elec361: signals and systems topic 9: the laplace transformamer/teach/elec361/slides/topic9... ·...

1

ELEC361: Signals And Systems

Topic 9: The Laplace Transform

o Introductiono Laplace Transform & Exampleso Region of Convergence of the Laplace Transformo Review: Partial Fraction Expansiono Inverse Laplace Transform & Exampleso Properties of the Laplace Transform & Exampleso Analysis and Characterization of LTI Systems Using the

Laplace Transformo LTI Systems Characterized by Linear Constant-Coefficient DEo SummaryDr. Aishy Amer

Concordia UniversityElectrical and Computer Engineering

Figures and examples in these course slides are taken from the following sources:

•A. Oppenheim, A.S. Willsky and S.H. Nawab, Signals and Systems, 2nd Edition, Prentice-Hall, 1997

•M.J. Roberts, Signals and Systems, McGraw Hill, 2004

•J. McClellan, R. Schafer, M. Yoder, Signal Processing First, Prentice Hall, 2003

•Web Site of Dr. Wm. Hugh Blanton, http://faculty.etsu.edu/blanton/

2

IntroductionTransforms: Mathematical conversion from one way of thinking to another to make a problem easier to solve

Reduces complexity of the original problem

Laplacetransform

solutionin

s domain

inverse Laplace

transform

solution in timedomain

problem in time domain

• Other transforms• Fourier Transform• z-transform

s = σ+jω

3

Introductiontime domain

lineardifferentialequation

timedomainsolution

Laplacetransformed

equation

Laplacesolution

Laplace domain orcomplex frequency domain

algebra

Laplace transforminverse Laplacetransform

x(t) y(t)

4

IntroductionCT Fourier Transform:

representation of signals as linear combination of complex exponentials est, s = jω

Laplace Transform: Representation of signals as linear combination of est,s = σ+jω

A generalization of CTFTCan be applied in contexts where the FT cannotInvestigation of stability/instability & causality of systems

Laplace transform applies to continuous-time signals

∫∫∞

∞−

∞

∞−

− == ωωπ

ω ωω dejXtxdtetxjX tjtj )(21)()()(

5

Introduction

Ω= j ez

Continuous-time analog signal

x(t)

Continuous-time analog signal

x(t)

Discrete-time analog sequence

x [n]

Discrete-time analog sequence

x [n]

Sample in timeSampling period = Ts

ω=2πfΩ = ω Ts,scale amplitude by 1/Ts

Sample in frequency,Ω = 2πn/N,N = Length

of sequence

ContinuousFourier Transform

X(f)

ContinuousFourier Transform

X(f)

∞≤≤∞

∫∞

∞−

f-

dt e x(t) ft2 j- π

Discrete Fourier Transform

X(k)

Discrete Fourier Transform

X(k)

10

e [n]x 1

0 =n

Nnk2j-

−≤≤

∑−

Nk

N π

Discrete-Time Fourier Transform

X(Ω)

Discrete-Time Fourier Transform

X(Ω)

π20

e [n]x - =n

j-

≤Ω≤

∑∞

∞

Ωn

LaplaceTransform

X(s)s = σ+jω

LaplaceTransform

X(s)s = σ+jω

∞≤≤∞

∫∞

∞−

−

s-

dt e x(t) st

z-TransformX(z)

z-TransformX(z)

∞≤≤∞−

∑∞

∞−

z =n

n- z [n]x

s = jωω=2πf

C CC

C

C D

D

DC Continuous-variable Discrete-variable

Ω= j ez

6

Introduction

Convert time-domain signals into frequency-domain x(t) → X(s) t∈R, s∈CLinear differential equations (LDE) →algebraic expression in complex plane

Graphical solution for key LDE characteristics(Discrete systems use the analogous z-transform)

ωσ js

dtetxsXtx st

+=

== ∫∞

∞−

−)()()]([L

7

Introduction: Complex Exponential e-st

part sinusoidal theoargument t theofpart is that thenoticingin apparent is This

sosillation theof rate thedeterminespart the th,decay/grow of rate thedetermines theWhile

)sin()cos()( )(

ωω

σσωσω ωσωσ

⇒

•=+++===• + tjttjst eAettAjttAAeAetx

σt is the phase

is the frequencyω

8

Introduction: the complex s-plane

•Any time s lies in the right half plane, the complex exponential will grow through time; any time s lies in the left half plane it will decay

Imaginary axis

rReal axis

ωσ js +=

σ

ωφφ−

ωσ js −=∗

(complex) conjugate

ω−

22||||

1tan

ωσ

σω

φ

+=∗

≡≡

−=≡∠

srs

s

r

Right Half PlaneLeft Half Plane

Axis tells how fast est

grows or decays

Axis tells how fast est oscillates (higher frequency)

9

Outline

o Introductiono Laplace Transform & Exampleso Region of Convergence of the Laplace Transformo Review: Partial Fraction Expansiono Inverse Laplace Transform & Examples o Properties of the Laplace Transform & Exampleso Analysis and Characterization of LTI Systems Using the Laplace

Transformo LTI Systems Characterized by Linear Constant-Coefficiento Summary

10

Laplace TransformAs mentioned earlier, the response of an LTI system with impulseresponse h(t) to a complex exponential of the form est is

where

If we let s = jw (pure imaginary), the integral above is essentially the Fourier transform of h(t)For arbitrary values of the complex variable s, this expression is referred to as the Laplace transform of h(t)Therefore, the Laplace transform of a general signal x(t) is defined as

Note that s is a complex variable, which can be expressed in general as s = σ+jω

11

Laplace TransformWhen s = jω, we get the Fourier transform of x(t)

Therefore, the Fourier transform is a special case of the Laplace transform

can be expressed as

X(σ+jω) is essentially the Fourier transform of x(t)e-σt

Properties of x(t)e-σt determine convergence of X(s)

Note: e-jωt sinusoidal, e.g., bounded

12

Laplace Transform

The Laplace transform X(s) for positive t≥0 typically exists for all complex numbers such that Res > a

where a is a real constant which depends on the growth behavior of x(t)

The subset of values of s for which the Laplace transform exists is called the region of convergence (ROC)Note: the two sided (-∞< t < ∞) Laplace transform is defined in a range a < Res < b

13

Laplace Transform

X(s) of expression algebraic thezecharacteri Completely :X(s) of Zeros& Poles

poles ofnumber theis ofOrder complex are zeroes and Poles

)0)( (So, 0)( :Zeroes))( (So, 0)( :ties)(signulari Poles

...)(

...)()2)(105(

252X(s);expression algebraic)()()(

spolynomial of rational aoften is transformLaplace

01

01

2

2

•=•

•==∋•

∞==∋•+++=•

+++=•

+++++

===

mX(s)

sXsNssXsDs

bsbsbsD

asasasNsss

sssDsNsX

mm

nn |X(s)| will be larger when it

is closer to the poles

|X(s)| will be smaller when it is closer to the zeros

14

Laplace Transform:Order of X(s)

KBsJsK

sDsN

sTK

sDsN

++=•

+=•

2)()(Order Second

1)()(Order First

Impulse response

Exponential

Step response

Step, exponential

Ramp response

Ramp, step, exponential

1 sT

K+

/1

2 TsKT-

sKT-

sK

+

/1

Ts

K-sK

+

15

Laplace Transform:Poles

The poles of a Laplace function are the values of s that make the Laplace function evaluate to infinityThe poles are therefore the roots of the denominator polynomial

has a pole at s = -1 and a pole at s = -3

Complex poles (e.g., s=-2+5j) always appear in complex-conjugate pairsThe response of a system is determined by the location of poles on the complex plane

)3)(1()2(10

+++ss

s

16

Laplace Transform: Zeros The zeros of a Laplace function are the values of s that make the Laplace function evaluate to zeroThe zeros are therefore the zeros of the numerator polynomial has a zero at s = -2

Complex zeros always appear in complex-conjugate pairs

Pole-Zero Cancellation: Do not eliminate poles as in

Think about what may happen if H(s) was a transfer function of a physical system where minor system (e.g., temperature) change could cause the pole or the zero to move

)3)(1()2(10

+++ss

s

)1()1)(3()(

−−+

=s

sssH

17

Laplace Transform:Visualization

FT: X(jω) a complex valued function of purely imagineryvariable jw

Visualize using 2D plot of real and imaginary part or magnitude and phase

LT: X(s) a complex valued function of a complex variable s=σ+j ω

Requires a 3D plot which is difficult to visualize or analyzeSolution: Poles (x) and Zeros (o) PlotExample:

Poles:s=(-1-3j)s=(-1+3j)s=-2

18

Laplace Transform: Example

x(t)of nature on the depends X(s) of eConvergenc-aRe(s)or 0a)Re(s ifonly 0lim

convergenot do power negative be NOT will zero,or negative is )( if converge power negative be be will positive, is )( if

notor converge transformLaplace he whether tdetermines hat apparent t becomesit bounded), (i.e., sinusoidal is that gRecognizin

1) OR

is x(t)of transformLaplace thehand,other On the

|)(|1)

is 0,a with ),X(j ansformFourier tr The

x(t)sidedright )(Let

)(

)(

)(

00

0

0

⇒>>+=•

⇒+

⇒+

•

+==+==

•

∞<+

===

>•

=•

+−

+−

+−

+−

−

+−∞

−+−∞

+−∞

∞−

−−

∞

∞−

∞−−

∞

∞−

−−

∞→

∫∫∫

∫∫∫

a)t(s

ta

ta

ta

jwt

a)t(sjwtσ)t(aa)t(sstat

jwtatjwtat

-at

eea

eae

e

eas

dteejωX(σdtedtu(t)eeX(s)

dttxajw

dteedtu(t)eeX(jω

tuex(t)

t

σ

σ

σ

σ

σ

ω

o

o

Recall: FT converges if

19


as

astue

asas

X(s)

js

aajw

dteejωX(σ

Lat

jwtσ)t(a

−>+

=

−>+

⎯→←•

−>+

=

+=•

>+++

==+•

=⇒•

−

∞−+−∫

Reas

1X(s)

Reas

1)( :Conclusion

Re1becomesequation last the, Since

0)( where)(

1)

Re(s) but to j torelatednot is ROC a- >Re(s) :eConvergencfor Condition

0

ωσ

σσ

σω

20

Laplace Transform: ExampleWe conclude from the above example that the Laplace transform exists for this particular x(t) only ifThe region in the complex plane in which the Laplace transform exists (or converges) is called region of convergence (ROC)The ROC for the above example is given in the following figure

For a single pole, the ROC lies to the right of this pole for right-sided signals; x(t) non zero for t≥0

21


left-sided x(t)

For a single pole, the ROC lies to the left of this pole for left-sided signals; x(t) non zero for t<0

22


two-sided h(t)

23


right-sided

24


1-01 when converges)3()1(

131

1

>→>+⇒−++

=−++

σσωσωσ jjj

a sum of real & complex exp.

25

Outline


Transformo LTI Systems Characterized by Linear Constant-Coefficient DEo Summary

26

Region of Convergence for Laplace Transform

Let X(s) be the Laplace transform of some signal x(t)The ROC of X(s), in general, has the following characteristics:1. The ROC of X(s) consists of strips parallel to

the jω-axis in the s-plane2. For rational Laplace transforms,

the ROC doesn’t contain any poles (since X(s)=∞)3. If x(t) is of finite duration and is absolutely integrable,

then the ROC is the entire s-plane

27


4. If x(t) is right-sided, and if the line Re s = σ0 is in the ROC, then all values of s for which Re s > σ0 will also be in the ROC

5. If x(t) is left-sided, and if the line Re s = σ0is in the ROC, then all values of s for which Re s < σ0 will also be in the ROC

6. If x(t) is two-sided, and if the line Re s = σ0 is in the ROC, then the ROC will consist of a strip in the s-plane that includes the line Re s = σ0

28


7. If the Laplace transform X(s) of x(t) is rational, then the ROC is bounded by poles or extends to infinity. In addition, no poles of X(s) are contained in the ROC

8. If the Laplace transform X(s) of x(t) is rational, then if x(t) is right-sided, the ROC is the region in the s-plane to the right of the rightmost pole. If x(t) is left sided, the ROC is the region in the s-plane to the left of the leftmost pole

29

Region of Convergence for Laplace Transform: Example

30

Common Laplace Transforms Pairs

number interger positive:numbers real are ,,

ntajsCs

•∞<<∞−••+=∈• τωσ

Name x(t) Graph X(s) ROC

Impulse 1 All s

Step R(s)>0

Delayed Impulse All s

Delayed step R(s)>0

Ramp R(s)>0

Power R(s)>0

ExponentialDecay

R(s)>-a

ExponentialApproach

R(s)>0

⎩⎨⎧

>=

==0001

)()(tt

ttx δ

)()( tutx =s1

)()( τδ −= ttxse τ−

)()( τ−= tutxs

e sτ−

2

1s

)()( ttutx =

1

!+ns

n)()( nuttx n=

as+1)()( tuetx at−=

)( assa+

)()1()( tuetx at−−=

31

Common Laplace Transforms Pairs

number interger positive:numbers real are ,,

ntajsCs

•∞<<∞−••+=∈• τωσ

Name x(t) Graph X(s) ROC

Left-sided step R(s)<0

Left-sided exponential R(s)<-a

Sine R(s)>0

Cosine R(s)>0

Exponentially Decaying Sine R(s)>-a

Exponentially Decaying Cosine R(s)>-a

Hyberbolic Sine R(s)>|ω|

Hyberbolic Cosine R(s)>|ω|

))(()( tutx −−= s1

as+1

)()( tuetx at −−= −

22 θθ+s)()sin()( tuttx θ=

22 θ+ss)()cos()( tuttx θ=

22)( θθ

++as)()sin()( tutetx at θ−=

22)( θ+++

asas

)()cos()( tutetx at θ−=

22 θθ−s

)()sinh()( tuttx θ=

22 θ−ss)()cosh()( tuttx θ=

32

Outline



33

Review:Partial Fraction Expansion

Partial fractions are several fractions whose sum equals a given fraction

Purpose: Working with transforms requires breaking complex fractions into simpler fractions to allow use of tables of transforms

1111

)1)(1()1(5)1(6

15

16

15

16

1111

2

2

−−

=−+

++−=

−+

+•

−+

+=

−−

•

ss

ssss

ss

ssss

34

Review:Partial Fraction Expansion

32)3()2(1

++

+=

+++

sB

sA

sss

( ))3()2(

2)3()3()2(

1++

+++=

+++

sssBsA

sss

32

21

)3()2(1

++

+−

=++

+ssss

s

1=+ BA 123 =+ BA

Expand into a term for each factor in the denominator.Recombine right hand side

Equate terms in s and constant terms. Solve.Each term is in a form so that inverse Laplace transforms can be applied.

35

Review:Partial Fraction Expansion: Different terms of 1st degree

To separate a fraction into partial fractions when its denominator can be divided into different terms of first degree, assume an unknown numerator for each fraction

561

11)1()111(

)1)(1()()(

)1)(1()1()1(

)1()1()1()111(

2

2

==⇒⎭⎬⎫

−=−=+

−−

=−+

−++=

−+++−

=

−+

+=

−−

BAABBA

ss

ssABsBA

sssBsA

sB

sA

ss

36

Review: Partial Fraction Expansion: Repeated terms of 1st degree

When the factors of the denominator are of the first degree but some are repeated, assume unknown numerators for each factor

If a term is present twice, make the fractions the corresponding term and its second powerIf a term is present three times, make the fractions the term and its second and third powers

2114

321

)()2()1()1(43)(

)()(

)1()1()1()1(

)1()1()1()1(43

2

22

33

2

323

2

===⇒⎪⎭

⎪⎬

⎫

=++=+

=

+++++=

++++=++=

=+

++

+++=

++

++

+=

+++

CBACBA

BAA

CBAsBAAsCsBsAsssN

sDsN

sC

ssBsA

sC

sB

sA

sss

37

Review: Partial Fraction Expansion: Different quadratic terms

When there is a quadratic term, assume a numerator of the form Bs + C

05.05.012

00

)2()()(1)1()1()2(1

)2()1()2)(1(1

2

2

22

===⇒⎪⎭

⎪⎬

⎫

=+=++

=+

++++++=

++++++=

+++

++

=+++

CBACACBA

BACAsCBAsBA

sCsBsssAss

CBss

Asss

38

Review:Partial Fraction Expansion: Repeated quadratic terms

When there is repeated quadratic term, assume two numerator of the form Bs + C and Ds+E

05.0025.025.0

1240324

0235022

0)1()1()2)(1(

)2)(1()2(1)2()2()1()2)(1(

1

2

222

22222

=−==−==

⇒

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

=++=++++

=+++=++

=+

++++++++

++++++=

+++

+++

++

+=

+++

EDCBA

ECAEDCBA

DCBACBA

BAsEsDssssC

sssBsssAss

EDsss

CBss

Asss

39

Outline

o Introductiono Laplace Transform & Exampleso Region of Convergence of the Laplace Transformo Review: Partial Fraction Expansiono Inverse Laplace Transform & Exampleso Properties of the Laplace Transform & Exampleso Analysis and Characterization of LTI Systems Using the Laplace


40

The Inverse Laplace Transform

Let X(s) be the Laplace transform of a signal x(t)The inverse Laplace transform is given by

for all values of s in the ROCA very useful technique in finding the inverse Laplace transform is to expandX(s) in the form

From the ROC of X(s), one can find the ROC for each individual term in the above expressionThe inverse Laplace transform can then be obtained for every term separately very easily

∫∞+

∞−

=j

j

stdsesXj

txσ

σπ)(

21)(

41

The Inverse Laplace Transform: Solving Using Tables

1. Write the function you wish to inverse transform, X(s), as a sum of other functions

Where each Xi(s) is known from the table2. Invert each Xi(s) to get xi(t)3. Sum up all xi(t) to get x(t)

∑=

=m

ii sXsX

1)()(

∑=

=m

ii txtx

1)()(

42

The Inverse Laplace Transform: Example

43

The Inverse Laplace Transform: Example

44

Outline



45

Properties of the Laplace Transform

46


47


48


49

Properties of the Laplace Transform: Examples

50


tof valuespositive from 0 approaches t as0 tof valuesnegative from 0 approaches t as0

+

−

•

•

51


52

Properties of the Laplace Transform:Examples

)4()2(2)(

++=

ssssYLaplace

transform of the function

Apply Final-Value theorem

Apply Initial-Value theorem

[ ]41

)40()20()0()0(2)(lim =

++=∞→ txt

[ ] 0)4()2()(

)(2)(lim 0 =+∞+∞∞

∞=→ txt

53


the initial value of x(t) as t approaches 0 from positive values of t is given by

The final value of x(t) as t approaches ∞is given by

11

lim)0(

11)(

)(

)(lim)0(

=+

=

+=

=

=

∞→

−

∞→

+

ssx

ssX

etxExample

ssXx

s

t

s)0( +x

01

lim)(lim

11)(

)(

)(lim)(lim

0

0

=+

=

+=

=

=

→∞→

−

→∞→

sstx

ssX

etxExample

ssXtx

st

t

st

54


55

Properties of the Laplace Transform: summary

x(t) X(s)

( ) )(txLsX =( ) sXLtx 1)( −=

)( asX −)(txeat

)()1( sXdsd

n

nn−

)0()( xssX

)(txt n

)(' tx −

)('' tx )0(')0()(2 xsxsXs −−

)(''' tx )0('')0(')0()( 23 xsxxssXs −−−

56

Properties of the Laplace Transform: Summary

57

Outline

o Introductiono Laplace Transform & Exampleso Region of Convergence of the Laplace Transformo Review: Partial Fraction Expansiono Inverse Laplace Transform & Examples o Properties of the Laplace Transform & Exampleso Analysis and Characterization of LTI Systems Using the

Laplace Transformo LTI Systems Characterized by Linear Constant-Coefficient DEo Summary

58

Analysis and Characterization of LTI Systems Using the Laplace Transform

Let x(t) be an input to some LTI system whose impulse response is h(t), the output y(t) of the system is given by

where ∗ denotes convolution In the s−domain, the above expression becomes

Y(s) is the Laplace transform (LT) of y(t)X(s) is the LT of x(t)H(s) is the LT of h(t)

It is customary to refer to H(s) as the transfer function of the LTI system

59

Transfer Function of an LTI system

A transfer function is an expression that relates the output to the input in the s-domain

H(s) = Y(s) / X(s)

H(s) relates the output of a linear system (or component) to its inputH(s) describes how a linear system responds to an impulseH(s) represents a normalized model of a process, i.e., can be used with any input

DifferentialEquation

x(t) y(t)

H(s):TransferFunction

X(s) Y(s)

60

Transfer Function of an LTI systemThe form of the transfer function indicates the dynamic behavior of the system

For a, b, and c positive constants, the transfer function terms indicate exponential decay and exponential oscillatory decay (first two terms)exponential growth (third term)The decay terms will reach zero with time but h(t) will continue to grow because of the growth term (third term)

ctbtat eCteBeAth

csC

bsB

asALth

csC

bbssB

asAsH

++=⎭⎬⎫

⎩⎨⎧

−+

+++

+=

−+

++++

+=

−−

−

)sin()(: table theUsing

)()()()(

)()2()()(

221

222

ω

ωω

ωω

61


Causality:A causal LTI system: the output at any time depends on present and past input values only (not on future value)A causal LTI system: h(t)=0 for t<0 (right-sided)

The ROC of H(s) of a causal system is in the right-half planeA system with rational transfer function is causal if and only if (iff) the ROC is the right-half plane to the right of the rightmost pole

62

Analysis and Characterization of LTI Systems Using the Laplace Transform: Example

Consider an LTI system whose H(s) is given by

Note that since the ROC is not specified and there are two poles at -1 and 2, then there are three possible solutions

63

Analysis and Characterization of LTI Systems Using the Laplace Transform: Example

1. Res < −1 : Then the solution is left-sided, and is given by

Clearly this system is non causal

2. Res > 2 : The solution here is given by

which is right-sided. In addition the ROC is extending towards ∞. Thus, the system is causal

3. -1 < Res < 2 : Then the solution must have a term that is left-sided and another that is right-sided; that is

The system now is non causal

64

Analysis and Characterization of LTI Systems Using the Laplace Transform:Example

65


StabilityA system is stable if bounded inputs produce bounded outputsAn LTI system is BIBO stable iif

An LTI system is stable iff the ROC of its transfer function includes the jω axisNote: A system maybe stable with a non-rational H(s)

Causality & StabilityClearly, a causal system with rational transfer function is stableiff all of the poles of the transfer function lie in the left-half plane

1Re,1

)(.,. −>+

= ssesHge

s

∞<∫∞

∞−

dtth |)(|

66

Analysis and Characterization of LTI Systems Using the Laplace Transform:Example

67

Unstable BehaviorIf the output of a system grows without bound for a bounded input, the system is referred to a unstableThe complex s-plane is divided into two regions depending on poles locations

1. the stable region, which is the left half of the s-plane2. the unstable region, which is the right half of the s-plane

If the real portion of any pole of a transfer function is positive and ROC lies in the right-half plane, the system corresponding to the transfer function isunstableIf any pole is located in the right half plane and the ROC lies in the right-half plane, the system is unstable

68

Outline



69

LTI Systems Characterized by Linear Constant-Coefficient DE

One of the great things about Laplace transform is that it can be used to solve fairly complicated linear differential equations very easilyConsider for example the following differential equation:Applying Laplace transform to both sides yieldsand the transfer function can be obtained as

Since this system has only one pole, there are two possible solutions:

1. Res > −3 : The system in this case is causal and is given by

2. Res < −3 : This results in the non-causal solution, namely,

70


The same procedure can be used to obtain H(s) from any differential equations with constant coefficientsA general linear constant-coefficient differential equation is of the form

Applying Laplace transform to both sides yields

For the transfer function, The zeros are the solutions ofThe poles are the solutions of

71


We remark here that the transfer function does not tell additional information about the ROC of the systemThe ROC is normally specified by additional information such as knowledge about the stability and causality of the system

72

LTI Systems Characterized by Linear Constant-Coefficient DE: Example

73


Now to determine the ROC of H(s), we know from the convolution property that the ROC of Y(s) must include at least the intersection of the ROCsof X(s) and H(s)

Since H(s) has two poles, then there are three choices for the ROCSince we have some knowledge about the ROC of Y(s), this limits our choices for the ROC of H(s) to one

Res > −1 which means that H(s) is stable and causalFrom H(s), one can obtain the differential equation that relates X(s) and Y(s)

74


textbook) theof 9.26 example (see)4()2(

4)(−+

=⇒ss

ssH

75

LTI Systems Characterized by Linear Constant-Coefficient DE: Solution Procedure

Any nonhomogeneous linear differential equation (LDE) with constant coefficients can be solved with the following procedure, which reduces the solution to algebraStep 1: Put LDE into standard form

y’’ + 2y’ + 2y = cos(t)y(0) = 1; y’(0) = 0

Step 2: Take the Laplace transform of both sidesLy” + L2y’ + L2y = Lcos(t)

76


Step 3: Use properties of transforms to express equation in s-domain

Ly” + L2y’ + L2y = Lcos(ω t)• Ly” = s2 Y(s) - sy(0) – y’(0)• L2y’ = 2[ s Y(s) - y(0)]• L2y = 2 Y(s)

• Lcos(t) =

s2 Y(s) - s + 2s Y(s) - 2 + 2 Y(s) =

)1( 2 +ss

)1( 2 +ss

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Step 4: Solve for Y(s)

)22)(1(222

)22(

2)1()(

2)1(

)()22(

)1()(22)(2)(

22

23

2

2

22

22

++++++

=++

+++=•

+++

=++

+=+−+−•

ssssss

ss

ss

s

sY

ss

ssYss

sssYssYssYs

78


Step 5: Expand equation into format covered by table

( )( ) ( ) ( )

2.1,8.04.0,2.0

22222

221

rmssimilar te Equate)2()22()2()(

221221222)(

23

2222

23

====

⎪⎪⎭

⎪⎪⎬

⎫

=+=++

=++=+

•+++++++++•

+++

+++

=++++++

=•

ECBA

EBCBAEBA

CA

EBsCBAsEBAsCAssECs

sBAs

sssssssY

1)1(4.0

1)1()1(8.0

222.18.0

)1(4.0

)1(2.0

)1(4.02.0

222

222

+++

+++

=++

+•

++

+=

++

•

sss

sss

sss

ss

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Step 6: Use table to convert s-domain to time domain

)sin(4.0)cos(8.0)sin(4.0)cos(2.0)(

)sin(4.01)1(

4.0

)cos(8.01)1()1(8.0

)sin(4.0)1(

4.0

)cos(2.0)1(

2.0

2

2

2

2

tetettty

tebecomess

tebecomess

s

tbecomess

tbecomess

s

tt

t

t

−−

−

−

+++=⇒

++•

+++

•

+•

+•

80

Outline



81

Summary

82

Summary: a quiz

A continuous-time LTI system has h(t) given by

a) Find the value of β such that the system is stableb) With the value a β found in part(a), find the range of

ROC such that the system is causalc) Is it possible to have the system stable and causal?

)(53)(

41)( 3 tuetueth tt −−= −β

83

Summary: quiz solution

A function of the form can be seen as the combination of the two exponentialsSince we know that andTherefore, we know that there will be a pole at s = –β and a pole at s = 3Knowing that the ROC is the intersection of the individual ROC’s

)(53)(

41)( 3 tuetueth tt −−= −β

ββ

β >+

=⇔= − Re,)()()( ss

AsHtuAeth t

3Re,3

)()( 3 <−

⇔−−= ss

BtuBeth t

84

Summary: quiz solution

a) β > 0 (not including 0), such that –β is in the left half of the s-plane. This way, the ROC (common intersection) will include the jω-axis, thus implying that the Fourier transform exists further implying that the system is stable

b) For the system to be causal, the ROC must extend outward to positive infinity. Furthermore, since an ROC cannot contain a pole, Res > 3, for the system to be causal

c) This system can only be causal or stable but not both. This is because, in order to be causal and stable, all poles must lie inthe left half of the s-plane such that the ROC can possibly extend from the rightmost pole to + infinity and include the jω-axis. However, due to the pole at s = 3. The ROC cannot extend toward infinity and include the jω-axis. Therefore, the system cannot be both stable and causal

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