Module 6

Introduction to Fourier series

Objective:To understand Fourier series representation of Periodic signals.

Introduction to Fourier Series:

A signal is said to be a continuous time signal if it is available at all instants of time. A real time

naturally available signal is in the form of time domain. However, the analysis of a signal is far more

convenient in the frequency domain. These are three important classes of transformation methods

available for continuous-time systems. They are:

1. Fourier series 2. Fourier Transform 3. Laplace Transform

Out of these three methods, the Fourier series is applicable only to periodic signals, i.e. signals

which repeat periodically over -∞<t<∞. Not all periodic signals can be represented by Fourier

series.

Description:

Basic idea of Fourier series is to project periodic signals onto a set of basis functions. The basis

functions are orthogonal and span the space of periodic functions. Any periodic signal can be

written as a weighted sum of these basis functions.

Representation of Fourier Series

The representation of signals over a certain interval of time in terms of the linear

combination of orthogonal functions is called Fourier Series. The Fourier analysis is also sometimes

called the harmonic analysis. Fourier series is applicable only for periodic signals. It cannot be

applied to non-periodic signals. A periodic signal is one which repeats itself at regular intervals of

time, i.e. periodically over -∞ to ∞. Three important classes of Fourier series methods are available.

They are:

1. Trigonometric form 2. Cosine form 3. Exponential form

In the representation of signals over a certain interval of time in terms of the linear combination

of orthogonal functions, if the orthogonal functions are exponential functions then it is called

exponential fourier series.

Similarly, in the representation of signals over a certain interval of time in terms of the linear

combination of orthogonal functions, if the orthogonal functions are trigonometric functions, then it

is called trigonometric Fourier series.

Existence of Fourier series

For the Fourier series to exist for a periodic signal, it must satisfy certain conditions.

The conditions under which a periodic signal can be represented by a Fourier series are

known as Dirichlet’s conditions named after the mathematician Dirichlet who first found them. They

are as follows:

In each period,

1. The function x(t) must be a single valued function.

2. The function x(t) has only a finite number of maxima and minima.

Example: This function x(t) violates Condition 2

3. The function x(t) has a finite number of discontinuities.

Example: This function violates condition 3

4. The function x(t) is absolutely integrable over one period, that is ⎹𝑥 𝑡 ⎸𝑑𝑡 < ∞𝑇

0.

These are the sufficient but not necessary conditions for the existence of the Fourier series of a

periodic function x(t). Condition 4 is known as the weak Dirichlet condition. If a function satisfied

the weak Dirichlet condition, the existence of Fourier series is guaranteed, but the series may not

converge at every point. Conditions 2 and 3 are known as strong Dirichlet conditions. If these are

satisfied, the convergence is also guaranteed.

Trigonometric form of Fourier series:

A sinusoidal signal, x(t)=A sin ⍵0t is a periodic signal with period T=2𝜋/⍵0. Also, the sum

of two sinusoids is periodic provided that their frequencies are integral multiples of a fundamental

frequency ⍵0. We can show that a signal x(t), a sum of sine and cosine functions whose frequencies

are integral multiples of ⍵0, is a periodic signal.

Let the signal x(t) be

x(t)= a0+a1cos⍵0t+a2cos2⍵0t+…..+akcosk⍵0t+ b1sin⍵0t+b2sin2⍵0t+…..+bksink⍵0t

i.e. x(t)= a0+ [an coskn=1 ⍵0nt + bnsin⍵0nt]

where a0, a1, a2,……ak and b0,b1, b2,……bk are constants, and⍵0 is the fundamental frequency.

For the signal x(t) to be periodic, it must satisfy the condition x(t) = x(t+T) for all t,

i.e. x(t+T)= a0+ [an coskn=1 ⍵0n(t + T) + bnsin⍵0n(t + T)]

= a0+ [an coskn=1 ⍵0n(t + 2π/⍵0) + bnsin⍵0n(t + 2π/⍵0)]

= a0+ [an coskn=1 (⍵0nt + 2nπ) + bnsin(⍵0nt + 2nπ)]

= a0+ [an coskn=1 ⍵0nt + bnsin⍵0nt]

=x(t)

This proves that the signal x(t), which is a summation of sine and cosine functions of

frequencies 0, ⍵0, 2⍵0, ….., k⍵0, is a periodic signal with period T. By changing ans and bns, we

can construct any periodic signal with period T. If k⟶∞ in the expression for x(t), we obtain the

Fourier series representation of any periodic signal x(t). That is, any periodic signal can be

represented as an infinite sum of sine and cosine functions which themselves are periodic signals of

angular frequencies 0, ⍵0 , 2⍵0 , ….., k⍵0 . This set of harmonically related sine and cosine

functions, i.e. sin⍵0𝑛𝑡 and cos⍵0𝑛𝑡, n=0,1,…. Forms a complete orthogonal set over the interval t0

to t0+T where T= 2𝜋/⍵0

The infinite series of sine and cosine terms of frequencies 0, ⍵0, 2⍵0, ….., k⍵0 is known as

trigonometric form of Fourier series and can be written as:

x(t)= [an cos∞n=0 ⍵0nt + bnsin⍵0nt]

or x(t)= a0+ [an cos∞n=0 ⍵0nt + bnsin⍵0nt]

where an and bn are constants; the coefficient a0 is called the dc component; a1cos⍵0t+b1sin⍵0t the

first harmonic, a2cos2⍵0t+b2sin2⍵0t the second harmonic and 𝑎𝑛 𝑐𝑜𝑠⍵0𝑛𝑡 + 𝑏𝑛𝑠𝑖𝑛⍵0𝑛𝑡 the nth

harmonic. The constant b0=0 because sin⍵0𝑛𝑡 = 0 for n=0.

Evaluation of Fourier coefficients of the trigonometric Fourier series

The constantsa0, a1, a2,……ak and b0, b1, b2,……bk are called Fourier coefficients. To evaluate a0, we

shall integrate both sides of the equation for x(t) over one period t0 to t0+T at an arbitrary time t0 .

Thus,

x t dtt0+T

t0 = a0 dt

t0+T

t0+ [

t0+T

t0 [an cos∞

n=1 ⍵0nt + bnsin⍵0nt] dt]

= a0T + an ∞n=1 cos ⍵0nt

t0+T

t0dt + bn

∞n=1 sin ⍵0nt dt

t0+T

t0

We know that cos ⍵0ntt0+T

t0dt=0 and sin ⍵0nt

t0+T

t0dt=0, since the net areas of sinusoids over

complete periods are zero for any nonzero integer n and any time t0. Hence, each of the integrals in

the above summation is zero.

Thus, we obtain

x t dtt0+T

t0 = a0T or a0 =

1

T x t dt

t0+T

t0

To evaluate an and bn, we can use the following results:

cos ⍵0nt cos ⍵0mtt0+T

t0dt =

0 for m ≠ n T

2 for m = n ≠ 0

sin ⍵0nt sin ⍵0mtt0+T

t0dt =

0 for m ≠ n T

2 for m = n ≠ 0

sin ⍵0nt cos ⍵0mtt0+T

t0dt = 0 for all m and n.

To find Fourier coefficients an, multiply the equation for x(t) by 𝑐𝑜𝑠 ⍵0𝑚𝑡 and integrate over one

period. That is,

x t cos ⍵0mtdtt0+T

t0= a0 cos ⍵0mt

t0+T

t0dt + an

∞n=1 cos ⍵0nt cos ⍵0mt

t0+T

t0dt

+ bn ∞n=1 sin ⍵0nt cos ⍵0mt

t0+T

t0dt

The first and third integrals in the above equation are equal to zero and the second integral is equal

to T/2 when m=n. Therefore,

x t cos ⍵0mtdtt0+T

t0 = am

T

2

an = 2

T x t cos ⍵0ntdt

t0+T

t0

To find bn, multiply both sides of equation for x(t) by sin⍵0mt and integrate over one period. Then

x(t)sin ⍵0mtt0+T

t0dt=a0 sin ⍵0mt

t0+T

t0dt + an

∞n=1 cos ⍵0nt sin ⍵0mt

t0+T

t0dt

+ bn ∞n=1 sin ⍵0nt sin ⍵0mt

t0+T

t0dt

The first and second integrals in the above equation are zero, and the third integral is equal to T/2

when m= n. Thus, we have

x(t)sin ⍵0mtt0+T

t0dt = bm

T

2

bm = 2

T x(t)sin ⍵0mt

t0+T

t0dt or bn =

2

T x(t)sin ⍵0nt

t0+T

t0dt

a0 , an and bn are called trigonometric Fourier series coefficients.

A periodic signal has the same Fourier series for the entire interval -∞ to ∞ as for the

interval t0 to t0+T, since the same function repeats after every T seconds. The Fourier series

expansion of a periodic function is unique irrespective of the location of t0 of the signal.

Cosine representation (Alternate form of the Trigonometric Representation):

The trigonometric Fourier series of x(t) contains sine and cosine terms of the same frequency.

i.e. x(t)= a0+ [an cos∞n=1 ⍵0nt + bnsin⍵0nt]

= a0+ an2 + bn

2∞n=1

an

an2 +bn

2cos⍵0nt +

bn

an2 +bn

2sin⍵0nt

Substituting the values A0= a0, An= an2 + bn

2 , cos θn = an

an2 +bn

2 and sin θn = −

bn

an2 +bn

2

i.e. θn= −tan−1 bn

an,

we have

x(t)= a0+ An[cos θncos∞n=1 ⍵0nt − sin θnsin⍵0nt]

x(t)= a0+ An[cos (⍵0nt + θn)]∞n=1

This is the cosine representation of x(t) which contains sinusoids of frequencies ⍵0, 2⍵0, ….

The term A0is called the dc component, and the term An[cos (⍵0nt + θn)] is called the nth

harmonic component of x(t). The first harmonic component A1[cos (⍵0nt + θ1)] is commonly

called the fundamental component as it has the same fundamental period as x(t).

The number An represents amplitude coefficients or harmonic amplitudes or spectral

amplitudes of the Fourier series and the number θn represents the phase coefficients or phase angles

or spectral phase of the Fourier series.

The cosine form is also called the Harmonic form Fourier series or Polar form Fourier series.

Wave Symmetry:

If the periodic signal x(t) has some type of symmetry, then some of the trigonometric Fourier

series coefficients may become zero and calculation of the coefficients becomes simple.

There are following four types of symmetry a function x(t) can have:

1. Even symmetry 2.Odd symmetry 3. Half wave symmetry

4. Quarter wave symmetry

If x(t) has even symmetry (also called mirror symmetry), then bn = 0 and only a0 and an are to be

evaluated.

On the other hand, if x(t) has odd symmetry (also called rotation symmetry), then a0 =0 and an =0.

Only bn is to be evaluated.

If x(t) has half wave symmetry, then a0 =0 and only odd harmonics exists.

If x(t) has quarter wave symmetry, then a0 =0 and either only an or only bn exists, then too only for

odd values of n.

A function x(t) is said to be even if 𝑥(𝑡) = 𝑥(−𝑡) A function x(t) is said to be odd if 𝑥(𝑡) = −𝑥(−𝑡) We know that any signal x(t) can be split into even and odd function. That is,

x(t) = xe(t) +xo(t)

so the even and odd parts of the signal x(t) can be obtained from the following relations:

xe(t) = 1

2[x t + x −t ]

xo(t) = 1

2[x t − x −t ]

These relations can be used to find the Fourier coefficients. To have a convenient form, we choose

the interval of integration from – (𝑇 2 ) to 𝑇 2 instead of from t0 to t0+T. also we know that

Odd function × Odd function = Even function

Even function × Even function =Even function

Odd function × Even function = Odd function

and for any even function xe(t),

xe(t) T 2

−T 2 dt = 2 xe (t)

T 2

0dt

and for any even function xo(t),

xo(t) T 2

−T 2 dt = 0

Based on these relations, we can easily calculated the Fourier series coefficients.

We have an= 2

T x t cos⍵0nt

T 2

−T 2 dt

Substituting x(t) = xe(t) +xo(t) in the above equations for an and bn, we have

an= 2

T xe t cos⍵0nt

T 2

−T 2 dt + xo t cos⍵0nt

T 2

−T 2 dt

and bn= 2

T xe t sin⍵0nt

T 2

−T 2 dt + xo t sin⍵0nt

T 2

−T 2 dt

1. Even or mirror symmetry

A function x(t) is said to have even or mirror symmetry, if

𝑥(𝑡) = 𝑥(−𝑡)

The waveforms of some even function are shown in figure below. The sum of two or more

even functions is always even. If a constant is added, the even nature of the function still persists.

These types of functions are always symmetrical with respect to the vertical axis. The product of two

even functions is always even.

If x(t) is an even function, then xo(t) =0 and x(t) = xe(t). therefore,

an= 2

T x t cos⍵0nt

T 2

−T 2 dt = an=

2

T xe t cos⍵0nt

T 2

−T 2 dt

=4

T x t cos⍵0nt

T 2

0 dt

ao= 1

T x t

T 2

−T 2 dt =

1

T xe t

T 2

−T 2 dt =

2

T x t

T 2

0dt

and bn = 2

T xe t sin⍵0nt

T 2

−T 2 dt = 0

because xe t sin⍵0nt is an odd function.

Thus, the Fourier series expansion of an even periodic function contains only cosine terms and a

constant.

When even or mirror symmetry exists, the trigonometric Fourier series coefficients are

ao= 2

T x t

T 2

0dt , an =

4

T x t cos⍵0nt

T 2

0dt and bn = 0

If the average value is 0, the coefficients ao will be absent, i.e. ao =0. The even function shows the

mirror symmetry with respect to the vertical axis. Thus, the waveform having even symmetry

contains only cosine terms and ao may be present.

Waveforms of Even or Mirror function

2. Odd or Rotation symmetry

A function x(t) is said to have odd or rotation symmetry, if

𝑥(𝑡) = −𝑥(−𝑡) The waveforms of some odd functions are shown in below figure. The sum of two or more odd

functions is always odd. If a constant is added, the odd nature of the function is removed. These

types of functions are always anti symmetrical with respect to the vertical axis. The product of two

odd functions is odd.

If x(t) is an odd function, then xe t = 0 and x(t) =xo t

∴ ao= 1

T x t

T 2

−T 2 dt =

1

T xo t

T 2

−T 2 dt =0

and an= 2

T x t cos⍵0nt

T 2

−T 2 dt = an=

2

T xo t cos⍵0nt

T 2

−T 2 dt =0

because xo t cos⍵0nt is an odd function.

bn =2

T x t sin⍵0nt

T 2

−T 2 dt =

2

T xo t sin⍵0nt

T 2

−T 2 dt =

4

T x t sin⍵0nt

T 2

0 dt

Thus, the Fourier series expansion of an odd periodic function contains only sine terms.

When odd or rotation symmetry exists, the trigonometric Fourier series coefficients are

ao= 0, an=0, and bn =4

T x t sin⍵0nt

T 2

0 dt

Waveforms of odd functions

3. Half wave symmetry

A periodic signal x(t) which satisfies the condition:

x(t) = −𝑥 𝑡 ±𝑇

2

is said to have half wave symmetry. Figure shown below a function having half wave symmetry.

This function is neither purely odd nor purely even . For such function, ao= 0. The Fourier series

expansion of such a type of periodic signal conditions odd harmonics only, that is ⍵0, 3⍵0, …. Etc.

As x(t) contains only odd harmonic terms, when n is even an=bn =0.

When n is odd,

an= 4

T x t cos⍵0nt

T 2

0dt

and bn = 4

T x t sin⍵0nt

T 2

0 dt

This can be proved as follows:

an= 2

T x t cos⍵0nt

T

0dt

= 2

T x t cos⍵0nt

T 2

0dt + x t cos⍵0nt

T

T 2 dt

Changing the variable from t to (t+T/2) in the second integral, we get

an= 2

T x t cos⍵0nt

T 2

0dt + x t + T/2 cos⍵0n(t + T/2)

T

T 2 dt

using the property x(t) = −x(t ±T

2), we get

an= 2

T [x t cos⍵0nt − x t cos⍵0nt cos nπ]

T 2

0dt

= 0 for even n 4

T x t cos⍵0nt

T 2

0dt for odd n

Similarly, it can be shown that

bn = 0 for even n 4

T x t sin⍵0nt

T 2

0dt for odd n

Half wave symmetry

4. Quarter wave symmetry:

A function x(t) is said to have quarter wave symmetry, if

x(t) = 𝑥(−𝑡) or x(t) = −𝑥(−𝑡) and also x(t) = −𝑥 𝑡 ±𝑇

2

i.e. a function x(t) which has either even symmetry or odd symmetry along with half wave symmetry

is said to have quarter wave symmetry. Below Figure shows the examples of such functions.

Quarter wave symmetry

The following two cases are discussed below for the waveforms having quarter wave symmetry.

Case 1: x(t) = −𝑥(−𝑡) and x(t) = −𝑥 𝑡 ±𝑇

2

ao= 0,

an= 0,

bn= 8

T x t sin⍵0nt

T 4

0dt for odd n

case 2: x(t) = 𝑥(−𝑡) and also x(t) = −𝑥 𝑡 ±𝑇

2

ao= 0,

an=8

T x t cos⍵0nt

T 4

0dt

bn = 0 for even n

Exponential Fourier series: The exponential Fourier series is the most widely used form of Fourier series. In this, the

function x(t) is expressed as a weighted sum of the complex exponential functions. Although the

trigonometric form and the cosine representation are the common forms of Fourier series, the

complex exponential form is more general and usually more convenient and more compact. So, it is

used almost exclusively, and it finds extensive application in communication theory.

The set of complex exponential functions

{ejn⍵0t, n = 0, ±1, ±2, ….}

forms a closed orthogonal set over an interval (t0, t0+T) where T= (2π/⍵0) for any value of t0, and

therefore it can be used as a Fourier series. Using Euler’s identity, we can write

An cos(n⍵0t +θn) = An ej(n⍵0t+θn )+ e−j(n⍵0t+θn )

2

Substituting this in the definition of the cosine Fourier representation, we obtain

x(t) = A0 + An

2 ej(n⍵0t+θn) + e−j(n⍵0t+θn) ∞

n=1

= A0 + An

2 ejn⍵0tejθn + e−jn⍵0te−jθn ∞

n=1

= A0 + An

2ejn⍵0tejθn + ∞

n=1 An

2e−jn⍵0te−jθn ∞

n=1

= A0 + An

2ejθn ejn⍵0t + ∞

n=1 An

2e−jθn ∞

n=1 e−jn⍵0t

Letting n = -K in the second summation of the above equation, we have

x(t) = A0 + An

2ejθn ejn⍵0t + ∞

n=1 Ak

2ejθk −∞

k=−1 ejk⍵0t

on comparing the above two equations for x(t), we get

An = Ak ; (-𝜃𝑛 ) = 𝜃𝑘 n > 0

k < 0

let us define

C0 = A0; Cn =An

2ejθn , n > 0

∴ x(t) = A0 + An

2ejθnejn⍵0t + ∞

n=1 An

2ejθn ejk⍵0t−∞

n=−1

i.e. x(t) = Cnejn⍵0t∞n=−∞

This is known as exponential form of Fourier series. The above equation is called the synthesis

equation.

So the exponential series from cosine series is:

C0 = A0

Cn =An

2ejθn

i. Determination of the coefficients of Exponential Fourier Series

We have x(t) = 𝐶𝑛𝑒𝑗𝑛 ⍵0𝑡∞

𝑛=−∞ ; ⍵0 = 2𝜋

𝑇

Multiplying both sides by 𝑒−𝑗𝑘⍵0𝑡 and integrating over one period, we get

x(t)e−jk⍵0tt0+T

t0dt = Cnejn⍵0t∞

n=−∞ e−jk⍵0tt0+T

t0dt

= Cn∞n=−∞ ejn⍵0te−jk⍵0tt0+T

t0dt

We know that

ejn⍵0te−jk⍵0tt0+T

t0dt =

0 for k ≠ nT for k = n

∴ x(t)e−jk⍵0tt0+T

t0dt = TCk

or Cn = 1

T x(t)e−jn⍵0tt0+T

t0dt

where Cn are the Fourier coefficients of the exponential series. The above equation is called the

analysis equation.

C0 = A0 = 1

T x(t)

t0+T

t0dt

The Fourier coefficients of x(t) have only a discrete spectrum because values of 𝐶𝑛 exist only for

discrete values of n. As it represents a complex spectrum, it has both magnitude and phase

spectra. The following points may be noted:

1. The magnitude line spectrum is always an even function of n.

2. The phase line spectrum is always an odd function of n.

ii. Trigonometric Fourier Series from Exponential Fourier Series

The complex exponential Fourier series is given by

x(t) = Cnejn⍵0t∞n=−∞ = C0+ Cnejn⍵0t−1

n=−∞ + Cnejn⍵0t∞n=1

= C0 + C−ne−jn⍵0t + Cnejn⍵0t ∞n=1

= C0 + C−n(cosn⍵0t − jsinn⍵0t) + Cn(cosn⍵0t + jsinn⍵0t) ∞n=1

x(t) = C0 + (Cn + C−n)cosn⍵0t + j(Cn − C− n)sinn⍵0t) ∞n=1

Comparing this x(t) with the standard trigonometric Fourier series to trigonometric Fourier series

as:

a0 = C0

an = Cn+C-n

bn = j(Cn - C-n)

Similarly, an exponential Fourier series can be derived from the trigonometric Fourier series

iii. Exponential Fourier series from Trigonometric Fourier series

From the exponential Fourier series, we know that

Cn = 1

T x(t)e−jn⍵0tT

0 dt =

1

T x(t)(cosn⍵0t − jsinn⍵0t)

T

0 dt

= 1

2

2

T x t cosn⍵0t dt − j

2

T x(t)

T

0sinn⍵0t

T

0 dt =

1

2[an - jbn]

C−n = 1

T x(t)ejn⍵0tT

0 dt =

1

T x(t)(cosn⍵0t + jsinn⍵0t)

T

0 dt

= 1

2

2

T x t cosn⍵0t dt + j

2

T x(t)

T

0sinn⍵0t

T

0 dt =

1

2[an + jbn]

C0 = 1

T x(t)

T

0 dt = a0

So, the formulae for conversion of trigonometric series to exponential series are:

C0 = a0

Cn = 1

2(an-jbn)

C-n = 1

2(an+jbn)

iv. Cosine Fourier Series from Exponential Fourier Series

We know that

A0 = a0 = C0

An= an2 + bn

2 = (Cn + C−n)2 + [j Cn − C−n ]2

= Cn2 + C−n

2 + 2CnC−n − (Cn2 + C−n

2 − 2CnC−n)

= 4CnC−n

= 2⎹Cn⎸

∴ A0 = C0

An = 2⎹Cn⎸

Solved Problems:

Problem 1:Find the Fourier series expansion of the half wave rectified sine wave shown in fig below

Solution :

The periodic waveform shown in fig with period 2π is half of a sine wave with period 2π.

x(t) = A sin⍵t = A sin

2π

2πt = A sin t 0 ≤ t ≤ π

0 π ≤ t ≤ 2π

Now the fundamental period T=2π

Fundamental frequency ⍵0=2π

T =

2π

2π =1

Let t0 = 0, t0+T =T = 2π

a0 = 1

T x(t)

T

0 dt =

1

2π x(t)

2π

0 dt =

1

2π A sin tπ

0 dt =

A

π

∴ a0 = A

π

an = 2

T x t cos ⍵0nt dt

T

0 =

2

2π x t cos nt dt

2π

0 =

1

π A sin tπ

0 cos nt dt

= −A

2π

(−1)n +1− 1

1+n +

(−1)n +1+ 1

1−n

For odd n, an = −A

2π

1− 1

1+n +

1+ 1

1−n = 0

For even n, an = −A

2π −1− 1

1+n +

−1− 1

1−n =−

2A

π(n2−1)

bn = 2

T x(t)sin ⍵0nt

T

0 dt = bn =

2

2π x t sin nt

2π

0 dt =

1

π A sin t sin ntπ

0 dt

= A

2π

sin n−1 t

n−1−

sin n+1 t

n+1

0

π

This is zero for all values of n except for n=1.

For n=1, b1 = A

2π

Therefore, the trigonometric Fourier series is:

x(t) = a0+ [an cos∞n=1 ⍵0nt + bnsin⍵0nt]

= a0 + b1sin t + an cos nt∞n=1

= A

π +

A

2πsin t −

2A

π(n2−1)cos nt∞

n=1

Problem 2: Obtain the trigonometric Fourier series for the waveform shown in below fig

Solution:

The waveform shown in fig above is periodic with a period T=2π

Let t0 = 0, t0+T =T = 2π

Fundamental frequency ⍵0=2π

T =

2π

2π =1

The waveform is described by

x(t) =

A

π t 0 ≤ t ≤ π

0 π ≤ t ≤ 2π

a0 = 1

T x(t)

T

0 dt =

1

2π x(t)

2π

0 dt =

1

2π

A

πt

π

0 dt =

A

2π2

t2

2

0

π

= A

4

an = 2

T x t cos ⍵0nt dt

T

0 =

2

2π

A

πt cos nt dt

π

0 =

A

π2n2(cos nπ - cos 0)

∴ an = −

2A

π2n2 for odd n

0 for even n

bn = 2

T x(t)sin ⍵0nt

T

0 dt =

2

2π x t sin nt

2π

0 dt =

1

2π

A

πt sin nt

π

0 dt =

A

nπ(−1)n+1

∴ bn =

A

πn for odd n

− A

πn for even n

The trigonometric Fourier series is :

x(t) = a0+ [an cos∞n=1 ⍵0nt + bnsin⍵0nt]

=A

4−

2A

π2 cos nt

n2∞n=odd +

A

π (−1)n+1 sin nt

n∞n=1

=A

4−

2A

π2 cos t +

1

32cos 3t +

1

52cos 5t + ⋯

+ A

π sin t −

1

2sin 2t +

1

3sin 3t −

1

4sin 4t +…

Problem 3: Find the trigonometric Fourier series for the waveform shown in below fig

Solution :

The waveform shown in figure is given by

x(t) = A

2π t, for 0 ≤ t ≤ 2π

Let 𝑡0 = 0, 𝑡0+T =T = 2𝜋

and Fundamental frequency ⍵0=2𝜋

𝑇 =

2𝜋

2𝜋 =1

The coefficients are evaluated as follows:

a0 = 1

𝑇 𝑥(𝑡)𝑇

0 𝑑𝑡 =

1

2𝜋

𝐴

2𝜋𝑡

2𝜋

0 𝑑𝑡 =

𝐴

(2𝜋)2 𝑡2

2

0

2𝜋

= 𝐴

2

an = 2

𝑇 𝑥 𝑡 𝑐𝑜𝑠 ⍵0𝑛𝑡𝑑𝑡𝑇

0 =

2

2𝜋

𝐴

2𝜋𝑡 𝑐𝑜𝑠 𝑛𝑡 𝑑𝑡

2𝜋

0

= 2𝐴

(2𝜋)2 0 − 0 +1

𝑛

(1−1)

𝑛 =0

bn = 2

𝑇 𝑥(𝑡)𝑠𝑖𝑛 ⍵0𝑛𝑡𝑇

0 𝑑𝑡 =

2

2𝜋

𝐴

2𝜋𝑡𝑠𝑖𝑛 𝑛𝑡

2𝜋

0 𝑑𝑡 = −

𝐴

𝜋𝑛

a0 = 𝐴

2 ; an = 0 ; bn = −

𝐴

𝜋𝑛

are the trigonometric Fourier series coefficients.

∴ x(t) = 𝐴

2 + −∞

𝑛=1𝐴

𝜋𝑛𝑠𝑖𝑛 𝑛𝑡=

𝐴

2−

𝐴

𝜋

𝑠𝑖𝑛 𝑛𝑡

𝑛∞𝑛=1 0≤ t ≤ 2𝜋

is the trigonometric Fourier series coefficients.

Problem 4: Find the Fourier series expansion for the waveform shown in figure below:

Solution :

The given waveform of figure is periodic with period T=2𝜋. For the computational convenience,

choose one cycle of the waveform from –𝜋 to 𝜋.

Let

𝑡0 = −𝜋, 𝑡0+T = 𝜋

and Fundamental frequency ⍵0=2𝜋

𝑇 =

2𝜋

2𝜋 =1

The waveform is described by

x(t) = −

𝐴

𝜋 𝑡𝑓𝑜𝑟 − 𝜋 ≤ 𝑡 ≤ 0

𝐴

𝜋 𝑡𝑓𝑜𝑟 0 ≤ 𝑡 ≤ 𝜋

The given waveform has even symmetry because x(t) =x(-t).

∴ a0 = 2

𝑇 𝑥(𝑡)𝑇 2

0 𝑑𝑡 , an =

4

𝑇 𝑥 𝑡 𝑐𝑜𝑠 ⍵0𝑛𝑡𝑑𝑡𝑇 2

0 and bn =0

Now, a0 = 2

𝑇 𝑥(𝑡)𝑇 2

0 𝑑𝑡

= 2

𝑇

𝐴

𝜋𝑡

𝑇 2

0 𝑑𝑡 =

𝐴

(𝜋)2 𝑡2

2

0

𝜋

= 𝐴

2

By observation also we can say that the average value a0 = 𝐴

2

an = 4

𝑇 𝑥 𝑡 𝑐𝑜𝑠 ⍵0𝑛𝑡𝑑𝑡𝑇 2

0

=4

2𝜋

𝐴

𝜋𝑡 𝑐𝑜𝑠 𝑛𝑡 𝑑𝑡

𝜋

0 =

2𝐴

𝜋2𝑛2(cos n𝜋 - cos 0) =2𝐴

𝜋2𝑛2 (−1)𝑛 − 1

∴ an = −

4𝐴

𝜋2𝑛2 𝑓𝑜𝑟𝑜𝑑𝑑𝑛

0 for even n

The trigonometric Fourier series is:

x(t) = a0+ [an cos∞n=1 ⍵0nt + bnsin⍵0nt]

=A

2 + −

4A

π2n2 cos nt∞

n=odd

= A

2−

4A

π2 cos nt

n2 ∞n=odd

Problem 5: Obtain the Fourier components of the periodic rectangular waveform shown in below

figure

Solution : The waveform shown in figure is a periodic waveform with period = T and can be expressed

as:

x(t) =

0 for

−T

2≤ t ≤

−T

4

A for −T

4≤ t ≤

T

4

0for T

4≤ t ≤

T

2

Let

t0 = −T

2, t0+T =

T

2

and Fundamental frequency ⍵0=2π

T

The given function has even symmetry because x(t) = x(-t).

∴ a0 = 2

T x(t)

T 2

0 dt , an =

4

T x t cos ⍵0nt dt

T 2

0 and bn =0

Now, a0 = 2

T x(t)

t0+T

t0dt

= 2

T

T 2

−T 2 A dt =

2

T A

T 4

0 dt =

2A

T t 0

T 4 =

A

2

an = 2

T x t cos ⍵0ntdt

t0+T

t0 =

2

T

T 2

−T 2 x t cos ⍵0nt dt

= 4

T

T 4

0 A cos n

2π

Tt dt =

2A

nπ sin

nπ

2

an =

2A

nπ for n = 1, 5, 9,…

−2A

nπ for n = 3,7, 11,…

0 for even n

The trigonometric Fourier series is:

x(t) = a0+ [an cos∞n=1 ⍵0nt + bnsin⍵0nt]

= = A

2+

2A

πn sin

nπ

2 cos

n2π

T ∞

n=odd t

Problem 6: Obtain the exponential Fourier Series for the wave form shown in below figure.

Solution: The periodic waveform shown in fig with a period T= 2π can be expressed as:

x(t) = A 0 ≤ t ≤ π

−A π ≤ t ≤ 2π

Let

t0 = 0, t0+T = 2π

and Fundamental frequency ⍵0=2π

T =

2π

2π =1

Exponential Fourier series

C0 = 1

T x(t)

T

0 dt

= 1

2π Aπ

0 dt +

1

2π −A

2π

π dt = 0

Cn = 1

T x(t)e−jn⍵0tT

0 dt

= 1

2π Ae−jntπ

0 dt +

1

2π −Ae−jnt2π

π dt

= −A

j2nπ (−1)n − 1 − 1 − (−1)n = −j

A

2nπ

Cn = −j

2A

πn for odd n

0 for even n

∴ x(t) = C0+ Cnejn⍵0t∞n=−∞

n≠0 = −j

A

2nπejnt∞

n=−∞

Problem 7: Find the exponential Fourier series for the full wave rectified sine wave given in below

figure.

Solution: The waveform shown in fig can be expressed over one period(0 to π) as:

x(t) = A sin ⍵t where ⍵= 2π

2π =1

because it is part of a sine wave with period = 2π

x(t) = A sin ⍵t 0≤ t ≤ π

The full wave rectified sine wave is periodic with period T = π

Let

t0 = 0, t0+T = 0+π = π

and Fundamental frequency ⍵0=2π

T =

2π

π = 2

The exponential Fourier series is

x(t) = Cnejn⍵0t∞n=−∞ = Cnej2nt∞

n=−∞

where Cn = 1

T x(t)e−jn⍵0tT

0 dt

= 1

π A sin t e−j2ntπ

0 dt =

A

π sin t e−j2ntπ

0 dt

= A

j2π

ej 1−2n t−e0

j 1−2n −

e−j 1−2n t−e0

−j 1−2n

∴Cn = 2A

π(1−4n2)

C0 = 1

T x(t)

T

0 dt

= 1

π Aπ

0 sin t dt =

A

π − cos 𝑡 0

𝜋 = 2A

π

The exponential Fourier series is given by

x(t) = 2A

π(1−4n2)ej2nt∞

n=−∞ = 2A

π +

2A

π ∞

n=−∞n≠0

ej2nt

1−4n2

Problem 8: Find the exponential Fourier series for the rectified sine wave shown in below figure

Solution:

The waveform shown in figure is a part of a sine wave with period =2

x(t) = A sin ⍵t 0≤ t ≤ 1, where ⍵= 2π

2 =π

hence x(t) = A sin πt for 0≤ t ≤ 1

The period of the rectified sine wave is T = 1

Let

t0 = 0, t0+T = 1

and Fundamental frequency ⍵0=2π

T =

2π

1 = 2π

The exponential Fourier series is

x(t) = Cnejn⍵0t∞n=−∞ = Cnej2πnt∞

n=−∞

where

Cn = 1

T x(t)e−jn⍵0tT

0 dt =

1

1 A

1

0 sinπt e−j2πnt dt

= A

j2π

ej 1−2n t−e0

j 1−2n −

e−j 1−2n t−e0

−j 1−2n

∴Cn = 2A

π(1−4n2)

C0 = 1

T x(t)

T

0 dt

= 1

1 A

1

0 sin t dt =

A

π − cos t 0

1 = 2A

π

The exponential Fourier series is:

x(t) =2A

π +

2A

π ∞

n=−∞n≠0

ej2nπ t

1−4n2

Problem 9: Derive the exponential Fourier series from the trigonometric Fourier series for the

waveform shown in figure.

Also find the exponential Fourier series directly.

Solution: Derivation of exponential series from trigonometric series:

From Example 2, we got the trigonometric Fourier coefficients as:

a0 =A

4, an =−

2A

π2n2 (for odd n) and bn =A

πn(−1)𝑛+1

C0 = a0 = A

4

Cn = 1

2(an−jbn) =

1

2 −

2A

π2n2 − jA

πn(−1)𝑛+1

Direct derivation of exponential Fourier series:

x(t) = C0 + Cnejn⍵0t∞n=−∞

n≠0 = C0 + Cnejnt∞

n=−∞n≠0

C0 = 1

T x t

T

0 dt =

1

2π x t

2π

0 dt =

1

2π

A

πt

π

0 dt =

A

4

Cn = 1

T x(t)e−jn⍵0tT

0 dt =

1

2π x t e−jnt2π

0 dt =

1

2π

A

πt e−jnt2π

0 dt

= A

2π2 −π

jn −1 n −

1

jn 2 −1 n − 1

∴ x(t) = A

4 +

A

2π2 −π

jn −1 n −

1

jn 2 −1 n − 1 e−jnt∞

n=−∞n≠0

Problem 10: Find the complex exponential Fourier series for the waveform shown in figure below

Solution:

The waveform shown in figure is periodic with period = T and can be expressed as

x(t) = A for 0 ≤ t ≤

T

2

0 for T

2≤ t ≤ T

fundamental frequency ⍵0= 2𝜋

𝑇

C0 = 1

T x t

T

0 dt =

1

T A

T 2

0 dt =

A

2

Cn = 1

T x(t)e−jn⍵0tT

0 dt =

1

T A

T 2

0 e−jn

2𝜋

𝑇tdt

= A

jn 2π 1 − −1 n

Cn =

A

jπn for odd n

0 for even n

∴ x(t) = A

2 +

A

jπn ∞

n=odd e−jn2𝜋

𝑇t

Assignment Problems:

1. Find the trigonometric Fourier series for the periodic signal x(t) shown in figure below:

2. Find the trigonometric Fourier series for the periodic signal x(t) shown in figure below:

3. Compute the exponential Fourier series of the signal shown in figure below:

4. Find the exponential series of the signal shown in the figure below:

5. Find the trigonometric Fourier series of x(t) =t2 over the interval (-1, 1).

6. Find the exponential series of the triangular waveform shown in figure below:

7. Find the exponential Fourier series coefficient of the signal x(t) = 2+cos 2𝜋𝑡

3 + 4 sin

5𝜋𝑡

3

8. Find the complex exponential Fourier series coefficient of the signal x(t) = 3cos 4⍵0t.

9. Find the complex exponential Fourier series coefficient of the signal x(t) = 2cos 3⍵0t.

10. Find the complex exponential Fourier series coefficient of the signal

(a) x(t) = 𝑠𝑖𝑛2𝑡 (b) x(t) = 𝑒2𝑡 with T=2

Simulation:

Approximating with Fourier Sine Series

A MATLAB code is used to plot the square wave function along with the Fourier sine series

in order to compare the accuracy and error between the approximation and the actual function.

Program: clc; clear all close all N = input( 'type the total number of harmonics' ); t = 0 : 0.001 : 1; y = square( 2 * pi * t ); plot( t , y , 'r' , 'linewidth' , 2 ) axis( [ 0 1 -1.5 1.5 ] ) hold; sq = zeros( size(t) ); for n = 1 : 2 : N sq = sq + (4 / (pi * n) * sin( 2 * pi * n * t)); end; plot( t , sq ) grid; xlabel( 'time' ); ylabel( 'amplitude' ); title('Synthesized Square Wave Using Fourier Series - Gibbs Phenomenon');

Output for N = 5 Harmonics:

Output for N = 25 Harmonics:

References:

[1] Alan V.Oppenheim, Alan S.Willsky and S.Hamind Nawab, “Signals & Systems”, Second

edition, Pearson Education, 8th

Indian Reprint, 2005.

[2] M.J.Roberts, “Signals and Systems, Analysis using Transform methods and MATLAB”, Second

edition,McGraw-Hill Education,2011

[3] John R Buck, Michael M Daniel and Andrew C.Singer, “Computer explorations in Signals and

Systems using MATLAB”,Prentice Hall Signal Processing Series

[4] P Ramakrishna rao, “Signals and Systems”, Tata McGraw-Hill, 2008

[5] Tarun Kumar Rawat, “Signals and Systems”, Oxford University Press,2011

[6] A.Anand Kumar, “Signals and Systems” , PHI Learning Private Limited ,2011