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 and state variables are
also discussed.
• Finally, circuit stability and network
synthesis will be covered.
2
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
v t   L
di  t 
dt
• Taking the Laplace Transform:
 
i 0
1
I s  V s 
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.
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Capacitors
• For a Capacitor
i t   C
dv  t 
dt
• Taking the Laplace Transform:
1
V s 
I s 
sC
 
v 0
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.
8
Initial Conditions
• One significant advantage of the Laplace
transform is that it includes both stead-state
and initial conditions.
• This allows for obtaining both the steadystate 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
Inductor
V  s   RI  s 
V  s   sLI  s 
Capacitor
V s 
1
I s
sC
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Impedances
• We can define the impedance in the sdomain as:
Z s 
V  s
I s
• Thus, the three elements will have the
following impedances:
Resistor
Inductor
Capacitor
Z s  R
Z  s   sL
1
Z s 
sC
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Admittance
• In the s-domain, the admittance is also
simply the reciprocal of the impedance:
Y s 
I s
1

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 opamps 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.
13
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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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.
H s 
Y s
X s
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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 sdomain, we have
u (t )

1H
1
F
3


1
s
sL  s
1 3

sC s
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Apply mesh analysis, on solving for V0(s)
V0 (s) 
3
2
2 (s  4) 2  ( 2 ) 2
Taking the inverse
transform give
v0 (t ) 
3  4t
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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20
Example 3:
Find v0(t) in the circuit shown below. Assume v0(0)=5V
.
Ans: v0 (t )  (10et  15e2t )u(t ) V
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Example 4:
The switch shown below has been in position b for a
long time. It is moved to position a at t=0. Determine
v(t) for t > 0.
Ans: v(t )  (V0  I0 R)et /  I0 R, t  0, where  RC
*
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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.
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Solution:
Transform the circuit from time-domain (a) into sdomain (b) using Laplace Transform. On rearranging
the terms, we have
V1 
35
30

s 1 s  2
By taking the inverse transform, we get
v1 (t )  (35et  30e2t )u(t ) V
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Example 6:
The initial energy in the circuit below is zero at t=0. Assume that
vs=5u(t) V. (a) Find V0(s) using the thevenin theorem. (b) Apply the
initial- and final-value theorem to find v0(0) and v0(∞). (c) Obtain v0(t).
Ans: (a) V0(s) = 4(s+0.25)/(s(s+0.3)) (b) 4,3.333V, (c) (3.333+0.6667e0.3t)u(t) V.
*
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Transfer Functions II
• There are four possible transfer functions:
H  s   Voltage gain 
H  s   Current gain 
H  s   Impedance 
H  s   Admittance 
Vo  s 
Vi  s 
Io  s 
Ii  s 
Vo  s 
Ii  s 
Io  s 
Vi  s 
• It is possible for a circuit to have many
transfer functions.
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Transfer Functions III
• Each transfer function can be found in two
ways:
• The first is to assume any convenient input
X(s).
• Then use any circuit analysis technique
(current or voltage division, nodal analysis,
etc…)
• Find the output Y(s)
• Take the ratio.
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Example 7:
The output of a linear system is y(t)=10e-tcos4t when the input is
x(t)=e-tu(t). Find the transfer function of the system and its impulse
response.
Solution:
Transform y(t) and x(t) into s-domain and apply H(s)=Y(s)/X(s), we
get
Y ( s)
10(s  1) 2
4
H ( s) 


10

40
X (s) (s  1) 2  16
( s  1) 2  16
Apply inverse transform for H(s), we get
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.
Ans:  2e3t  4e6t , t  0; 2 (t)-12e-6t u(t )
*Refer to in-class illustration, textbook
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State Variables
• So far we have only looked at techniques for
analyzing system with only one input and
only one output.
• Many systems have more than one input or
output.
• The state variable method is a very important
tool for analyzing such highly complex
systems.
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State Variables
• In the state variable model, we specify a
collection of variables that describe the
internal behavior of the system.
• These variables are known as the state
variables of the system.
• They are variables that determine the future
behavior of a system.
• They are a reflection of the current state of
the system, without requiring the history of
the system.
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State Variables
• The standard way to represent a state
variable is to arrange them as a set of first
order differential equations.
x  Ax  Bz
• And the output equation is:
y  Cx  Dz
• The transfer equation is found by taking the
Laplace transform of the first equation.
H  s   C  sI  A  B
1
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State Variables
• There are three steps to applying the state variable
method:
1. Select the inductor current i and capacitor voltage v
as the state variables, making sure they are
consistent with the passive sign convention.
2. Apply KCL and KVL to the circuit and obtain circuit
variables (voltages and currents) in terms of the
state variables. This should lead to a set of firstorder differential equations necessary and sufficient
to determine all state variables.
3. Obtain the output equation and put the final result in
state-space representation.
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16.4 State Variables (2)
Example 9:
Obtain the state variable model for the circuit shown
below. Let R1=1, R2=2 ,C=0.5F and L=0.2H and
obtain the transfer function.
 1
v 
Ans     R1C
 i  1
 L
1 
 1 

v


C
     R1C  vs ,

 R2   i  
0 

L 
v 
v0  0 R2   
i 
; H(s) 
20
s 2  12s  30
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*Refer to in-class illustration, textbook
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Applications
• The Laplace transform certainly plays an
important role in circuit analysis, however
that is not the only area where it has
applications.
• It also has applications in signal processing
and control systems.
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