(!T=2)ej!td!

(10)

=1

¼

BZ¡B

sin(!T=2)

!ej!td! only the even part of ej!t contributes (11)

=1

¼

BZ¡B

sin(!T=2)

!cos(!t) d! =

2

¼

BZ0

sin(!T=2)

!cos(!t) d! (12)

Use the trig identity 2 sinX cosY = sin(X ¡ Y ) + sin(X + Y ) so that

y(t) =1

¼

BZ0

©sin[!(T=2¡ t)] + sin[!(T=2 + t)]ª d!

!: (13)

Let » = !(T=2¨ t) so thatBZ0

sin[!(T=2¨ t)] d!!=

B(T=2¨t)Z0

sin »

»d» = Si[B(T=2¨ t)] (14)

whereupon (13) is

y(t) =1

¼

©Si[B(T=2¡ t)] + Si[B(T=2 + t)]ª

=1

¼

©Si[B(t+ T=2)]¡ Si[B(t¡ T=2)]: (15)

92

Linear Combination of Step-Responses. Recognize that our particular input can be

written as

x(t) = ¦(t=T ) = u(t+ T=2)¡ u(t¡ T=2) (16)

so that the response is

y(t) = s(t+ T=2)¡ s(t¡ T=2) (17)

where the step response is

s(t) =

tZ¡1

h(¿) d¿ =B

¼

tZ¡1

sinc(B¿) d¿ =1

¼

BtZ¡1

sin »

»d» (18)

=1

¼

24 0Z¡1

sin »

»d» +

BtZ0

sin »

»d»

35 = 1

¼

h¼2+ Si(Bt)

i=1

2+1

¼Si(Bt): (19)

The linear combination (17) of shifted step responses now yields

y(t) =1

¼

©Si[B(t+ T=2)]¡ Si[B(t¡ T=2)]: (20)

The MATLAB routine sinint.m calls the symbolic toolbox, built upon MAPLE, and does

an extremely slow and ine±cient numerical integration to compute the function Si(x).

It doesn't work at all on my computer, so here is an alternate computation that uses

MATLAB's routine expint.m for the exponential integral, a di®erent but related special

function.

E1(z) =

1Zz

e¡t

tdt

Si(x) =¼

2+ ImfE1(jx)g

The routine for E1(jx) will not work for (real) values of x · 0, but that's ok since we knowthat Si(0) = 0 and it is an odd function Si(¡x) = ¡Si(x). Therefore Si(x) = Si(jxj) sgn(x).

function f=sinintRWS(x)

% f=sinintRWS(x) sine integral Si(x) for real x

f=zeros(size(x));

k=find(x»=0);f(k)=(pi/2+imag(expint(i*abs(x(k))))).*sign(x(k));

93

¡2:0 ¡1:5 ¡1:0 ¡0:5 0.0 0.5 1.0 1.5 2.0¡0:2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

t=T

y(t)

dashed curve: BT = 10

solid curve: BT = 50

...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..........................................................................................................................................................................................

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

......................................................................................................

....................................................................................................

..................................................................................................

..................................................................................................

............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.........

........

........

........

........

.................................................................................................

............................................................................................................

..............................................................................................................................................................................................................................

..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..................................................................................................................

.................................................................................................................................................................................................................................................................................................................................................................................

..........................................................................................................................................................................................................................................................................................................................................................................

.........................................................................................................................

...........................................................................

Fig. 1. Rectangular pulse response of ideal low-pass ¯lter.

94

Ideal Band-Pass Filter

..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

!......................................................................................................................................

H(!)

..............................

¡!0 !00

B..................................................................................................... ....................................................................................... ..........

.... B ....................................................................................... ...................................................................................................................

H(!) = ¦

μ! ¡ !0B

¶+¦

μ! + !0

B

¶

..........................................................................................................................................................................................................................................................................................................................................................................

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

......

f(t) F (!)

B

2¼sinc

μBt

2

¶¦³ !B

´

ej!0tf(t) F (! ¡ !0)

h(t) =B

2¼sinc(Bt=2)

£ej!0t + e¡j!0t

¤=B

¼sinc(Bt=2) cos(!0t)

.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

......

t

h(t)

!0 = 100, B = 20

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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95

Complex Phasors for Time-Harmonic Circuit Analysis. Consider the sinusoidal-

steady-state response of the given RLC circuit, that is ¯nd vo(t) if vi(t) = Vp cos(!0t+Á).

The time-domain integro-di®erential equation (Kirchho®'s voltage law) that is appropriate

for this network is

Ldi(t)

dt+1

C

tZ¡1

i(¿) d¿ +Ri(t) = vi(t)

and the output voltage is simply a scaled version of the mesh current, that is vo(t) = Ri(t).

.......

........................................................................................................................................................................................................................................

................................

................................

................................

................................

..................................................................i(t)

.....................................................................................................................

.....................................................................................................................

.........................................................................................................................................................................................................................................................................................................................................................................................................

................................................... ................................................................ ..................................................................................................................................................

.......

.......

.......

.......

.......

.......

.......

..

.......

.......

.......

.......

.......

.......

.......

..

...................................................

.......

.......

.......

.......

.......

.......

.......

..

...............................................

................................................................................

.................................

²

²

..........................................................

..........................................................

..........................................................

.

+

¡

vi(t)

+

¡

vo(t)

LC

R...............................................................................................................................................................................

(1) Solve for vo(t) by representing the time-domain signals in terms of their Fourier

transforms.

(2) Show that the Fourier transform analysis is equivalent to, and is therefore the mo-

tivation for, the (simple and compact!) complex phasor analysis of time-harmonic

circuits such as this one. Recall that the time-harmonic signal i(t) is represented

in terms of the complex phasor I via

i(t) = RefIe+j!tg:

The Fourier transform of the input or generator voltage is

Vi(!) = F [vi(t)] = F [Vp cos(!0t+ Á)] = F½Vp

2

£ej(!0t+Á) + e¡j(!0t+Á)

¤¾=Vp

2

£2¼ejÁ±(! ¡ !0) + 2¼e¡jÁ±(! ¡ !0)

¤:

The Fourier transform of the entire (KVL) equation above (both the LHS and RHS) is

j!LI(!) +1

C

·I(!)

j!+ ¼I(0)±(!)

¸+RI(!) = ¼Vp

£ejÁ±(! ¡ !0) + e¡jÁ±(! + !0)

¤:

Physically argue that I(0) = 0 here, since we expect zero steady-state DC current through

the capacitor, whereuponμR+ j!L+

1

j!C

¶| {z }

Z(!)

I(!) = ¼Vp£ejÁ±(! ¡ !0) + e¡jÁ±(! + !0)

¤:

96

Now solve this algebraic equation for the unknown spectrum

I(!) = ¼VpejÁ±(! ¡ !0) + e¡jÁ±(! + !0)

Z(!):

Fourier inversion proceeds as

i(t) =1

2¼

1Z¡1

I(!)ej!td!

=Vp

2

24ejÁ 1Z¡1

±(! ¡ !0)Z(!)

ej!td! + e¡jÁ1Z

¡1

±(! + !0)

Z(!)ej!td!

35=Vp

2

·ejÁ

ej!0t

Z(!0)+ e¡jÁ

e¡j!0t

Z(¡!0)¸

note Z(¡!0) = Z¤(+!0)

=1

2

·Vpe

jÁ

Z(!0)ej!0t +

Vpe¡jÁ

Z¤(!0)e¡j!0t

¸=1

2

·Vpe

jÁ

Z(!0)ej!0t + c.c.

¸an expression plus its complex conjugate

= Re·Vpe

jÁ

Z(!0)ej!0t

¸= Re

£Iej!0t

¤where the complex phasor

I =Vpe

jÁ

Z(!0)=

Vi

Z(!0)

introduced itself! If you still want it,

Vo = RI and vo(t) = Ri(t):

It looks like the two steps alluded to in the problem statement are more naturally done

together.

97

Homework Problem on Uncertainty Principle

Consider the signal

x(t) =1pT¦(t=T )

where T > 0 and the total energy in x(t) is

Ex =1Z

¡1jx(t)j2 dt = 1:

(The multiplying factor T¡1=2 ensures that Ex is independent of T .) If x(t) is the inputvoltage to the given electric circuit, with ¯xed values of R and L, consider the energy in

the output voltage y(t)

Ey =1Z

¡1jy(t)j2 dt:

Again, note that changing T has no e®ect on the total input energy. If T is increased, will

the output energy Ey increase, decrease, or remain the same? Explain.

..................................................................... ................................................................

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.................................................. .................................................................................................................................................................... .......................

.......................................................................................................................................................................................................................................................................................................................................................................................... ..................................................................

........................................................................

........................................................................

......................................................................................

........

.......

.......

.......

.

²

²L

R

+

¡

x(t)

+

¡

y(t)

98

Approximate Analysis of Dispersion

x(t) y(t)H(!)

............................................................................................................................................................................................................................................................................................................................................................................

........................................................... ............. ........................................................... .............

Fig. 1. System Block Diagram

If the modulating signal or envelope of a high frequency sinusoid of frequency !0 is

the baseband signal a(t), having bandwidth considerably less than !0 (say 10%), then the

spectrum of the the modulated carrier

x(t) = a(t) cos(!0t) (1)

is

X(!) = 12 [A(! ¡ !0) +A(! + !0)]: (2)

An example spectrum is shown in Fig. 2.

...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

0.............

¡!0.............

!0

!

jX(!)j

........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

Fig. 2. Typical Narrow-Band Spectrum

For real signals x(t) and h(t), the spectra satisfy X(¡!) = X¤(!) and H(¡!) = H¤(!).The response of the system is the inverse Fourier transform ofX(!)H(!), with the positive

and negative frequency contributions written separately as

99

y(t) =1

2¼

1Z0

X(!)H(!)ej!td! +1

2¼

0Z¡1

X(!)H(!)ej!td!

=1

2¼

1Z0

X(!)H(!)ej!td! +1

2¼

1Z0

X(¡−)H(¡−)e¡j−td−

=1

2¼

1Z0

X(!)H(!)ej!td! +1

2¼

1Z0

£X(−)H(−)ej−t

¤¤d−

=1

2¼

1Z0

X(!)H(!)ej!td! + c.c.

= 12 [y+(t) + y

¤+(t)] (3)

The signal y+(t) is called the pre-envelope of y(t), and consists of positive frequencies only

y+(t) =1

2¼

1Z0

[A(! ¡ !0) +A(! + !0)]H(!)ej!td!

¼ 1

2¼

1Z¡1

A(! ¡ !0)H(!)ej!td! = 1

2¼

1Z¡1

A(−)H(−+ !0)ej(−+!0)td− (4)

where A(−) is the original, baseband or low-pass signal.

The phase or argument of the transfer function H(!) in a neighborhood of the carrier

frequency !0 is of most interest. Write

H(!) = exp[j£(!)]: (5)

A Taylor series about !0 of the phase function

£(!0 + −) = £(!0) + −£0(!0) + 1

2−2£00(!0) + : : : (6)

gives the approximate transfer function

H(−+ !0) ¼ ej£(!0)ej−£0(!0): (7)

An approximation for the pre-envelope (4) is therefore

y+(t) ¼ 1

2¼ej[!0t+£(!0)]

1Z¡1

A(−)ej−[£0(!0)+t]d−

= ej[!0t+£(!0)]a[t+£0(!0)] (8)

and so the ¯nal approximation for the output of the dispersive ¯lter (5) is

y(t) ¼ a[t+£0(!0)] cos[!0t+£(!0)]: (9)

The phase delay (of the carrier) is due to £(!0) and the group delay (of the envelope) is

due to £0(!0).

100

Numerical Example. The phase function of a length of waveguide transmission line is

£(!) = ¡p!2 ¡ !2c t0

where !c is the cuto® frequency of the guide and t0 = d=c where d is the length and c is

the free-space phase velocity.

If !c = 0, then £0(!0) = ¡t0.

If !c = 25, !0 = 30, t0 = 3, then

£0(!0) = ¡ (30)(3)p302 ¡ 252 = ¡5:43

The following ¯gures are the result of an inverse FFT approximation for the signal

x(t) = e¡t2

cos(!0t)

applied to the input of the above dispersive waveguide section. The 4096 point FFT uses

¢! = 0:3.

101

0 1 2 3 4 5 6 7 8 9 10

t

¡1:0

¡0:5

0.0

0.5

1.0

y(t)

!c = 0, t0 = 3, !0 = 30

........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................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0 1 2 3 4 5 6 7 8 9 10

t

¡1:0

¡0:5

0.0

0.5

1.0

y(t)

!c = 25, t0 = 3, !0 = 30

........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................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Fig. 3. Dispersive Filter Output

102

Fourier Integral Solution of Potential Problem

Laplace's equation

r2Ã(x; y) =μ@2

@x2+@2

@y2

¶Ã(x; y) = 0

boundary conditions

Ã(x; 0) =

½1; jxj < 10; jxj > 1

separate variables

Ã(x; y) = X(x)Y (y)

X 00(x)X(x)| {z }=¡k2

+Y 00(y)Y (y)| {z }=+k2

= 0

solution form

Ã(x; y) = cos(kx)e¡kjyj

Ã(x; y) =

1Z0

A(k) cos(kx)e¡kjyj dk

Ã(x; 0) =

1Z0

A(k) cos(kx) dk

inverse Fourier transform

A(k) =2

¼

1Z0

Ã(x; 0) cos(kx) dx =2

¼

1Z0

cos(kx) dx =2

¼

sin k

k

Ã(x; y) =2

¼

1Z0

sin k cos kx

ke¡kjyjdk

=1

¼

1Z0

sin[k(x+ 1)]¡ sin[k(x¡ 1)]k

e¡kjyjdx

=1

¼

·tan¡1

μx+ 1

jyj¶¡ tan¡1

μx¡ 1jyj

¶¸using G&R 3.941(1).

103

−3 −2 −1 0 1 2 30

1

2

3

4

5

6

Fig. 1. Equipotentials

104

Fourier Integral Solution of Potential Problem

M.J. Ablowitz and A.S. Fokas, Complex Variables, Cambridge, 1997, page 60 prob 9.

Laplace's equation

r2Ã(x; y) = 0boundary conditions

Ã(x; 0) = sgn(x) = §1 (x ? 0)

separate variables

Ã(x; y) = X(x)Y (y)

X 00(x)X(x)| {z }=¡k2

+Y 00(y)Y (y)| {z }=+k2

= 0

solution form

Ã(x; y) = sin(kx)e¡kjyj

Ã(x; y) =

1Z0

B(k) sin(kx)e¡kjyj dk

Ã(x; 0) =

1Z0

B(k) sin(kx) dk

inverse Fourier transform

B(k) =2

¼

1Z0

Ã(x; 0) sin(kx) dx =2

¼

1Z0

sin(kx) dx =2

¼k

a bit tricky! recall ECE 370 and sgn(x) = u(x)¡ u(¡x)

Ã(x; y) =2

¼

1Z0

sin kx

ke¡kjyjdk

=2

¼tan¡1

μx

jyj¶= 1¡ 2

¼tan¡1

μ jyjx

¶using G&R 3.941(1).

105

The Fourier Transform of combT (t) is2¼Tcomb2¼=T (!)

Recall our notation F (!) for the Fourier transform

F (!) = F [f(t)] =1Z

¡1f(t)e¡j!t dt F() f(t) = F¡1[F (!)] = 1

2¼

1Z¡1

F (!)ej!t d!: (1)

The Fourier transform of Woodward's comb function, or impulse train

f(t) = combT (t) =

1Xn=¡1

±(t¡ nT ) (2)

is therefore

F [combT (t)] =1X

n=¡1e¡jn!T = F (!) = F (! + 2¼=T ); (3)

periodic in ! with period 2¼=T . Let's integrate F (!) over the range a < ! < b, where

¡¼=T < a < b < ¼=T . The minimum and maximum restrictions cover one period, centeredabout the origin (! = 0). We will make use of the discontinuous sum

1Xn=1

sin(nx)

n=¼ ¡ jxj2

sgn(x) (¡2¼ < x < 2¼) (4)

where, in this context, we de¯ne

sgn(x) =

8><>:¡1; x < 0

0; x = 0

+1; x > 0

(5)

in order to ensure the vanishing of the sum for x = 0. The integral of interest is

bZa

1Xn=¡1

e¡jn!Td! = b¡ a+ 21Xn=1

bZa

cos(n!T ) d!

= b¡ a+ 2

T

1Xn=1

1

n[sin(nbT )¡ sin(naT )]

= b¡ a+³ ¼T¡ jbj

´sgn(b)¡

³ ¼T¡ jaj

´sgn(a)

=

½2¼=T if ¡ ¼=T < a < 0 < b < ¼=T0 if a < b < 0 or 0 < a < b

: (6)

Therefore, we see that F [combT (t)] = 2¼Tcomb2¼=T (!): ¥

Now we are in a position to derive the Fourier series from the Fourier integral. And recall

that we essentially derived the Fourier integral from the \spectral representation of the

Dirac-delta" function. We are in excellent shape to now do some real damage.

106

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t

combT (t)

¡2T ¡T 0 T 2T

: : : : : :

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!

¡4¼T

¡2¼T

02¼

T

4¼

T

: : : : : :

2¼

Tcomb2¼

T

(!)F()

Fourier Transform of a Periodic Signal

...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.......

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.......

...

¡T 0 T 2T

t

g(t)

............................................................................................................................................................................................................................................................................................................

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.......

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...

¡T 0 T 2T

t

: : : : : :

f(t) = f(t+ T )

............................................................................................................................................................................................................................................................................................................

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Consider the arbitrary periodic signal f(t) = f(t+T ), and let's focus our attention on the

particular period 0 · t · T . Any period would work, but since all periods are equivalentwe might as well work with the selected one. (The function f does not have to go to zero

at t = nT ; it was simply easier to draw one like this. And of course the function f does

not have to be positive, or even real-valued.) Introduce the new aperiodic function g(t),

also de¯ned for all t according to

g(t) =

½f(t); 0 · t · T0; t < 0 or t > T:

(1)

We can easily reproduce the entire periodic signal f(t) by replicating g(t) all up and down

the t-axis. That is, we write f(t) as the periodic extension of g(t) in the form

f(t) =

1Xn=¡1

g(t¡ nT ): (2)

107

Recognize that (2) can be written as

f(t) = g(t)~1X

n=¡1±(t¡ nT ) = g(t)~ combT (t): (3)

The Fourier transform of (3) is

F (!) = G(!) ¢ 2¼Tcomb2¼

T

(!)

=2¼

TG(!)

1Xn=¡1

±(! ¡ n!0) where !0 ,2¼

T

=2¼

T

1Xn=¡1

G(n!0)±(! ¡ n!0): (4)

The spectrum F (!) of the periodic signal f(t) consists of discrete spectral lines at integer

multiples n!0 of the fundamental frequency !0 = 2¼=T . The nth multiple is called the

nth harmonic. Note that negative n are just as important as positive n.

!

!0¡10 10

¡8 8¡6 6

4¡4 2¡2 0..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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................................................................................................

108

The inverse Fourier transform of F (!) should give us our original periodic f(t). It does.

In fact, it gives us a whole new representation for a periodic signal

f(t) =1

2¼

1Z¡1

F (!)ej!td! =

1Xn=¡1

G(n!0)

T

1Z¡1

±(! ¡ n!0)ej!td! =1X

n=¡1

G(n!0)

Tejn!0t

(5)

where

G(n!0) = G(!)¯̄!=n!0

=

1Z¡1

g(t)e¡jn!0tdt =

TZ0

g(t)e¡jn!0tdt =

TZ0

f(t)e¡jn!0tdt: (6)

We usually call the Fourier coe±cients cn instead of G(n!0)=T and write the whole

package as

f(t) =

1Xn=¡1

cnejn!0t cn =

1

T

TZ0

f(t)e¡jn!0tdt: (7)

Of course any period will work for the integration limits, since the total integrand is

periodic with period T :

cn =1

T

t0+TZt0

f(t)e¡jn!0tdt: (8)

You now have the complete and uncluttered derivation of the Fourier series representation

of a periodic signal, starting from the Fourier integral. The two key components are the

correspondingly appropriate spectral representations for the aperiodic Dirac-delta function

and for its periodic version, Woodward's comb function. All that was required on our part

was a reasonable °uency with the Fourier transform properties and a few fundamental

signals.

F Exercise 8. Find an interested (or at least willing) fellow student or more who are at

or above your present level with respect to Fourier analysis, and explain to them, in detail,

the above derivation starting from the top of three pages back. If you explain it correctly

and with some degree of coherence, and your entire audience has not deserted you, buy the

survivor(s) a Coke. (They can split it to ¯t any ¯nancial boundary conditions, applied, of

course, to your total solution.) You deserve one, too. Otherwise go back and do any of the

homework exercises (including enough of the 16 Fourier transform pairs on page 59) that

strike your fancy. If none ¯t that description, see me for more.

109

Periodic Signals and Fourier Series

The period of a periodic signal is the smallest positive number T such that f(t) = f(t+T )

for all t. It will be frequently convenient to de¯ne !0 , 2¼=T to be the fundamental

frequency of the periodic signal, with n!0 the nth harmonic frequency for integer n =

0;§1;§2; : : : . An important property of a periodic signal is that the integralt0+TZt0

f(t) dt =

TZ0

f(t) dt =

T=2Z¡T=2

f(t) dt =

3T=4Z¡T=4

f(t) dt = : : :

is independent of the time t0. That is, integration over any complete period yields the same

result. Another useful observation is that if f(t) = f(t + T ), then the product f(t)ejn!0t

is also periodic with period T , so that the integral

t0+TZt0

f(t)ejn!0t dt

is also independent of the time t0. F Be sure you can show/verify this important property

for yourself.

DEFN: Orthogonality. Two functions (generally complex) are said to be orthogonal on

the domain a · t · b if the inner product

hf(t); g(t)i¯̄̄ba,

bZa

f(t)g¤(t) dt = 0:

In any given situation, let's assume the interval of interest a · t · b is known and ¯xed, sothat we can condense the notation on the inner product angle brackets. The inner product

of a signal and itself is sometimes called the energy in the signal

hf(t); f(t)i ,bZa

jf(t)j2 dt = E

and the square-root of this necessarily non negative quantity is called the norm of the

signal

kf(t)k = +phf(t); f(t)i =

24 bZa

jf(t)j2 dt351=2 :

When dealing with periodic signals so that the interval of interest is any period t0 · t ·t0 + T , the e®ective or root-mean-square value is

frms =

24 1T

t0+TZt0

jf(t)j2 dt351=2 :

110

Kronecker delta. This simple function is a function of two integers m and n and is

de¯ned as

±m;n =

½1; m = n

0; m6= n:It is unrelated to the Dirac-delta function ±(t) that is explicitly written as a function of

the continuous time variable t. The operational advantage of introducing this notational

convenience will be clear as we progress.

Orthogonality of the Fourier basis.

hejn!0t; ejm!0tiT =t0+TZt0

ej(n¡m)!0tdt = T±n;m (n;m = 0;§1;§2; : : : )

hcos(n!0t); cos(m!0t)iT =t0+TZt0

cos(n!0t) cos(m!0t) dt =T

2±n;m (n;m = 1; 2; : : : )

hsin(n!0t); sin(m!0t)iT =t0+TZt0

sin(n!0t) sin(m!0t) dt =T

2±n;m (n;m = 1; 2; : : : )

hcos(n!0t); sin(m!0t)iT =t0+TZt0

cos(n!0t) sin(m!0t) dt = 0 (n;m = 0; 1; 2; : : : )

When n = 0 the nth cosine is unity, and we have

h1; cos(m!0t)iT =t0+TZt0

cos(m!0t) dt = T±0;m (m = 0; 1; 2; : : : )

h1; sin(m!0t)iT =t0+TZt0

sin(m!0t) dt = 0 (m = 1; 2; : : : )

Note that sin(m!0t) is trivial, and therefore not used, when m = 0.

F Homework: Verify/derive each of the above orthogonality relations. You will need to

use trig identities to simplify the products in the integrands.

111

Derivation of Fourier Coe±cients by Minimizing the Mean Square Error

The concept of the Fourier series arose naturally back on page 108 when we wrote a

periodic signal as the inverse Fourier integral of its Fourier transform. In doing so, the

Fourier coe±cients appeared naturally. Now let's start over by assuming that the periodic

signal f(t) = f(t+ T ) can be represented by a Fourier series, call it

ef(t) = 1Xn=¡1

cnejn!0t

to temporarily avoid any premature claims that ef(t) = f(t). One approach to derive theexpression for the Fourier coe±cients cn is to ask the question: What values of the cnminimize the mean square error between the original signal f(t) and its proposed Fourier

series representation ef(t) ?Firstly, write the mean square error

E2 = 1

T

t0+TZt0

jf(t)¡ ef(t)j2 dt = 1

T

t0+TZt0

£f(t)¡ ef(t)¤£f¤(t)¡ ef¤(t)¤ dt

=1

T

t0+TZt0

£jf(t)j2 + j ef(t)j2 ¡ f(t) ef¤(t)¡ f¤(t) ef(t)¤ dt:We need the following component integrals.

1

T

t0+TZt0

j ef(t)j2 dt = 1

T

t0+TZt0

1Xn=¡1

cnejn!0t

| {z }ef(t)

1Xm=¡1

c¤me¡jm!0t

| {z }ef¤(t)dt

=

1Xn=¡1

cn

1Xm=¡1

c¤m1

T

t0+TZt0

ej(n¡m)!0tdt

=

1Xn=¡1

cn

1Xm=¡1

c¤m±n;m

=

1Xn=¡1

cnc¤n

112

1

T

t0+TZt0

f¤(t) ef(t) dt = 1Xn=¡1

cn1

T

t0+TZt0

f¤(t)ejn!0tdt

1

T

t0+TZt0

f(t) ef¤(t) dt = 1Xn=¡1

c¤n1

T

t0+TZt0

f(t)e¡jn!0tdt

Now the mean square error is

E2 = 1

T

t0+TZt0

jf(t)j2 dt +1X

n=¡1cnc

¤n ¡

1Xn=¡1

cn1

T

t0+TZt0

f¤(t)ejn!0tdt

¡1X

n=¡1c¤n1

T

t0+TZt0

f(t)e¡jn!0tdt

To minimize E2 with respect to the Fourier coe±cients cn, set to zero the partial derivativewith respect to each of the ck (for all integer k)

@

@ckE2 = c¤k ¡

1

T

t0+TZt0

f¤(t)ejk!0tdt = 0

or

c¤k =1

T

t0+TZt0

f¤(t)ejk!0tdt

the conjugate of which gives the required Fourier coe±cient formula

ck =1

T

t0+TZt0

f(t)e¡jk!0tdt

where k can now be replaced by n

cn =1

T

t0+TZt0

f(t)e¡jn!0tdt:

113

Fourier Coe±cients by a Direct Inner Product

Write

f(t) =

1Xn=¡1

cnejn!0t

and then take the inner product of both sides with ejm!0t

t0+TZt0

f(t)e¡jm!0tdt =1X

n=¡1cn

t0+TZt0

ej(n¡m)!0tdt

=

1Xn=¡1

cnT±n;m = Tcm

and therefore

cm =1

T

t0+TZt0

f(t)e¡jm!0tdt:

Trigonometric Form of the Fourier Series

f(t) = a0 +

1Xn=1

£an cos(n!0t) + bn sin(n!0t)

¤To ¯nd the trig coe±cients an and bn (n = 0; 1; 2; : : : ) in terms of the complex exponential

coe±cients cn (n = : : : ;¡1; 0; 1; 2; : : : ) set1X

n=¡1cne

jn!0t = a0 +

1Xn=1

£an cos(n!0t) + bn sin(n!0t)

¤

c0 +

1Xn=1

hcne

jn!0t + c¡ne¡jn!0ti= a0 +

1Xn=1

£an cos(n!0t) + bn sin(n!0t)

¤

c0 +

1Xn=1

hcn©cos(n!0t) + j sin(n!0t)

ª+ c¡n

©cos(n!0t)¡ j sin(n!0t)

ªi= a0 +

1Xn=1

£an cos(n!0t) + bn sin(n!0t)

¤

c0+

1Xn=1

h(cn+c¡n) cos(n!0t)+j(cn¡c¡n) sin(n!0t)

i= a0+

1Xn=1

£an cos(n!0t)+bn sin(n!0t)

¤114

Therefore

a0 = c0 =1

T

t0+TZt0

f(t) dt

an = cn + c¡n =2

T

t0+TZt0

f(t) cos(n!0t) dt (n = 1; 2; : : : )

bn = j(cn ¡ c¡n) = 2

T

t0+TZt0

f(t) sin(n!0t) dt (n = 1; 2; : : : )

F Homework. Derive the trig Fourier coe±cients by taking the direct inner product of

f(t) = a0 +

1Xn=1

£an cos(n!0t) + bn sin(n!0t)

¤with each of the trig Fourier basis functions cos(m!0t), sin(m!0t), and unity.

115

Two Standard Notations for the Trigonometric Fourier Series

f(t) = a0 +

1Xn=1

£an cos(n!0t) + bn sin(n!0t)

¤a0 =

1

T

t0+TZt0

f(t) dt

an =2

T

t0+TZt0

f(t) cos(n!0t) dt (n = 1; 2; : : : )

bn =2

T

t0+TZt0

f(t) sin(n!0t) dt (n = 1; 2; : : : )

f(t) =a0

2+

1Xn=1

£an cos(n!0t) + bn sin(n!0t)

¤an =

2

T

t0+TZt0

f(t) cos(n!0t) dt (n = 0; 1; 2; : : : )

bn =2

T

t0+TZt0

f(t) sin(n!0t) dt (n = 1; 2; : : : )

The second form uses the general form for an to also calculate the DC term a0, avoiding

the need for a separate formula for a0.

It is silly, but observe that this captures the spirit:

f(t) =|7+

1Xn=1

£an cos(n!0t) + bn sin(n!0t)

¤| = 7

T

t0+TZt0

f(t) dt

F Homework. In some applications, the Fourier series is written in the form

f(t) = A0 +

1Xn=1

An cos(n!0t+ Án):

Find an expression for An and Án in terms of an and bn.

116

Riemann-Lebesgue Lemma - Fourier Coe±cients

f(t) = f(t+ T ) =

1Xn=¡1

cnejn!0t; !0 = 2¼=T

Tcn =

t0+TZt0

f(t)e¡jn!0tdt

=f(t)e¡jn!0t

¡jn!0

¯̄̄̄¯t0+T

t=t0

+1

jn!0

t0+TZt0

f 0(t)e¡jn!0tdt

by integration-by-parts. The boundary terms are zero by the periodicity of the integrand.

The remaining integral has¯̄̄̄¯̄t0+TZt0

f 0(t)e¡jn!0tdt

¯̄̄̄¯̄ ·

t0+TZt0

jf 0(t)j dt · K <1

if f 0(t) is absolutlely integrable. Therefore

jcnj · K

2n¼:

If f 0(t) is absolutlely integrable, then cn ! 0 as n!1.

Corollary.

If f (k)(t) is absolutely integrable, but f (k+1)(t) is not, then cn = O(n¡k) as n!1.

117

Gibb's Phenomenon - Example Square wave

0.0 0.5 1.0

t=T

¡1:0

¡0:5

0.0

0.5

1.0

f(t)

S5(t)

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Fourier series of the above square wave is

f(t) =4

¼

1Xk=1

sin[(2k ¡ 1)!0t]2k ¡ 1 ; !0 = 2¼=T

Truncated Fourier series

SN (t) =4

¼

NXk=1

sin[(2k ¡ 1)!0t]2k ¡ 1

Di®erentiate to ¯nd its extrema

S0N (t) =8

T

NXk=1

cos[(2k ¡ 1)!0t]

write

cos[(2k ¡ 1)!0t] = sin(2k!0t)¡ sin[(2k ¡ 1)!0t]2 sin!0t

All terms cancel except

S0N (t) =4

T

sin(2N!0t)

sin!0t

118

Smallest zero is

t0 =¼

2N!0

SN (t0) =4

¼

NXk=1

sinh(2k¡1)¼2N

i2k ¡ 1 =

4

¼

NXk=1

¼

2N

sinh(2k¡1)¼2N

i(2k¡1)¼2N

Riemann sumaZ0

sinx

xdx ¼

NXk=1

sin£2k¡12 ¢x

¤2k¡12 ¢x

¢x

where N¢x = a. Here ¢x = ¼=N so a = ¼.

SN (t0) ¼ 2

¼

¼Z0

sinx

xdx =

2

¼Si(¼) ¼ 1:18

overshoot is1:18¡ 1

total jump of 2£ 100% ¼ 9%

Pointwise Convergence of the Fourier Series

The Dirichlet kernel is

DN (x) =

NXn=¡N

ejnx = e¡jNx2NXn=0

ejnx = e¡jNx1¡ ej(2N+1)x1¡ ejx =

sin[(N + 1=2)x]

sin(x=2);

an even function DN (¡x) = DN (x). Consider a periodic function f(t) = f(t+ T ) havinga single jump discontinuity at the point ¡T=2 < t < T=2. De¯ne the right and left-handlimits

f(t+) = lim²!0+

f(t+ ²) and f(t¡) = lim²!0+

f(t¡ ²):

With the Fourier coe±cients

cn =1

T

T=2Z¡T=2

f(¿)e¡j2n¼¿=T d¿;

the truncated Fourier series efN (t) = NXn=¡N

cnej2n¼t=T

119

is

efN (t) = T=2Z¡T=2

f(¿)1

T

NXn=¡N

ej2n¼(t¡¿)=Td¿ =

T=2Z¡T=2

f(¿)1

TDN [2¼(t¡ ¿)=T ] d¿

=

tZ¡T=2

f(¿)1

TDN [2¼(¿ ¡ t)=T ] d¿

| {z }I¡

+

T=2Zt

f(¿)1

TDN [2¼(¿ ¡ t)=T ] d¿| {z }

I+

I¡ =

tZ¡T=2

[f(¿)¡ f(t¡)] 1TDN [2¼(¿ ¡ t)=T ] d¿ + f(t¡)

tZ¡T=2

1

TDN [2¼(¿ ¡ t)=T ] d¿

Introduce the change-of-integration variable

x =2¼

T(N + 1=2)(¿ ¡ t)

such that1

TDN [2¼(¿ ¡ t)=T ] d¿ = sinx

sin

μx

2N + 1

¶ dx

(2N + 1)¼

¿ = ¡T=2 =) x = ¡(2N + 1)¼(1=2 + t=T ) = ¡X¿ = t =) x = 0

¿ ¡ t = T

(2N + 1)¼

I¡ =1

¼

0Z¡X

½f

·T

(2N + 1)¼x+ t

¸¡ f(t¡)

¾sinx

(2N + 1) sin

μx

2N + 1

¶dx

+f(t¡)¼

0Z¡X

sinx

(2N + 1) sin

μx

2N + 1

¶dxSince ¡X < x < 0 , as N !1,

f

·T

(2N + 1)¼x+ t

¸¡¡¡¡!N!1

f(t¡)

120

so that the ¯rst integral above vanishes. The denominator becomes

(2N + 1) sin

μx

2N + 1

¶¡¡¡¡!N!1

x

and with X !1, the second integral gives

I¡ ¡¡¡¡!N!1

f(t¡)¼

0Z¡1

sinx

xdx =

f(t¡)2

:

The same philosophy yields

I+ ¡¡¡¡!N!1

f(t+)

¼

1Z0

sinx

xdx =

f(t+)

2

so thatNX

n=¡Ncne

j2n¼t=T ¡¡¡¡!N!1

f(t+) + f(t¡)2

: ¥

F Homework Consider the simpler case where f(t) is continuous everywhere, and there-

fore demonstrate thatNX

n=¡Ncne

j2n¼t=T ¡¡¡¡!N!1

f(t):

121

ECE 370 QUIZ 3 16 January 1978 NAME

Take Home - Due Wednesday January 18

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.......................................................................................................................................................

t

f(t)

.......

.......

.....

¡T...................

¡T=2 0...................

T=2...................

T...................

3T=2

................... 1

: : : : : :

1.) Find a Fourier series representation of the signal f(t).

2.) Use MATLAB to graph the Nth partial sum of this Fourier series. For example, if

f(t) =a0

2+

1Xn=1

an cos(n!0t);

then let

fN (t) =a0

2+

NXn=1

an cos(n!0t):

Pick several N to demonstrate (pictorially) the convergence of this Fourier series.

There are no sines (bn ´ 0) in this even function, and the DC or average value is

a0

2=1

T

TZ0

f(t) dt =1

2; (1)

as the graph shows (perhaps after a little thought). The cosine coe±cients (for n = 1; 2; : : : )

are

an =2

T

t0+TZt0

f(t) cos(n!0t) dt =2

T

T=2Z¡T=2

f(t) cos(n!0t) dt =2

T¢ 2

T=2Z0

f(t) cos(n!0t) dt (2)

122

by the even symmetry of the product of f(¡t) = f(t) and the nth cosine. The last integralabove requires an expression for f(t) in the range 0 · t · T=2, which is the line of slope2=T through the origin: f(t) = 2t=T . Required now is

an =2

T¢ 2

T=2Z0

2t

Tcos(n!0t) dt =

8

T 2

T=2Z0

t cos(n!0t) dt: (3)

In terms of di®erent variables, we need the integral

I =

dZ0

x cos(®x) dx: (4)

It is perfectly acceptable to use a table of integrals (perhaps in your calculus book, or

a table such as the CRC). But here are two di®erent techniques to evaluate it. The

¯rst method is based on the observation that the parameter ® is independent from the

integration variable x, so that we can write

I =

dZ0

x cos(®x) dx =d

d®

dZ0

sin(®x) dx =d

d®

1¡ cos(®d)®

=d sin(®d)

®¡ 1¡ cos(®d)

®2: (5)

The second method is integration by parts:

d(uv) = u dv + v du =)bZa

udv = uv¯̄̄ba¡

bZa

v du:

The choice

u = x

dv = cos(®x) dx=)

du = dx

v =1

®sin(®x)

gives

I =

dZ0

x cos(®x) dx =x

®sin(®x)

¯̄̄dx=0

¡ 1

®

dZ0

sin(®x) dx

=d

®sin(®d) +

1

®2cos(®x)

¯̄̄dx=0

=d

®sin(®d) +

cos(®d)¡ 1®2

; (6)

123

in agreement with the ¯rst result (5). The needed integral in (3) has upper limit d = T=2

and frequency (the factor multiplying the integration variable t in the argument of the

cosine) ® = n!0, whereupon ®d = n!0T=2 = n¼ (recall !0T = 2¼). Note that the values

of the trig functions are thus sin(n¼) = 0 and cos(n¼) = (¡1)n. The Fourier coe±cients(3) are now

an = ¡ 8

T 21¡ (¡1)n(n!0)2

= ¡ 2

(n¼)2[1¡ (¡1)n] =

8<: ¡μ2

¼

¶21

n2; n odd

0; n even:

(7)

Together with the DC value (1), the entire trigonometric Fourier series for our even, peri-

odic function f(t) is

f(t) =1

2¡μ2

¼

¶2 1Xn=1;3;5;:::

1

n2cos(2n¼t=T ): (8)

The sum is reasonably clear (at least to me), but it would more commonly be written in

the (probably nicer looking) completely equivalent form

f(t) =1

2¡μ2

¼

¶2 1Xn=1

1

(2n¡ 1)2 cos[(2n¡ 1)2¼t=T ]: (9)

And it would be perfectly acceptable to write it in terms of the fundamental frequency as

f(t) =1

2¡μ2

¼

¶2 1Xn=1

1

(2n¡ 1)2 cos[(2n¡ 1)!0t];

as long as it is perfectly clear that we have de¯ned !0 = 2¼=T .

This series converges nicely, since janj » n¡2 as n!1 (in fact, these an are 1=n2 for all

odd n. In fact, Fourier series representations of known functions are used to sum certain

series. In this case, note that f(T=2) = 1 and so (9) gives

1 =1

2¡μ2

¼

¶2 1Xn=1

1

(2n¡ 1)2 cos[(2n¡ 1)¼]

or 1Xn=1

1

(2n¡ 1)2 =¼2

8:

Note that the sum over both odd and even integers is

1Xn=1

1

n2=

1Xn=1

1

(2n¡ 1)2 +1Xn=1

1

(2n)2

=

1Xn=1

1

(2n¡ 1)2 +1

4

1Xn=1

1

n2

124

or

3

4

1Xn=1

1

n2=

1Xn=1

1

(2n¡ 1)2

=¼2

8

and therefore 1Xn=1

1

n2=¼2

6:

This is the Riemann zeta function

³(x) =

1Xn=1

1

nx

of argument x = 2.

125

On the Convergence and Summation of Some Series

The basic form to consider is

S =

1Xn=1

an = a1 + a2 + a3 + : : :

where the indexing can just as easily start at n = 0

1Xn=0

an = a0 + a1 + a2 + : : :

and negative indices are also commonly encountered

1Xn=¡1

an = a0 +

1Xn=1

an +

1Xn=1

a¡n:

If the series1Pn=1

janj converges, then we say that1Pn=1

an is absolutely convergent. Any series

that is absolutely convergent is convergent, since¯̄̄̄¯1Xn=1

an

¯̄̄̄¯ ·

1Xn=1

janj:

A convergent series that is not absolutely convergent is called conditionally convergent.

Here is an important class of series: With p real

1Xn=1

1

npconverges i® p > 1.

This can be shown by using the integral argument (Do it!) on the next page.

126

If all of the terms of the original series are positive (true for exp(z) if z > 0), then a related

integral can provide error bounds by the following reasoning.

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g(x)........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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............................................

N+1 N+2 N+3 : : :

1Xn=N+1

g(n) >

1ZN+1

g(x) dx

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x

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N N+1 N+2 N+3 : : :

1Xn=N+1

g(n) <

1ZN

g(x) dx

Therefore, we have both a lower and an upper bound, so that the error or \tail" sum is

1ZN+1

g(x) dx <

1Xn=N+1

g(n) <

1ZN

g(x) dx:

127

If the series1Pn=1

janj converges, then we must have limn!1 an = 0: The converse is not true,

as we demonstrate by example, since the series

1Xn=1

1

np

is divergent for p = 1=2 but1pn¡¡¡!n!1

0:

Comparison Test. Let1Pn=1

an and1Pn=1

bn be two series with nonnegative terms. If there

exists an index N such that an · bn for all n > N , then the convergence of the series1Pn=1

bn implies the convergence of the series1Pn=1

an, and the divergence of the series1Pn=1

an

implies the divergence of1Pn=1

bn.

Weierstrass M-Test for Absolute Convergence. Let1Pn=1

an and1Pn=1

bn be series.

Suppose there exists an index N such that janj · bn for all n > N . Then a su±cient

condition for absolute convergence of the series1Pn=1

an is that the series1Pn=1

bn converges.

Ratio (d'Alembert's) Test. Suppose that the limit limn!1

¯̄̄̄an+1

an

¯̄̄̄= ® exists for the

series1Pn=1

an.

(1) if ® < 1 then the series1Pn=1

an converges absolutely

(2) if ® > 1 then the series1Pn=1

an diverges

(3) there exist both absolutely convergent and divergent series for which ® = 1.

128

Acceleration of series convergence by subtracting out the asymptotic form

(Kummer transform):

Consider the series1Pn=1

f(n) where the large n form of f(n) is

f(n) ¡¡¡!n!1

fasy(n):

If the original terms f(n) decay so slowly with n as to be numerically disagreeable, but

the asymptotic form can be summed by some alternate method, then write

1Xn=1

f(n) =

1Xn=1

[f(n)¡ fasy(n)] +1Xn=1

fasy(n):

The terms of the di®erence f(n)¡fasy(n) will approach zero faster than the original f(n),so that truncation of the di®erence sum can be made at a smaller n to achieve a given

level of accuracy.

Example. One series that can be summed exactly using the Poisson sum formula (on

pages 150-151) is1X

n=¡1

1

n2 + b2=¼

bcoth¼b:

We choose an example series where we know the exact value of the sum simply to be able

to check our result. Let's get this series into the form1Pn=1

f(n), where f(¡n) = f(n) is aneven function of the summation index,

1Xn=¡1

1

n2 + b2= 2

1Xn=1

1

n2 + b2+1

b2

and therefore

S =

1Xn=1

1

n2 + b2=¼b coth¼b¡ 1

2b2:

Now let's compare the brute-force summation of S as it stands, with a simple Kummer

transform. Having the exact value of S allows us to gauge our accuracy. The asymptotic

form of the summand f(n) is fasy(n) = n¡2 so that the Kummer transform is

S =

1Xn=1

·1

n2 + b2¡ 1

n2

¸+

1Xn=1

1

n2:

The asymptotic or \tail" sum is the Riemann zeta function of argument 2 (a well-known,

actually famous sum) that we encountered back on page 124

³(2) =

1Xn=1

1

n2=¼2

6:

129

The di®erence terms

1

n2 + b2¡ 1

n2=n2 ¡ (n2 + b2)n2(n2 + b2)

=¡b2

n2(n2 + b2)= O(n¡4)

decay much faster than the original terms of O(n¡2). The Kummer transform of the

original sum S is now

S = ¡b21Xn=1

1

n2(n2 + b2)+¼2

6:

To do the comparison, label the truncated brute-force sum

S0(N; b) =

NXn=1

1

n2 + b2

and the truncated Kummer-accelerated sum

SK(N; b) = ¡b2NXn=1

1

n2(n2 + b2)+¼2

6:

With the exact value

S(b) =¼b coth¼b¡ 1

2b2

let's compare the numerical performance of

S0(N; b)

S(b)with

SK(N; b)

S(b)

as a function of the truncation index N for one or more values of the parameter b.

NS0(N; 10)

S(10)

SK(N; 10)

S(10)

101 0:499726021056655 1:125492496191760

102 0:934787298579118 1:000214644029801

103 0:993428003029169 1:000000218841745

104 0:999342482848786 1:000000000219156

Exercise. Using ³(4) =1Pn=1

n¡4 = ¼4=90, perform a second Kummer transform on the

series in the example and compare its convergence to the data above.

130

Homework Problem: Convergence of Fourier Series. Consider the three periodic

functions x(t), y(t), and z(t) as graphed below. Find appropriate Fourier series expansions

for each of the three functions. Study and discuss the convergence of each series, including

graphical display.

For example, say the odd function y(t) is most conveniently represented by the Fourier

sine series

y(t) =

1Xn=1

bn sin(2n¼t=T )

where the coe±cients bn are known. Truncation of this Fourier series yields the approxi-

mation

yN (t) =

NXn=1

bn sin(2n¼t=T ):

Compare these truncated series yN (t) with the original function y(t) for several (you decide

which range of values has signi¯cance to your convergence study) values of the ¯nal or

truncation index N . Certainly, the rate of decay of jbnj as n ! 1 is the important

feature.

Note: How are x(t), y(t), and z(t) related? How then are their Fourier series related?

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................

......

................

¡T=2 0 T=2 Tt

................... +1

x(t)

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t

+1

¡1

y(t)

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...N

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...Nz(t)

(+1)

(¡1)

131

Fourier Series of Full-Wave Recti¯ed Sine Wave

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.......

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.....

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.....

¡T 0 T

t

sin( 12!0t) = sin(¼t=T )

.......

.......

.......

.....

.......

.......

.......

.....

.......

.......

.......

.....

¡T 0 T

t

f(t) = j sin(12!0t)j = j sin(¼t=T )j

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................................................................................................................

The even function f(t) = f(¡t) is expanded in a Fourier series

f(t) =a0

2+

1Xn=1

an cos(n!0t) (1)

where its fundamental frequency is !0 = 2¼=T , in terms of its period T . Note that the

period of the sine wave prior to recti¯cation is 2T . Equivalently, the frequency of the

un-recti¯ed sine wave is half the frequency of the recti¯ed signal f(t).

The DC or average value of f(t) is

a0

2=1

T

TZ0

f(t) dt =1

T

TZ0

sin³!02t´dt =

2

!0Tcos³!02t´¯̄̄̄¯0

t=T

=2

2¼[1¡ cos¼] = 2

¼: (2)

For n = 1; 2; : : :

an =2

T

t0+TZt0

f(t) cos(n!0t) dt =2

T¢ 2

T=2Z0

sin³!02t´cos(n!0t) dt (3)

132

since

f(t) =¯̄̄sin³!02t´¯̄̄= sin

³!02t´

for 0 · t · T=2: (4)

Use of the trig identity

2 sinA cosB = sin(A¡B) + sin(A+B)yields

an =2

T

T=2Z0

½sin£(n+ 1

2 )!0t¤¡ sin£(n¡ 1

2 )!0t¤¾dt: (5)

Let's economize the algebra and do both forms at once by considering

2

T

T=2Z0

sin£(n§ 1

2)!0t¤dt =

2

T

1

(n§ 12 )!0

cos£(n§ 1

2 )!0t¤¯̄̄̄¯0

t=T=2

=1¡ cos£(n§ 1

2 )!0T=2¤

(n§ 12 )!0T=2

=1¡ cos£(n§ 1

2 )¼¤

(n§ 12 )¼

=1

(n§ 12)¼

:

(6)

Insertion of (6) into (5) gives

an =1

¼

·1

n+ 12

¡ 1

n¡ 12

¸=1

¼

(n¡ 12 )¡ (n+ 1

2 )

(n+ 12 )(n¡ 1

2)= ¡ 1

¼(n2 ¡ 14 ): (7)

Together with the DC term (2), the full Fourier series for f(t) is now

f(t) =2

¼¡ 1

¼

1Xn=1

1

n2 ¡ 14

cos(2n¼t=T ); (8)

and its truncated version (Nth partial sum) is

fN (t) =2

¼¡ 1

¼

NXn=1

1

n2 ¡ 14

cos(2n¼t=T ): (9)

..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.......

.......

.....

0 1

t=T

solid curve: N = 3

dashed curve: N = 20

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.........................................................................................

133

Parseval's Theorem for the Fourier Series

Consider a periodic signal f(t) = f(t+ T ) and its Fourier series representation

f(t) =

1Xn=¡1

cnejn!0t (!0 = 2¼=T ):

The average power (per period) \in" or \of" f(t) is

Pav =1

T

t0+TZt0

jf(t)j2 dt

=1

T

t0+TZt0

f(t)f¤(t) dt =1

T

t0+TZt0

1Xn=¡1

cnejn!0t

1Xm=¡1

c¤me¡jm!0tdt

=

1Xn=¡1

cn

1Xm=¡1

c¤m1

T

t0+TZt0

ej(n¡m)!0tdt =1X

n=¡1cn

1Xm=¡1

c¤m ±mn =1X

n=¡1cnc

¤n

=

1Xn=¡1

jcnj2

F Homework: Find the comparable expression in terms of the trigonometric Fourier

series coe±cients.

134

Homework Problem: Power Supply Performance.

..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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.....

¡T 0 T

t

sin(12!0t) = sin(¼t=T )

.......

.......

.......

.....

.......

.......

.......

.....

.......

.......

.......

.....

¡T 0 T

t

x(t) = j sin(12!0t)j = j sin(¼t=T )j

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................................................................................................................

x(t) =2

¼¡ 1

¼

1Xn=1

1

n2 ¡ 14

cos(2n¼t=T )

Consider a real power supply, consisting of a full-wave recti¯er followed by a simple low-

pass ¯lter. Find an expression for, and graph, the time-domain output y(t) of the power

supply for several values of the important parameter !b=!0. De¯ne the percent ripple as

the ratio of average power in the unwanted harmonics to the desired DC power. Graph

percent ripple versus !b=!0. Recall Parseval's theorem for Fourier series.

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..

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..

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............................................

²

²R

C

+

¡

x(t)

+

¡

y(t)

135

Solution. The transfer function of the low-pass ¯lter is

H(!) =

1

j!C

R+1

j!C

=1

1 + j!RC=

1

1 + j!=!bwith !b ,

1

RC:

In polar form

H(!) = jH(!)jej£(!)

the magnitude of H(!) is

jH(!)j = 1p1 + (!=!b)2

and the phase of H(!) is

£(!) = ¡ tan¡1(!=!b):Note that the ¯lter impulse response is necessarily real, since the ¯lter is an actual, physical

electric circuit. Also note that H(0) = 1 is real; that is, £(0) = 0. With the ¯lter input

written as the particular Fourier series

x(t) =

1Xn=0

an cos(n!0t);

the response of the ¯lter is

y(t) =

1Xn=0

anjH(n!0)j cos[n!0t+£(n!0)]:

Note that the DC term is now called a0, instead of a0=2. Recall that !0 , 2¼=T is the

fundamental frequency of the recti¯ed sine, which is twice the frequency of the pure sine

wave that is the initial input to the full-wave recti¯er. Speci¯cally, the above expression is

y(t) =2

¼¡ 1

¼

1Xn=1

1

n2 ¡ 14

1p1 + (n!0=!b)2

cos£n!0t¡ tan¡1(n!0=!b)

¤:

Note that the single (dimensionless!) numerical parameter required to characterize the

complete power supply output is !0=!b. Time t is always multiplied by !0 = 2¼=T , so

that in terms of normalized time t=T , we have

y

μt

T

¶=2

¼¡ 1

¼

1Xn=1

1

n2 ¡ 14

1p1 + (n!0=!b)2

cos

·2n¼

t

T¡ tan¡1(n!0=!b)

¸: (1)

The form of the Fourier series for y(t) is, with An = anjH(n!0)j and Án = £(n!0),

y(t) =

1Xn=0

An cos(n!0t+ Án)

136

and the total average power in the output y(t) is

Ptot =1

T

t0+TZt0

jy(t)j2 dt

=1

T

t0+TZt0

" 1Xn=0

An cos(n!0t+ Án)

#" 1Xm=0

Am cos(m!0t+ Ám)

#¤dt

=

1Xn=0

1Xm=0

AnA¤m

1

T

t0+TZt0

cos(n!0t+ Án) cos(m!0t+ Ám) dt:

The DC terms (n = 0 and/or m = 0) are a little di®erent, but I included them in the series

(starting at n = 0 instead of starting at n = 1) for notational convenience. Recall or note

that Á0 ´ 0. The inner products of interest are:

1

T

t0+TZt0

1 ¢ 1 dt = 1 (n = m = 0)

1

T

t0+TZt0

1 ¢ cos(m!0t+ Ám) dt = 0 (n = 0;m = 1; 2; : : : )

1

T

t0+TZt0

cos(n!0t+ Án) cos(m!0t+ Ám) dt =1

2±nm (n;m = 1; 2; : : : ):

Therefore, the total average power in the signal y(t) is

Ptot = jA0j2 + 12

1Xn=1

jAnj2;

which is the form of Parseval's theorem that applies for our particular representation of

y(t). The DC (n = 0) term is our desired power supply output, while all of the higher

harmonics (n = 1; 2; : : : ) give rise to the unwanted \ripple." Therefore, one de¯nition of

percent ripple is

r , unwanted power in the ripples

total power output£ 100%

or

r =

12

1Pn=1

jAnj2

jA0j2 + 12

1Pn=1

jAnj2£ 100%: (2)

137

However, the problem statement evidently de¯ned percent ripple as

r =

12

1Pn=1

jAnj2

jA0j2 £ 100%:

These two r expressions will be nearly identical when the majority of the output power is

in the DC component, but for a very poor power supply I suppose some marketing/sales

people would prefer to use (2). The second form might be greater than 100%!

¡0:75 ¡0:50 ¡0:25 0.00 0.25 0.50 0.75

t=T

¡0:2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

y(t)

!i=!b = 0:1

10

1

input frequency of unrecti¯ed sine is !i = !0=2

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...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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................................................................................................................................................

0 1 2 3 4 5

!i=!b = freq of unrecti¯ed sine/¯lter break

0

2

4

6

8

10

12

14

16

18

20

r (%)

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138

Periodic Excitation of a Simple GLC Electric Circuit

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................ig(t)

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L

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...

² ²

² ²

G C

v(t)

.................................................................................

If the source signal ig(t) = ig(t + T ) is a periodic signal, then the steady-state solution

for all of the linear signals (common voltage v(t), all three branch currents) will also be

periodic with the same fundamental frequency !0 = 2¼=T of the excitation. If we know

the waveform ig(t), then we can ¯nd its Fourier series representation

ig(t) =

1Xn=¡1

cnejn!0t: (1)

Here is a case where it is more convenient and natural to use the complex exponential

form of the Fourier series: The trigonometric form would su±ce in principle, but carrying

around two sets of coe±cients an and bn is more awkward. Plus di®erentiating exp(jn!0t)

is much cleaner than di®erentiating cos(n!0t) and sin(n!0t).

Kirchho®'s current law supplies the equation-of-motion for the unknown voltage signal

v(t) in our circuit

Gv(t) + Cdv(t)

dt+1

L

tZ¡1

v(¿) d¿ = ig(t): (2)

This integrodi®erential equation is converted to a pure di®erential equation by di®erenti-

ating once to eliminate the integral operator in the inductive current

Cd2v(t)

dt2+G

dv(t)

dt+1

Lv(t) = i0g(t): (3)

The order of the terms was rearranged to put the di®erential equation in standard form,

and the prime on the generator current denotes di®erentiation with respect to its argument,

which is our independent variable t. The periodic voltage signal is written as a Fourier

series

v(t) =

1Xn=¡1

°nejn!0t (4)

139

where the Fourier coe±cients °n are unknown, so far. Insertion of both series (1) and (4)

into the di®erential equation (3) gives

1Xn=¡1

°n

·(jn!0)

2C + (jn!0)G+1

L

¸ejn!0t =

1Xn=¡1

(jn!0)cnejn!0t: (5)

Since the above two Fourier series in (5) are equal for all t and they are written in terms

of the same Fourier basis functions exp(jn!0t), the corresponding coe±cients are equal

°n

·(jn!0)

2C + (jn!0)G+1

L

¸= (jn!0)cn: (6)

Therefore the °n are given explicitly in terms of the known cn

°n =jn!0

(jn!0)2C + (jn!0)G+1

L

cn =jn!0L

1¡ (n!0)2LC + jn!0LG cn; (7)

and our particular solution (4) for the voltage is

v(t) =

1Xn=¡1

jn!0L

1¡ (n!0)2LC + jn!0LG cnejn!0t: (8)

Note that the (SI or mksC) units of the numerator are those of !0L, which are (H/s) or

(−). All three terms in the denominator are dimensionless (check for yourself), and the

Fourier coe±cients cn de¯ned in (1) are in (A). The units of the °n are therefore (V), as

required.

Problems

1. Find an expression for the average power dissipated as heat in this circuit.

2. Give a physical explanation for the absence (zero value) of the n = 0 term in the voltage.

3. If the lossy element is removed from the circuit (G = 0 or G¡1 = R =1), what is thephysical signi¯cance of a harmonic having n!0 = 1=

pLC ?

4. Write down an approximation for the higher-order terms in the series for v(t), that

is as n ! +1. (The terms with negative indices behave similarly.) Interpret this highfrequency approximation in terms of the circuit.

5. If f(t) is a real function of time, show that the coe±cients cn in its complex exponential

Fourier series

f(t) =

1Xn=¡1

cnejn!0t

must satisfy the requirement cn = c¤¡n. One way to see this is to examine f

¤(t) = f(t).

6. If ig(t) is a real-valued source current, show that the expression (8) is also real.

7. Graph the signal v(t) when the current source is a square wave, say of amplitude §I0.Show the e®ect of varying the circuit element values. Use normalized parameters, as much

as possible, to simplify the results.

140

Simple Low-Pass and Hi-Pass Filter Response to a Periodic Input

Problem: Find expressions for and graph the signals ya(t) and yb(t) for several values of

the important (nondimensional) parameters that appear naturally.

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¡T 0 T 2Tt

................... Ax(t)

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..

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............................................

²

²R

C

+

¡

x(t)

+

¡

ya(t)

............................

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..

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..

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.......

......

.......

..

²

²C

R

+

¡

x(t)

+

¡

yb(t)

Note that the excitation x(t) has both even and odd parts, but the even part is simply

A=2. The graph of ex(t) = x(t)¡A=2 is clearly an antisymmetric (odd) function, so that itsFourier series is most logically written in the trigonometric form with only sine functions.

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.........................................................................................................................................................................................................................................................................................................

.........................................................................................¡T 0 T 2Tt

...................A=2

...................¡A=2

ex(t) = x(t)¡A=2

ex(t) = ¡ex(¡t) ex(t) = Aμ tT¡ 12

¶(0 < t < T=2)

ex(t) = 1Xn=1

bn sin(n!0t) (!0 , 2¼=T )

bn =2

T

T=2Z¡T=2

ex(t) sin(n!0t) dt = 4

T

T=2Z0

ex(t) sin(n!0t) dt=4A

T

T=2Z0

μt

T¡ 12

¶sin(n!0t) dt

141

T=2Z0

sin(n!0t) dt =cos(n!0t)

n!0

¯̄̄̄0t=T=2

=1¡ cos(n!0T=2)

n!0=1¡ cos(n¼)

n!0=1¡ (¡1)nn!0

=

8<:2

n!0n odd

0 n even

(but this form is not as useful as it might appear)

Zu dv = uv¡

Zv du u = t du = dt dv = sin(n!0t) dt v = ¡cos(n!0t)

n!0

T=2Z0

t sin(n!0t) dt =t cos(n!0t)

n!0

¯̄̄̄0t=T=2

+1

n!0

T=2Z0

cos(n!0t) dt

= ¡ T

2n!0cos(n!0T=2) +

sin(n!0t)

(n!0)2

¯̄̄̄T=2t=0

= ¡ T

2n!0cos(n¼) +

sin(n¼)

(n!0)2

=¡T (¡1)n2n!0

Now go back and assemble the required Fourier coe±cient

bn =4A

T

·¡(¡1)n2n!0

¡ 1¡ (¡1)n

2n!0

¸=

¡4A2n!0T

=¡An¼:

The Fourier series representation of the input signal is therefore

x(t) =A

2¡ A¼

1Xn=1

sin(n!0t)

n:

The transfer functions of our two ¯lters, from a simple frequency domain voltage divider,

are

Ha(!) =

1

j!C

R+1

j!C

=1

1 + j!RC= jHa(!)jej£a(!)

Hb(!) =R

R+1

j!C

=j!RC

1 + j!RC= jHb(!)jej£b(!):

Let's not write out the polar representations explicitly: MATLAB will compute those for

us. That is, assume the magnitude function jHa(!)j and £a(!) are known (they are!), andobviously the same for the b circuit. Note that Ha(0) = 1 and Hb(0) = 0.

142

The Fourier series for the two output signals are therefore

ya(t) =A

2¡ A¼

1Xn=1

jHa(n!0)jn

sin£n!0t+£a(n!0)

¤

yb(t) = ¡A¼

1Xn=1

jHb(n!0)jn

sin£n!0t+£b(n!0)

¤Let's use normalized time t=T and normalize by the amplitude A of the input to write

ya(t=T )

A=1

2¡ 1

¼

1Xn=1

jHa(n!0)jn

sin£2n¼t=T +£a(n!0)

¤yb(t=T )

A= ¡ 1

¼

1Xn=1

jHb(n!0)jn

sin£2n¼t=T +£b(n!0)

¤We need

Ha(n!0) =1

1 + jn!0RC=

1

1 + jn!0=!b¡¡¡!n!1

!b

jn!0

Hb(n!0) =jn!0RC

1 + jn!0RC=

jn!0=!b

1 + jn!0=!b¡¡¡!n!1

1

where

!b ,1

RC

is the break frequency (or cut-o® frequency or half-power frequency) of (both of) the ¯lters.

Therefore, the important nondimensional parameter of our problem is !0=!b = !0RC.

Recall that !0 = 2¼=T is the fundamental frequency of the periodic excitation.

As n!1, the terms in the series for the low-pass ¯lter output ya(t) vary like n¡2 whilethe higher-order terms in the high-pass ¯lter output yb(t) vary like n

¡1. Therefore, thenumerical convergence of the signal ya(t) is much faster. That is, truncation of the ya(t)

series at some ¯nite value of n will give a much better approximation for the in¯nite series

than will the same truncation of the series for yb(t). Note that the outputs ya(t) and yb(t)

are intimately related since

Ha(!) +Hb(!) = 1 =) ya(t) + yb(t) = x(t):

Therefore, we can numerically evaluate the series for ya(t) (because its series converges

faster), and then avoid computing the poorly convergent series for yb(t) simply by using

the known relationship

yb(t) = x(t)¡ ya(t):

143

...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................t=T

!0=!b = 0:1

ya(t)

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........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................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...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................t=T

yb(t)

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144

...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................t=T

!0=!b = 1

ya(t)

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...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................t=T

yb(t)

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145

...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................t=T

!0=!b = 10

ya(t)

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...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................t=T

yb(t)

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146

The Vibrating String

This is one of the most celebrated problems of mathematical physics. And since the early

1950's, the unchallenged lead instrument in the sound track of Western culture has been

constructed around a set of six such steel strings. Let's see what our Fourier analysis can

tell us about this simple and important wave equation.

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.......

.......

.......

.....

x

y(x; t0)

0 a

.................................................................................................................................................................................................................................................................................................................................................................................................................................................

...........................................................................................................................................................................................................................................................................................................................................

.......................................................................................................................................................................

....................................................

Fig. 1. Snapshot of string displacement at time t = t0.

Fig. 1 shows the pro¯le (at some particular time t0) of the vertical de°ection of a string, of

total length a, about its equilibrium position y = 0. The string is ¯xed at each end; hence

y(0; t) = y(a; t) = 0 (1)

are the boundary conditions imposed on our unknown function y(x; t), for all t.

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........................................

........................................

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.....................................................................

.......................................................................................................

..........................................................................................................................................

T

T

μ

μ0

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.......................................................

............................................................................................................................................................

x x+¢x

Fig. 2. Free-body diagram of elemental string length.

147

Consider the free-body diagram (Fig. 2) of a di®erential length of string. If the entire

string is subject to a tensile force T (N), then the sum of the forces in the y-direction,

according to Newton's second law, must be

T sin μ0 ¡ T sin μ = ½¢x@2y

@t2(2)

where ½ is the lineal mass density (kg/m) of the string. For small displacements about

equilibrium

sin μ0 ¼ tan μ0 = @

@xy(x+¢x) and sin μ ¼ tan μ = @

@xy(x) (3)

resulting in@

@xy(x+¢x; t)¡ @

@xy(x; t)

¢x=½

T

@2

@t2y(x; t): (4)

In the limit as ¢x! 0, we have the wave equation·@2

@x2¡ 1

c2@2

@t2

¸y(x; t) = 0 (5)

where c =pT=½ will be the wave speed in (m/s).

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.

.......

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.......

..

0 a

x

Á(x)

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

Fig. 3. Initial displacement of string.

Let's excite our string by holding it in the initial rest position

y(x; 0) = Á(x) (one possibility shown in Fig. 3) (6)

@

@ty(x; 0) = 0 (initially at rest) (7)

and releasing it. We can solve our partial di®erential equation (5) by expanding y(x; t)

into a Fourier series in x, which is equivalent to periodically extending the function y(x)

beyond our original physical domain 0 · x · a. Since y vanishes at the end points x = 0

148

and x = a, we can use a sine series, which is the odd periodic extension of period 2a. So

the form of our sought-after solution is

y(x; t) =

1Xn=1

Bn(t) sin(n¼x=a); (8)

where the functions Bn(t) (Fourier coe±cients for the x-expansion) are unknown at this

point. We need the appropriate Fourier series representation of our initial displacement

(6); write it as

Á(x) =

1Xn=1

bn sin(n¼x=a) (9)

where the bn are known, in principle and in reality, since we know the initial displacement

Á(x):

bn =2

a

aZ0

Á(x) sin(n¼x=a) dx: (10)

Insertion of our solution form (8) into the wave equation (5) yields

1Xn=1

·¡³n¼a

´2¡ 1

c2d2

dt2

¸Bn(t) sin(n¼x=a) = 0: (11)

This is itself a Fourier series representation of zero, so each coe±cient must be zero; that

is ·d2

dt2+³n¼ca

´2 ¸Bn(t) = 0: (12)

The solution of this homogeneous di®erential equation is a linear combination of½cos

sin

¾μn¼ct

a

¶:

We select the cosine (discard the sine) in accordance with the initial condition (7). Now

we have Bn(t) = ¯n cos(n¼ct=a) in terms of (unknown) constants ¯n, which gives

y(x; t) =

1Xn=1

¯n cos(n¼ct=a) sin(n¼x=a): (13)

Evaluation at t = 0, according to the initial condition (6), provides ¯n = bn, and so

y(x; t) =

1Xn=1

bn cos(n¼ct=a) sin(n¼x=a): (14)

149

The physical interpretation of our solution(!) (this is it!!!) is much clearer upon rewriting

it (via a trig identity) as

y(x; t) =1

2

1Xn=1

bn

nsinhn¼a(x¡ ct)

i+ sin

hn¼a(x+ ct)

io(15)

or, ¯nally,

y(x; t) =1

2

h~Á(x¡ ct) + ~Á(x+ ct)

i(16)

where ~Á(¢) means the odd periodic extension (Fig. 4) of the initial displacement Á(x) asgiven in the sine series (9):

~Á(x) =

1Xn=¡1

Á(x¡ 2na)¡1X

n=¡1Á[¡x¡ (2n+ 1)a]: (17)

In (17) we are taking Á(») to be zero outside of the original domain 0 · » · a. Also notethat ~Á(x¨ ct) is a wave propagating with speed c in the §x-direction.

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....................................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................................................................................................................................................................................................................................

...

¡a ..........

0..........

a..........

2ax

~Á(x)

Fig. 4. Odd periodic extension of initial displacement of string.

Problems

1.) Verify that any functions f(x + ct) and g(x ¡ ct) satisfy the wave equation (5).Certainly, then, (16) is a solution. Such forms are called the d'Alembert solution.

2.) Verify that (16) satis¯es the initial conditions (6) and (7), and the homogeneous

boundary conditions (1).

3.) Why do we say that f(x + ct) and g(x ¡ ct) are propagating in the ¨x-directions,respectively, with speed c?

150

Poisson Sum Formula

Let f(t) be any function that is Fourier transformable. Then its periodic extension can be

represented by a Fourier series:

g(t) =

1Xn=¡1

f(t+ nT ) =

1Xk=¡1

ckej2k¼t=T

with Fourier coe±cients

ck =1

T

TZ0

g(t)e¡j2k¼t=Tdt

=1

T

1Xn=¡1

TZ0

f(t+ nT )e¡j2k¼t=T dt

with ¿ = t+ nT

=1

T

1Xn=¡1

(n+1)TZnT

f(¿)e¡j2k¼¿=T ej2kn¼d¿

=1

T

1Z¡1

f(¿)e¡j2k¼¿=Td¿ =1

TF

μ2k¼

T

¶:

The resulting Poisson sum formula

1Xn=¡1

f(t+ nT ) =1

T

1Xk=¡1

F

μ2k¼

T

¶ej2k¼t=T

is often written in the common form, with t = 0 and T = 1 such that

1Xn=¡1

f(n) =

1Xk=¡1

F (2k¼):

F Homework. Sum the slowly converging series

1Xn=¡1

1

n2 + b2=¼

bcoth¼b:

151

Solution. Recall that the Fourier transform of g(t) = exp(¡®jtj) with ® > 0 is

G(!) =

0Z¡1

e(®¡j!)tdt+

1Z0

e¡(®+j!)tdt =1

®¡ j! +1

®+ j!=

2®

®2 + !2:

Duality says that the Fourier transform pair

g(t)F() G(!) also gives G(t)

F() 2¼g(¡!):so that the Fourier transform of

f(t) =1

t2 + ®2

is

F (!) =¼

®e¡®j!j:

We will also use the geometric series

1Xk=1

zk =z

1¡ z (jzj < 1):

The Poisson sum formula 1Xn=¡1

f(n) =

1Xk=¡1

F (2k¼)

therefore gives (assume b > 0 and note that the desired sum is even in b)

1Xn=¡1

1

n2 + b2=¼

b

1Xk=¡1

e¡2jkj¼b =¼

b

"1 + 2

1Xk=1

e¡2¼bk#=¼

b

·1 +

2e¡2¼b

1¡ e¡2¼b¸

=¼

b

1 + e¡2¼b

1¡ e¡2¼b =¼

b

e¼b + e¡¼b

e¼b ¡ e¡¼b =¼

bcoth(¼b): ~

It is interesting to look at the limiting case of very small b. If jbj ¿ 1, then

1Xn=¡1

1

n2 + b2¡¡¡!b!0

1

b2+ 2

1Xn=1

1

n2=1

b2+ 2³(2) =

1

b2+¼2

3:

Formula 4.5.67 of Big Red shows that, for jzj < ¼,

coth z =1

z+z

3¡ z3

45+ : : :

so that¼

bcoth(¼b) ¡¡¡!

b!0

¼

b

·1

¼b+¼b

3¡ (¼b)

3

45+ : : :

¸=1

b2+¼2

3+O(b2):

How can you see the behavior of the sum for very large b, that is for b!1 ?

152

Sampling and Reconstruction

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combT (t)

x(t) xs(t)h(t)

y(t)

Consider a continuous-time (analog) signal x(t) that is uniformly sampled at the times

tn = nT (n = : : : ;¡2;¡1; 0; 1; 2; 3; : : : ) so that the sampling frequency is

!s =2¼

T(1)

in terms of the sampling period or interval T . Once x(t) has been sampled, all we have is

a table (list) of numbers x(nT ). But a useful representation of the sampled signal xs(t) in

continuous-time notation is

xs(t) = x(t) combT (t) =

1Xn=¡1

x(nT )±(t¡ nT ): (2)

From our table of Fourier transform properties we need

F£f(t)g(t)¤ = 1

2¼F (!)~G(!): (3)

The easiest way to get this is to take the inverse Fourier transform of the frequency-domain

convolution

F¡1£F (!)~G(!)¤ = 1

2¼

1Z¡1

1Z¡1

F (−)G(! ¡ −) d− ej!t d!

=

1Z¡1

F (−)1

2¼

1Z¡1

G(! ¡ −)ej!td! d−:

The inner integral (in !) is

1

2¼

1Z¡1

G(! ¡ −)ej!td! = 1

2¼

1Z¡1

G(¸)ej(¸+−)td¸ = ej−tg(t)

and therefore

F¡1£F (!)~G(!)¤ = g(t) 1Z¡1

F (−)ej−td− = 2¼f(t)g(t)

which is identical to (3).

Typeset by AMS-TEX

153

The Fourier fransform of the time-domain comb function is

F£combT (t)¤ = 2¼

Tcomb2¼

T

(!) (4)

and thus the spectrum of the sampled signal xs(t) is

Xs(!) =1

2¼X(!)~ 2¼

Tcomb2¼

T

(!) =1

TX(!)~

1Xk=¡1

±(! ¡ k!s)

=1

T

1Xk=¡1

X(! ¡ k!s): (5)

If x(t) is a \low-pass" signal, then

X(!) = 0 for j!j > !h: (6)

This means that the spectrum of x(t) has a highest frequency !h. An example low-pass

spectrum appears in the ¯rst row of the ¯gure below.

X(!)

Xs(!) when !s=2 > !h

Xs(!) when !s=2 < !h

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.....

¡2!s...........

¡!s...........

0...........

2!s

......

.....

!s.............

¡!s=2.............

!s=2

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.....

¡!s...........

!s...........

0.............

¡!s=2.............

!s=2

............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ !...................................................................................................................................................................................................................................................

0...........

¡!h...........

!h

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The middle curve showsXs(!) for the case !s > 2!h and the bottom curve is for !s < 2!h.

If we pass xs(t) through an ideal low-pass ¯lter having transfer function

H(!) = T¦

μ!

!s

¶; (7)

154

then we see that, for all !,

Y (!) = X(!) if !s > 2!h: (8)

In this case, y(t) = x(t) and we have exactly recovered the low-pass signal x(t) from its

sample values! The minimum sampling frequency is called the Nyquist rate or Nyquist

frequency 2!h. If we sample at too low a frequency, as shown in the bottom curve of the

previous ¯gure, then

Y (!)6= X(!) if !s < 2!h (9)

and of course then y(t)6= x(t). The error or distortion that results from the overlap in the

adjacent shifted transforms X(! ¡ k!s) is called aliasing. The whole concept is usuallycalled:

Shannon's sampling theorem. If the low-pass signal x(t) having spectrum X(!) = 0

for j!j > !h is sampled uniformly at the rate !s > 2!h (the Nyquist rate), then x(t) is

exactly recoverable by passing the sampled signal through the ideal reconstruction ¯lter

having transfer function H(!) = T¦(!=!s).

The frequency domain renders this derivation clear and simple. Now let's examine the

output y(t) in the time-domain to get another perspective. The impulse response of the

ideal reconstruction ¯lter is

h(t) = F¡1£T¦(!=!s)¤ = sinc(!st=2) = sinc(¼t=T ); (10)

and this ideal low-pass ¯lter is also called the sinc interpolation ¯lter. Independent of x(t)

being a low-pass signal or not, and for an arbitrary sampling frequency (that is, ignoring

both Nyquist and Shannon), the output of the ¯lter in response to the sampled signal (2)

is always

y(t) = xs(t)~ h(t) =1X

n=¡1x(nT )±(t¡ nT )~ h(t) =

1Xn=¡1

x(nT )h(t¡ nT )

=

1Xn=¡1

x(nT ) sinc

·¼

μt

T¡ n

¶¸: (11)

If k is an integer, observe that

sinc(k¼) =sin(k¼)

k¼=

½1; k = 0

0; k6= 0 = ±k0; (12)

in terms of the Kronecker-delta. Therefore, we see that at the sample times t = mT , the

output is always equal to the input in that

y(mT ) =

1Xn=¡1

x(nT ) sinc [¼ (m¡ n)] =1X

n=¡1x(nT )±mn = x(mT ): (13)

155

Dissection of the Time-Domain Output of the sinc Interpolation Filter: Consider

a small number (four) of samples x[n] = x(nT ) for n = 0; 1; 2; 3. If all other (n = 4; 5; : : :

and n = ¡1;¡2; : : : ) of the x[n] are zero (or absent), then the output (11) of the sincinterpolation ¯lter is

y(t) = x[0] sinc

μ¼t

T

¶+ x[1] sinc

·¼

μt

T¡ 1¶¸

+ x[2] sinc

·¼

μt

T¡ 2¶¸+ x[3] sinc

·¼

μt

T¡ 3¶¸:

Each of these individual terms are graphed below for the example data

x[0] = 0:5 x[1] = 1:0 x[2] = 0:8 x[3] = 0:6

along with the sum of all four terms. Do you see which curve is which?

¡2 ¡1 0 1 2 3 4 5

t=T

¡0:4

¡0:2

0.0

0.2

0.4

0.6

0.8

1.0

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²

²

²

²

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156

DISCRETE-TIME SIGNALS, SYSTEMS, AND TRANSFORMS

A common source for a discrete-time signal is to uniformly sample a continuous-time signal,

as in

x[n] = x(nT ) n = : : : ;¡2;¡1; 0; 1; 2; 3; : : :The discrete-time signal x[n] is also called a sequence; it is a function of the integers

n. Usually we are not concerned with any underlying sample period T that was used to

obtain our sequence from some original x(t). Once we're in the discrete-time world, the

independent time variable is simply the integers n. Our sequence x[n] is simply a list or

table of ordered numbers. Note that the sequence x[n] can take on real or complex values.

Many of our familiar continuous-time signals have discrete-time analogues or cousins, but

be aware that their behavior is often (usually? always?) quite di®erent. For starters, there

is no such thing as continuity in discrete-time. The continuous-time calculus operators of

di®erentiation and integration have analogous operators of di®erencing and accumulation

in discrete-time. Sometimes the discrete-time version of a common signal, for example

the Heaviside unit-step function u[n], goes by the same name and similar symbol as its

continuous-time counterpart. In other cases, di®erent names are used. I suppose the

Heaviside unit-step sequence is a better name that draws attention to its di®erence with

the continuous-time signal u(t).

Heaviside unit-step sequence.

u[n] ,½0; n = ¡1;¡2; : : :1; n = 0; 1; 2; : : :

..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................² ² ²

² ² ² ² ²

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.

n

u[n]: : : : : :

¡3 ¡2 ¡1 0 1 2 3 4

Unlike u(t) that is not well-de¯ned at t = 0, the unit-step sequence clearly takes on the

value u[0] = 1. Note, however that u[0:5] is complete notational nonsense.

Kronecker delta sequence. (also called unit impulse sequence)

±[n] ,½0; n = §1;§2;§3; : : :1; n = 0

..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................² ² ²

²

² ² ² ²................................................................

n

±[n]: : : : : :

¡3 ¡2 ¡1 0 1 2 3 4

Unlike the continuous-time Dirac-delta function ±(t) that is illegal at t = 0, the discrete-

time Kronecker-delta sequence clearly takes on the value ±[0] = 1. The Kronecker-delta

possesses the sampling property

x[n] =

1Xk=¡1

x[k]±[k ¡ n]:

Relationship.

±[n] = u[n]¡ u[n¡ 1] u[n] =

nXk=¡1

±[k]

157

Complex exponential. First, recall that for any positive real number ! (called the

frequency), the continuous-time complex exponential

x(t) = ej!t

is always periodic in time t with period T = 2¼=!, since

x(t+ T ) = ej!(t+T ) = ej!T ej!t = ej!t = x(t)

whenever !T = 2¼, since e2¼j = 1.

The discrete-time complex exponential is substantially di®erent. Consider the discrete-

time sequence

x[n] = ej−n

where the frequency − is any real number: Take it to be positive for starters. Is x[n]

periodic in the independent time variable n ? Let's denote its candidate period by the

speci¯c integer N , and examine

x[n+N ] = ej−(n+N) = ej−Nej−n:

If x[n] is periodic with the period N , then x[n+N ] = x[n] which requires that

−N = 2k¼

where k is any integer. This only happens if

− =k

N2¼ = a rational multiple of 2¼:

However, the discrete-time complex exponential is a periodic function of the frequency

variable −, since

ejn(−+2¼) = e2n¼jejn− = ejn−:

Therefore, in discrete-time work, we only have to consider frequencies − in the primary

range ¡¼ < − · ¼, for example. The highest frequency possible is − = ¼ (or equivalently,¡¼), and the lowest frequency is − = 0 (which is identical with − = 2¼).

158

Exercises.

1. Write an expression, in terms of n, N ,

and u[¢], for the given sequence x[n].

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................² ² ² ²

²

²

²

²

²

²

² ²

1

2

3

N¡1

N

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n

x[n]

: : : : : :

¡3 ¡2 ¡1 0 1 2 3 : : : N¡1 N N+1

2. Graph the exponential (or geometric) sequence x[n] = anu[n].

Look at the following speci¯c cases:

a = 1=2; a = ¡1=2; a = 2; a = rej− where r = 1=2 and − = 1

Discrete-Time Systems

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................................................... ............. ...................................................................................... .............x[n] y[n]

The equation-of-motion, that describes the relationship between the input x[n] and the

output y[n], for a discrete-time system is often a di®erence equation. An example of a

second-order, constant-coe±cient di®erence equation is

c0y[n] + c1y[n¡ 1] + c2y[n¡ 2] = b0x[n] + b1x[n¡ 1]:When accompanied by two (the di®erence equation is second order) initial conditions of

the form

y[¡1] = y¡1 and y[¡2] = y¡2;we have a discrete-time initial value problem. Classical solution techniques are analogous

to the classical methods used to solve ordinary di®erential equations in continuous-time.

Di®erence equations can sometimes be solved recursively:

y[n] = ¡c1c0y[n¡ 1]¡ c2

c0y[n¡ 2] + b0

c0x[n] +

b1

c0x[n¡ 1]:

The discrete-time version of the Laplace transform (or Fourier transform) is called the

Z-transform, and it, too, provides an attractive path to solve discrete-time di®erenceequations by converting them into purely algebraic equations in the transform (z) domain.

The inverse Z-transform gets our solution back to the time [n] domain.

159

Linear Time-Invariant Discrete-Time Systems

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................................................... ............. ...................................................................................... .............x[n] y[n] = Tfx[n]g

Assume that the zero-state response sequence y[n] to the exciting sequence x[n] is given

by the operator or system transformation rule y[n] = Tfx[n]g. If the operator T satis¯esthe superposition property

Tfc1x1[n] + c2x2[n]g = c1Tfx1[n]g+ c2Tfx2[n]g

for any values of the constants c1 and c2 and for arbitrary sequences x1[n] and x2[n], then

the discrete-time system is linear.

Given that the zero-state response to the input sequence x[n] is y[n], if the response to the

delayed input x[n¡n0] is the delayed response y[n¡n0], then the system is time-invariant.

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................................................... ............. ...................................................................................... .............x[n] y[n] = Tfx[n]g

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................................................... ............. ...................................................................................... .............x[n¡ n0] y[n¡ n0] = Tfx[n¡ n0]g

Input/Output relationship. Using the sampling property of the Kronecker-delta se-

quence, write the input sequence as

x[n] =

1Xk=¡1

x[k]±[n¡ k]:

The system operator T operates only on functions of the running time variable n, treating

all other variables as constants, so that the T operator can be moved inside the summation

in

y[n] = Tfx[n]g = T( 1Xk=¡1

x[k]±[n¡ k])=

1Xk=¡1

x[k]Tf±[n¡ k]g:

The response to a Kronecker-delta or unit impulse sequence is the system impulse response

h[n] = Tf±[n]g. By the (assumed or de¯ned) time-invariance of the linear system, the

160

response due to a delayed unit impulse ±[n¡ k] is the delayed h[n¡ k] = Tf±[n¡ k]g, andtherefore the zero-state response is

y[n] =

1Xk=¡1

x[k]h[n¡ k] = x[n]~ h[n]:

This is the discrete-time convolution sum between the two signals x[n] and h[n].

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................................................... ............. ...................................................................................... .............x[n] y[n] = x[n]~ h[n]h[n]

Example of a LTI Discrete-Time System. Let's look at one particular discrete-time

system in detail and study its input/output dynamics from several perspectives. One

common way to specify a discrete-time system is via a system block diagram. We need to

¯rst introduce several basic components of such system block diagrams.

.......

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.....

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Pf [n]

g[n]

f [n] + g[n]summer

.....................................................................................

............................

................................................................

.............................................................................. .............. .............................................................................. ..........

....®f [n] ®f [n]ampli¯er/attenuator

............................................................................................................................................................................................................................

.............................................................................. .............. .............................................................................. ..........

....z¡1f [n] f [n¡ 1]unit delay

The reason that the unit delay is represented by the symbol z¡1 will be clear when westudy the Z-transform.

161

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.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

z¡1®

Px[n] y[n]²

The output of the summer is

y[n] = x[n] + ®y[n¡ 1]

and so the discrete-time system depicted by the block diagram above is governed by an

equation-of-motion that is a linear, constant coe±cient, ¯rst-order di®erence equation.

The impulse response h[n] is the response y[n] to the input sequence x[n] = ±[n], and being

a zero-state response, its initial condition is h[¡1] = 0. Sometimes di®erence equations

can be solved recursively, and this is one

h[n] = ®h[n¡ 1] + ±[n]h[¡1] = 0h[0] = ®h[¡1] + 1 = 1h[1] = ®h[0] + 0 = ®

h[2] = ®h[1] = ®2

h[3] = ®h[2] = ®3

...

h[n] = ®nu[n]

With the unit impulse response h[n] in hand, the zero-state response due to any input

sequence is readily computed via the convolution sum. For example, the unit step response

is

s[n] = u[n]~ h[n] =1X

k=¡1h[k]u[n¡ k] =

nXk=¡1

h[k] =

nXk=¡1

®ku[k] =

nXk=0

®k

=1¡ ®n+11¡ ® u[n]:

The required ¯nite geometric series is detailed two pages ahead.

162

The Z-Transform

In order to derive the Z-transform from the Laplace transform, we need to represent our

sequence x[n] in continuous-time notation: Take a step backwards, so to speak, and write

a sampled (discretized) continuous-time signal as

xd(t) = x(t) combT (t) =

1Xn=¡1

x(nT )±(t¡ nT ) =1X

n=¡1x[n]±(t¡ nT ):

t...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

²

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²

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T............................................. ................ .............................................................

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The bilateral Laplace transform of a continuous-time signal f(t) is de¯ned as the integral

Lff(t)g = F (s) =1Z

¡1f(t)e¡stdt

where s = ¾+j! is a complex frequency variable. The Laplace-transform of the discretized

signal xd(t) is therefore

Xd(s) =

1Xn=¡1

x[n]

1Z¡1

±(t¡ nT )e¡stdt =1X

n=¡1x[n]e¡nsT :

Introduce the transformed variable

z = esT

so that the above Laplace-transform is now called the Z-transform

Zfx[n]g = X(z) =1X

n=¡1x[n]z¡n:

The Z-transform converts a function of the integers n (a sequence) to a function of a

complex variable z.

163

Geometric series. Firstly, consider the ¯nite geometric series

S =

NXn=0

an

where a is an arbitrary complex number. There is no question about the convergence

of this series, since it is simply the sum of a ¯nite number of terms. We can obtain a

closed-form expression for the sum S by this neat trick:

S = 1+a+ a2 + a3 + ¢ ¢ ¢+ aN¡1 + aNaS = a+ a2 + a3 + ¢ ¢ ¢+ aN¡1 + aN + aN+1

S ¡ aS = 1¡aN+1

S(1¡ a) = 1¡ aN+1 =) S =

NXn=0

an =1¡ aN+11¡ a

Note that when a = 1, we have S = N +1, which is the limit as a! 1 of the above result.

Similarly, the expression for the in¯nite geometric series

1Xn=0

an =1

1¡ a (jaj < 1)

explicitly displays its region-of-convergence jaj < 1.

Z-transform example: causal exponential sequence. Consider the sequence

x[n] = anu[n]

where a is an arbitrary complex number. Then the Z-transform of x[n] is

X(z) =

1Xn=0

³az

´n=

1

1¡ az

=z

z ¡ a (jzj > jaj):

The region-of-convergence in the complex z-plane for X(z) to be a valid transform of x[n]

is the exterior of the circle of radius jzj = jaj.

164

Fibonacci Sequence - Di®erence Equation and Z-Transform

f [n] = f [n¡ 1] + f [n¡ 2] initial conditions: f [¡2] = 0; f [¡1] = 1

Zff [n]g = F (z) =1Xn=0

f [n]z¡n

Zff [n¡ 1]g = f [¡1] + z¡1F (z); Zff [n¡ 2]g = f [¡2] + z¡1f [¡1] + z¡2F (z)F (z) = 1 + z¡1F (z) + z¡1 + z¡2F (z)

F (z) =1 + z¡1

1¡ z¡1 ¡ z¡2 =z(z + 1)

z2 ¡ z ¡ 1

F (z)

z=

z + 1

z2 ¡ z ¡ 1 =z + 1³

z ¡ 1+p5

2

´³z ¡ 1¡p5

2

´=

12

³1 + 3p

5

´z ¡ 1+

p5

2

+

12

³1¡ 3p

5

´z ¡ 1¡p5

2

via partial fraction expansion

f [n] =1

2

μ1 +

3p5

¶Ã1 +

p5

2

!n+1

2

μ1¡ 3p

5

¶Ã1¡p52

!nf = 0; 1; 1; 2; 3; 5; 8; 13; 21; 34; 55; 89; 144; : : : for n = ¡2;¡1; 0; 1; 2; : : :

f [n+ 1]

f [n]¡¡¡!n!1

1 +p5

2

165

Convolution Theorem of the Z-Transform

Zff [n]~ g[n]g =1X

n=¡1

( 1Xk=¡1

f [k]g[n¡ k])z¡n

=

1Xk=¡1

f [k]

" 1Xn=¡1

g[n¡ k]z¡n#

(let m = n¡ k) =

1Xk=¡1

f [k]

" 1Xm=¡1

g[m]z¡(m+k)#

=

1Xk=¡1

f [k]z¡k1X

m=¡1g[m]z¡m

= F (z)G(z)

Z-Transform of a Delayed Sequence.

Zff [n¡ n0]g =1X

n=¡1f [n¡ n0]z¡n =

1Xm=¡1

f [m]z¡(m+n0) = z¡n01X

m=¡1f [m]z¡m

= z¡n0F (z)

System Transfer Function.

y[n] = x[n]~ h[n] Z() Y (z) = X(z)H(z)

H(z) =Y (z)

X(z)

Recall the system block diagram for the example four pages back. The Z-transform of thegoverning di®erence equation

y[n]¡ ®y[n¡ 1] = x[n]

is

(1¡ ®z¡1)Y (z) = X(z)and therefore the system transfer function is

Zfh[n]g = H(z) = 1

1¡ ®z¡1 =z

z ¡ ®:

166

Discrete-Time Fourier Transform. With z written in polar form as z = rej−, if we

restrict z to lie on the unit-circle (jzj = r = 1), then we have the discrete-time Fourier

transform

X(z)¯̄̄jzj=1

= X(ej−) =

1Xn=¡1

x[n]e¡jn−:

The discrete-time Fourier transform (DTFT) is a mapping from a function of the integers

n to a periodic function of the continuous real variable −, with period 2¼. Observe that

the DTFT is essentially a Fourier series representation of the periodic function X(ej−),

where x[n] are the Fourier coe±cients! Therefore, the inverse DTFT is simply obtained

by applying the operator

1

2¼

¼Z¡¼

d− ejm−

to the DTFT above

1

2¼

¼Z¡¼X(ej−)ejm−d− =

1Xn=¡1

x[n]1

2¼

¼Z¡¼

ej(m¡n)−d−

| {z }±mn

= x[m]:

The discrete-time Fourier transform pair is therefore

DTFTfx[n]g = X(ej−) =1X

n=¡1x[n]e¡jn−

DTFT¡1fX(ej−)g = x[n] = 1

2¼

¼Z¡¼X(ej−)ejn−d−:

Discrete Fourier Transform. Consider a ¯nite or partial sequence x[n], where our

attention is centered on the indices n = 0; 1; 2; : : : ; N ¡ 1. Its DTFT is

X(ej−) =

N¡1Xn=0

x[n]e¡jn−:

If we now sample the DTFT at N values of −

−k =2k¼

N(k = 0; 1; 2; : : : ; N ¡ 1)

that are uniformly distributed around the unit circle, we have

X(ej−k) =

N¡1Xn=0

x[n]e¡j2nk¼=N :

167

This is a mapping from a function x[n] of the integers n = 0; 1; 2; : : : ; N ¡ 1 to anotherfunction of the same integers k = 0; 1; 2; : : : ; N¡1. Rather than write it as the cumbersomeX(ej−k), we focus our attention on the discrete frequency parameter k and write it as the

discrete Fourier transform or DFT

X[k] =

N¡1Xn=0

x[n]e¡j2nk¼=N (k = 0; 1; 2; : : : ; N ¡ 1):

The inverse DFT is most easily derived by ¯rst looking at the comparable discrete-time

spectral representation of the Kronecker-delta sequence

±[n] =1

N

N¡1Xk=0

e§j2nk¼=N :

The truth of this last equation can be established in a number of ways. Observe that

the sum is a ¯nite geometric series (page 163): The result follows at once. Therefore,

application of the operator

1

N

N¡1Xk=0

ej2mk¼=N (m = 0; 1; 2; : : : ; N ¡ 1)

to the DFT above gives

1

N

N¡1Xk=0

X[k]ej2mk¼=N =

N¡1Xn=0

x[n]1

N

N¡1Xk=0

ej2(m¡n)k=N| {z }±mn

= x[m]:

Hence, the DFT transform pair is

DFTfx[n]g = X[k] =N¡1Xn=0

x[n]e¡j2nk¼=N (k = 0; 1; 2; : : : ; N ¡ 1)

DFT¡1fX[k]g = x[n] = 1

N

N¡1Xk=0

X[k]ej2nk¼=N (n = 0; 1; 2; : : : ; N ¡ 1):

The DFT (FFT is the fast algorithm that e±ciently computes the DFT) is therefore a

transformation that converts N numbers into N other numbers. But each of the sets of

N numbers, x[n] and X[k], are a single period of periodic sequences, each with period N .

To see this, observe that

x[n+N ] =1

N

N¡1Xk=0

X[k]ej2(n+N)k¼=N =1

N

N¡1Xk=0

X[k]ej2nk¼=Nei2k¼

=1

N

N¡1Xk=0

X[k]ej2nk¼=N = x[n]

168

and

X[k +N ] =

N¡1Xn=0

x[n]e¡j2(k+N)n¼=N =N¡1Xn=0

x[n]e¡j2nk¼=Ne¡j2n¼

=

N¡1Xn=0

x[n]e¡j2nk¼=N = X[k]:

Exercises.

1. Find the Z-transform and its region-of-convergence for the ¯nite, rectangular pulse

sequence x[n] = u[n]¡ u[n¡N ], where N > 0.

2. Given the sequence x[n] = ±[n] (n = 0; 1; 2; : : : ; N ¡ 1), ¯nd its DFT X[k].

3. Given the sequence x[n] = 1 (n = 0; 1; 2; : : : ; N ¡ 1), ¯nd its DFT X[k].

4. The Bessel function of order n and argument t is de¯ned as the integral

Jn(t) =1

2¼

2¼Z0

ej(nÁ¡t sinÁ)dÁ:

Approximate this integral using the DFT. Use the fft algorithm in MATLAB to generate

values of Jn(1) and Jn(50) for n = 0; 1; 2; : : : ; 20. Compare with Table 9.4 of Big Red:

M. Abramowitz and I.A. Stegun, Handbook of Mathematical Functions with Formulas,

Graphs, and Mathematical Tables. National Bureau of Standards (also reprinted by Dover).

5. Recall from page 165 that the Z-domain transfer function of the system having block

diagram on page 161 is

H(z) =1

1¡ ®z¡1 :

Graph the magnitude and phase response H(ej−) as a function of − for these values of the

constant ®:

(a) ® = 1 and ® = ¡1 (graph on the same axes)(b) ® = 1=2 and ® = ¡1=2 (also graph on common axes)

Graph 20 log10fjH(j−)jg (dB) for the magnitude characteristic. Explain the frequencybehavior with respect to the sign of ®.

169

DFT APPROXIMATION OF THE CONTINUOUS-TIME

FOURIER TRANSFORM

F (!) =

1Z¡1

f(t)e¡j!t dt ¼1X

n=¡1f(n¢t)e¡j!n¢t¢t ¼ ¢t

N=2Xn=¡N=2+1

f(n¢t)e¡jn!¢t;

subject to appropriate restrictions. Evaluation at speci¯c frequencies gives

F

μk2¼

N ¢t

¶¼ ¢t

N=2Xn=¡N=2+1

f(n¢t)e¡j2¼nk=N :

De¯ne the periodic sequence

x[n] =

½f(n¢t); n = 0; 1; : : : ; N=2

f((n¡N)¢t); n = N=2 + 1; : : : ; N ¡ 1

where x[n +N ] = x[n]. Also note that exp(¡j2¼nk=N) is periodic in both n and k withperiod N . The approximate Fourier transform above is now

F

μk2¼

N ¢t

¶¼ ¢t

N¡1Xn=0

x[n]e¡j2¼nk=N ¼ ¢tX[k]

in terms of the DFT X[k] of the sequence x[n]. If F (!k) is desired for negative k, employ

the periodicity of X[k] = X[k +N ]. Also note that the range of available frequencies is

¡ ¼

¢t

μ1¡ 2

N

¶· ! · ¼

¢t:

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t

170

Numerical Example of Using the FFT to Evaluate the Fourier Integral:

Time-Domain Response of an RC Filter to a Gaussian Pulse

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.

²

²

CR

+

¡

x(t)

+

¡

y(t)

y(t) = x(t)~ h(t) F() Y (!) = X(!)H(!) (1)

The time-domain input to the simple RC ¯lter is a Gaussian pulse

x(t) = e¡®t2

(2)

having spectrum

X(!) =

r¼

®e¡!

2=4®: (3)

The circuit transfer function is

H(!) =Y (!)

X(!)=

R

R+1

j!C

=j!RC

1 + j!RC=

j!=!b

1 + j!=!b(4)

where the ¯lter break frequency is de¯ned to be

!b =1

RC: (5)

The time-domain response is the inverse Fourier transform

y(t) =1

2¼

1Z¡1

H(!)X(!)ej!td! =1

2¼

1Z¡1

j!=!b

1 + j!=!b

r¼

®e¡!

2=4®ej!td!: (6)

Let ® = ¿¡2 so that (6) becomes a little clearer

y(t) =1

2p¼

1Z¡1

j!¿

!b¿

1 +j!¿

!b¿

exp£¡1

4(!¿)2¤exp

·j!¿

t

¿

¸¿ d!: (7)

171

Now introduce the normalized variables

z = !¿; zb = !b¿; v =t

¿(8)

so that the circuit input and output signals are

x(v) = e¡v2

(9)

y(v) =1

2p¼

1Z¡1

jz=zb

1 + jz=zbe¡z

2=4 ejzv dz: (10)

Normalized time is v = t=¿ and the single parameter requiring a numerical value is

zb = !b¿ =¿

RC=

1p®RC

: (11)

This parameter is essentially the ratio of the input \pulse width" to the circuit time-

constant. Note that as the cut-o® frequency of the high-pass ¯lter approaces zero, the

output approaches the input

y(v) ¡¡¡!zb!0

1

2p¼

1Z¡1

e¡z2=4 ejzv dz = x(v): (12)

x(v) =1

2p¼

1Z¡1

e¡z2=4 ejzv dz ¼ 1

2p¼

N=2Xn=¡N=2+1

e¡(n¢z)2=4ejn¢z v¢z

Let

vk =2¼k

N¢z(k = ¡N=2 + 1; : : : ; N=2)

x

μ2¼k

N¢z

¶¼ ¢z

2p¼

N=2Xn=¡N=2+1

e¡(n¢z)2=4ej2¼nk=N

x[k] =N¢z

2p¼FFT¡1 fX[n]g

with

x[n] = e¡(n¢z)2=4:

172

¡4 ¡3 ¡2 ¡1 0 1 2 3 4¡0:50

¡0:25

0.00

0.25

0.50

0.75

1.00

t=¿

¿=RC = 10

1

0:1

x(t)

y(t)

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Fig. 1. Response of RC High-Pass Filter to a Gaussian Pulse x(t) = exp[¡(t=¿)2]Cases: ¿=RC = 0:1; 1; 10

173

Example: DFT Approximation of a Fourier Sine Integral

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...................

...................

...................

0 T=2 T N¢t

t

...................1

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f(t) f(t) =

8>>>>><>>>>>:

0; t < 0

2t

T; 0 · t < T=2

2

T(T ¡ t); T=2 · t < T

0; t ¸ T

exact Fourier sine transform for comparison (never start with one you don't know!)

Fs(!) =

1Z0

f(t) sin(!t) dt =2T

(!T )2

£2 sin(!T=2)¡ sin(!T )¤

Fs(!) ¼N¡1Xn=0

f(n¢t) sin(n!¢t)¢t

Fs

μ2¼k

N¢t

¶¼ ¢t

N¡1Xn=0

f(n¢t) sin(2¼nk=N)

=j¢t

2

N¡1Xn=0

f(n¢t)£e¡j2¼nk=N ¡ ej2¼nk=N¤

discrete Fourier transform (DFT) of a sequence x[n] is the sequence

X[k] =

N¡1Xn=0

x[n]e¡j2¼nk=N (k = 0; 1; 2; : : : ; N ¡ 1)

both sequences are periodic with period N , that is X[k +N ] = X[k]

let x[n] = f(n¢t) for n = 0; 1; 2; : : : ; N ¡ 1

Fs

μ2¼k

N¢t

¶¼ j¢t

2

nX[k]¡X[N ¡ k]

ohopefully good for k < N=2 or slightly less

174

In this particular run, T = 1, ¢t = 0:02, and N = 1024. In the range 0 < ! < 20

the approximate and exact curves are indistinguishable to the naked eye. Therefore, the

percentage error

E(!) = Fap(!)¡ Fex(!)Fex(!)

£ 100%

is also graphed.

0 2 4 6 8 10 12 14 16 18 20¡0:1

0.0

0.1

0.2

0.3

0.4

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!

Fs(!)

0 2 4 6 8 10 12 14 16 18 200.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

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!

E(!)

175

% DFTsineFT.m DFT for a sine Fourier Transform

% 9 November 2008 R.W.Scharstein

% Practice for the Army 3-dim slotted cylinder problem

% and a good example for ECE 370.

%

% Original f(t) is causal, and an isosoles triangular waveform of total

% pulse width T.

T=1; N=1024;dt=0.02;

n=0:(N-1); t=n*dt;

x=(2*t/T).*(t<=(T/2))+(2*(T-t)/T).*(((T/2)<t)&(t<=T))+0*(t>T);

%plot(t,x)

X=fft(x); Y=[X(2:end) X(1)]; XX=fliplr(Y);

%[n' X.' XX.']

Fs=(i*dt/2)*(X-XX); w=2*pi*n/(N*dt);

% Fexact has a divide by zero at w=0, but so what!

Fexact=2*T*(2*sin(w*T/2)-sin(w*T))./(w*T).̂ 2;

% look at the first N/4 data points

m=1:N/4; wm=w(1:N/4); Fsm=Fs(1:N/4);

Fexactm=[0 Fexact(2:N/4)];

[m' wm' Fsm.' Fexactm.']

figure

plot(wm,Fexactm,wm,real(Fsm))

axis([0 20 -0.1 0.5])

error=100*(Fsm-Fexactm)./Fexactm;

figure

plot(wm,error)

axis([0 20 0 2])

176

Discrete-Time Approximation for the Step Response

of an Underdamped RLC Series Circuit

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+

¡V0u(t)

R L

C

+

¡vC(t)

....................................................................................................................................

i(t)

Exact continuous-time solution for veri¯cation:

Kirchho® voltage law gives the equation of motion as the integro-di®erential equation

Ldi(t)

dt+Ri(t) +

1

C

tZ¡1

i(¿) d¿ = V0u(t):

The inductor forces the initial condition i(0) = 0. For t > 0 application of L¡1d=dt toeliminate the integral yields the homogeneous second-order di®erential equation

d2i(t)

dt2+R

L

di(t)

dt+

1

LCi(t) = 0 (t > 0)

A second initial condition i0(0) = V0=L comes from the ¯rst integro-di®erential equation

evaluated at t = 0+ with the capacitor initially uncharged. A nice, clean statement of the

pertinent initial value problem is now

d2i(t)

dt2+ 2®

di(t)

dt+ !20i(t) = 0

with initial conditions

i(0) = 0 and i0(0) = V0=L;

where the introduction of

® , R

2Land !20 ,

1

LC

simpli¯es the mathematics by clarifying the physically-important combinations of the orig-

inal R, L, and C circuit values. The exact solution to this well-posed initial value problem

is

i(t) =V0

¯Le¡®t sin¯t

where

¯ ,q!20 ¡ ®2:

This form is most natural for the subject underdamped case where R < 2pL=C.

177

Discrete-time approximation:

The discretization of the derivative derives from the de¯ning limit as

df(t)

dt, lim

¢t!0

f(t)¡ f(t¡¢t)¢t

=) f(nT )¡ f((n¡ 1)T )T

=) f [n]¡ f [n¡ 1]T

:

A successive application of this gives the second-derivative

d

dt

df(t)

dt=)

f [n]¡ f [n¡ 1]T

¡ f [n¡ 1]¡ f [n¡ 2]T

T=f [n]¡ 2f [n¡ 1] + f [n¡ 2]

T 2:

The discretized forms of the initial conditions i(0) = 0 and i0(0) = V0=L are thus

i[0] = 0 and i[¡1] = ¡V0T=L:

The di®erence equation that approximates the continuous-time di®erential equation is

similarly ©1 + 2®T + (!0T )

2ªi[n]¡ 2(1 + ®T )i[n¡ 1] + i[n¡ 2] = 0:

Using the two initial conditions at n = ¡1 and n = 0, a recursive solution can be con-

structed for n = 1; 2; : : : according to

i[n] =2(1 + ®T )

1 + 2®T + (!0T )2i[n¡ 1]¡ 1

1 + 2®T + (!0T )2i[n¡ 2]:

Numerical Results:

The circuit values

R = 200 (−) L = 20 (H) C = 40 (¹F)

give

® = 5 (s¡1) and ¯ = 35 (s¡1):

The MATLAB code discreteRLC.m compares the normalized exact solution

¯L

V0i(t) = e¡®t sin¯t

to the recursive solution of the discrete-time problem.

178

0.0 0.1 0.2 0.3 0.4 0.5

t (s)

¡0:5

0.0

0.5

1.0

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Normalized Current

solid curve: exact solutiondashed curve: sampling period T = 0:010 (s)

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0.0 0.1 0.2 0.3 0.4 0.5

t (s)

¡0:5

0.0

0.5

1.0

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Normalized Current

solid curve: exact solutiondashed curve: sampling period T = 0:001 (s)

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179