Download Chapter 4 - UniMAP Portal

CHAPTER 4
Laplace
Transform.
School of Computer and Communication Engineering,
UniMAP
Hasliza A Rahim @ Samsuddin
EKT 230
1
4.0 Laplace Transform.
4.1
4.2
4.3
4.4
4.5
Introduction.
The Laplace Transform.
The Unilateral Transform and Properties.
Inversion of the Unilateral.
Solving Differential Equation with Initial
Conditions.
4.6 Laplace Transform Methods in Circuit Analysis.
4.7 Properties of the Bilateral Laplace Transform.
4.8 Properties of the Region of Convergence.
4.9 Inversion of the Bilateral Laplace Transform.
4.10 The Transfer Function
4.11 Causality and Stability.
4.12 Determining the Frequency Response from Poles2
and Zeros.
4.1 Introduction.
 In Chapter 3 we developed representation of signal and LTI by
using superposition of complex sinusoids.
 In this Chapter 4 we are considering the continuous-time signal and
system representation based on complex exponential signals.
 The Laplace transform can be used to analyze a large class of
continuous-time problems involving signal that are not absolutely
integrable, such as impulse response of an unstable system.
 Laplace transform come in two varieties;
(i) Unilateral (one sided); is a tool for solving differential equations
with initial condition.
(ii) Bilateral (two sided); offer insight into the nature of system
characteristic such as stability, causality, and frequency response.
3
4.2 Laplace Transform.
 Let est be a complex exponential with complex frequency
s = s +jw. We may write,
e st  est coswt   jest sin wt .
 The real part of est is an exponential damped cosine
 And the imaginary part is an exponential damped sine as
shown in Figure 6.1.
 The real part of s is the exponential damping factor s.
 And the imaginary part of s is the frequency of the cosine and
sine factor, w.
4
Cont’d…
Figure 4.1: Real and imaginary parts of the complex exponential est,
where s = s + jw.
5
4.2.1 Eigen Function Property of
st
e .
 Apply an input to the form x(t) =est to an LTI system with impulse
response h(t). The system output is given by,
Derivation:
y t   H xt 
 ht * xt 


 h xt   d

 We use x(t) =est to obtain
y t  

s t  


h

e
d



 We define transfer function as
 e st  h e s d

H s  

 st


h

e
d


6
Cont’d…
 We can write


y t   H e st   H s e st


 An eigen function is a signal that passes through the system
without being modified except by multiplication by scalar.
 The equation below indicates that,
- est is the eigenfunction of the LTI system.
- H(s) is the eigen value.
H s   H s  e
j  s 
7
Cont’d…
Express complex-value transfer function in Polar Form
 Where |H(s)| and (s) are the magnitude and phase of H(s)
y t   H s  e j  s e st
substitute
s  s  jw
y t   H s  jw  est e jwt  (s  jw )
 H s  jw  est coswt   s  jw   j H s  jw  est sin wt   s  jw .
8
4.2.2 Laplace Transform
Representation.
 H(s) is the Laplace Transform of h(t) and the h(t) is the inverse
Laplace transform of H(s).
 The Laplace transform of x(t) is

X s  

x t e  st dt

 The Inverse Laplace Transform of X(s) is
s  j
xt  
1
2j

s
X ( s )e st ds
 j
 We can express the relationship with the notation
x t    X s 
L
9
4.2.3 Convergence.
 The condition for convergence of the Laplace transform is the
absolute integrability of x(t)e-at ,

 xt e
st
dt  

 The range of s for which the Laplace transform converges is
termed the region of convergence (ROC)
Figure 4.2: 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-st is absolutely integrable for s > 1.
10
4.2.4 The s-Plane.
 It is convenience to represent the complex frequency s graphically
in termed the s-plane.
(i) the horizontal axis represents the real part of s
(exponential damping factor s).
(ii) The vertical axis represents the imaginary part of s
(sinusoidal frequency w)
X  jw   X s  |s 0
 In s-plane, s =0 correspond to imaginary axis.
 Fourier transfrom is given by the Laplace transform evaluated along
the imaginary axis.
11
Cont’d…
 The jw-axis divides the s-plane in half.
(i) The region to the left of the jw-axis is termed the left half of
the s-plane.
(ii) The region to the right of the jw-axis is termed the right half
of the s-plane.
 The real part of s is negative in the left half of the s-plane and
positive in the right half of the s -plane..
Figure 4.3: The s-plane. The horizontal axis is Re{s}and the vertical axis is Im{s}.
Zeros are depicted at s = –1 and s = –4  2j, and poles are depicted at
s = –3, s = 2  3j, and s = 4.
12
4.2.5 Poles and Zeros.
 Zeros. The ck are the root of the numerator polynomial and are
termed the zeros of X(s). Location of zeros are denoted as “o”.
 Poles. The dk are the root of the denominator polynomial and are
termed the poles of X(s). Location of poles are denoted as “x”.
 The Laplace transform does not uniquely correspond to a signal
x(t) if the ROC is not specified.
 Two different signal may have identical Laplace Transform, but
different ROC. Below is the example.
Figure 6.4a
Figure 6.4b
Figure 4.4a. The ROC for x(t) = eatu(t) is depicted by the shaded region. A pole is
located at s = a.
Figure 4.4b. The ROC for y(t) = –eatu(–t) is depicted by the shaded region. A pole is
located at s = a.
13
Example 4.1: Laplace Transform of a Causal
Exponential Signal.
Determine the Laplace transform of x(t)=eatu(t).
Solution:
Step 1: Find the Laplace transform.
X s  

 st


x
t
e
dt



  e ( s  a )t dt
0
 1 ( s  a ) t

e
sa

0
To evaluate e-(s-a)t, Substitute s=s + jw
X s  
1
e (s  jw a )t
s  jw  a

0
14
Cont’d…
If s > 0, then e-(s-a)t goes to zero as t approach infinity,
1
(0  1),
s  jw  a
1

,
Re( s )  a.
sa
X s  
s a
*The Lapalce transform does not exist for s=<a because the
integral does not converge.
*The ROC is at s>a, the shade region of the s-plane in Figure
below. The pole is at s=a.
4.5: The ROC for x(t) = eatu(t) is depicted by the shaded region. A pole
 Figure
.
is located at s = a.
15
4.3 The Unilateral Laplace
Transform and Properties.
 The Unilateral Laplace Transform of a signal x(t) is defined by

X s  

xt e  st dt
0
 The lower limit of 0- implies that we do not include discontinuities
and impulses that occur at t = 0 in the integral. H(s) depends on
x(t)for t >= 0.
 The relationship between X(s) and x(t) as
xt   
 X s 
Lu
 The unilateral and bilateral Laplace transforms are equivalent for
signals that are zero for time t<0.
16
Cont’d…
Properties of Unilateral Laplace Transform.
Scaling
 s 
1

xat 

X


a
a
Lu
Linearity,
Lu




ax t  by t 
 aX s   bY s 
For a>0
Time Shift
xt   
 e  s X s 
Lu
for all t such that x(t - )u(t) = x(t - )u(t - )
 A shift in  in time correspond to multiplication of the Laplace
transform by the complex exponential e-s.
17
Cont’d…
s-Domain Shift
Lu
s
t
o
e
xt 
 X s  s
o
 Multiplication by a complex exponential in time
introduces a shift in complex frequency s into the Laplace
transform.
Figure 4.6: 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.
18
Cont’d…
Convolution.
xt * y t 
 X s Y s .
Lu
 Convolution in time corresponds to multiplication of
Laplace transform. This property apply when x(t)=0 and
y(t) = 0 for t < 0.
Differentiation in the s-Domain.
d
 tx t 

X s .
ds
Lu
 Differentiation in the s-domain corresponds to
multiplication by -t in the time domain.
19
Cont’d…
Differentiation in the Time Domain.
 
d
Lu

xt   sX s   x 0
dt
Initial and Final Value Theorem.
 
Lim sX s   x 0 .


s 
 The initial value theorem allow us to determine the initial
value, x(0+), and the final value, x(infinity, of x(t) directly from
X(s).
 The initial value theorem does not apply to rational functions
X(s) in which the order of the numerator polynomial is greater
than or equal to hat of the denominator polynomial.
20
Example 4.2: Applying Properties.
Find the unilateral Laplace Transform of x(t)=(-e3tu(t))*(tu(t)).
Solution:
Find the Unilateral Laplace Transform.
1
 e u (t ) 
( s  3)
Lu
3t
And
1
u
(
t
)


Apply s-domain differentiation property,
s
Lu
1
tu(t )  2
s
Lu
Use the convolution property,
1
x (t )  (e u (t ) * (tu(t ))  X ( s )  2
1
s ( s 213)
.
3t
Lu
4.4 Inversion of the Unilateral
Laplace Transform.
 We can determine the inverse Laplace transforms using one-toone relationship between the signal and its unilateral Laplace
transform.
Appendix D1 consists of the table of Laplace Transform.
 X(s) is the sum of simple terms,
N
~
X ( s )  k 1
Ak
s  dk
 Using the residue method, solve for a system linear equation.
Ak
Ak e u t  
s  dk
dkt
Lu
 Then sum the Inverse Laplace transform of each term.
At n1 d k t
A
Lu
e u t  
.
n
n  1!
s  d k 
22
Example 4.3: Inversion by Partial-Fraction Expansion.
Find the Inverse Laplace Transform of
Solution:
3s  4
X s  
( s  1)( s  2) 2
Step 1: Use the partial fraction expansion of X(s) to write
A
B
C
X s  


( s  1) ( s  2) ( s  2) 2
Solving the A, B and C by the method of residues
(3s  4)
A( s  2) 2
B( s  1)( s  2)
C ( s  1)



( s  1)( s  2) 2 ( s  1)( s  2) 2 ( s  2) 2 ( s  1) ( s  2) 2 ( s  1)
23
Cont’d…
(3s  4)  A( s  2) 2  B ( s  2)( s  1)  C ( s  1)
 A( s 2  4 s  4)  B ( s 2  3s  2)  C ( s  1)
 ( A  B ) s 2  ( 4 A  3B  C ) s  ( 4 A  2 B  C )
so, compare coefficient ,
A B  0
       (1)
4 A  3B  C  3      ( 2)
4 A  2 B  C  4      (3)
(3)  ( 2);
 B 1
B  1.
From(1)
A B  0
A 1
Substitute B and
A, int o( 2)
4(1)  3( 1)  C  3
C  2.
24
Cont’d…
X s  
A=1, B=-1 and C=2
1
1
2


( s  1) ( s  2) ( s  2) 2
Step 2: Construct the Inverse Laplace transform from the above
partial-fraction term above.
- The pole of the 1st term is at s = -1, so
1
e u (t ) 
( s  1)
t
Lu
- The pole of the 2nd term is at s = -2, so
1
e u (t ) 
( s  2)
-The double pole of the 3rd term is at s = -1, so
 2t
Lu
Lu
2te  2t u (t ) 
2
( s  2) 2
Step 3: Combining the terms.
t
2t
2t
x(t )  e u(t )  e u(t )  2te u(t ).
 .
25
Example 4.4: Inversion An Improper Rational Laplace
Transform.
Find the Inverse Laplace Transform of
Solution:
2s 3  9s 2  4s  10
X s  
s 2  3s  4
Step 1: Use the long division to espress X(s) as sum of rational
polynomial function.
2 s  3 __________
s 2  3s  4 2s 3  9s 2  4s  10
2 s 3  6 s 2  8s
 3s 2  12s  10
We can write,
 3s 2  9 s  12
3s  2
X s   2s  3 
3s  2
s 2  3s  4
26
Cont’d…
Use partial fraction to expand the rational function,
1
2
X s   2s  3 

s 1 s  4
Step 2: Construct the Inverse Laplace transform from the above
partial-fraction term above. Refer to the Laplace transform
Table.
x(t )  2
 .
(1)
(t )  3 (t )  e u(t )  2e u(t ).
t
4t
27
4.5 Solving Differential Equation
with Initial Condition.
 Primary application f unilateral Laplace transform in system
analysis, solving differential equations with nonzero initial
condition.
 Refer to the example.
28
Example 4.5: RC Circuit Analysis (Initial condition)
Use the Laplace transform to find the voltage across the capacitor , y(t),
for the RC circuit shown in Figure 4.7 in response to the applied voltage
x(t)=(3/5)e-2tu(t) and the initial condition y(0-) = -2.
Solution:
Figure 4.7: RC circuit for
Examples 6.4 and 6.10. Note
that RC = 1/5.
Step 1: Derive differential equation from the circuit.
KVL around the loop.
 d

xt   R C
y t   y t   0
 dt

 d

R C
y t   y t   xt 
divide
by RC
 dt

d
1
1
y t  
y t  
xt 
       (1)
dt
RC
RC
RC  1K * 200mF  0.2 s
From
(1)
d
29
y t   5 y t   5 xt 
dt
Cont’d…
Step 2: Get the unilateral Laplace Transform.
 
d
Lu
xt  
sX s   x 0 
dt
Apply the differential property,
sY ( s )  y (0  )  5Y ( s )  5 X ( s )
solving for Y ( s ),
1
Y (s) 
5 X ( s )  y (0  )
       (2)
s5


Step 3: Substitute Unilateral Laplace Transform of x(t) into Y(s).
Given initial condition y(0-)=-2.
3e 2t u (t )
x(t ) 
5
(2) Into (3),
Lu
x(t ) 
X ( s) 
3/ 5
s2
     (3)
3
2
Y ( s) 

( s  2)( s  5) ( s  5)
30
Cont’d…
Step 4: Expand Y(s) into partial fraction.
1
1
2
Y ( s) 


s2 s5 s5
1
3
Y ( s) 

s2 s5
Step 5: Take Inverse Unilateral Laplace Transform of Y(s).
2t
5t
y(t )  e u(t )  3e u(t ).
31
 .
4.6 LT Methods in Circuit
Analysis.
 We can use the differentiation and integration properties to
transform circuits involving capacitive and inductive elements so
that we can solve using Laplace transforms.
 Replace resistive, capacitive, and inductive elements by the
Laplace transform equivalent.
Figure 4.8: Laplace transform circuit models for use with Kirchhoff ’s voltage law.
(a) Resistor. (b) Inductor with initial current iL(0–).
32
(c) Capacitor with initial voltage vc(0–).
Cont’d…
(a) Resistor
v

t   Ri t 
R
R

V

s   RI s 
R
R
Eq (4.29)
 Transforming the above equation
 Representation of the transform resistor element.
d
v t L
i t 
L
dt L

(b) Inductor.
 Transforming the above equation
V
 



sLI
s

Li
 0 .
L s
L
L


Eq (4.30)
 Representation of the transform inductor element.
33
Cont’d…
(c) Capacitor.
t
vc t  
1
C

 
ic  dt  vc  0  .
 
0
 Transforming the above equation


vc  0  .
1


Vc s  
I c s  
sC
s
Eq (6.31)
 Representation of the transform capacitor element.
34
Cont’d…
 When apply Kirchhoff ’s voltage law to solve a circuit, use
equation (6.29),(6.30) and (6.31).
 When apply Kirchhoff ’s current law to solve a circuit, rewrite
equation (6.29),(6.30) and (6.31) to express current as a function of
voltage.
Figure 4.9: Laplace transform circuit models for use wit Kirchhoff ’s current law.
(a) Resistor. (b) Inductor with initial current iL(0–).
(c) Capacitor with initial voltage vc(0–).
35
Example 4.6: Solving the Second Order Circuit.
Use the Laplace transform circuit model to determine the voltage y(t) in
the circuit of Figure 4.10 for the applied voltage x(t)=3e-10tu(t) V. the
voltage across the capacitor at time t=0- is 5V.
Solution:
Figure 4.10 Electrical circuit (a) Original circuit.
Step 1: Transform the circuit (a) to (b).
I1(s) and I2(s) are the currents through the branch.
Figure 4.11: Electrical circuit (a) Original circuit. (b) Transformed circuit.
36
Cont’d…
Step 2: Derive equation from the Transformed circuit.
Y ( s )  1000( I ( s )  I ( s ))
           (1)
KVL
1
2
1
5
I1 (s)   0
4
s
s (10 )
1
5
X (s)  Y (s) 
I1 ( s) 
4
s
s (10 )
 X (s)  Y (s) 
I 1 ( s )  s (10  4 ) X ( s )  s (10  4 )Y ( s )  5(10  4 )
       (2)
X ( s )  Y ( s )  1000 I 2 ( s )
I 2 (s) 
X (s) Y (s)

1000 1000
           (3)
Y ( s )  1000I 1 ( s )  I 2 ( s ) 
(2) and (3) into(1),
5
X ( s) Y ( s) 
 sX ( s ) sY ( s )
Y ( s )  1000





 10000 10000 10000 1000 1000 
 sX ( s ) sY ( s ) 5

Y (s)  

  X (s)  Y (s) 
10
10
 10

sY ( s ) sX ( s )
5
2Y ( s ) 

 X ( s) 
10
10
10
s 

 s
 5
Y ( s ) 2    X ( s )  1 
10 

 10  10
Y (s) 
Y (s) 
Y (s) 
u sin g
.
Y (s) 
 s
 5
X ( s )  1 
10

 10
s


2  
10 

 s  10  5
X ( s )

 10  10
 s  20 


 10 
( s  10)
5
X (s)

( s  20) ( s  20)
3
X (s) 
we
( s  10)
2
( s  20)
obtain,
y(t )  2e 20t u(t )V .
37
4.7 Properties of Bilateral
Laplace Transform.
 The Bilateral Lapalace Transform is suitable to the problems
involving no causal signals and system.
 The properties of linearity, scaling, s-domain shift, convolution and
differentiation in the s-domain is identical fort the bilateral and
unilateral LT, the operations associated y these properties may
change the ROC.
 Example; a linearity property.
xt  X s .
with
ROC
Rx
y t  Y s .
with
ROC
Ry
L
L
then
ax t   by t  aX s   bY s 
L
with
ROC
Rx  Ry
 ROC of the sum of the signals is an intersection of the individual
38
ROCs.
Cont’d…
Time Shift
xt   
 e s X s 
L
 The bilateral Laplace Transform is evaluated over both
positive and negative values of time. ROC is unchanged by a
time shift.
Differentiation in the Time Domain.
L
d
xt 
 sX s 
dt
with
ROC
at
least
Rx ,
 Differentiation in time corresponds to multiplication by s.
39
Cont’d…
Integration with Respect to Time.
t

x d 
L
X s 
,
s
with
ROC R x  Re s  0.

 Integration corresponds to division by s
 Pole is at s=0, we ae integrating to the right the ROC must
lie to the right of s=0.
40
4.8 Properties of Region of
Converges.
 The Bilateral Lapalace Transform is suitable to the problems
involving no causal signals and system.
 The properties of linearity, scaling, s-domain shift, convolution and
differentiation in the s-domain is identical for the bilateral and
unilateral LT, the operations associated y these properties may
change the ROC.
 Example; a linearity property.
 ROC of the sum of the signals is an intersection of the individual
41
ROCs.
4.9 Inversion of Bilateral Laplace
Transform.
 The inversion of Bilateral Laplace transforms are expressed as a
ratio of polynomial in s.
 Compare to the unilateral, in the bilateral Laplace transform we
must use the ROC to determine the unique inverse transform in
bilateral case.
Ak
Ak e u (t ) 

, with
s  dk
dkt
L
Ak
 Ak e u (t ) 

, with
s  dk
dkt
L
ROC
ROC
Re( s )  d k
Re( s )  d k
42
Example 4.7: Inverting a Proper Rational Laplace
Transform.
Find the Inverse bilateral Laplace Transform of
With ROC -1<Re(s)<1.
 5s  7
X s  
( s  1)( s  1)( s  2)
Solution:
Step 1: Use the partial fraction expansion of X(s) to write
A
B
C
X s  


( s  1) ( s  1) ( s  2)
Solving the A, B and C by the method of residues
1
2
1
X s  


( s  1) ( s  1) ( s  2)
43
Cont’d…
Step 2: Construct the Inverse Laplace transform from the above
partial-fraction term above.
- The pole of the 1st term is at s = -1, the ROC lies to the right of this
pole, choose the right-sided inverse Laplace Transform.
1
e u (t ) 

( s  1)
t
L
- The pole of the 2nd term is at s = 1, the ROC is to the left of the pole,
choose the right-sided inverse Laplace Transform.
L
2et u (t ) 


2
( s  1)
-The pole of the 3rd term is at s = -2, the ROC is to the right of the
pole, choose the left-sided inverse Laplace Transform.
1
e u (t ) 

( s  2)
 2t
L
44
Cont’d…
Step 3: Combining the terms.
 Combining this three terms we obtain,
x(t )  et u(t )  2et u(t )  e2t u(t ).
Figure 4.12 : Poles and ROC for Example 6.17.
45
4.10 Transfer Function.
 The transfer function of an LTI system is defined as the Laplace
transform of the impulse response.
 Take the bilateral Laplace transform of both sides of the equation
and use the convolution properties result in,
Y ( s)  H ( s) X ( s)
 Rearrange the above equation result in the ratio of Laplace
transform of the output signal to the Laplace transform of the
input signal. (X(s) is nonzero)
Y ( s)
H ( s) 
X ( s)
46
4.10.1 Transfer Function and
Differential-Equation System
Description.
Given a differential equation.
M
dk
dk
ak k y(t )   bk k x(t )

dt
dt
k 0
k 0
N
Step 1: Substitute y(t) = estH(s) into the equation.
y(t) = estH(s), substitute to the above equation result in,
M
 N
d k st 
d k st
  ak k e  H s    bk k e
dt
k 0
 k 0 dt

 
Step 2: Solve for H(s).
H(s) is a ratio of polynomial and
s is termed a rational transfer function.
 
M
H s  
k
b
s
 k
k 0
N
k
a
s
 k
k 0
47
Example 4.8: Find the Transfer Function.
Find the transfer function of the LTI system described by the
differential equation below
d2
d
d
yt   3 yt   2 yt   2 xt   3x(t )
2
dt
dt
dt
Solution:
Step 1: Substitute y(t) = estH(s) into the equation.
 
 
 
   
d 2 st
d st
d st
st
st
e
H
(
s
)

3
e
H
(
s
)

2
e
H
(
s
)

2
e

3
e
dt 2
dt
dt
 
d k st
k st
e

s
e
k
dt
Step 2: Solve for H(s).
 d 2 st
d
H ( s) 2 e  3 e st  2 e st
dt
 dt
 

d
  2 e st  3 e st
dt

   
   
48
Cont’d…
   
d st
2 e  3 e st
dt
H ( s) 
 d 2 st
d
 2 e  3 e st  2 e st
dt
 dt
 



   
 
d k st
k st
e

s
e
k
dt
The transfer function is,
2s  3
H (s)  2
s  3s  2
.
49
Reference Table.
50