Transcript Chapter 13 – Circuit Analysis using Laplace Transforms

Chapter 13 EGR 272 – Circuit Theory II
Read: Ch. 13, Sect. 1-5, 7 in Electric Circuits, 9th Edition by Nilsson
Solving Differential Equations using Laplace Transforms
Laplace transforms can be used to transform differential equations into algebraic
equations. The idea is to avoid solving a differential equation by:
• taking a Laplace transform of the Differential Equation (D.E.) for x(t)
• solving for X(s) in the resulting algebraic equation
• finding x(t), which is the inverse Laplace transform of X(s)
This procedure is illustrated below:
Solve D.E.
Take
Laplace
Transform
of D.E.
Given:
D.E. for x(t)
Algebraic
equation
for X(s)
Solve
algebraic
equation
Find:
x(t) = ?
Take
Inverse
Laplace
Transform
X(s) =
1
Chapter 13
EGR 272 – Circuit Theory II
A Laplace transform of the differential equation (D.E.) is found by
making use of the time-differentiation property of Laplace transforms.
Recall:
d

L  f(t)   sF(s) - f(0)
 dt

 d2

2
L  2 f(t)   s F(s) - sf(0) - f '(0)
 dt

 d3

3
2
L  3 f(t)   s F(s) - s f(0) - sf '(0) - f ''(0)
 dt

etc.
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EGR 272 – Circuit Theory II
Example: Solve the following differential equation using Laplace transforms.
2
d x
dt
2
 3
dx
 2x(t)  e
dt
and x(0)  2,
dx(0)
dt
 -3
-t
for t  0
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Circuit Analysis using the Laplace-transformed circuit
Although Laplace transforms can be used effectively to solve differential
equations, this is not the preferred approach when analyzing circuits. Instead it
is easier to analyze the circuit directly in the s-domain.
Advantages to working directly in the s-domain include:
1) It eliminates the need to write or solve a differential equation
2) It eliminates the need for all initial conditions except initial capacitor
voltages and initial inductor currents (vC(0) and IL(0)) . Recall that the
differential equations approach would require finding N initial conditions (x(0),
x’(0), x’’(0), ….) for an Nth order circuit.
Representation of components in the Laplace-transformed Circuit
How do we draw a circuit in the s-domain? The circuit models for the sdomain can be developed by taking the Laplace transform of the time-domain
relationships, as shown on the following pages.
Chapter 13
EGR 272 – Circuit Theory II
Resistors - Develop the s-domain model
Inductors - Develop the s-domain model
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Chapter 13
EGR 272 – Circuit Theory II
Capacitors - Develop the s-domain model
Sources - Show the s-domain models for various types of sources
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Procedure: Circuit Analysis using the Laplace-transformed Circuit
1) Draw the time-domain circuit at t = 0-.
• Assume that the circuit is in steady state.
• Draw inductors as short circuits and capacitors as open circuits.
• Find vC(0-) and iL(0-).
• These values are needed for the s-domain circuit (step 2).
2) Draw the s-domain circuit for t > 0.
3) Analyze the circuit as you might analyze a DC circuit (using any circuit
analysis method). Recall that the s-domain impedances sL and 1/(sC) act
essentially like resistors. Determine the desired result in the s-domain (V(s),
I(s), etc).
4) Convert the result back to the time domain. In other words, use inverse
Laplace transforms to find v(t) or i(t) from V(s) or I(s).
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Example: Find v(t) for t  0 if v(0) = 20V and i(0) = 2A.
i(t)
5H
50 u(t) V
_+
30
1
F
125
+
v(t)
_
Chapter 13
EGR 272 – Circuit Theory II
Review: Methods of Circuit Analysis
Since the Laplace-transformed circuit (or s-domain circuit) can be
analyzed like a DC resistive circuit, any of the circuit analysis methods
covered in EGR 260 can be used, including:
• KVL and KCL
• Combining impedances in series and parallel
• Voltage and current division
• Source transformations
• Superposition
• Node and mesh equations
• Operational amplifiers
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Voltage Division
• Applies to series circuits only.
• Voltage divides proportionally between series impedances with the largest
impedance getting the most voltage.
• The general form for voltage division is:
V i (s)  VSource
 Z (s) 
(s )  i

Z
(s)

 eq
where Z eq (s)  Z 1 (s)  Z 2 (s)  ...  Z N (s)
+ Vi(s)
Z1(s)
VSource(s) +
_
Z2(s)
…
Zi(s)
_
.
..
ZN(s)
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Example: Use voltage division in the s-domain circuit to find v(t) for t  0.
t=0
25V
+
_
5H
50 
1
F
120
+
v(t)
_
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EGR 272 – Circuit Theory II
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Current Division
• Applies to parallel circuits only.
• Current divides between parallel impedances with the largest impedance
getting the least current.
• The general form for current division is:
 Z eq (s) 
I i (s)  I Source ( s ) 

Z
(s)
 i

where Z eq (s)  Z 1 (s) Z 2 (s) Z 3 (s)
... Z N (s)
…
…
Ii(s)
ISource(s)
Z1(s)
Z2(s)
… Zi(s)
…
…
…
ZN(s)
Note: Also discuss
the form for current
division with two
parallel impedances.
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Example: Use current division (after finding the source current) in the sdomain circuit to find i(t) for t  0.
t=0
12 V +
_
5
i(t)
0.5 H
1
F
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EGR 272 – Circuit Theory II
Chapter 13
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Source Transformations
A real voltage source is defined as a voltage source in series with an impedance.
A real current source is defined as a current source in parallel with an
impedance.
Recall that circuits can often be simplified by replacing a real voltage source
with a real current source (and vice versa) using the relationships shown below.
I(s)
+
_
Vs(s)
+
V(s)
_
Zs(s)
I(s)
Load
IP(s)
Real Voltage Source
a real voltage
source
to a real current source :
I P (s) 
V s (s)
Z s (s)
and
Z P (s)  Z s (s)
Load
Real Current Source
(subscript s indicates “series”)
Converting
+
ZP(s) V(s)
_
(subscript P indicates “parallel”)
Converting
a real current source
to a real voltage
source :
V s (s)  I P (s)Z P (s)
and
Z s (s)  Z P (s)
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Notes about source transformations:
1) All sources are not transformable. Note that a voltage source MUST have a
SERIES impedance to be transformable. Note that a current MUST have a
PARALLEL impedance to be transformable.
2) Dependent sources can be transformed also.
3) Transformed sources are equivalent in that they provide the same terminal
voltage and terminal current (V and I) to any connected load.
4) Transformed sources are not equivalent internally. For example, the current
through or the voltage across ZS and ZP is not the same. To assume that they
are the same is a common error.
Simple Example:
I(s)
I(s)
4s
20
s
+
_
Convert V-source to I-source
+
V(s)
_
5
s2
Convert I-source to V-source
4s
+
V(s)
_
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Chapter 13
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Example: Use source transformations to find i(t) for t  0.
t=0
4
12V
i(t)
2H
12
1/4 F
+
v(t)
_
8V
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Mesh Equations - Mesh equations are a set of simultaneous KVL equations.
Refer to the text or to EGR 271 notes for procedures and examples.
Example: Use mesh equations to find i(t) for t  0 if v(0) = 5V and iL (0) = 3A.
i(t)
1
10 u(t) V _+
iL
1H
+
1 F
4
V
_
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Node Equations - Node equations are a set of simultaneous KCL equations.
Refer to the text or to EGR 271 notes for procedures and examples.
Example: Use node equations to find v(t) for t  0.
2H
4
t=0
+
48 V _
12
+
1
4
F
1
v(t)
_
EGR 272 – Circuit Theory II
Chapter 13
Example: Find v(t) for t  0 if v(0) = 4V and i(0) = 2A.
2e-t A
i(t)
1
3
2H
1
F
2
+
v(t)
_
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EGR 272 – Circuit Theory II
Chapter 13
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Operational Amplifiers
Recall that operational amplifiers can be analyzed using the following rules:
• Assume that V+ = V- (the voltage is the same at the two inputs)
• Assume that I+ = I- = 0 (no input current enters the op amp)
• Use node equations
Example: Find Vo (DC circuit review).
R2
Vin
_
R1
Vo
+
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EGR 272 – Circuit Theory II
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Example: Find Vo using Laplace transforms (assume zero initial conditions)
0.05F
40
Vin(t)
_
10
0.1F
+
Vo(t)
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Transfer Functions (Network Functions)
A transfer function, H(s), can be used to describe a system or circuit in the sdomain in terms of its input and output as illustrated below.
Input = x(t)
Circuit
or
System
y(t) = Output
Time-domain representation of a circuit or system
Input = X(s)
H(s)
Y(s) = Output
s-domain representation of a circuit or system using a transfer function H(s)
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The transfer function H(s) is defined more specifically as:
H (s) 
Y (s)
X (s)
Notes:
• Transfer functions are always defined with zero initial condition (or zero
initial stored energy). Therefore, VC(0) = IL(0) = 0, so the voltage sources in
the models for capacitors and inductors disappear.
• Y(s) and X(s) typically represent voltages or currents.
• The input and the output must be designated by the user (there might be many
possible transfer functions for a given circuit or system). However, most
circuits or systems have well-defined inputs and outputs.
• H(s) completely characterizes the circuit or system. Once H(s) is known, you
can calculate circuit outputs for various inputs without ever seeing the circuit
or system again.
Chapter 13
EGR 272 – Circuit Theory II
Example: Find H(s) for the circuit shown below if: H (s)
24

V o (s)
V i (s)
Vi
_+
1 F
24
+
2H
12
Vo
_
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Chapter 13
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Example: Find H(s) for the circuit shown below if:
H (s) 
V o (s)
V i (s)
1H
1
Vi
+
_
1F
+
1
Vo
_
EGR 272 – Circuit Theory II
Chapter 13
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Finding the output, y(t) using H(s) and the input, x(t)
S ince H (s) 
Y (s)
w e can see that
X (s)
Y(s)  H(s)  X(s)  (transfer
y(t)  L
-1
H(s)
 X(s)
function)
 (Laplace transform
of the input)

So we can determine the output of a circuit by taking the inverse Laplace
transform of the product of the transfer function and the Laplace transform of
the input to the circuit.
Example: Find the output, y(t) for the circuit below if the input is x(t) = 25u(t)V.
x(t)
+
-
2s
(s + 1)
2
+
y(t)
-
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Example: Find the output, Vo, for the circuit below if the input is Vi = 10 u(t)V
using the transfer function H(s) found in a previous example.
Vi
_+
1 F
24
+
2H
12
Vo
_
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Unit Step Response and Impulse Response
Although H(s) can be used to find the output for any given input, there are two
special cases that are often of interest:
1) Impulse response, h(t) - the output when the input is (t)
2) Unit step response (USR) - the output when the input is u(t)
Impulse response, h(t) - the output to a circuit when the input x(t) = (t)
If x(t)   (t), then X(s)  1, so Y(s)  H(s)  1  H(s), so y(t)  L - 1H(s)
h(t)  impulse
response  L
-1
  h(t)
H(s) 
Unit step response (USR) - the output to a circuit when the input x(t) = u(t)
If x(t)  u(t), then X(s) 
1
, so Y(s) 
s
y(t)  USR  unit step response
H(s)
s
 H(s) 
L 

s


-1
Chapter 13
EGR 272 – Circuit Theory II
Example: For the circuit shown:
A) Find the transfer function H(s) = I(s)/V(s)
1
v(t)
+
_
i(t)
4
1F
4
2H
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Chapter 13
EGR 272 – Circuit Theory II
B) Use the transfer function to determine the output if v(t) = 10e-2t V
C) Use the transfer function to determine the impulse response, h(t)
D) Use the transfer function to determine the unit step response, USR
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