[PPT] - Topic 9: The Laplace Transform o Introduction o Laplace Transform PowerPoint Presentation

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ELEC361: Signals And Systems

Topic 9: The Laplace Transform

Introduction
Laplace Transform & Examples
Region of Convergence of the Laplace Transform
Review: Partial Fraction Expansion
Inverse Laplace Transform & Examples
Properties of the Laplace Transform & Examples
Analysis and Characterization of LTI Systems Using the

Laplace Transform

LTI Systems Characterized by Linear Constant-Coefficient DE
Summary
Dr. Aishy Amer

Concordia University Electrical 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/

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Introduction

Transforms: Mathematical conversion from

ne way of thinking to another to make a

problem easier to solve

Reduces complexity of the original problem

Laplace transform solution in s domain inverse Laplace transform solution in time domain problem in time domain

Other transforms
Fourier Transform
z-transform

s = σ+jω

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Introduction

time domain linear differential equation time domain solution Laplace transformed equation Laplace solution Laplace domain or complex frequency domain algebra Laplace transform inverse Laplace transform

x(t) y(t)

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Introduction

CT 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 CTFT Can be applied in contexts where the FT cannot Investigation of stability/instability & causality of systems Laplace transform applies to continuous-time signals

∫ ∫

∞ ∞ − ∞ ∞ − −

= = ω ω π ω

ω ω

d e j X t x dt e t x j X

t j t j

) ( 2 1 ) ( ) ( ) (

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Introduction

Ω

=

j

e z

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 time Sampling period = Ts ω=2πf Ω = ω Ts, scale amplitude by 1/Ts Sample in frequency,

Ω = 2πn/N,

N = Length

f sequence

Continuous Fourier Transform

X(f)

Continuous Fourier Transform

X(f)

∞ ≤ ≤ ∞

∫

∞ ∞ −

f

dt

e x(t)

ft 2 j

π

Discrete Fourier Transform

X(k)

Discrete Fourier Transform

X(k) 1 e [n] x

1 = n N nk 2 j

−

≤ ≤

∑

−

N k

N π

Discrete-Time Fourier Transform

X(Ω)

Discrete-Time Fourier Transform

X(Ω)

π 2 e [n] x

=

n j

≤

Ω ≤

∑

∞ ∞ Ωn

Laplace Transform X(s)

s = σ+jω

Laplace Transform X(s)

s = σ+jω

∞ ≤ ≤ ∞

∫

∞ ∞ − −

s

dt

e x(t)

st

z-Transform X(z) z-Transform X(z) ∞ ≤ ≤ ∞ −

∑

∞ ∞ −

z

= n n

z

[n] x s = jω ω=2πf

C C C C C D D D C

Continuous-variable Discrete-variable

Ω

=

j

e z

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Introduction

Convert time-domain signals into frequency-domain x(t) → X(s) t∈R, s∈C Linear differential equations (LDE) →

algebraic expression in complex plane

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

ω σ j s dt e t x s X t x

st

+ = = =

∫

∞ ∞ − −

) ( ) ( )] ( [ L

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Introduction:

Complex Exponential e-st

part sinusoidal the

argument t

the

f

part is that the noticing in apparent is This s

sillation

the

f

rate the determines part the th, decay/grow

f

rate the determines the While ) sin( ) cos( ) (

) (

ω ω σ σ ω σ ω

ω σ ω σ

⇒

=

+ + + = = =

+

t j t t j st

e Ae t t Aj t t A Ae Ae t x

σt is the phase is the frequency

ω

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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

r

Real axis

ω σ j s + =

σ ω

φ

φ −

ω σ j s − =

∗

(complex) conjugate

ω −

2 2 | | | | 1 tan ω σ σ ω φ + = ∗ ≡ ≡ − = ≡ ∠ s r s s

r

Right Half Plane Left Half Plane Axis tells how fast est grows or decays Axis tells how fast est oscillates (higher frequency)

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Outline

Introduction
Laplace Transform & Examples
Region of Convergence of the Laplace Transform
Review: Partial Fraction Expansion
Inverse Laplace Transform & Examples
Properties of the Laplace Transform & Examples
Analysis and Characterization of LTI Systems Using the Laplace

Transform

LTI Systems Characterized by Linear Constant-Coefficient
Summary

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Laplace Transform

As mentioned earlier, the response of an LTI system with impulse

response 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

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Laplace Transform

When s = jω, we get the Fourier transform of x(t) Therefore, the Fourier transform is a special case

f 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

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Laplace Transform

The Laplace transform X(s) for positive t≥0

typically exists for all complex numbers such that Re{s} > 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 < Re{s} < b

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Laplace Transform

X(s)

f

expression algebraic the ze characteri Completely : X(s)

f

Zeros & Poles poles

f

number the is

f

Order complex are zeroes and Poles ) ) ( (So, ) ( : Zeroes ) ) ( (So, ) ( : ties) (signulari Poles ... ) ( ... ) ( ) 2 )( 10 5 ( 2 5 2 X(s) ; expression algebraic ) ( ) ( ) ( s polynomial

f

rational a

ften

is transform Laplace

1 1 2 2

=
=

= ∋

∞

= = ∋

+

+ + =

+

+ + =

+

+ + + + = = = m X(s) s X s N s s X s D s b s b s b s D a s a s a s N s s s s s s D s N s X

m m n n

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

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Laplace Transform: Order of X(s)

K Bs Js K s D s N sT K s D s N + + =

+

=

2

) ( ) ( Order Second 1 ) ( ) ( Order First

Impulse response Exponential Step response Step, exponential Ramp response Ramp, step, exponential

1 sT K + / 1

2

T s KT

s

KT

s

K + / 1 T s K

s

K +

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Laplace Transform: Poles

The poles of a Laplace function are the values of s that

make the Laplace function evaluate to infinity

The 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 pairs

The response of a system is determined by the location

f poles on the complex plane

) 3 )( 1 ( ) 2 ( 10 + + + s s s

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Laplace Transform: Zeros

The zeros of a Laplace function are the values of s that make

the Laplace function evaluate to zero

The 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

f a physical system where minor system (e.g., temperature)

change could cause the pole or the zero to move

) 3 )( 1 ( ) 2 ( 10 + + + s s s

) 1 ( ) 1 )( 3 ( ) ( − − + = s s s s H

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Laplace Transform: Visualization

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

variable 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 analyze
Solution: Poles (x) and Zeros (o) Plot
Example:

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

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Laplace Transform: Example

x(t)

f

nature

n the

depends X(s)

f

e Convergenc

a

Re(s)

r

a) Re(s if

nly

lim converge not do power negative be NOT will zero,

r

negative is ) ( if converge power negative be be will positive, is ) ( if not

r

converge transform Laplace he whether t determines hat apparent t becomes it bounded), (i.e., sinusoidal is that g Recognizin 1 ) OR is x(t)

f

transform Laplace the hand,

ther

On the | ) ( | 1 ) is 0, a with ), X(j ansform Fourier tr The x(t) sided right ) ( Let

) ( ) ( ) (

⇒ > > + =

⇒

+ ⇒ +

+

= = + = =

∞

< + = = = >

=
+

− + − + − + − − + − ∞ − + − ∞ + − ∞ ∞ − − − ∞ ∞ − ∞ − − ∞ ∞ − − −

∞ →

∫ ∫ ∫ ∫ ∫ ∫

a)t (s t a t a t a jwt a)t (s jwt σ)t (a a)t (s st at jwt at jwt at

at

e e a e a e e e a s dt e e jω X(σ dt e dt u(t)e e X(s) dt t x a jw dt e e dt u(t)e e X(jω t u e x(t)

t

σ σ σ

σ σ ω

Recall:

FT converges if

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Laplace Transform: Example

a s a s t u e a s a s X(s) j s a a jw dt e e jω X(σ

L at jwt σ)t (a

− > + = − > + ⎯→ ←

−

> + = + =

>

+ + + = = +

=

⇒

−

∞ − + −

∫

} Re{ a s 1 X(s) } Re{ a s 1 ) ( : Conclusion } Re{ 1 becomes equation last the , Since ) ( where ) ( 1 ) Re(s) but to j to related not is ROC a

>

Re(s) : e Convergenc for Condition ω σ σ σ σ ω

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Laplace Transform: Example

We conclude from the above example that the Laplace transform

exists for this particular x(t) only if

The 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

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Laplace Transform: Example

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

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Laplace Transform: Example

two-sided h(t)

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Laplace Transform: Example

right-sided

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Laplace Transform: Example

1

1

when converges ) 3 ( ) 1 ( 1 3 1 1 > → > + ⇒ − + + = − + + σ σ ω σ ω σ j j j a sum of real & complex exp.

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Outline

Introduction
Laplace Transform & Examples
Region of Convergence of the Laplace Transform
Review: Partial Fraction Expansion
Inverse Laplace Transform & Examples
Properties of the Laplace Transform & Examples
Analysis and Characterization of LTI Systems Using the Laplace

Transform

LTI Systems Characterized by Linear Constant-Coefficient DE
Summary

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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-plane

2. 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

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Region of Convergence for Laplace Transform

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} = σ0 is 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

f a strip in the s-plane that includes the

line Re {s} = σ0

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Region of Convergence for Laplace Transform

7.

If the Laplace transform X(s) of x(t) is rational, then the ROC is bounded by poles

r 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

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Region of Convergence for Laplace Transform: Example

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Common Laplace Transforms Pairs

number interger positive : numbers real are , , n t a j s C s

∞

< < ∞ −

+

= ∈

τ

ω σ

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 Exponential Decay R(s)>-a Exponential Approach R(s)>0

⎩ ⎨ ⎧ > = = = 1 ) ( ) ( t t t t x δ

) ( ) ( t u t x =

s 1 ) ( ) ( τ δ − = t t x

s

e τ

−

) ( ) ( τ − = t u t x

s e

s τ − 2

1 s

) ( ) ( t tu t x =

1

!

+ n

s n

) ( ) ( n u t t x

n

=

a s+ 1

) ( ) ( t u e t x

at −

=

) ( a s s a +

) ( ) 1 ( ) ( t u e t x

at −

− =

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Common Laplace Transforms Pairs

number interger positive : numbers real are , , n t a j s C s

∞

< < ∞ −

+

= ∈

τ

ω σ

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|

)) ( ( ) ( t u t x − − =

s 1 a s+ 1

) ( ) ( t u e t x

at −

− =

−

2 2 θ

θ + s

) ( ) sin( ) ( t u t t x θ =

2 2 θ

+ s s

) ( ) cos( ) ( t u t t x θ =

2 2

) ( θ θ + +a s

) ( ) sin( ) ( t u t e t x

at

θ

−

=

2 2

) ( θ + + + a s a s

) ( ) cos( ) ( t u t e t x

at

θ

−

=

2 2 θ

θ − s

) ( ) sinh( ) ( t u t t x θ =

2 2 θ

− s s

) ( ) cosh( ) ( t u t t x θ =

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Outline

Introduction
Laplace Transform & Examples
Region of Convergence of the Laplace Transform
Review: Partial Fraction Expansion
Inverse Laplace Transform & Examples
Properties of the Laplace Transform & Examples
Analysis and Characterization of LTI Systems Using the Laplace

Transform

LTI Systems Characterized by Linear Constant-Coefficient DE
Summary

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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

1 1 11 ) 1 )( 1 ( ) 1 ( 5 ) 1 ( 6 1 5 1 6 1 5 1 6 1 1 11

2 2

− − = − + + + − = − + +

−

+ + = − −

s

s s s s s s s s s s s

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Review: Partial Fraction Expansion

3 2 ) 3 ( ) 2 ( 1 + + + = + + + s B s A s s s

( )

) 3 ( ) 2 ( 2 ) 3 ( ) 3 ( ) 2 ( 1 + + + + + = + + + s s s B s A s s s

3 2 2 1 ) 3 ( ) 2 ( 1 + + + − = + + + s s s s s

1 = + B A

1 2 3 = + B A

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.

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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

5 6 1 11 ) 1 ( ) 1 11 ( ) 1 )( 1 ( ) ( ) ( ) 1 )( 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 11 (

2 2

= = ⇒ ⎭ ⎬ ⎫ − = − = + − − = − + − + + = − + + + − = − + + = − − B A A B B A s s s s A B s B A s s s B s A s B s A s s

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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 power
If a term is present three times, make the fractions the term and its second and third

powers

2 1 1 4 3 2 1 ) ( ) 2 ( ) 1 ( ) 1 ( 4 3 ) ( ) ( ) ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( 4 3

2 2 2 3 3 2 3 2 3 2

= = = ⇒ ⎪ ⎭ ⎪ ⎬ ⎫ = + + = + = + + + + + = + + + + = + + = = + + + + + + = + + + + + = + + + C B A C B A B A A C B A s B A As C s B s A s s s N s D s N s C s s B s A s C s B s A s s s

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Review: Partial Fraction Expansion: Different quadratic terms

When there is a quadratic term, assume a numerator of

the form Bs + C

5 . 5 . 1 2 ) 2 ( ) ( ) ( 1 ) 1 ( ) 1 ( ) 2 ( 1 ) 2 ( ) 1 ( ) 2 )( 1 ( 1

2 2 2 2

= = = ⇒ ⎪ ⎭ ⎪ ⎬ ⎫ = + = + + = + + + + + + + = + + + + + + = + + + + + = + + + C B A C A C B A B A C A s C B A s B A s C s Bs s s A s s C Bs s A s s s

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Review: Partial Fraction Expansion: Repeated quadratic terms

When there is repeated quadratic term, assume two numerator of the form

Bs + C and Ds+E

5 . 25 . 25 . 1 2 4 3 2 4 2 3 5 2 2 ) 1 ( ) 1 ( ) 2 )( 1 ( ) 2 )( 1 ( ) 2 ( 1 ) 2 ( ) 2 ( ) 1 ( ) 2 )( 1 ( 1

2 2 2 2 2 2 2 2 2

= − = = − = = ⇒ ⎪ ⎪ ⎪ ⎭ ⎪ ⎪ ⎪ ⎬ ⎫ = + + = + + + + = + + + = + + = + + + + + + + + + + + + + + + = + + + + + + + + + = + + + E D C B A E C A E D C B A D C B A C B A B A s E s Ds s s s C s s s Bs s s A s s E Ds s s C Bs s A s s s

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Outline

Introduction
Laplace Transform & Examples
Region of Convergence of the Laplace Transform
Review: Partial Fraction Expansion
Inverse Laplace Transform & Examples
Properties of the Laplace Transform & Examples
Analysis and Characterization of LTI Systems Using the Laplace

Transform

LTI Systems Characterized by Linear Constant-Coefficient DE
Summary

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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 ROC

A very useful technique in finding the inverse Laplace transform is to expand

X(s) in the form

From the ROC of X(s), one can find the ROC for each individual term in the

above expression

The inverse Laplace transform can then be obtained for every term separately

very easily

∫

∞ + ∞ −

=

j j stds

e s X j t x

σ σ

π ) ( 2 1 ) (

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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 table

2.

Invert each Xi(s) to get xi(t)

3.

Sum up all xi(t) to get x(t)

∑

=

m i i s

X s X

1

) ( ) (

∑

=

m i i t

x t x

1

) ( ) (

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The Inverse Laplace Transform: Example

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The Inverse Laplace Transform: Example

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Outline

Introduction
Laplace Transform & Examples
Region of Convergence of the Laplace Transform
Review: Partial Fraction Expansion
Inverse Laplace Transform & Examples
Properties of the Laplace Transform & Examples
Analysis and Characterization of LTI Systems Using the Laplace

Transform

LTI Systems Characterized by Linear Constant-Coefficient DE
Summary

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Properties of the Laplace Transform

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Properties of the Laplace Transform

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Properties of the Laplace Transform

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Properties of the Laplace Transform

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Properties of the Laplace Transform: Examples

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Properties of the Laplace Transform

t

f

values positive from approaches t as t

f

values negative from approaches t as

+ −

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Properties of the Laplace Transform

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Properties of the Laplace Transform: Examples

) 4 ( ) 2 ( 2 ) ( + + = s s s s Y

Laplace

transform of the function

Apply Final-

Value theorem

Apply Initial-

Value theorem

[ ]

4 1 ) 4 ( ) 2 ( ) ( ) ( 2 ) ( lim = + + =

∞ →

t x

t

[ ]

) 4 ( ) 2 ( ) ( ) ( 2 ) ( lim = + ∞ + ∞ ∞ ∞ =

→

t x

t

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Properties of the Laplace Transform: Examples

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

1 1 lim ) ( 1 1 ) ( ) ( ) ( lim ) ( = + = + = = =

∞ → − ∞ → +

s s x s s X e t x Example s sX x

s t s

) (

+

x

1 lim ) ( lim 1 1 ) ( ) ( ) ( lim ) ( lim = + = + = = =

→ ∞ → − → ∞ →

s s t x s s X e t x Example s sX t x

s t t s t

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Properties of the Laplace Transform: Examples

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Properties of the Laplace Transform: summary

x(t) X(s) ( ) { }

) (t x L s X = ( ) { }

s X L t x

1

) (

−

=

) ( a s X −

) (t x eat

) ( ) 1 ( s X ds d

n n n

−

) ( ) ( x s sX

) (t x t n

) ( ' t x −

) ( ' ' t x

) ( ' ) ( ) (

2

x sx s X s − −

) ( ' ' ' t x

) ( ' ' ) ( ' ) ( ) (

2 3

x sx x s s X s − − −

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Properties of the Laplace Transform: Summary

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Outline

Introduction
Laplace Transform & Examples
Region of Convergence of the Laplace Transform
Review: Partial Fraction Expansion
Inverse Laplace Transform & Examples
Properties of the Laplace Transform & Examples
Analysis and Characterization of LTI Systems Using the

Laplace Transform

LTI Systems Characterized by Linear Constant-Coefficient DE
Summary

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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

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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 input H(s) describes how a linear system responds to an impulse

H(s) represents a normalized model of a process, i.e., can be used

with any input

Differential Equation x(t) y(t) H(s): Transfer Function X(s) Y(s)

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Transfer Function of an LTI system

The 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)

ct bt at

e C t e B e A t h c s C b s B a s A L t h c s C b bs s B a s A s H + + = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − + + + + + = − + + + + + + =

− − −

) sin( ) ( : table the Using ) ( ) ( ) ( ) ( ) ( ) 2 ( ) ( ) (

2 2 1 2 2 2

ω ω ω ω ω

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Analysis and Characterization of LTI Systems Using the Laplace Transform

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 plane

A 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

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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

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

1.

Re{s} < −1 : Then the solution is left-sided, and is given by Clearly this system is non causal

2.

Re{s} > 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 < Re{s} < 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

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64

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

SLIDE 65

65

Analysis and Characterization of LTI Systems Using the Laplace Transform

Stability A system is stable if bounded inputs produce bounded outputs An LTI system is BIBO stable iif

An LTI system is stable iff the ROC of its transfer function includes the jω axis

Note: A system maybe stable with a non-rational H(s) Causality & Stability Clearly, a causal system with rational transfer function is stable

iff all of the poles of the transfer function lie in the left-half plane

1 } Re{ , 1 ) ( ., . − > + = s s e s H g e

s

∞ <

∫

∞ ∞ −

dt t h | ) ( |

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66

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

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67

Unstable Behavior

If the output of a system grows without bound for a bounded input, the system

is referred to a unstable

The complex s-plane is divided into two regions depending on poles locations
1. the stable region, which is the left half of the s-plane
2. 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 is unstable

If any pole is located in the right half plane and the ROC lies in the right-half

plane, the system is unstable

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Outline

Introduction
Laplace Transform & Examples
Region of Convergence of the Laplace Transform
Review: Partial Fraction Expansion
Inverse Laplace Transform & Examples
Properties of the Laplace Transform & Examples
Analysis and Characterization of LTI Systems Using the Laplace

Transform

LTI Systems Characterized by Linear Constant-Coefficient DE
Summary

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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 easily

Consider for example the following differential equation: Applying Laplace transform to both sides yields

and the transfer function can be obtained as

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

solutions:

1. Re{s} > −3 : The system in this case is causal and is given by
2. Re{s} < −3 : This results in the non-causal solution, namely,

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LTI Systems Characterized by Linear Constant-Coefficient DE

The same procedure can be used to obtain H(s) from

any differential equations with constant coefficients

A 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 of The poles are the solutions of

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LTI Systems Characterized by Linear Constant-Coefficient DE

We remark here that the transfer function does

not tell additional information about the ROC of the system

The ROC is normally specified by additional

information such as knowledge about the stability and causality of the system

SLIDE 72

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LTI Systems Characterized by Linear Constant-Coefficient DE: Example

SLIDE 73

73

LTI Systems Characterized by Linear Constant-Coefficient DE: Example

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 ROCs

f X(s) and H(s)

Since H(s) has two poles, then there are three choices

for the ROC

Since we have some knowledge about the ROC of

Y(s), this limits our choices for the ROC of H(s) to one

Re{s} > −1 which means that H(s) is stable and causal From H(s), one can obtain the differential equation

that relates X(s) and Y(s)

SLIDE 74

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LTI Systems Characterized by Linear Constant-Coefficient DE: Example

textbook) the

f

9.26 example (see ) 4 ( ) 2 ( 4 ) ( − + = ⇒ s s s s H

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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 algebra

Step 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 sides L{y”} + L{2y’} + L{2y} = L{cos(t)}

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

Step 3: Use properties of transforms to

express equation in s-domain

L{y”} + L{2y’} + L{2y} = L{cos(ω t)}

L{y”} = s2 Y(s) - sy(0) – y’(0)
L{2y’} = 2[ s Y(s) - y(0)]
L{2y} = 2 Y(s)
L{cos(t)} =

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

) 1 (

2 +

s s

) 1 (

2 +

s s

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

Step 4: Solve for Y(s)

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

2 2 2 3 2 2 2 2 2 2

+ + + + + + = + + + + + =

+

+ + = + + + = + − + −

s

s s s s s s s s s s s Y s s s s Y s s s s s Y s sY s s Y s

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78

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

Step 5: Expand equation into format covered by table

( )( ) ( ) ( )

2 . 1 , 8 . 4 . , 2 . 2 2 2 2 2 2 2 1 rms similar te Equate ) 2 ( ) 2 2 ( ) 2 ( ) ( 2 2 1 2 2 1 2 2 2 ) (

2 3 2 2 2 2 2 3

= = = = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = + = + + = + + = +

+

+ + + + + + + +

+

+ + + + + = + + + + + + =

E

C B A E B C B A E B A C A E B s C B A s E B A s C A s s E Cs s B As s s s s s s s Y

1 ) 1 ( 4 . 1 ) 1 ( ) 1 ( 8 . 2 2 2 . 1 8 . ) 1 ( 4 . ) 1 ( 2 . ) 1 ( 4 . 2 .

2 2 2 2 2 2

+ + + + + + = + + +

+

+ + = + +

s

s s s s s s s s s s

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

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

) sin( 4 . ) cos( 8 . ) sin( 4 . ) cos( 2 . ) ( ) sin( 4 . 1 ) 1 ( 4 . ) cos( 8 . 1 ) 1 ( ) 1 ( 8 . ) sin( 4 . ) 1 ( 4 . ) cos( 2 . ) 1 ( 2 .

2 2 2 2

t e t e t t t y t e becomes s t e becomes s s t becomes s t becomes s s

t t t t − − − −

+ + + = ⇒ + +

+

+ +

+
+

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Outline

Introduction
Laplace Transform & Examples
Region of Convergence of the Laplace Transform
Review: Partial Fraction Expansion
Inverse Laplace Transform & Examples
Properties of the Laplace Transform & Examples
Analysis and Characterization of LTI Systems Using the Laplace

Transform

LTI Systems Characterized by Linear Constant-Coefficient DE
Summary

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Summary

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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 stable

b)

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

c)

Is it possible to have the system stable and causal?

) ( 5 3 ) ( 4 1 ) (

3

t u e t u e t h

t t

− − =

−β

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83

Summary: quiz solution

A function of the form

can be seen as the combination of the two exponentials

Since we know that

and

Therefore, we know that there will be a pole at

s = –β and a pole at s = 3

Knowing that the ROC is the intersection of the

individual ROC’s

) ( 5 3 ) ( 4 1 ) (

3

t u e t u e t h

t t

− − =

−β

β β

β

> + = ⇔ =

−

} Re{ , ) ( ) ( ) ( s s A s H t u Ae t h

t

3 } Re{ , 3 ) ( ) (

3

< − ⇔ − − = s s B t u Be t h

t

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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, Re{s} > 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 in the 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