Fundamentals of

Electric Circuits
Chapter 15

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
 Laplace Transform
• 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

2
 The Laplace transform is an integral transformation of a function f (t )
from the time domain into the complex frequency domain, giving F (s).


L  f ( t )  = F ( s ) =  f ( t ) e − st dt
0−

Example 1
Determine the Laplace transform of each of the
following functions shown below:

3
Solution:
a) The Laplace Transform of unit step, u(t) is given by

Lu (t ) = F ( s) = 
 1
− st
1e dt =
0 s

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Solution:
b) The Laplace Transform of exponential function,
e-atu(t),a>0 is given by

Lu (t ) = F ( s) = 
 1
at − st
e e dt =
0 s +a

5
Solution:
c) The Laplace Transform of impulse function,
δ(t) is given by

Lu (t ) = F ( s ) =   (t )e − st dt = 1


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7
8
Summary of Properties

9
Summary of Properties II

10
Transform Pairs

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12
Inverse Laplace Transform

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 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:
N (s)
F (s) =
D (s)

• Where N(s) and D(s) are both polynomials.

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 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.

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 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:
k1 k k kn
F (s) = + 2 + 3 + +
s + p1 s + p2 s + p3 s + pn

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 Simple Poles II
• With
ki = ( s + pi ) F ( s ) s =− p
i

• Yielding:
(
f ( t ) = k1e − p1t + k2 e − p2t + k3e − p3t + )
+ kn e − pnt u ( t )

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Example 2
Find the inverse Laplace transform of

3 5 6
F (s) = − + 2
s s +1 s + 4
Solution:

−1  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
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 Repeated Poles
• If F(s) has n repeated poles at s=-p, then:
kn kn −1 k2 k1
F (s) = + + + + + F1 ( s )
(s + p) (s + p) (s + p)
n −1
n 2
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 kn 
f ( t ) =  k1e + k2e + t e +
− pt
+ t e  u ( t ) + f1 ( t )
n −1 − pt
 2! ( n − 1)! 
 

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21
Fundamentals of
Electric Circuits
Chapter 16

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
 Overview
• This chapter will apply the Laplace transform
to circuit analysis.
• The equivalent models for the resistor,
capacitor, and inductor will be introduced.
• Setting up proper initial conditions will be
covered.
• Transfer functions

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 Circuit Element Models
• We will now look at how to apply Laplace
transforms to circuit.
1. Transform the circuit from time domain to
the s-domain.
2. Solve the circuit using nodal analysis, mesh
analysis, source transformation,
superposition, or any circuit analysis
technique with which we are familiar.
3. Take the inverse transform of the solution
and thus obtain the solution in the time
domain.
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 Circuit Element Models II
• Only the first step in the process is new.
• As was done in phasor analysis, we
transform a circuit from time domain to the
frequency or s-domain.
• For a resistor, the voltage current
relationship in the time domain is:
v ( t ) = Ri ( t )

• Taking the Laplace transform:

V ( s ) = RI ( s )

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 Inductors
• For an inductor
di ( t )
v (t ) = L
dt

• Taking the Laplace Transform:

1
I (s) = V (s) +
i 0− ( )
sL s

• Note here that the initial conditions can be

represented either as a voltage or as a
current source
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 Inductors
• Here are the equivalent circuit elements
including the sources for the initial
conditions.

1
I (s) = V (s) +
( )
i 0−
sL s

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 Capacitors
• For a Capacitor
dv ( t )
i (t ) = C
dt

• Taking the Laplace Transform:

V (s) =
1
I (s) +
( )
v 0−
sC s

• Note here that the initial conditions can also

be represented either as a voltage or as a
current source

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 Capacitors
• Here are the equivalent circuit elements
including the sources for the initial
conditions.

V (s) =
1
I (s) +
( )
v 0−
sC s

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 Initial Conditions
• One significant advantage of the Laplace
transform is that it includes both steady-
state and initial conditions.
• This allows for obtaining both the steady-
state response as well as the transient
response.
• The s-domain equivalent elements can be
readily used in first and second order
circuits.

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 Zero Initial Conditions
• In many applications, we will
assume the initial conditions
are zero.
• The resistor of course stays
the same.
• For the inductor and
capacitor, the equations are
simplified:
Resistor V ( s ) = RI ( s )
Inductor V ( s ) = sLI ( s )
Capacitor 1
V (s) = I (s)
sC 31
 Impedances
• We can define the impedance
in the s-domain as:
V (s)
Z (s) =
I (s)

• Thus, the three elements will

have the following
impedances:
Resistor Z (s) = R
Inductor Z ( s ) = sL
Capacitor 1
Z (s) =
sC
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 Admittance
• In the s-domain, the admittance is also
simply the reciprocal of the impedance:
1 I (s)
Y (s) = =
Z (s) V (s)

• Use of the Laplace transform in circuit

analysis facilitates the use of various signals
sources, such as:
– Impulse
– Step
– Ramp
– Exponential
– Sinusoidal
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 Dependent Sources
• The models for dependent sources and op-
amps are easy to develop.
• Dependent sources have a linear output
based on the input.
• In the Laplace transform, multiplying the
time-domain equation by a constant results
in the same constant multiplied against the
Laplace transform.
• Op amps get treated like resistors; they
multiply a voltage times a constant.
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 Circuit Analysis
• Using the s-domain to analyze circuits is
relatively easy.
• Converting from time domain to s-domain
results in complicated equations based on
derivatives and integrals into simple
multiples of s and 1/s.
• One need only use algebra to set-up and
solve the circuit equations.
• All theorems and relationships developed for
DC work in the s-domain.
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Example 1:
Find v0(t) in the circuit shown below, assuming zero
initial conditions.

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Solution:
Transform the circuit from the time domain to the s-
domain, we have

1
u (t ) 
s
1H  sL = s
1 1 3
F  =
3 sC s

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Apply mesh analysis, on solving for V0(s)

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Apply mesh analysis, on solving for V0(s)

3 2
V0 ( s) =
2 ( s + 4) 2 + ( 2 ) 2

Taking the inverse

transform give

3 − 4t
v0 (t ) = e sin( 2t ) V, t  0
2

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Example 2:
Determine v0(t) in the circuit shown below, assuming
zero initial conditions.

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41
Example 3:
Find v0(t) in the circuit shown below. Assume v0(0)=5V
.

42
Ans : v 0 (t ) = (10 e − t + 15e −2t )u (t ) 43
V
 Circuit analysis
• Circuit analysis is again relatively easy to do when we are in the s-domain.
• We merely need to transform a complicated set of mathematical relationships in
the time domain into the s-domain where we convert operators (derivatives and
integrals) into simple multipliers of s and 1/s .
• This now allows us to use algebra to set up and solve our circuit equations.
• The exciting thing about this is that all of the circuit theorems and relationships
we developed for dc circuits are perfectly valid in the s-domain.

• Remember, equivalent circuits, with capacitors and inductors, only exist in the
s-domain;
• they cannot be transformed back into the time domain.

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Example 5:
Consider the circuit below. Find
the value of the voltage across
the capacitor assuming that the
value of vs(t)=10u(t) V and
assume that at t=0, -1A flows
through the inductor and +5 is
across the capacitor.

45
46
 Transfer Functions
• The transfer function is a key concept in signal
processing because it indicates how a signal is
processed as it passes through a network.
• The transfer function relates the output of a circuit to a
given input, assuming zero initial conditions.
• The transfer function H(s) is the ratio of the output response
Y(s) to the input excitation X(s), assuming all initial conditions
are zero.

Y (s)
H (s) =
X (s)

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 Transfer Functions II
• There are four possible transfer functions:
Vo ( s )
H ( s ) = Voltage gain =
Vi ( s )
Io ( s )
H ( s ) = Current gain =
Ii ( s )
Vo ( s )
H ( s ) = Impedance =
Ii ( s )
Io ( s )
H ( s ) = Admittance =
Vi ( s )

• It is possible for a circuit to have many

transfer functions.

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49
 Y ( s) 10( s + 1) 2 4
H ( s) = = = 10 − 40
X ( s) ( s + 1) 2 + 16 ( s + 1) 2 + 16

h(t ) = 10 (t ) − 40e−t sin( 4t )u(t )

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Example 8:
The transfer function of a linear system is

2s
H ( s) =
s+6

Find the output y(t) due to the input e-3tu(t) and its
impulse response.

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52
53
Poles and Zeros of H(s)

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Definition

◼H(s) can be expressed as the ratio of two

factored polynomials N(s)/D(s).
◼The roots of the denominator D(s) are called
poles and are plotted as Xs on the complex s-
plane.
◼The roots of the numerator N(s) are called zeros
and are plotted as os on the complex s-plane.

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56
 Example
N (s) (s + 5)[s − (−3 + j4)][s − (−3 − j4)]
H(s) = = .
D(s) s(s + 10)[s − (−6 + j8)][s − (−6− j8)]

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