Signal and Systems Prof. H. Sameti Chapter 9: Laplace Transform Motivatio n and Definition of the...

Signal and SystemsProf. H. Sameti

Chapter 9: Laplace Transform

Motivation and Definition of the (Bilateral) Laplace Transform Examples of Laplace Transforms and Their Regions of Convergence

(ROCs) Properties of ROCs Inverse Laplace Transforms Laplace Transform Properties The System Function of an LTI System Geometric Evaluation of Laplace Transforms and Frequency Responses

Book Chapter#: Section#

2

Motivation for the Laplace Transform

CT Fourier transform enables us to do a lot of things, e.g.• Analyze frequency response of LTI systems • Sampling• Modulation

Why do we need yet another transform? One view of Laplace Transform is as an extension of the

Fourier transform to allow analysis of broader class of signals and systems

In particular, Fourier transform cannot handle large (and important) classes of signals and unstable systems, i.e. when

Computer Engineering Department, Signal and Systems


3

Motivation for the Laplace Transform (continued)

In many applications, we do need to deal with unstable systems, e.g.• Stabilizing an inverted pendulum• Stabilizing an airplane or space shuttle• Instability is desired in some applications, e.g. oscillators and lasers

How do we analyze such signals/systems? Recall from Lecture #5, eigenfunction property of LTI systems:

is an eigenfunction of any LTI system can be complex in general



4

The (Bilateral) Laplace Transform

s = + σ j is a complex variable – Now we explore the full ωrange of

Basic ideas:

1. A critical issue in dealing with Laplace transform is convergence:—X(s) generally exists only for some values of s, located in what is called the region of convergence(ROC): so that

2. If = is in the ROC (i.e. = 0), then σ


absolute integrability needed

absolute integrability condition


5

Example #1: (a – an arbitrary real or complex number)

This converges only if Re(s+a) > 0, i.e. Re(s) > -Re(a)



6

Example #2:

This converges only if Re(s+a) < 0, i.e. Re(s) < -Re(a) Same as (s), but different ROC Key Point (and key difference from FT): Need both X(s)

and ROC to uniquely determine x(t). No such an issue for FT.



7

Graphical Visualization of the ROC Example1: Example2:



8

Rational Transforms Many (but by no means all) Laplace transforms of interest to us are

rational functions of s (e.g., Examples #1 and #2; in general, impulse responses of LTI systems described by LCCDEs), where

X(s) = N(s)/D(s), N(s),D(s) – polynomials in s

Roots of N(s)= zeros of X(s)

Roots of D(s)= poles of X(s)

Any x(t) consisting of a linear combination of complex exponentials for t > 0 and for t < 0 (e.g., as in Example #1 and #2) has a rational Laplace transform.



9

Example #3



10

Laplace Transforms and ROCs Some signals do not have Laplace Transforms (have no ROC) for all t since for all

for all t for all X(s) is defined only in ROC; we don’t allow impulses in LTs



11

Properties of the ROCThe ROC can take on only a small number of

different forms1. 1) The ROC consists of a collection of lines

parallel to the j -axis in the ω s-plane (i.e. the ROC only depends on ).Why? σdepends only on

2. If X(s) is rational, then the ROC does not contain any poles. Why?

Poles are places where D(s) = 0 ⇒ X(s) = N(s)/D(s) = ∞ Not convergent.



12

More Properties If x(t) is of finite duration and is absolutely integrable, then the ROC is

the entire s-plane.



13

ROC Properties that Depend on Which Side You Are On - I

If x(t) is right-sided (i.e. if it is zero before some time), and if Re(s) = is in the ROC, then all values of s for which Re(s) > are also in the ROC.


ROC is a right half plane (RHP)


14

ROC Properties that Depend on Which Side You Are On -II

If x(t) is left-sided (i.e. if it is zero after some time), and if Re(s) = is in the ROC, then all values of s for which Re(s) < are also in the ROC.


ROC is a left half plane (LHP)


15

Still More ROC Properties If x(t) is two-sided and if the line Re(s) = is in the ROC,

then the ROC consists of a strip in the s-plane



16

Example:


Intuition?

Okay: multiply by constant () and will be integrable

Looks bad: no will dampen both sides


17

Example (continued):

Overlap if , with ROC:

What if b < 0? No overlap No Laplace Transform⇒ ⇒



18

Properties, Properties If X(s) is rational, then its ROC is bounded by poles or extends to

infinity. In addition, no poles of X(s) are contained in the ROC. Suppose X(s) is rational, then

a) If x(t) is right-sided, the ROC is to the right of the rightmost pole.b) If x(t) is left-sided, the ROC is to the left of the leftmost pole.

If ROC of X(s) includes the j -axis, then FT of x(t) exists.ω



19

Example: Three possible ROCs

x(t) is right-sided ROC: III No x(t) is left-sided ROC: I Nox(t) extends for all time ROC: II Yes


Fourier Transform exists?

20

The z-Transform

• Last time:

• Unit circle (r = 1) in the ROC ⇒DTFT exists

• Rational transforms correspond to signals that are linear combinations of DT exponentials


-depends only on r = |z|, just like the ROC in s-plane only depends on Re(s)

nxZznxzXnxn

n

)(

n

nj rnxrez ||at which ROC

( )jX e

Book Chapter10: Section 1

21

Some Intuition on the Relation between ZT and LT


The (Bilateral) z-Transform

Can think of z-transform as DTversion of Laplace transform with

( ) ( ) ( ) { ( )}stx t X s x t e dt L x t

TenTx nsT

n nxT

)()(lim

0

n

nsT

TenxT )(lim

0

}{)( nxzznxzXnxn

n

sTez

Let t=nT


22

More intuition on ZT-LT, s-plane - z-plane relationship

LHP in s-plane, Re(s) < 0 |⇒ z| = | esT| < 1, inside the |z| = 1 circle.

Special case, Re(s) = -∞ ⇔|z| = 0. RHP in s-plane, Re(s) > 0 |⇒ z| = | esT| > 1, outside the |z| = 1 circle.

Special case, Re(s) = +∞ ⇔|z| = ∞. A vertical line in s-plane, Re(s) = constant⇔| esT| = constant, a circle in z-plane.


zesT )( plane-sin axis jsj plan-zin circleunit a1 Tjez


23Computer Engineering Department, Signal and SystemsComputer Engineering Department, Signal and Systems 23

Inverse Laplace Transform

})({

,)()(

t

st

etxF

ROCjsdtetxsX

Fix σ ROC and apply the inverse Fourier ∈

dejXetx tjt )(2

1)(

dejXtx tj )()(2

1)(

But s = σ + jω (σ fixed) ds =⇒ jdω

j

j

stdsesXj

tx )(2

1)(

Book Chapter 9 : Section 2

24Computer Engineering Department, Signal and Systems

Inverse Laplace Transforms Via Partial FractionExpansion and Properties

Example:

3

5 ,

3

2

21)2)(1(

3)(

BA

s

B

s

A

ss

ssx

Three possible ROC’s — corresponding to three different signals

Recall sided-left )(}{ ,1

tueaseas

at

1, { } ( ) right-sidedate s a e u t

s a



ROC I: — Left-sided signal.

ttuee

tuBetuAetx

tt

tt

as es Diverg)(3

5

3

2

)()()(

2

2

ROC II: — Two-sided signal, has Fourier Transform.

ttuetue

tuBetuAetx

tt

tt

as erges Div)(3

5)(

3

2

)()()(

2

2

ROC III: — Right-sided signal.

tu(t)ee

tuBetuAetx

tt

tt

as erges Div3

5

3

2

)()()(

2

2



26

Many parallel properties of the CTFT, but for Laplace transforms we need to determine implications for the ROC

For example:


Properties of Laplace Transforms

Linearity

)()()()( 2121 sbXsaXtbxtax

ROC at least the intersection of ROCs of X1(s) and X2(s)

ROC can be bigger (due to pole-zero cancellation)

0)(0)()( Then

and )()( E.g.

21

21

sXtbxtax

batxtx

⇒ ROC entire s-


Time Shift

)( as ROCsame ),()( sXsXeTtx sT

? 2}{ ,2

:Example3

ses

e s

Ttttueses

e tsT

|)( 2}{ ,2

2

3 T

)3( 2}{ ,2

)3(23

tueses

e ts



Time-Domain Differentiation

j

j

j

j

stst dsessXjdt

tdxdsesX

jtx )(

2

1)( ,)(

2

1)(

ROC could be bigger than the ROC of X(s), if there is pole-zero cancellation. E.g.,

)( of ROC thecontaining ROC with),()(

sXssXdt

tdx

plane-s entire ROC1

1)( )(

0}{ ,1

)( )(

sst

dt

tdx

ses

tutx

s-Domain Differentiation

X(s)ds

sdXttx as ROCsame with,

)()(

(Derivation is

similar to )d

sdt

aseasasds

dtute at

}{ ,)(

11)( E.g.

2



Convolution Property

x(t) y(t)=h(t)＊ x(t)h(t)

For

Then )()()(

)()(),()(),()(

sXsHsY

sHthsYtysXtx

• ROC of Y(s) = H(s)X(s): at least the overlap of the ROCs of H(s) & X(s)

• ROC could be empty if there is no overlap between the two ROCs

E.g.

x(t)=etu(t),and h(t)=-e-tu(-t)

• ROC could be larger than the overlap of the two.

)()(*)( tthtx



The System Function of an LTI System

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

function system the)()( sHth

The system function characterizes the system⇓

System properties correspond to properties of H(s) and its ROC

A first example:

dtth )( stable is System

axis theincludes

of ROC

jω

H(s)



Geometric Evaluation of Rational Laplace Transforms

Example #1: A first-order zeroassX )(1



Example #2: A first-order pole

)(

11)(

12 sXas

sX

)( )(

)|)(|log log(or )(

1 )(

12

121

2

sXsX

sX(s)||XsX

sX

Still reason with vector, but remember to "invert" for poles

Example #3: A higher-order rational Laplace transform

)(

)( )(

1

1

jPj

iRi

s

sMsX

jPj

iRi

s

sMsX

1

1 )(

R

i

p

jji ssMsX

1 1

)()( )(



First-Order System

Graphical evaluation of H(jω)

1

}{,/1

/1

1

1)(

se

sssH

)(1

)( / tueth t

)(1)( / tuets t

/1

11

/1

/1 )(

jjjH



Bode Plot of the First-Order System

1/ 2/

1/ 4/

0 0

)(tan )(

1/ /1

1/ 2/1

0 1

)/1(

/1 )(

1/j

1/ )(

1

22

jH

jH

jH



Second-Order System

e(pole)e{s} ROC2

)(22

2

nn

n

sssH

10 complex poles— Underdamped

1

1

double pole at s = −ωn

— Critically damped

2 poles on negative real axis— Overdamped



Demo Pole-zero diagrams, frequency response, and step response of first-order and second-order CT causal systems



Bode Plot of a Second-Order System

Top is flat whenζ= 1/√2 = 0.707⇒a LPF for

ω < ωn



Unit-Impulse and Unit-Step Response of a Second- Order System

No oscillations when ≥ 1ζ

⇒ Critically (=) and over (>) damped.



First-Order All-Pass System

1. Two vectors have the same lengths

2.

)0( }{ ,)(

aaseas

assH

a

a

jH

0~

2/

0

2

)(

)(

2

22

21


CT System Function Properties

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

)()()( sXsHsY H(s) = “system function”

1) System is stable ROC of H(s) includes jω axis

2) Causality => h(t) right-sided signal => ROC of H(s) is a right-half plane

dtth )(

Question:If the ROC of H(s) is a right-half plane, is the system causal?

E.x. sided-right )(1}{ ,1

)( thees

esH

eT

Tttt

Ttt

eT

tues

Ls

eLth

|)(

1

1

1 )( 11

0at 0)( )( tTtue Tt Non-causal40

Properties of CT Rational System Functions

a) However, if H(s) is rational, then

The system is causal ⇔ The ROC of H(s) is to the right of the rightmost pole

b) If H(s) is rational and is the system function of a causal system, then

The system is stable ⇔ jω-axis is in ROC ⇔ all poles are in

41

Checking if All Poles Are In the Left-Half Plane

)(

)()(

sD

sNsH

Poles are the root of D(s)=sn+an-1sn-1+…+a1s+a0

Method #1: Calculate all the roots and see!Method #2: Routh-Hurwitz – Without having to solve for roots.

210

01201

22

3

01012

00

and

0,0,0order-Third

0,0order-Second

0order-First LHP thein are roots

all that so ConditionPolynomial

aaa

aaaasasas

aaasas

aas

42

Initial- and Final-Value Theorems

If x(t) = 0 for t < 0 and there are no impulses or higher order discontinuities at the origin, then

)(lim )0( ssXxs

)(lim )(0

ssXxs

Initial value

Final value

If x(t) = 0 for t < 0 and x(t) has a finite limit as t → ∞, then

43

Applications of the Initial- and Final-Value Theorem

For

n-order of polynomial N(s), d – order of polynomial D(s)

• Initial value：

• Final value：

1

1 0 finite

1 0

)(lim )0(

nd

nd

nd

ssXxs

?)0( 1

1)( E.g.

x

ssX

)(

)()(

sD

sNsX

0at poles No

)(lim0)(lim)( If00

s

sXssXxss

44

LTI Systems Described by LCCDEs

N

k

M

kk

k

kk

k

k dt

txdb

dt

tyda

0 0

)()(

N

k

M

k

kk

kk

kk

k

sXsbsYsa

sdt

ds

dt

d

0 0

)()(

,property ationdifferenti of use Repeated ：

)( where

)()()(

Rational

0

0

N

k

kk

M

k

kk

sa

sbsH

sXsHsY

roots of numerator ⇒ zeros

roots of denominator ⇒ poles

ROC =? Depends on: 1) Locations of all poles. 2) Boundary conditions, i.e. right-, left-, two-sided signals.

45

System Function Algebra

Example: A basic feedback system consisting of causal blocks

)()()()()()( 2 sYsHsXsZsXsE

)()()()()()()( 211 sYsHsXsHsEsHsY

)()(1

)(

)(

)()(

21

1

sHsH

sH

sX

sYsH

More on this later in feedback

ROC: Determined by the roots of 1+H1(s)H2(s), instead of H1(s)

46

Block Diagram for Causal LTI Systems with Rational System Functions

Example： )()()( sXsHsY

23

642)(

2

2

ss

sssH )642(

23

1 22

ss

ss

— Can be viewed as cascade of two systems.

)(23

1)( Define

2sX

sssW

：

)(2)(

3)()(

or

restat initially ),()(2)(

3)(

2

2

2

2

twdt

tdwtx

dt

twd

txtwdt

tdw

dt

twd

)(6)(

4)(

2)(

)()642()(

Similarly

2

2

2

twdt

tdw

dt

twdty

sWsssY

47

Example (continued)

Instead of

)(64223

1)(

)(

22

tyssss

tx

sH

We can construct H(s) using:

)(6)(

4)(

2 )(

)(2)(

3)( )(

2

2

2

2

twdt

tdw

dt

twdty

twdt

tdwtx

dt

twd

Notation: 1/s—an integrator48

Note also that

Cascade 1

)1(2

2

3

1

3

2

12 )(

s

s

s

s

s

s

s

)(ssH

connection parallel 1

8

2

62

ss

PFE

Lesson to be learned： There are many different ways to construct a system that performs a certain function.

49

The Unilateral Laplace Transform(The preferred tool to analyze causal CT systems described by LCCDEs with initial conditions)

Note:

1) If x(t) = 0 for t < 0,

2) Unilateral LT of x(t) = Bilateral LT of x(t)u(t-)

3) For example, if h(t) is the impulse response of a causal LTI

system then,

4) Convolution property: If x1(t) = x2(t) = 0 for t < 0,

Same as Bilateral Laplace transform

0

)()()( txdtetxs st ULX

)()( ssX X

)()( ssH H

)()()()( 2121 sxtxtx XXUL

50

Differentiation Property for Unilateral Laplace Transform

)0()()(

)()(

xssdt

tdx

stx

X

X

Initial condition!

Derivation：

Note：

0

)()(dte

dt

tdx

dt

tdx stUL

0

)(

0|)()( st

s

st etxdtetxs

X

)0()( xssX

integration by parts∫f． dg=fg－∫ g． df

)0(')0()(

)0('

)(

))0()(()()(

2

2

2

xsxss

x

dt

tdx

xsssdt

tdx

dt

d

dt

txd

X

UL

X

51

Use of ULTs to Solve Differentiation Equations with Initial Conditions

Example：

Take ULT：

)()(,)0(',)0(

)()(2)(

3)(

2

2

tutxyy

txtydt

tdy

dt

tyd

ssssss

dt

dy

dt

yd

)(2))((3)(

2

2

2 YYY

ULUL

ZSR

sss

ZIR

ssss

ss

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

)3()(

Y

ZIR — Response for zero input x(t)=0

ZSR — Response for zero state,β=γ=0, initially at rest

52

Example (continued)

• Response for LTI system initially at rest ( = = 0 )β γ

• Response to initial conditions alone ( = 0). For αexample:

)()2)(1(

1

)(

)()( sH

sss

ss

Y

H

)0,1( 0)0(' ,1)0( ),input no(0)( yytx

0 ,2)(

2

1

1

2

)2)(1(

3)(

2

teety

ssss

ss

tt

Y

53

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