continuous-time signals and lti systems

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Continuous-Time Signals and LTI Systems
At the start of the course both continuous and discrete-time sig-nals were introduced. In the world of signals and systems model-ing, analysis, and implementation, both discrete-time andcontinuous-time signals are a reality. We live in an analog world,is often said. The follow-on courses to ECE2610, Circuits andSystems I (ECE2205) and Circuits and Systems II (ECE3205)focus on continuous-time signals and systems. In particular cir-cuits based implementation of systems is investigated in greatdetail. There still remains a lot to discuss about continuous-timesignals and systems without the need to consider a circuit imple-mentation. This chapter begins that discussion.

Continuous-Time Signals

• To begin with signals will be classified by their support inter-val

Two-Sided Infinite-Length Signals

• Sinusoids are a primary example of infinite duration signals,that are also periodic

Chapt

9

ECE 2610 Signal and Systems 9–1


(9.1)

• The period for both the real sinusoid and complex sinusoidsignals is

• The signal may be any periodic signal, say a pulse train orsquarewave

• A two-sided exponential is another example

(9.2)

x t( ) A ω0t φ+( ) ∞– t ∞< <,cos=

x t( ) Aejφejω0t ∞– t ∞< <,=

T0 2π ω0⁄=

x t( ) Ae β t– ∞– t ∞< <,=

�10 �5 5 10

0.5

1.0

1.5

2.0

�4 �2 2 4

�4

�2

2

4

�4 �2 2 4

0.5

1.0

1.5

2.0

2.5

3.0

t

t

t

x t( ) 5 2π t2---⎝ ⎠

⎛ ⎞cos=

x t( ) 2e t 2⁄–=

x t( ) Pulse Train=

Period = 2sPulse Width = 0.5s

Two-sided exponential

ECE 2610 Signals and Systems 9–2


One-Sided Signals

• Another class of signals are those that exist on a semi-infiniteinterval, i.e., are zero for (support )

• The continuous-time unit-step function, , is useful fordescribing one-sided signals

(9.3)

• When we multiply the previous two-side signals by the step-function a one-side signal is created

t t0< t [0 ∞),∈

u t( )

u t( ) 1, t 0≥0, otherwise⎩

⎨⎧

=

�2 2 4 6 8 10

0.5

1.0

1.5

2.0

�1 1 2 3 4

0.2

0.4

0.6

0.8

1.0

�1 1 2 3 4

�4

�2

2

4x t( ) 5 2π t

2--- π

4---–⎝ ⎠

⎛ ⎞ u t( )cos=

x t( ) 2e t 2⁄– u t( )=

x t( ) u t( )=

One-sided exponential

t

t

t



• The start time can easily be changed by letting

(9.4)

Finite-Duration Signals

• Finite duration signals will have support over just a finitetime interval, e.g.,

• A convenient way of crating such signals is via pulse gatingfunction such as

(9.5)

t t t0–→

x t( ) u t 2–( ) 1, t 2≥0, otherwise⎩

⎨⎧

= =

t [4 10),∈

p t( ) u t 4–( ) u t 10–( )–1, 4 t 10<≤0, otherwise⎩

⎨⎧

= =

t

t

p t( )

x t( ) 5 2π t2--- π

4---–⎝ ⎠

⎛ ⎞ p t( )cos=

2 4 6 8 10 12

�4

�2

0

2

4

2 4 6 8 10 12

�4

�2

2

4


The Unit Impulse

The Unit Impulse

• The topics discussed up to this point have all followed logi-cally from our previous study of discrete-time signals andsystems

• The unit impulse signal, , however is more difficult todefine than the unit impulse sequence,

• Recall that

• The unit impulse signal is defined as

(9.6)

and

(9.7)

• What does this mean?

– It would seem that must have zero width, yet havearea of unity

• A test function, , can be defined that in fact becomes as

(9.8)

δ t( )δ n[ ]

δ n[ ] 1, n 0=

0, otherwise⎩⎨⎧

=

δ t( ) 0 t 0≠,=

δ t( ) td∞–

∞

∫ 1=

δ t( )

δΔ t( )δ t( ) Δ 0→

δΔ t( )1

2Δ------- , Δ– t Δ< <

0, otherwise⎩⎪⎨⎪⎧

=


The Unit Impulse

• The claim is that

(9.9)

• Check (9.6) and (9.7)

• In plotting a scaled unit-impulse signal, e.g., , we plot avertical arrow with the amplitude actually corresponding tothe area

t

δΔ t( )1

2Δ1---------

12Δ2---------

Δ1 Δ20Δ– 1Δ– 2

δΔ t( )Δ 0→lim δ t( )=

δΔ t( )Δ 0→lim 0 t 0≠,=

δΔ t( ) td∞–

∞

∫ 1=

Aδ t( )

t

Aδ t( )A( )

0


The Unit Impulse

Sampling Property of the Impulse

• A noteworthy property of is that

(9.10)

• Discussion

– Since is zero everywhere except , only thevalue is of interest

– Using the test function we also note that

(9.11)

so as the only value of that matters is

– Also observe that

(9.12)

δ t( )

f t( )δ t t0–( ) f t0( )δ t t0–( )=Sampling Property

δ t t0–( ) t t0=f t0( )

δΔ t( )

f t( )δΔ t( )f t( ) 2Δ( ),⁄ Δ– t Δ< <0, otherwise⎩

⎨⎧

=

Δ 0→ f t( ) f 0( )

t t

f t( )f t( )

f t( )δΔ t( ) f 0( )δΔ t( )≈ f t( )δ t( ) f 0( )δ t( )≈

f t( )δΔ t( )

f 0( )2Δ---------

f 0( )( )

f t( )δ t( ) td∞–

∞

∫ f 0( )δ t( ) td∞–

∞

∫=

f 0( ) δ t( ) td∞–

∞

∫ f 0( )==


The Unit Impulse

• Integral Form

(9.13)

Example:

• The sampling property of results in

• When integrated we have

Operational Mathematics and the Delta Function

• The impulse function is not a function in the ordinary sense

• It is the most practical when it appears inside of an integral

• From an engineering perspective a true impulse signal doesnot exist

– We can create a pulse similar to the test function aswell as other test functions which behave like impulsefunctions in the limit

• The operational properties of the impulse function are veryuseful in continuous-time signals and systems modeling, aswell as in probability and random variables, and in modelingdistributions in electromagnetics

f t( )δ t t0–( ) td∞–

∞

∫ f t0( )=

Sampling/Sifting Property

2πt( )δ t 1.2–( )cos u t( )δ t 3–( )+

δ t( )

2π 1.2( )( )δ t 1.2–( )cos u 3( )δ t 3–( )+

2πt( )δ t 1.2–( )cos u t( )δ t 3–( )+[ ] td∞–

∞

∫2.4π( )cos u 3( )+ 2.4π( )cos 1+==

δΔ t( )


The Unit Impulse

Derivative of the Unit Step

• A case in point where the operational properties are veryvaluable is when we consider the derivative of the unit stepfunction

• From calculus you would say that the derivative of the unitstep function, , does not exist because of the discontinu-ity at

• Consider

(9.14)

• The area property of states that

(9.15)

• Invoking the area property we have

(9.16)

which says that this integral behaves like the unit step func-tion

(9.17)

u t( )t 0=

δ τ( ) τd∞–

t

∫δ t( )

δ t( ) tda

b

∫1, a 0 and b 0≥<0, otherwise⎩

⎨⎧

=

δ τ( ) τd∞–

t

∫1, t 0≥0, otherwise⎩

⎨⎧

=

u t( ) δ τ( ) τd∞–

t

∫=


The Unit Impulse

• From calculus we recognize that (9.17) implies also that

(9.18)

• Similarly,

(9.19)

• If we now consider situations where a product exits, i.e., , we can invoke the product rule for derivatives to

obtain

(9.20)

Example:

• The derivative of is

δ t( ) ddt-----u t( )=

δ t t0–( ) ddt-----u t t0–( )=

x t( )f t( )u t( )=

ddt-----f t( )u t( ) d

dt-----f t( )⎝ ⎠⎛ ⎞ u t( ) f t( ) ddt

-----u t( )⎝ ⎠⎛ ⎞+=

f ′ t( )u t( ) f t( )δ t( )+=

x t( ) e 4t– u t( ) u t 1–( )+=

x t( )

x′ t( ) ddt-----x t( ) 4e 4t–

– u t( ) e 4t– δ t( ) δ t 1–( )+ += =

4e 4t– u t( )– δ t( ) δ t 1–( )+ +=

-1 -0.5 0.5 1 1.5 2

-4

-3

-2

-1

1

-1 -0.5 0.5 1 1.5 2

0.2

0.4

0.6

0.8

1dx t( )dt

------------x t( ) (1) (1)


Continuous-Time Systems

Continuous-Time Systems

• A continuous-time system operates on the input to producean output

(9.21)

Basic System Examples

(9.22)

(9.23)

(9.24)

(9.25)

• In all of the above we can calculate the output given the inputand the definition of the system operator

• For linear time-invariant systems we are particularly inter-ested in the impulse response, that is the output, ,when , for the system initially at rest

y t( ) T x t( ){ }=

T { }x t( ) y t( )

y t( ) x t( )[ ]2=Squarer

y t( ) x t td–( )=Time Delay

y t( ) dx t( )dt

------------=

Differentiator

y t( ) x τ( ) τd∞–

t

∫=

Integrator

y t( ) h t( )=x t( ) δ t( )=


Linear Time-Invariant Systems

Example: Integrator Impulse Response

• Using the definition


• In the study of discrete-time systems we learned the impor-tance of systems that are linear and time-invariant, and howto verify these properties for a given system operator

Time-Invariance

• A time invariant system obeys the following

(9.26)

for any

• Both the squarer and integrator are time invariant

• The system

(9.27)

is not time invariant as the gain changes as a function of time

y t( ) h t( ) δ τ( ) τd∞–

t

∫ u t( )= = =

x t t0–( ) y t t0–( )→

t0

y t( ) ωct( )x t( )cos=



Linearity

• A linear system obeys the following

(9.28)

where the inputs are applied together or applied individuallyand combined via and later

• The squarer is nonlinear by virtue of the fact that

produces a cross term which does not exist when the twoinputs are processed separately and then combined

• The integrator is linear since

The Convolution Integral

• For linear time-invariant (LTI) systems the convolution inte-gral can be used to obtain the output from the input and thesystem impulse response

(9.29)

αx1 t( ) βx2 t( )+ αy1 t( ) βy2 t( )+→

α β

y t( ) αx1 t( ) βx2 t( )+[ ]2=

α2x12 t( ) 2αβx1 t( )x2 t( ) β2x2 t( )+ +=

y t( ) αx1 τ( ) βx2 τ( )+[ ] τd∞–

t

∫=

α x1 τ( ) τd∞–

t

∫ β x2 τ( ) τd∞–

t

∫+=

y t( ) x τ( )h t τ–( ) τd∞–

∞

∫ x t( )*h t( )= =

Convolution Integral



• The notation used to denote convolution is the same as thatused for discrete-time signals and systems, i.e., the convolu-tion sum

• Evaluation of the convolution integral itself can prove to bevery challenging

Example:

• Setting up the convolution integral we have

or simply

,

which is known as the unit ramp

y t( ) x t( )*h t( ) u t( )*u t( )= =

y t( ) u τ( )u t τ–( ) τd∞–

∞

∫=

τ

u τ( )u t τ–( )

t 0 t

1

y t( )0, t 0<

τ,d0

t

∫ t 0≥⎩⎪⎨⎪⎧

=

0, t 0<t, t 0≥⎩

⎨⎧

=

y t( ) tu t( ) r t( )≡=


Impulse Response of Basic LTI Systems

Properties of Convolution

• Commutativity:

(9.30)

• Associativity:

(9.31)

• Distributivity over Addition:

(9.32)

• Identity Element of Convolution:

(9.33)

What is ?

– It turns out that

proof

Impulse Response of Basic LTI Systems

• For certain simple systems the impulse response can befound by driving the input with and observing the output

• For complex systems transform techniques, such as theLaplace transform, are more appropriate

x t( )*h t( ) h t( )*x t( )=

x t( )*h1 t( )[ ]*h2 t( ) x t( )* h1 t( )*h2 t( )[ ]=

x t( )* h1 t( )*h2 t( )[ ] x t( )*h1 t( ) x t( )*h2 t( )+=

x t( )*h t( ) h t( )=

x t( )

x t( ) δ t( )= δ t( )*h t( )⇒ h t( )=

δ τ( )h t τ–( ) τd∞–

∞

∫ δ τ( )h t 0–( ) τd∞–

∞

∫=

h t( ) δ τ( ) τd∞–

∞

∫ h t( )==

δ t( )


Convolution of Impulses

Integrator

(9.34)

Ideal delay

(9.35)

• Note that this means that

(9.36)

Convolution of Impulses

• Basic Theorem:

(9.37)

Example:

• Using the time shift property (9.36)

Evaluating Convolution Integrals

Step and Exponential

• Consider and

• We wish to find

h t( ) x τ( ) τd∞–

t

∫x τ( ) δ τ( )=

u t( )= =

h t( ) x t td–( )x t( ) δ t( )=

δ t td–( )= =

x t( )*δ t td–( ) x t td–( )=

δ t t1–( )*δ t t2–( ) δ t t1 t2+( )–( )=

δ t( ) 2δ t 3–( )–[ ]*u t( )

δ t( )*u t( ) 2δ t 3–( )*u t( )– u t( ) 2u t 3–( )–=

x t( ) u t 2–( )= h t( ) e 3t– u t( )=

y t( ) x t( )*h t( )=



(9.38)

• To evaluate this integral we first need to consider how thestep functions in the integrand control the limits of integra-tion

• For or there is no overlap in the product thatcomprises the integrand, so

• For or there is overlap for , sohere

(9.39)

y t( ) e 3τ– u τ( )u t τ– 2–( ) τd∞–

∞

∫=

τe 3τ– u τ( )

u t τ– 2–( ) t 2 0<–

t 2– 0 t 2–

1u t τ– 2–( ) t 2 0>–

t 2 0<– t 2<y t( ) 0=

t 2 0>– t 2> τ [0 t 2)–,∈

y t( ) e 3τ– τd0

t 2–

∫=

e 3τ–

3–----------

0

t 2–

=

13--- 1 e 3 t 2–( )–

–[ ]u t 2–( )=

t20

13---

y t( )



• Note: The use of the exponential impulse response in exam-ples is significant because it occurs frequently in practice,e.g., an RC lowpass filter circuit

Example: and

• Find by evaluating the convolution integral

• Suppose that and

y t( )x t( )

R

C

h t( ) 1RC--------e

tRC--------–

u t( )=

x t( ) e at– u t( )= h t( ) e bt– u t( )=

y t( ) x t( )*h t( )=

y t( ) e aτ– u τ( )e b t τ–( )– u t τ–( ) τd∞–

∞

∫=

e aτ– e b t τ–( )– τd0

t

∫=

e bt– e a b–( )τ– τd0

t

∫=

e bt–

a b–------------ e a b–( )τ–

a b–( )–--------------------

0

t

⋅ e bt–

a b–------------ 1 e a b–( )t–

–[ ]u t( )==

1a b–------------ e bt– e at–

–[ ]u t( ) a b≠,=

a 2= b 3=



Square-Pulse Input

• Consider a pulse input of the form

(9.40)

where is the pulse width and

• The output is

(9.41)

• From the step response analysis we know that

, (9.42)

so

�1 1 2 3 4 5 6

0.02

0.04

0.06

0.08

0.10

0.12

0.14y t( ) a 2 b 3–,=

t

x t( ) u t( ) u t T–( )–=

Tx t( )

T0 t

1

h t( ) e at– u t( )=

y t( ) u t( )*h t( ) u t T–( )*h t( )–=

u t( )*h t( )1a--- 1 e at–

–[ ]u t( )=


Properties of LTI Systems

(9.43)

• Plot the results for and


Cascade and Parallel Connections

• We have studied cascade and parallel system earlier

• For a cascade of two LTI systems having impulse responses and respectively, the impulse response of the cas-

cade is the convolution of the impulse responses

(9.44)

y t( ) 1a--- 1 a at–

–[ ]u t( ) 1a--- 1 a a t T–( )–

–[ ]u t T–( )–=

T 5= a 1=

5 10 15

0.2

0.4

0.6

0.8

1.0y t( )

a 1 T, 5= =

t

h1 t( ) h2 t( )

hcascade t( ) h1 t( )*h2 t( )=

h1 t( ) h2 t( )

h t( ) h1 t( )*h2 t( )=

x t( ) y t( )

x t( ) y t( )Cascade



• For two systems connected in parallel, the impulse responseis the sum of the impulse responses

(9.45)

Differentiation and Integration of Convolution

• Since the integrator and differentiator are both LTI systemoperations, when used in combination with another systemhaving impulse response, , we find that the cascade prop-erty holds

• What this means is that performing differentiation or integra-tion before a signal enters and LTI system, gives the sameresult as performing the differentiation or integration after thesignal passes through the system

hparallel t( ) h1 t( ) h2 t( )+=

h1 t( )

h2 t( )

x t( ) y t( )

h t( ) h1 t( ) h2 t( )+=x t( ) y t( )

Parallel

h t( )

h t( )

h t( )

( )∫ or d ( ) dt⁄

( )∫ or d ( ) dt⁄

x t( )

x t( ) y t( )

y t( )



Example: Step Response from

• Knowing the impulse response of a system we can find therespond to a step input by just integrating the output, since

at the input is obtained by integrating

• Thus we can write that

• This result is consistent with earlier analysis

Stability and Causality

• Definition: A system is stable if and only if every boundedinput produces a bounded output. A bounded input/output is asignal for which for all values of t.

• A theorem which applies to LTI systems states that a system(LTI system) is stable if and only of

(9.46)

– if and only if holds in either direction

h t( ) e at– u t( )=

u n( ) δ t( )

y t( ) u t( )*h t( ) h τ( ) τd∞–

t

∫= =

e aτ– u τ( ) τd∞–

t

∫ e aτ– τd0

t

∫==

e aτ–

a–----------

0

t1a--- 1 e at–

–[ ]u t( )==

x t( ) or y t( ) ∞<

h t( ) td∞–

∞

∫ ∞<Stability for LTI Systems



Example: LTI with

• For stability

• We must have for stability

• Note that result in , which is an integra-tor system, hence an integrator system is not stable

• Definition: A system is causal if and only if the output at thepresent time does not depend upon future values of the input

• A theorem which applies to LTI systems is

(9.47)

• This definition and LTI theorem also holds for discrete-timesystems

Example: Simulate an LTI System using Matlab lsim()

• As a final example we consider how we can use MATLAB tosimulate LTI systems

• The function we use is lsim(), which has behavior similarto that of filter(), which is used for discrete-time sys-tems

h t( ) e at– u t( )=

e at– u t( ) td∞–

∞

∫ e at– td0

∞

∫=

e at–

a–---------

0

∞1a--- a 0>,==

a 0>

a 0= h t( ) u n( )=

h t( ) 0 for t 0<=Causal for LTI Systems



>> t = -1:0.01:15; % create a time axis>> x = zeros(size(t)); % next 3 lines create a pulse>> i_pulse = find(t>=0 & t<=5); % duration is 5s>> x(i_pulse) = ones(size(i_pulse));>> subplot(211)>> plot(t,x)>> axis([-1 15 0 1.1]); grid>> ylabel('Input x(t)')>> subplot(212)>> y = lsim(tf([1],[1 1]),x,t);% h(t) = e^(-1*t) u(t)Warning: Simulation will start at the nonzero initial time T(1).> In lti.lsim at 100>> plot(t,y); grid>> ylabel('Output y(t)')>> xlabel('Time (s)')

0 5 10 150

0.5

1

Inpu

t x(t

)

0 5 10 150

0.5

1

Out

put y

(t)

Time (s)

Input pulse of duration 5s

Impulse response = e-t u(t)


continuous-time signals and lti systems

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