Fundamentals of
Electric Circuits
Chapter 15
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Overview
• This chapter introduces the Laplace
Transform.
• The properties of the transform will be
discussed.
• The inverse transform and convolution
integral will also be covered.
• Finally, the general approach to using the
Laplace transform in circuit analysis will be
introduced.
2
Laplace Transform
• We have seen that a powerful technique for
circuit analysis is to perform a time domain
to frequency domain transformation.
• Now we will look at a more generalized
approach to this, called the Laplace
transformation.
• It has the advantage of making solutions
simpler.
• It also is capable of providing the total
response of the circuit with both the natural
and forced responses.
3
Laplace Transform II
• Given a function f(t) its Laplace transform is:
L  f  t    F  s  


f  t  e  st dt
0
• Where s is a complex variable given by:
s    j
• This parameter has the dimensions of
frequency.
• The lower limit is denoted as 0- to include
any discontinuities present at t=0
4
Example 1
Determine the Laplace transform of each of the
following functions shown below:
5
Solution:
a) The Laplace Transform of unit step, u(t) is given by
Lu (t )  F ( s)  

0
1
1e dt 
s
 st
6
Solution:
b) The Laplace Transform of exponential function,
e-atu(t),a>0 is given by
Lu (t )  F ( s )  

0
1
e e dt 
s a
at  st
7
Solution:
c) The Laplace Transform of impulse function,
δ(t) is given by
Lu(t )  F ( s)    (t )e dt  1

 st
0
8
Laplace Transform III
• The Laplace transform is considered a
transformation from t-domain to s-domain.
• s can be seen as the complex frequency
domain.
• In circuit analysis, the differential equations
that describe the circuit behavior are
converted to algebraic equations in the sdomain
• The complementary function is the inverse
Laplace transform
  j
1
1
st
L  f  s    f  t  
F
s
e
ds



2 j 1  j
1
9
Linearity:
If F1(s) and F2(s) are, respectively, the Laplace
Transforms of f1(t) and f2(t)
La1 f1 (t )  a2 f 2 (t )  a1F1 (s)  a2 F2 (s)
Example:


s
 1 jt

 jt
Lcos(t )u(t )  L  e  e
u(t )  2
2
2
 s 
10
Scaling:
If F (s) is the Laplace Transforms of f (t), then
1
s
L f (at )  F ( )
a a
Example:
2
Lsin( 2t )u (t )  2
s  4 2
11
Time Shift:
If F (s) is the Laplace Transforms of f (t), then
L f (t  a)u(t  a)  e F (s)
 as
Example:
Lcos(  (t  a)) u (t  a)  e
 as
s
s2   2
12
Frequency Shift:
If F (s) is the Laplace Transforms of f (t), then

Le
 at

f (t )u(t )  F (s  a)
Example:

Le
 at

sa
cos(t )u (t ) 
( s  a) 2   2
13
Time Differentiation:
If F (s) is the Laplace Transforms of f (t), then the
Laplace Transform of its derivative is
 df

L  u (t )  sF ( s)  f (0 )
 dt

Example:

Lsin( ωt )u(t)  2
s 2
14
Time Integration:
If F (s) is the Laplace Transforms of f (t), then the
Laplace Transform of its integral is
t
1

L  f (t ) dt   F ( s )
 0
 s
Example:
 
n!
L t  n 1
s
n
15
Frequency Differentiation:
If F(s) is the Laplace Transforms of f (t), then the
derivative with respect to s, is
Ltf (t )  
dF ( s)
ds
Example:


1
L te u (t ) 
( s  a) 2
 at
16
Properties of the Transform II
• Frequency shift:
L e at f  t  u  t    F  s  a 
• Time differentiation:
 
L  f   t    sF  s   f 0
• Or
 
 
L  f   t    s 2 F  s   sf 0  f  0
• Or for the n’th derivative:
 d n f t  
n
n 1

n2


L

s
F
s

s
f
0

s
f
0




n
 dt 
 
 
 
 s 0 f  n 1 0
17
Properties of the Transform III
• Time integration:
t
 1
L   f  x  dx   F  s 
0
 s
• Frequency differentiation:
L tf  t    
dF  s 
ds
• Time periodicity: The function can be
considered to be composed of a sum of time
shifted functions
F1  s 
F s 
Ts
1 e
18
Initial and Final Values
• It is possible to get the initial value f(0) and
final value f() of a function directly from the
Laplace transform:
f  0   lim sF  s 
s 
f     lim sF  s 
s 0
19
Summary of Properties
20
Summary of Properties II
21
Transform Pairs
22
Inverse Laplace Transform
• Although it is always possible to use the
formula for the inverse Laplace transform, it
is preferable to use the table of transform
pairs.
• Suppose that F(s) has the form:
F s 
N s
D s
• Where N(s) and D(s) are both polynomials.
23
Inverse Transform
• The steps to finding the inverse transform
are:
1. Decompose F(s) into simple terms using
partial fraction expansion.
2. Find the inverse from the transform pairs.
• Next we will consider three possible forms
F(s) may take and apply these two steps to
them.
24
Simple Poles
• If F(s) has only simple poles, then it can be
expressed as:
N s
F s 
 s  p1  s  p2   s  pn 
• Where s=-p1, -p2,…,-pn are the simple poles.
• This can be expanded as:
F s 
k
k1
k
 2  3 
s  p1 s  p2 s  p3

kn
s  pn
25
Simple Poles II
• With
ki   s  pi  F  s  s  p
i
• Yielding:

f  t   k1e  p1t  k2e  p2t  k3e  p3t 

 k n e  pnt u  t 
26
Example 2
Find the inverse Laplace transform of
3
5
6
F ( s)  
 2
s s 1 s  4
Solution:
 3  1  5  1  6 
f (t )  L    L 
 L  2

s
 s 1
 s 4
 (3  5e t  3 sin(2t )u(t ), t  0
1
27
Repeated Poles
• If F(s) has n repeated poles at s=-p, then:
F s 
kn
 s  p
n

kn1
s  p

n 1

k2
s  p
2

k1
 F1  s 
s p
• Where F1(s) is the remaining part of F(s) that
does not have a pole at s=-p.
• The transform will be:
  pt
k3 2  pt
 pt
f  t    k1e  k2e  t e 

2!


kn
n 1  pt

t e  u  t   f1  t 

 n  1!

28
The Convolution Integral
• Convolution means folding.
• It is a valuable tool as it provides a means of
viewing and characterizing physical systems.
• For example it is used in finding the
response y(t) of a system to an excitation
x(t), knowing the impulse response h(t).
• This is done using the convolution integral:
y t  

 x    h t    d 
or
y t   x t  * h t 

29
The Convolution Integral II
• In the integral,  is a dummy variable.
• The asterisk denotes convolution.
• The convolution process is commutative,
meaning the order in which the two functions
are convolved does not matter.
• The convolution of two signals consists of
time-reversing one of the signals, shifting it,
and multiplying it point by point with the
second signal and integrating the product.
30
Simplified Integral
• If x(t)=0 for t<0, and if the system’s impulse
response is causal (h(t)=0 for t<0), then:
t
y t   h t  * x t    x    h t    d 
0
• Here are some properties:
1
x t  * h t   h t  * x t 
2
f  t  *  x  t   y  t   f  t  * x  t   f  t  * y t 
3
f  t  *  x  t  * y  t    f  t  * x  t  * y  t 
31
Integral Properties
• Here are some more properties:
f t  * t  

 f     t    d   f t 

f  t  *   t  t0   f  t  t 0 
f t  *  t  

 f      t    d   f  t 

f t  * u t  

t


 f    u t    d    f    d 
32
Relationship to Laplace
• The convolution integral and the Laplace
transformation can be related to each other:
• Given two functions f1(t) and f2(t) with
Laplace transforms F1(s) and F2(s).
t
f  t   f1  t  * f 2  t    f1    f 2  t    d 
0
• Taking the Laplace transform gives:
F  s   L  f1  t  * f 2  t    F1  s  F2  s 
33
• It is defined as

y(t )   x( )h(t   )d or y(t )  x(t ) * h(t )

• Given two functions, f1(t) and f2(t) with Laplace
Transforms F1(s) and F2(s), respectively
F1 (s) F2 (s)  L f1 (t ) * f 2 (t )
• Example:
y(t )  4et and h(t )  5e2t
 5  4 
t
 2t
h(t ) * x(t )  L1H (s) X (s)  L1 

  20(e  e ), t  0
 s  2  s  1 
34
t-domain vs. s-domain
• From this we can see that many cases of
convolution are simply multiplication in the
s-domain.
• But sometimes the product is too
complicated to do the inverse transform of.
• Or there may only be numerical data for the
input and no Laplace transform exists. In
these cases, the convolution is better to be
done in the time domain.
35
Steps to Evaluate
1. Folding: Take the mirror image of h() about
the ordinate axis to obtain h(-).
2. Displacement: Shift or delay h(-) by t to
obtain h(t-).
3. Multiplication: Find the product of h(t-) and
x().
4. Integration: For a given time t, calculate the
area under the product h(t-)x() for 0< < t
to get y(t) at t.
36
Application to
Integrodifferential Equations
• The Laplace transform is useful in solving
linear integrodifferential equations.
• Using the differential and integral properties
of Laplace transforms, each term in the
equation is transformed.
• Initial conditions are automatically taken into
account.
• The resulting algebraic equation is solved in
the s-domain.
37
Application to
Integrodifferential Equations
II
• The solution is then converted back to time
domain by using the inverse transform.
38
Example 3:
Use the Laplace transform to solve the differential
equation
d 2v(t )
dv(t )
6
 8v(t )  2u (t )
2
dt
dt
Given: v(0) = 1; v’(0) = -2
39
Solution:
Taking the Laplace transform of each term in the given
differential equation and obtain
s V (s)  sv(0)  v' (0) 6sV (s)  v(0)  8V (s)  2s
2
Substituting v(0)  1; v' (0)  2, we have
1
1
1
2 s 2  4s  2
4
2
( s  6 s  8)V ( s )  s  4  
 V (s)  
 4
s
s
s s2 s4
By theinverseLaplaceT ransform,
1
v(t )  (1  2e  2t  e  4t )u (t )
4
2
40