Signal & Systems - KOCWelearning.kocw.net/KOCW/document/2015/chungnam/kimjin... · 2016. 9. 9. ·...

Kim, J. Y.

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Signal & Systems

Chonnam National University

Dept. of Electronics Engineering

Kim, Jinyoung

http://www.chonnam.ac.kr/

Kim, J. Y.

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6. Representation of Signals Using Continuous-Time Complex Exponentials :

Laplace Transform


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Laplace, Pierre Simon

1749~1827 프랑스의 수학자, 물리학자, 천문학자, 노르망디의 빈농출신. 어릴 때부터 비상한 재능이 있었고, 1784년에는 에콜 노르말의 교수가 되었다. 나폴레옹 1세 하에서 내상(1799)과 백작이 되었고, 또 왕정복고 시기에 후작의 지휘를 받았다. (1817). 정치적으로는 입장이 불명확했다. 해석학에 뛰어나, 이것을 천체역학이나 확률론에 응용하여 많은 성과를 얻었다. 명저[천체역학](Mecanique celeste, 5권 1799~1825)는 뉴튼 이래의 천체역학을 집대성하여, 태양의 천체세게 관련된 많은 현상을 해명했다. 특히 그 섭동론은 천왕성 운행의 이론적 계산치의차이를 이용하여 해왕성의 크기와 위치를 예언하고, 그 발견에 기여한 르베리에(Urbain J.J. Le Verrier, 1811~77, 프랑스의 천문학자. 해왕성은 J.Galle에 의해 1846년 9월에 발견됨)등의 사업의 기초를 마련했다. 또 태양계의 기원에 대해 칸트-라플라스설인 성운성을 완성시켰다. 그 외에도, 수학, 물리학 연구에 뛰어난 업적을 남겼다.


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Shortage of Fourier Transform

1) Damped Signal

2) Non-stationary

t j te e

http://www.chonnam.ac.kr/http://kr.youtube.com/watch?v=0dPS-EHl-FE

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6.1 Introduction

Laplace Transform

- A representation in terms of complex

exponential signals

- Generalization of FT


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6.2 The Laplace Transform

LTI system response to complex

exponential function

Definition of Laplace transform

S-plane


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Complex exponential : est

cos( ) sin( )st t te e t je t


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6.2.1 Eigenfunction Property of est

H x(t)=est y(t)

( )

( ) { }

( )* ( )

( ) ( )

( )

( )

st

s t

st s

y t H e

h t x t

h x t d

h e d

e h e d

{ } ( )

, where ( ) ( )

st st

s

H e H s e

H s h e d


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Express in polar form

LTI System Response to Complex Exponential Function

( )( ) ( ) j s sty t H s e e

Substitute s j

{ ( )}( ) ( )

( ) cos( ( ))

( ) sin( ( ))

t j t j

t

t

y t H j e e

H j e t j

j H j e t j


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6.2.2 Laplace Transform Representation : Derivation of Laplace Transform

A representation for arbitrary signals as a

weighted superposition of eigenfunction est.

( )( ) ( ) ( )

( )

1( ) ( )

2

j t

t j t

t j t

H s H j h e d

h e e d

h t e H j e d

( )

1( ) ( )

2

1( )

2

t j t

j t

h t e H j e d

H j e d


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Derivation of Laplace Transform

2

Substituting and js jdsd /

Laplace Transform

1( ) ( )

2

j

st

j

h t H s e dsj

( ) ( )

1( ) ( )

2

st

j

st

j

X s x t e dt

x t X s e dsj


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6.2.3 Convergence

A necessary condition for convergence of

the Laplace transform is absolute

integrability of x(t)e-t.

ROC(region of convergence) : the range

of for which the Laplace transform

converges.

( ) tx t e dt


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The Laplace transform applies to more general

signals than the Fourier transform does. (a) Signal for

which the Fourier transform does not exist.

(b) Attenuating factor associated with Laplace

transform. (c) The modified signal x(t)e-t is absolutely

integrable for > 1.


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Right half plane

Left half plane

6.2.4 S-plane

6.2.5 Pole-Zero

Represent the complex frequency s in

terms of a complex plane termed the s-

plane

j

X(j)=X(s)|=0

Zero:X(s)=0

Pole:X(s)=


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General Form of Laplace

Transform

A ratio of two polynomials in s

ck : zeros, dk :poles

1

1 1 0

1

1 1 0

1

1

( )

( )( )

( )

M M

M M

N N

N

M

M kk

N

kk

b S b S b s bX s

S a S a s a

b s cX s

s d


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Examples 1

Example6.1 Determine the Laplace

transform of x(t)=eatu(t)

j

a

( )

0

( )

0

( ) ( )

1

1, Re( )

at st

s a t

s a t

X s e u t e dt

e dt

es a

s as a


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Examples 2

Example6.2 Determine the Laplace

transform of y(t)=-eatu(-t)

a

j

0

( )

0

( )

( ) ( )

1

1,Re( )

at st

s a t

s a t

Y s e u t e dt

e dt

es a

s as a


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Examples 3

Comment

In the previous examples the Laplace

transforms X(s) and Y(s) are equal even

though the signals x(t) and y(t) are

different.

The ROC must be specified for the

Laplace transform to be unique.


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6.3 The Unilateral Laplace

Transform

Definition

Properties

Examples


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Definition

Definition : unilateral Laplace transform

Example

0

( ) ( )

( ) ( )U

st

L

X s x t e dt

x t X s

1( )

1( ) with ROC { }

ULat

Lat

e u ts a

e u t re s as a


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6.4 Properties of the Unilateral

Laplace Transform

Linearity

Scaling

Time Shift

( ) ( ) ( ) ( )UL

ax t by t aX s bY s

1( )

| |

UL sx at X

a a

( ) ( )

for all such that ( ) ( ) ( ) ( )

ULsx t e X s

τ x t u t x t u t


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Time shifts for which the unilateral Laplace

transform time-shift property does not apply. (a) A

nonzero portion of x(t) that occurs at times t 0 is

shifted to times t < 0. (b) A nonzero portion x(t) that

occurs at times t < 0 is shifted to times t 0.


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Properties 2

s-Domain Shift

Convolution

Differentiation in the s-domain

0

0( ) ( )UL

s te x t X s s

( )* ( ) ( ) ( )UL

x t y t X s Y s

( ) ( )UL d

tx t X sds


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Properties 3

Differentiation in the time domain

General form for the differentiation

property

00 0

( ) ( ) ( ) ( )

( ) ( ) (0 )

U

U

Lst st st

L

d dx t x t e dt x t e s x t e dt

dt dt

dx t sX s x

dt

1 2

1 20 0

2 1

0

( ) ( ) ( ) | ( ) |

... ( ) | (0 )

Un n nLn

n n nt t

n n

t

d d dx t s X s x t s x t

dt dt dt

ds x t s x

dt


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Properties 4

Initial and final value theorems

Integration property

0

lim ( ) (0 )

lim ( ) ( )

s

s

sX s x

sX s x

( 1)

0

( 1)

(0 ) ( )( ) ,

where (0 ) ( )

Ut L x X s

x ds s

x x d


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Examples 1

Example6.3 Find the unilateral Laplace

transform of x(t)=(-e3tu(t))*(tu(t))

(sol)

Apply the s-domain differentiation property

3 1( )3

1( )

U

U

Lt

L

e u ts

u ts

2

1( )

UL

tu ts

3

2

1( ) ( ( ))*( ( ))

( 3)

ULtx t e u t tu t

s s


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Example 2

Example 6.4 RC filter output x(t)=te2tu(t)

/( )

2

2

1 5( ) ( )

5

1( )

( 2)

5( )

( 2) ( 5)

uLt RCh t e u t

RC s

X ss

Y ss s


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Examples 3

Example6.5 Verify the differentiation

property for the signal x(t)=eatu(t)

(sol)

, 0

( ) ( )U

at at

Lat

de ae t

dt

d ax t ae u t

dt s a

1( ) ( ) (0 ) 1

ULdx t sX s x s

dt s a

a

s a


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Examples 4

Example6.5 Determine the initial and

final values of a signal x(t) whose

unilateral LT is

(sol)

7 10( )

( 2)

sX s

s s

0

7 10(0 ) lim 7

( 2)

7 10( ) lim 5

( 2)

s

s

sx s

s s

sx s

s s


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6.5 Inversion of the Laplace

Transform

Direct inversion of Laplace transform

given by

requires an understanding of contour

integration .

Determine inverse Laplace transform

using the one-to-one relationships

between a signal and its unilateral

Laplace transform.

1( ) ( )

2

j

st

j

x t X s e dsj


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Inverse Lapace Transform based on Several Basic Transform Pairs 1

Laplace transforms that are a ratio of

polynomials in s.

1

1 1 0

1

1 1 0

1

1 1 0

1

1

( )

( )

( )

( )

M M

M M

N N

N

M M

M M

N

k

k

Nk

k k

b S b S b s bX s

S a S a s a

b S b S b s bX s

s d

AX s

s d


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Inverse Lapace Transform based on Several Basic Transform Pairs 2

Basic Laplace transform pair

(CF) If a pole di is repeated r times, then

there are r terms in thr partial fraction

expansion associated with this pole.

( )U

k

Ld t k

k

k

AA e u t

s d

1 2

2

1

, , ,

( )( 1)!

r

U

k

i i i

r

i i i

n Ld t

n

k

A A A

s d s d s d

At Ae u t

n s d


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Example

Example6.7

(sol)

2

3 4( )

( 1)( 2)

sX s

s s

31 2

2

2

( )1 2 2

1 1 2( )

1 2 2

AA AX s

s s s

X ss s s

2 2( ) ( ) ( ) 2 ( )t t tx t e u t e u t te u t


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A pair of Complex-Conjugate

Poles 1

+j0 and -j0 are a pair of complex-

conjugate poles.

1 21 2

0 0

1 2

0 0

2 01 2 1

2 2 2 2 2 2

0 0 0

1 21 1 2 1 2 0 2 2

0

1 2

( * )

( )( )

( )

( ) ( ) ( )

( , ( ) /( )

( , : )

A AA A

s j s j

B s B

s j s j

CB s B C s

s s s

B s BC B C B B

s

B B real


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A pair of Complex-Conjugate

Poles 2

Basic Laplace transform pairs

11 0 2 2

0

2 01 0 2 2

0

( )cos( ) ( )

( )

sin( ) ( )( )

U

U

Lt

Lt

C sC e t u t

s

CC e t u t

s


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Example

Example6.9

(sol)

2

3 2

4 6( )

2

sX s

s s

1 2

2

2

2 2

( )1 ( 1) 1

2 2 2

1 ( 1) 1

2 1 12 4

1 ( 1) 1 ( 1) 1

B s BAX s

s s

s

s s

s

s s s

( ) 2 ( ) 2 cos( ) ( ) 4 sin( ) ( )t t tx t e u t e t u t e t u t


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6.6 Solving Differential Equations with Initial Conditions

Primary application of the unilateral

Laplace transform in systems : solving

differential equations with nonzero initial

conditions.


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Example 6.10 : RC circuit analysis

2( ) 5 ( ) ( ), ( ) 3 ( ), (0 ) 2td

y t y t x t x t e u t ydt

2 5

( ) (0 ) 5 ( ) ( )

1( ) ( ) (0 )

5

3( )

2

3 2( )

( 2)( 5) 5

1 3

2 2

( ) ( ) 3 ( )t

sY s y Y s X s

Y s X s ys

X ss

Y ss s s

s s

y t e tu t e u t


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General Description of Differential Equation in Laplace Transform Domain 1

1

1 1 01

1

1 1 01

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

N N

N NN N

M M

M NM M

d d da y t a y t a y t a y t

dt dt dt

d d db y t b x t b x t b x t

dt dt dt

1

1 1 0

1

1 1 0

11

1 0 0

11

1 0 0

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

( )

( )

( ) ( )

( ) ( )

N N

N N

M M

M N

lN kk l

k lk l t

lM kk l

k lk l t

A s Y s C s B s X s D s where

A s a s a s a s a

B s b s b s b s b

dC s a s y t

dt

dD s b s x t

dt


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General Description of Differential Equation in Laplace Transform Domain 2

Clear separation between

the natural response of the system to initial conditions and

the forced response of the system associated with the input

( ) ( )

( )

( )

( ) ( ) ( ) ( )( )

( ) ( )

( ) ( )

( ) ( ) ( )( )

( )

( )( )

( )

f n

f

n

B s X s D s C sY s

A s A s

Y s Y s

B s X s D sY s

A s

C sY s

A s


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Example 6.11

2

2

0

( ) 5 ( ) 6 ( ) 2 ( ) ( )

where, ( ) ( ), (0 ) 1, ( ) | 2t

d d dy t y t y t x t x t

dt dt dt

dx t u t y y t

dt

2

0

2

0

2

( 5 6) ( ) (0 ) ( ) | 5 (0 ) (2 1) ( ) 2 (0 )

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

5 6

(0 ) ( ) | 5 (0 )

5 6

t

t

ds s Y s sy y t y s X s x

dt

s X s xY s

s s

dsy y t y

dt

s s


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Example 6.11 : continued

( )

( )

( ) 2 3

( ) 2 3

1 1/ 6 1/ 2 1/ 3( )

( 2)( 3) 2 3

7 5 4( )

( 2)( 3) 2 3

1 1 1( ) ( ) ( ) ( )

6 2 3

( ) 5 ( ) 4 ( )

f

n

f t t

n t t

Y ss s s s s s

sY s

s s s s

y t u t e u t e u t

y t e u t e u t


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6.7 Laplace Transform Method

in Circuit Analysis - 생략

Basic electrical circuit elements

- Resistance

- Inductance

- Capacitance

( ) ( )

( ) ( )

R R

R R

v t Ri t

V s RI s

( )

( ) ( ) (0 )

L L

L L L

dv L i t

dt

V s sLI s Li

0

1( ) ( ) (0 )

(0 )1( ) ( )

t

C C C

CC C

v t i d vC

vV s I s

sC sC


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Laplace Transform Circuit

Model 2

Laplace transform circuits models for use

with Kirchhoff’s voltage law


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Model 3

Laplace transform circuits models for use

with Kirchhoff’s current law


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Model 4

Example6.13 : use Laplace transform circuit

modes to determine the voltage y(t) in the

circuit of the following figure.

x(t)=3e-10tu(t) and vC(0-)=5


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Model 6

Example6.11 : (sol) continued

1 2

14

2

20

( ) 1000( ( ) ( ))

1 5( ) ( ) ( )

10

( ) ( ) 1000 ( )

10 5( ) ( )

20 20

2 1( ( ) 3 )

20 10

( ) 2 ( )t

Y s I s I s

X s Y s I ss s

X s Y s I s

sY s X s

s s

X ss s

y t e u t


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6.8 The Bilateral Laplace

Transform

The bilateral Laplace Transform

( ) ( ) ( )L

stx t X s x t e dt


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Properties of the Bilateral Laplace

Transform 1

The linearity, scaling, s-domain, convolution, and differentiation is the s-domain properties are identical for the bilateral and unilateral Laplace transform, although operations associated with these properties may change the ROC.

The bilateral Laplace transform properties involving time shift and differentiation in time domain, the time integration differ slightly from their unilateral counterparts.


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Properties of the Bilateral Laplace

Transform 2

Time shift

Differentiation in the time domain

Time Integration

Integration introduces a pole at s=0

( ) ( )L

sx t e X s

( ) ( ) with ROC at least L

x

dx t sX s R

dt

( )( ) with ROC Re( ) 0

s

t L

x

X sx d R s


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Example 6.15 Find the Laplace

transform

x(t)

(sol)

2

3( 2)

2( ) ( 2)t

dx t e u t

dt

3

3( 2) 2

2 23( 2) 2

2

1( ) , with ROC Re( ) 3

3

1( 2) , with ROC Re( ) 3

3

( ) ( 2) , with ROC Re( ) 33

Lt

Lt s

Lt s

e u t ss

e u t e ss

d sx t e u t e s

dt s


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6.9 Properties of the Region of

Convergence

Convergence

- ROC cannot contain any poles

- Convergence of the Laplace transform for a signal x(t) implies that

-The quantity is the real part of s, so the ROC depends only on the real part of s. ->ROC consists of strips parallel to the j-axis in s-plane

( ) | ( ) | tI x t e dt


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Properties of the Region of

Convergence 2

ROC for x(t) that is finite duration signal.

: ROC for a finite-duration signal

includes the entire s-plane

( ) 0 for and x t t a t b

( ) | ( ) |

( / ) , 0

( ), 0

b

t

a

bt

a

I Ae dt x t A

A e

A b a


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Convergence 3

ROC for x(t) that is infinite duration

signal.

0

0

( ) | ( ) |

( ) ( )

, ( ) | ( ) |

( ) | ( ) |

t

t

t

I x t e dt

I I

where I x t e dt

I x t e dt


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Convergence 4


signal. : continued

np


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Convergence 5


signal. : continued

| ( ) | , 0| ( ) | , 0

p

n

t

t

x t Ae t

x t Ae t

00

( ) ( )

( ) ( )

0 0

( )

( )

n n

p p

t t

n

t t

p

AI A e dt e

AI A e dt e

and n p

p n


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Convergence 6

Change in ROC : example

( ) ( ) with ROC

( ) ( ) with ROC

( ) ( ) ( ) ( )

with ROC at least

L

x

L

y

L

x y

x t X s R

y t Y s R

ax t by t aX s bY s

R R


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Convergence 7

Relationship between the time extent of a signal

and the ROC


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Example 6.16

Identify the ROC associated with the

bilateral Laplace transform

2

1

2

2

( ) ( ) ( )

( ) ( ) ( )

t t

t t

x t e u t e u t

x t e u t e u t

( ) ROC Re(s)

( ) ROC Re(s)

at

at

e u t a

e u t a

2

1

2

2

1( ) ( ) ( ) , 2 Re( ) 1

( 1)( 2)

( ) ( ) ( ) X

Lt t

Lt t

x t e u t e u t ROC ss s

x t e u t e u t


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ROCs for signals in Example 6.16. (a) The shaded regions denote

the ROCs of each individual term, e–2tu(t) and e–tu(–t). The doubly

shaded region is the intersection of the individual ROCs and

represents the ROC of the sum. (b) The shaded regions represent

the individual ROCs of

e–2tu(–t) and e–tu(t). In this case there is no intersection, and the

Laplace transform of the sum does not converge for any value of s.


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Inversion of The Bilateral Laplace

Transform 1

Use the ROC to determine a unique

inverse transform in the bilateral case.

The ROC of a right-sided exponential

signal lies to the right of a pole, while the

ROC of a left-sided exponential signal lies

to the left of a pole


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Transform 2

Example 6.17 Inverting a Laplace

transform

(sol)

5 7( ) with ROC 1 Re( ) 1

( 1)( 1)( 2)

sX s s

s s s

2

1 2 1( )

1 1 2

( ) ( ) 2 ( )

( )

t t

t

X ss s s

x t e u t e u t

e u t


Kim, J. Y.

IC & DSP

Research

Group


Transform 3

Example 6.18 : Inverting an improper

rational Laplace transform

3 2

2

2 9 4 10( ) , with ROC Re( ) 1

s 3 4

s s sX s s

s

(1) 4

1 2( ) 2 3

1 4

( ) 2 ( ) 3 ( ) ( ) 2 ( )t t

X s ss s

x t t t e u t e u t


Kim, J. Y.

IC & DSP

Research

Group


Transform 4

Partial fraction expansion

1

( )N

k

k k

AX s

s d

( ) with ROC Re( )

( ) with ROC Re( )

k

k

Ld t k

k k

k

Ld t k

k k

k

AA e u t s d

s d

AA e u t s d

s d


Kim, J. Y.

IC & DSP

Research

Group


Transform 5

Some Laplace transform pairs

1

1

( ) ; right sided signal( 1)! ( )

( ) ; left sided signal( 1)! ( )

k

k

n Ld t

n

k

n Ld t

n

k

At Ae u t

n s d

At Ae u t

n s d

11 0 2 2

0

11 0 2 2

0

( )cos( ) ( )

( )

; right sided signal

( )cos( ) ( )

( )

; left sided signal

Lt

Lt

C sC e t u t

s

C sC e t u t

s


Kim, J. Y.

IC & DSP

Research

Group

6.11 Transform Function

The transfer function of a system : the

Laplace transform of the impulse

response. ( ) ( )* ( )

( ) ( ) ( )

( ) ( ) / ( )

y t h t x t

Y s H s X s

H s Y s X s


Kim, J. Y.

IC & DSP

Research

Group

The Transfer Function and

Differential Equations 1

Differential-equation description for a

LTI system

x(t)=est is an eigenfunction of the system

0 0

( ) ( )k kN M

k kk kk k

d da y t b x t

dt dt

0 0

0

0

,

( )

k k kN Mst st st k st

k kk k kk k

Mk

k

k

Nk

k

k

d d da e b e e s e

dt dt dt

b s

H s

a s


Kim, J. Y.

IC & DSP

Research

Group



Example6.19

Pole-zero form

2

2( ) 3 ( ) 2 ( ) 2 ( ) 3 ( )

d d dy t y t y t x t x t

dt dt dt

2

2 3( )

3 2

sH s

s s

1

1

( / ) ( )( )

( )

M

M N kk

N

kk

b a s cH s

s d


Kim, J. Y.

IC & DSP

Research

Group



Example6.18

1

2

Torque produced by the motor : ( ) ( )

Back electromortive fource : ( ) ( )

t K i t

dv t K y t

dt


Kim, J. Y.

IC & DSP

Research

Group



Example6.20:solution

2

12

2

1 122

2

1 2 1

2

1( ) ( ), ( ) [ ( ) ( )]

( ) [ ( ) ( )] [ ( ) ( )]

( ) ( ) ( )

dJ y t K i t i t x t v t

dt R

K Kd dJ y t x t v t x t K y t

dt R R dt

K K Kd dJ y t y t x t

dt R dt R

11

2 1 2 1 2

( )

KK

RJRH sK K K K

Js s s sR R


Kim, J. Y.

IC & DSP

Research

Group

6.12 Causality and Stability

Causality : Impulse response of a causal

system is zero for t

Kim, J. Y.

IC & DSP

Research

Group

The relationship between the locations of poles and the impulse

response in a causal system. (a) A pole in the left half of the s-plane

corresponds to an exponentially decaying impulse response. (b) A

pole in the right half of the s-plane corresponds to an exponentially

increasing impulse response. The system is unstable in this case.


Kim, J. Y.

IC & DSP

Research

Group

Stability

Stability : The impulse response is

absolutely integrableThe Fourier

transform exists and thus the ROC

includes the j-axis in the s-plane.

This knowledge is sufficient to uniquely

determine the inverse Laplace transform

of the transfer function.


Kim, J. Y.

IC & DSP

Research

Group

The relationship between the locations of poles and the impulse

response in a stable system. (a) A pole in the left half of the s-

plane corresponds to a right-sided impulse response. (b) A pole

in the right half of the s-plane corresponds to an left-sided

impulse response. In this case, the system is noncausal.


Kim, J. Y.

IC & DSP

Research

Group

A system that is both stable and causal must have a

transfer function with all of its poles in the left half of

the s-plane, as shown here.


Kim, J. Y.

IC & DSP

Research

Group

Example 6.21

Transfer function

(a) stable (b) causal

-3 2 -3 2

2 1( )

3 2H s

s s

3 2( ) 2 ( ) ( )t th t u t e u t 3 2( ) 2 ( ) ( )t th t u t e u t


Kim, J. Y.

IC & DSP

Research

Group

Inverse Systems 1

Inverse system

Minimum phase : H(s) has all of its poles and

zeros in the left half of the s-plane : A

nonminimum phase system cannot have a

stable and causal inverse system.

1

1 1

( )* ( ) ( )

1( ) ( ) 1 or ( )

( )

h t h t t

H s H s H sH s

1 1

1

( )( )

( / ) ( )

N

kk

M

M N kk

s dH s

b a s c


Kim, J. Y.

IC & DSP

Research

Group

Inverse Systems 2

Example6.21 Find the inverse system

(sol)

:The inverse system cannot be both stable and causal

2

2( ) 3 ( ) ( ) ( ) 2 ( )

d d dy t y t x t x t x t

dt dt dt

2

2

1

2

( )( 3) ( )( 2)

( ) 2( )

( ) 3

1 3 3( )

( ) 2 ( 1)( 2)

Y s s X s s s

Y s s sH s

X s s

s sH s

H s s s s s


Kim, J. Y.

IC & DSP

Research

Group

6.13 Determining the Frequency Response from Poles and Zeros

s j

at some fixed , 0

1

1

( / ) ( )( )

( )

M

M N kk

N

kk

b a j cH j

j d

010

01

/( )

M

M N kk

N

kk

b a j cH j

j d

0 0

1

0

1

arg ( ) arg / arg

arg

M

M N k

k

N

k

k

H j b a j c

j d


Kim, J. Y.

IC & DSP

Research

Group

The function |j – g| corresponds to the lengths of

vectors from g to the j-axis in the s-plane. (a) Vectors

from g to j for several frequencies. (b) |j – g| as a

function of j.


Kim, J. Y.

IC & DSP

Research

Group

Components of the magnitude response.

(a) Magnitude response associated with a zero.

(b) Magnitude response associated with a pole.


Kim, J. Y.

IC & DSP

Research

Group

Example 6.23

Sketch the magnitude and phase response of

the system having the transfer function

(sol)

( 0.5)( )

( 0.1 5 )( 0.1 5 )

sH s

s j s j


Kim, J. Y.

IC & DSP

Research

Group

하나의 꿈, 다양한 지식

Sagres castle

바스코 다 가마 항로 (1497)

조선 지도

천문학

항해

탐험

수학자 고문서 학자

신학자

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