CHAPTER 4
Laplace
Transform.
School of Computer and Communication Engineering,
UniMAP
Pn. Nordiana Mohamad Saaid
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 Forced and Natural Responses in Unilateral
Laplace Transform.
4.7 Properties of the Bilateral Laplace Transform
4.8 Inversion of the Bilateral Laplace Transform.
4.9 The Transfer Function
4.10 Causality and Stability
2
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 the input, x(t) =est to obtain
the output, y(t) as:

y t    h e s t  d


 e st  h e  s d

 Thus, transfer function is:
H s  

 s


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.
 The transfer function expressed in terms of magnitude and phase;
H s   H s  e
j  s 
7
Cont’d…
Express complex-value transfer function in Rectangular Form
 Where |H(s)| and (s) are the magnitude and phase of H(s)
The output;
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.. thus 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 terms of 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 4.4a
Figure 4.4b
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
13
located at s = a.
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, while as t is
zero, e-(s-a)t goes to minus one, hence
1
(0  1),
s  jw  a
1

,
Re( s )  a.
sa
X s  
s a
*The Laplace transform does not exist for s=<a because the
integral does not converge.
*The ROC is at s>a, the shaded region of the s-plane in figure
below. The pole is at s=a.

Figure 4.5: The ROC for x(t) = eatu(t) is depicted by the shaded region. A pole
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 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
e xt   X s  s0 
s0 t
Lu
 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 domain 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 that of the denominator
20
polynomial.
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)
3t
Lu
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  3 B  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 = -2, so
 2t
Lu
Lu
2te  2t u (t ) 
Step 3: Combining the terms.
t
2t
2
( s  2) 2
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 2 s 3  9 s 2  4 s  10
2 s 3  6 s 2  8s
 3s 2  12 s  10
 3s 2  9 s  12
3s  2
We can write,
X s   2 s  3 
3s  2
s 2  3s  4
26
Cont’d…
Use partial fraction to expand the rational function,
1
2
X s   2 s  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 of 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
where RC  1K * 200mF  0.2 s
substitute into (1),
d
y t   5 y t   5 xt 
dt
29
Cont’d…
Step 2: Get the unilateral Laplace Transform.
Apply the properties of differential in time domain,
 
d
Lu
xt  
sX s   x 0 
dt
sY ( s )  y (0  )  5Y ( s )  5 X ( s )
Rearrange the equation in terms of Y ( s );
Y ( s) 

1
5 X ( s )  y (0  )
s5

       (2)
Step 3: Substitute Unilateral Laplace Transform of x(t) into Y(s).
3e 2t u (t )
Laplace transform of the
x(t ) 
5
applied voltage, x(t);
3 1 
x(t )  X ( s )  

5 s2
Lu
Given initial condition,
substitute y(0-)=-2 into (2),
And substitute (3) into (2), Y ( s) 
     (3)
3
2

( 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 ).
by referring to the Laplace transform table.
 .
31
QUIZ 1
Use the basic Laplace transforms & Laplace transform
properties in Appendix D to determine the following
signals:
a)
b)
c)
x(t )  tu(t ) * cos( 2t )u (t )
2t
x(t )  u(t  1) * e u(t  1)
d t
x(t )  t e cos(t )u (t ) 
dt
32
QUIZ 2
 Use the method of partial fractions to find the time
signals corresponding to the following Laplace
transforms:
 a)
s3
X ( s)  2
s  3s  2
 b)
2s  1
X (s)  2
s  2s  1
 C)
5s  4
X (s)  3
2
s  3s  2 s
33
4.6 Forced and Natural Responses
in Unilateral Laplace Transform.
 Primary application of the unilateral Laplace transform is to
solve differential equations with non-zero initial conditions.
 The initial conditions are incorporated into the solution as the
values of the signal and its derivatives that occur at time zero
in the differentiation property.
 General form of differentiation property:n 1
dn
d
Lu
n
x
(
t
)


S
X ( s )  n 1 x(t )
n
dt
dt
n2 d
n 1
0
 ...  s
x(t ) t 0  s x( ).
dt
t 0
d n2
 s n  2 x(t )
dt
t 0
Eq. 6.19 
34
4.6 Forced and Natural Responses
in Unilateral Laplace Transform.
 The forced response of a system represents the component of
the response associated entirely with the input, denoted as Y ( f ) (s) .
This response represents the output when the initial conditions
are zero.
 The natural response represents the component of the output
due entirely to the initial conditions, denoted as Y ( n ) ( s) . This
response represents the system output when the input is zero.
35
Example 4.6: Finding forced and natural
responses.
Use the unilateral Laplace transform to determine the output of a
system represented by the differential equation,
d2
d
d
y
(
t
)

5
y
(
t
)

6
y
(
t
)

x(t )  6 x(t )
2
dt
dt
dt
in response to the input x(t )  u (t ) . Assume that the initial
conditions on the system are;
d
y (0 )  1 and
y (t )
dt

t 0
2
Solution.
Using differentiation property in Eq. 6.19 and taking Laplace
transform of both sides of the differential equation,
s 2Y ( s ) 
d
y (t )
dt
t 0

 sy (0  )  5sY ( s )  5 y (0  )  6Y ( s )  sX ( s )  6 X ( s )
36
Contd…
Rearranging the terms;
s
2

 5s  6 Y ( s ) 
d
y (t )
dt
t 0

 sy (0  )  5 y (0  )  s  6 X ( s )
Solving for Y(s), we get;

s  6X ( s )
Y ( s) 

sy (0  ) 
s 2  5s  6
d
y (t ) t 0  5 y (0  )
dt
s 2  5s  6
The first term is associated with the forced response , Y ( f ) (s) . The
second term corresponds to the natural response, Y ( n) (s) .
Using the transform of input, X ( s)  1 / s and the initial conditions,
we obtain;
s6
Y ( f ) ( s) 
and
Y
(n)
s( s  2)( s  3)
s7
( s) 
( s  2)( s  3)
37
Contd…
Next, taking the inverse Laplace transforms of Y ( f ) (s) and Y ( n ) ( s) ,
we obtain;
y
(f)
2t
3t
(t )  u(t )  2e u(t )  e u(t )
and
2t
3t
y (t )  5e u (t )  4e u (t )
( n)
Hence, the output of the system is;
y(t )  y (t )  y (t )
(f)
(n)
38
Contd…

a) Forced response of the system, y(f)(t).
(b) Natural response of the system, y(n)(t). (c) System output.
39
4.7 Properties of Bilateral
Laplace Transform.
 The Bilateral Laplace Transform is suitable to the problems
involving non-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 by these properties may
change the ROC.
 Example; a linearity property.
L
xt  

X s .
with
ROC
L
y t  

Y s .
Rx
with
ROC
Ry
then
L
axt   by t  

aX s   bY s 
with
ROC
R x  Ry
 ROC of the sum of the signals is an intersection of the individual
40
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.
d
L
xt   sX s 
dt
with
ROC
at
least
 Differentiation in time domain corresponds to
multiplication by s.
Rx ,
41
Cont’d…
Integration with Respect to Time.
X s 
x d  s ,
t
L
with
ROC
Rx  Re s   0.
 Integration corresponds to division by s
 Pole is at s=0, we are integrating to the right, hence the
ROC must lies to the right of s=0.
42
4.8 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
ROC
Ak
 Ak e u (t ) 

, with ROC
s  dk
dkt
L
Re( s )  d k
Re( s)  d k
right  sided
left - sided
43
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)
44
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 left-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 right-sided inverse Laplace Transform.
1
e u (t ) 

( s  2)
 2t
L
45
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.
46
4.9 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, it results in;
Y ( s)  H ( s) X ( s)
 Rearrange the above equation will result in the ratio of Laplace
transform of the output signal to the Laplace transform of the
input signal. (Note: X(s) is non-zero)
Y ( s)
H ( s) 
X ( s)
47
4.9.1 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 as rational transfer function. H s  
 
M
k
b
s
 k
k 0
N
k
a
s
 k
k 0
48
Example 4.8: Find the Transfer Function.
Find the transfer function of the LTI system described by the
differential equation below,
Solution:
d2
d
d
yt   3 yt   2 yt   2 xt   3x(t )
2
dt
dt
dt
Step 1: est is an eigenfunction of LTI system. If input, x(t)=est
Then, y(t) = estH(s). Hence, substitute 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 st
H ( s) 2 e  3
e  2 e st
dt
 dt
 

d st
  2
e  3 e st
dt

   
   
49
Cont’d…
   
d st
2
e  3 e st
dt
H ( s) 
 d 2 st
d st
 2 e  3 e  2 e st
dt
 dt
 
k



   
 
d
st
k st
e

s
e
k
dt
Hence, the transfer function is,
2s  3
H ( s)  2
s  3s  2
50
4.10 Causality and Stability
4.10.1 Causality :
 Impulse response of a causal system is zero for t < 0.
 A system pole at s = dk in the left half plane [Re(dk) < 0]
contributes an exponentially decaying term to the impulse
response (See Figure 4.13 (a)).
 A pole in the right half plane [Re(dk) > 0] contributes an
increasing exponential term to the impulse response (See
Figure 4.13(b)).
51
Figure 4.13: 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.
52
4.10.2 Stability :
 If the system is stable, the impulse response is absolutely
integrable.
 A pole of the system transfer function that locates in the right half
plane contributes a left sided decaying exponential term to the
impulse response (See Figure 4.14 (a)).
 A pole in the left half plane contributes a right-sided decaying
exponential term to the impulse response (See Figure 4.14 (b)).
53
Figure 4.14: 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.
54
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.
55
Reference Table.
56