Transcript No Slide Title

APPLICATION OF THE LAPLACE TRANSFORM
TO CIRCUIT ANALYSIS
LEARNING GOALS
Laplace circuit solutions
Showing the usefulness of the Laplace transform
Circuit Element Models
Transforming circuits into the Laplace domain
Analysis Techniques
All standard analysis techniques, KVL, KCL, node,
loop analysis, Thevenin’s theorem are applicable
Transfer Function
The concept is revisited and given a formal meaning
Pole-Zero Plots/Bode Plots
Establishing the connection between them
Steady State Response
AC analysis revisited
LAPLACE CIRCUIT SOLUTIONS
We compare a conventional approach to solve differential equations with a
technique using the Laplace transform
Comple “Take Laplace transform” of the equation
mentary
di
(t )
dt
 di 
VS ( s )  RI ( s )  LL  
i  iC  i p
Initial conditions
 dt 
di
KVL : v S (t )  Ri (t )  L (t )
are automatically
di 

P
dt
L    sI ( s )  i (0)  sI ( s ) included
a
 dt 
Complementary equation
r
1
1
I
(
s
)



RI
(
s
)

LsI
(
s
)
diC
t
RiC (t )  L
(t )  0  iC (t )  K C e t i
s ( R  Ls)
s
dt
1/ L
K
K2
Only algebra
c I ( s) 
R
 1
t
t
RKC e  LKC (e )  0   
u
s ( R / L  s ) s s  R / L is needed
L
l
1
Particular solution for this case
a K1  sI ( s ) |s 0 
No need to
R
i p (t )  K p  v S  1  RK p
r
search for
1
R
particular
K 2  ( s  R / L) I ( s ) | s   R / L  
 t Use boundary conditions
1
L
R
i (t )   K C e
or complev S (t)  0 for t  0  i( 0 )  0
R
R
mentary

 t
1
R
L
 t
i (t )  1  e ; t  0
1 
solutions
L



i (t ) 
1 e
;t 0
R



R 

v S (t )  Ri (t )  L
L
LEARNING BY DOING
Find v (t ), t  0
v  vS
R
vS
RC
1
s
1
1 / RC
V ( s) 

s ( RCs  1) s ( s  1 / RC )
RCsV ( s)  V ( s) 
C
Model using KCL
In the Laplace domain the differential
equation is now an algebraic equation
C
dv
dt
dv v  v S

0
dt
R
dv
 v  vS
dt
V ( s) 
1 / RC
K
K2
 1
s ( s  1 / RC ) s s  1 / RC
K1  sV ( s) |s 0  1
 dv 
RCL    V ( s )  VS ( s )
 dt 
L
 dv 
L    sV ( s )  v (0)  sV ( s )
 dt 
v S (t )  0, t  0  v (0)  0 Initial condition
v S  u(t )  VS ( s) 
Use partial fractions to determine inverse
1
s
given in implicit
form
K 2  ( s  1 / RC )V ( s) |s 1 / RC  1
v (t )  1  e

t
RC ,
t 0
CIRCUIT ELEMENT MODELS
The method used so far follows the steps:
1. Write the differential equation model
2. Use Laplace transform to convert the model to an algebraic form
For a more efficient approach:
1. Develop s-domain models for circuit elements
2. Draw the “Laplace equivalent circuit” keeping the interconnections and replacing
the elements by their s-domain models
3. Analyze the Laplace equivalent circuit. All usual circuit tools are applicable and all
equations are algebraic.
Independent sources
v S ( t )  VS ( s )
Resistor
i S (t )  I S ( s)
Dependent sources
v D (t )  AiC (t )  VD ( s)  AIC ( s )
i D (t )  BvC (t )  I D ( s)  BVC ( s)
...
v (t )  Ri (t )  V ( s)  RI ( s)
Capacitor: Model 1
Source transformation
t
 I ( s)
L   i ( x )dx  
s
0

v (0)
I eq  s  Cv (0)
1
Cs
1t
1
v (0)
v (t )   i ( x )dx  v (0) V ( s) 
 Cs
I ( s) 
C0
Cs
s
Capacitor: Model 2
1t
v (t )   i ( x )dx  v (0)
C0
Impedance in series
with voltage source
I ( s)  CsV ( s)  Cv(0)
Impedance in parallel
with current source
Inductor Models
 di 
L    sI ( s )  i (0)
 dt 
v (t )  L
di
(t )  V ( s)  L( sI ( s)  i (0))
dt
V ( s) i (0)
I ( s) 

Ls
s
V ( s)  LsI ( s)  Li (0)
I ( s) 
V ( s) i (0)

Ls
s
Mutual Inductance
di1
di
(t )  M 2 (t )
dt
dt
di
di
v2 (t )  M 1 (t )  L2 2 (t )
dt
dt
v1 (t )  L1
V1 ( s)  L1sI1 ( s)  L1i1 (0)  MsI 2 ( s)  Mi 2 (0)
V2 ( s)  MsI 1 ( s)  Mi1 (0)  LsI 2 ( s)  Li2 (0)
Combine into a single source in the primary
Single source in the secondary
LEARNING BY DOING
Determine the model in the s-domain and the expression for
the voltage across the inductor
Steady state for t<0
i (0)  1A Inductor with
initial current
KVL : 1  (1  s ) I ( s )
Ohm's Law
V ( s)  1 I ( s)  V ( s)  
Equivalent circuit in
s-domain
1
s 1
ANALYSIS TECHNIQUES
All the analysis techniques are applicable in the s-domain
LEARNING EXAMPLE
I S ( s) 
3
s 1
Draw the s-domain equivalent and find the voltage in both
s-domain and time domain
One needs to determine the initial voltage
across the capacitor
i S (t )  0, t  0  vo (0)  0
RC  (10 103 )(25 106 )  0.25
120
K
K
 1  2
( s  4)( s  1) s  4 s  1
K1  ( s  4)Vo ( s) |s 4  40
Vo ( s ) 
1 

Vo ( s )   R ||  I S ( s )
Cs 

R
1/ C
3 103
Cs
Vo ( s) 
I ( s) 

1 S
s  1 / RC
s 1
R
Cs
K 2  ( s  1)Vo ( s) |s 1  40


vo (t )  40 e t  e 4t u(t )
LEARNING EXAMPLE
Write the loop equations in the s-domain
Do not increase
number of loops
Loop 1
VA ( s) 
R1I1 ( s ) 
v1 (0) v 2 (0)

 L1i1 (0) 
s
s
1
1
I1 ( s ) 
( I1 ( s )  I 2 ( s ))  L1s ( I1 ( s )  I 2 ( s ))
C1s
C2 s
Loop 2
L1i1 (0) 
L1s ( I 2 ( s )  I1 ( s )) 
v2 (0)
 L2i2 (0)  VB ( s) 
s
1
( I 2 ( s )  I1 ( s ))  ( L2 s  R2 ) I 2 ( s )
C2 s
LEARNING EXAMPLE
Write the node equations in the s-domain
Node V1
Do not increase number
of nodes
i1 (0)
i (0)
 C1v1 (0)  2

s
s


 1

1
1
 G1 

 C1s V1 ( s)  
 C1s V2 ( s)
L1s L2 s


 L2 s

Node V2
I A ( s) 
i2 (0)

s


1 
1 
 G2  C2 s  C1s 
V2 ( s)   C1s 
V1 ( s)
L2 s 
L2 s 


I B ( s)  C2v2 (0)  C1v1 (0) 
LEARNING EXAMPLE
Find vo (t ) using node analysis, loop analysis, superposition,
source transformation, Thevenin' s and Norton's theorem.
Assume all initial conditions are zero
V1 ( s )
KCL @ V1
1
2
12
s
V1 ( s) 
4
s  V1 ( s)  Vo ( s)  0  s
 
1
s
s
s
KCL@Vo
Could have
Node Analysis
Vo ( s ) Vo ( s )  V1 ( s )

 02
1
2
s
used voltage
divider here
4 s  12
 2s
s
 (1  s 2 )
 2 sV1 ( s)  (1  2 s)Vo ( s)  0
(1  s 2 )V1 ( s)  s 2Vo ( s) 
Vo ( s ) 
8( s  3)
(1  s ) 2
Loop Analysis
Loop 1
I1 ( s) 
4
s
Loop 2
1
12
s( I 2 ( s)  I1 ( s))  I 2 ( s)  2 I 2 ( s) 
s
s
4( s  3)
I 2 ( s) 
( s  1) 2
Vo ( s )  2 I 2 ( s ) 
8( s  3)
( s  1) 2
Source Superposition
Applying current source
I 2'
Current divider
Vo' ( s )  2 
s
4

1
2 2 s
s
Applying voltage source
Voltage divider
8( s  3)
Vo ( s )  V o ( s )  V o ( s ) 
( s  1) 2
'
"
Vo" ( s ) 
2
12

1
2  s s
s
Source Transformation
Combine the sources and use current
divider
Vo ( s)  2 
Vo ( s ) 
The resistance is redundant
s
 4 12 
  2
1
s   2 s s 
s
8( s  3)
( s  1) 2
Using Thevenin’s Theorem
Reduce this part
1
s2  1
ZTh   s 
s
s
Voltage
divider
VOC ( s) 
12
4 4 s  12
s 
s
s
s
Only independent sources
Vo ( s ) 
2
4 s  12
s2  1 s
2
s
Vo ( s ) 
8( s  3)
( s  1) 2
Using Norton’s Theorem
Reduce this part
ZTh  s
Current
division
s
4 s  12
2
1
s
s 2
s
8( s  3)
Vo ( s ) 
( s  1) 2
Vo ( s )  2 
4 12 / s 4 s  12
I SC ( s)  

s
s
s2
LEARNING EXAMPLE
Determine the voltage vo (t ). Assume all initial conditions to be zero
Selecting the analysis technique:
Transforming the circuit to s-domain
. Three loops, three non-reference nodes
. One voltage source between non-reference
nodes - supernode
. One current source. One loop current known
or supermesh
. If v_2 is known, v_o can be obtained with a
voltage divider
Doing the algebra : V1 ( s)  V2 ( s)  12 / s
I ( s )  V2 ( s ) / 2  6 / s
(1 / 2)( s  1)V2 ( s)  12 / s   2(V2 ( s) / 2  6 / s)
 V2 ( s) /( s  1)  0
12
s
V ( s) V ( s)
V ( s)
KCL@ supernode : 1  1  2 I ( s)  2
0
2
2/ s
s 1
V ( s)
Controlling variable : I ( s)   1
2
1
Voltage divider : V0 ( s) 
V2 ( s)
s 1
Supernode constraint : V2 ( s)  V1 ( s) 
V2 ( s ) 
12( s  1)( s  3)
s ( s 2  4 s  5)
Vo ( s ) 
12( s  3)
s ( s 2  4 s  5)
Continued ...
Compute Vo (s) using Thevenin' s theorem
-keep dependent source and controlling
variable in the same sub-circuit
-Make sub-circuit to be reduced as simple
as possible
-Try to leave a simple voltage divider after
reduction to Thevenin equivalent

VOC  12 / s
VOC  12 / s VOC  12 / s

 2I ' 0
2
2/ s
12
VOC  12 / s
V
(
s
)

OC
I ' 
s
2
 I '(2 I ' ) /( 2 / s)  2 I '  0  I '  0
12 / s

I " 6 / s
I SC  2 I " I "2 I " /( 2 / s )  0
6( s  3) Z  VOC ( s )  2
TH
I SC 
I SC ( s ) s  3
s
Vo ( s ) 
1
1 s 
2
s3

12
s
Continued …
Computing the inverse Laplace transform
Analysis in the s-domain has established that the Laplace transform of the
output voltage is
2
2
12( s  3)
s

4
s

5

(
s

2

j
1
)(
s

2

j
1
)

(
s

2
)
1
Vo ( s ) 
2
s ( s  4 s  5)
12( s  3)
Ko
K1
K1*
Vo ( s) 



s( s  2  j1)( s  2  j1)
s ( s  2  j1) ( s  2  j1)
K o  sVo ( s) |s 0
 36
5
K1
K1*

 2 | K1 | e t cos( t  K1 )u(t )
( s    j ) ( s    j )
12(1  j1)
12 245
One can also use
K  ( s  2  j1)V ( s) |


1
o
s  2  j1 (2  j1)( j 2)
5153.43(290)
quadratic factors...
 3.79  198.43  3.79161.57
 36

2t
12( s  3)
Co C1 ( s  2)
C2
v
(
t
)


7
.
59
e
cos(
t

161
.
57


 u( t )
o
Vo ( s ) 



2
2
2
 5

s ( s  2)  1 ( s  2)  1
s s  2  1


C1 ( s   )
C2 

 e t [C1 cos t  C 2 sin t ]u(t )
2
2
2
2
(s   )  
(s   )  
12( s  3)  Co (( s  2)2  1)  s[C1( s  2)  C2 ] s  2  12  Co  2C 2  C 2  36 / 10  6  12 / 5
Co  sVo ( s) |s 0  36 / 5
Equating coefficients of s2 : 0  Co  C1  C1  36 / 5
12
 36

vo ( t )   (1  e  2 t cos t )  e  2 t sin t  u(t )
5
5

LEARNING EXTENSION Find io (t ) using node equations
VS
Vo  VS
KCL at supernode
supernode
Vo
Assume zero initial conditions
Implicit circuit transformation to s-domain
K1
K1*

 2 | K1 | e t cos( t  K1 )u(t )
( s    j ) ( s    j )

1
15 
 I o ( s) | 1
K1   s   j
s   j
4
4 

4
 1
15 

1  6   j
4
4



15
15
4
2j
4
6.33  66.72
2 Vo ( s) Vo ( s)
K

 6.53  156.72
Cs(Vo ( s)  VS ( s))  

0
1
0.9790
s
Ls
2
12
V ( s)
VS ( s)  , I o ( s)  o
t



s
2
4 cos 15 t  156.72 
i
(
t
)

13
.
06
e
o
 4

Doing the algebra


1  6s
1  6s
I o ( s)  2

2
s  0.5s  1 
1  15
s   
4  16

K 1*
1  6s
K1
I o ( s) 



1
15 
1
15  
1
15  
1
15 
 s   j
 s   j
  s   j
  s   j

4
4
4
4
4
4
4
4


 
 

LEARNING EXTENSION Find vo (t ) using loop equations
I 2 ( s) 
K 0  sI 2 ( s ) |s 0  2
supermesh
constraint due to source
2
 I 2  I1
s
KVL on supermesh
K1  ( s  0.27) I 2 ( s ) |s  0.27 
K 2  ( s  3.73) I 2 ( s ) |s  3.73 
1
12
I1  2 I1  sI 2   2 I 2  0
s
s
Solve for I2
I 2 ( s) 
16 s  2
16 s  2

s ( s 2  4 s  1) s ( s  0.27)( s  3.73)
Determine inverse transform
K0
K1
K2


s s  0.27 s  3.73

16(0.27)  2
 2.48
(0.27)( 0.27  3.73)
16(3.73)  2
 4.47
(3.73)( 3.73  0.27)

i2 (t )  2  2.48e 0.27t  4.47e 3.73t u(t )
v o ( t )  2i2 ( t )
TRANSIENT CIRCUIT ANALYSIS USING LAPLACE TRANSFORM
For the study of transients, especially transients due to switching, it is important
to determine initial conditions. For this determination, one relies on the properties:
1. Voltage across capacitors cannot change discontinuously
2. Current through inductors cannot change discontinuously
LEARNING EXAMPLE
Determine vo (t ), t  0
i L( 0  )

vC (0  )

Assume steady state for t<0 and determine
voltage across capacitors and currents
through inductors
For DC case capacitors are open circuit
inductors are shortcircuit
vC (0)  1V , i L (0)  1A
Circuit for t>0
Vo ( s) 
2s  7
2 s 2  3s  2
Now determine the inverse transform
b2  4ac  0  complex conjugate roots
Laplace
K 1*
K1
Vo ( s ) 

3
7
3
7
s  j
s  j
4
4
4
4
Circuit for t>0

3
7
 2.14  76.5
Vo ( s )
K1   s   j
4
4 
3
7

s   j
4
K1
K1*

 2 | K1 | e t cos( t  K1 )u(t )
( s    j ) ( s    j )
Use mesh analysis
4
( s  1) I1  sI 2   1  s
s
Solve for I2
2
1
 sI1  ( s  1  ) I 2    1  ( s  1)
s
s
I 2 ( s) 
4
2s  1
2
1
V
(
s
)

I
(
s
)

o
2
s
s
2 s 2  3s  2
vo (t )  4.28 cos(
7
t  76.5)
4
LEARNING EXTENSION
Determine i1 (t ), t  0
Initial current through inductor
i L (0)  i L (0  )  1 A
12
6
I1 ( s )
I1 ( s) 
2s
2s
1

2 s  18 s
1
s
Current
divider
I1 ( s) 
s
 i1 (t )  e  9t u(t )
s9
LEARNING EXTENSION
Determine vo (t ), t  0
2s


8
V
3

Vo (s )
Determine initial current through inductor
i L (0)
Use source
superposition
i12V  2 A
Vo ( s ) 
2
 12 8 
    (voltage divider)
4  2s  s 3 
Vo ( s ) 
(8 s  36) K1 K 2


3 s ( s  2)
s s2
K1  sVo ( s ) |s 0  6
2
i4V   A
3
K 2  ( s  2)Vo ( s) |s 2  
4
i L (0)  A
3
8


v o ( t )   6  e  2 t  u( t )
3


10
3
TRANSFER FUNCTION
X (s)
System with all
initial conditions
set to zero
Y (s )
bn s n  ...  b1s  b0
H ( s) 
am s m  ...  a1s  a0
For the impulse function
Y ( s)
x (t )   (t )  X ( s )  1
X ( s)
H(s) can also be interpreted as the Laplace
If the model for the system is a differential transform of the output when the input is
H ( s) 
equation
dny
d n1 y
dy
bn n  bn1 n1  ...  b1  bo y
dt
dt
dt
dmx
d m 1 x
dx
 am m  am 1 m 1  ...  a1  ao x
dt
dt
dt
If all initial conditions are zero
d k y 
L  k   s kY (s )
 dt 
bn s nY ( s )  ...  b1sY ( s )  b0Y ( s )
 am s m X ( s )  ...  a1sX ( s )  a0 X ( s )
bn s n  ...  b1s  b0
Y ( s) 
X ( s)
m
am s  ...  a1s  a0
an impulse and all initial conditions are zero
The inverse transform of H(s) is also
called the impulse response of the system
If the impulse response is known then one
can determine the response of the system
to ANY other input
LEARNING EXAMPLE
A network has impulse response h(t )  e t u(t)
Determine the response, vo (t ), for the input vi (t )  10e 2t u(t )
In the Laplace domain, Y(s)=H(s)X(s)
Vo ( s)  H ( s)Vi ( s)
1
s 1
10
vi (t )  10e 2t u(t )  Vi ( s) 
s2
h(t )  e t u(t )  H ( s) 
Vo ( s ) 
10
K
K
 1  2
( s  1)( s  2)
s 1 s  2
K1  ( s  1)Vo ( s) |s 1  10
K 2  ( s  2)Vo ( s) |s 2  10


vo (t )  10 e t  e 2t u(t )
Impulse response of first and second order systems
t
Case 2 :   1 : Underdamped network

K

First order system H ( s ) 
 h(t )  Ke
s  1
poles : s1, 2   0  j 0 1   2
h(t )  Ke ot cos(o 1   2 t   )
Normalized second order system
02
H ( s)  2
s  20 s  02
poles : s1, 2   0   0  2  1
Case 1 :   1 : Overdamped network
h(t )  K1e ( 0  0
 2 1) t
 K 2e ( 0  0
 2 1) t
Case 3 :   1 : Critically damped network
h(t )  K1teot  K 2e ot
LEARNING EXAMPLE
Vo ( s )
Vi ( s )
a) C  8F  poles : s1, 2  0.25  j 0.25
Determine the transfer function H ( s ) 
Transform the circuit to the Laplace
domain. All initial conditions set to zero
b) C  16F  poles : s1, 2  0.25
c) C  32F  poles : s1, 2  0.427,  0.073
Vi (s )
Mesh analysis
Vi ( s )  2 I1  I 2
Vo ( s) 
1 

0   I1  1  s 
I2
sC 

(1 / 2C )
Vo ( s )  2
s  (1 / 2) s  1 / C
1
I 2 ( s)
sC
LEARNING EXAMPLE
Determine the transfer function, the type of damping and the
unit step response
Transform the circuit to the Laplace
domain. All initial conditions set to zero
V1 ( s)  VS ( s) V1 ( s) V1 ( s) V1 ( s)  V0 ( s)



0
1
1
1
1
s
1
Vo ( s )
32

VS ( s ) s 2  1 s  1
o2  0  0.25
2 16
2o    1
Unit step response  VS ( s) 
(1 / 32)
Vo ( s ) 
2

1
s
Ko
K11
K12


s s  0.25 ( s  0.25) 2
1

s s  
4

K o  sVo ( s ) |s 0  0.5
V  0
V1 ( s) Vo ( s)

0
1
V1 ( s )   sVo ( s )  1
s
K12  ( s  0.25)2Vo ( s) |s0.25  0.125

d s 2Vo ( s )
K11 
ds


 0.5
s  0.25

vo (t )  0.5  0.125t  0.5e 0.25t u(t )
Determine the pole-zero plot, the type of damping and the
unit step response
LEARNING EXTENSION
s  10
s 2  4s  8
zero : z  -10
poles :
H ( s) 
1
s  10
Y ( s)  H ( s) 
s s ( s 2  4 s  8)
s 2  4s  8  ( s  2  j 2)( s  2  j 2)
s 2  4 s  8  0  s1, 2  2  j 2
x
j
j2
2
O

 10
x
s  4s  8
2
2o
o2
 
2
2
K1
K2
K 2*
Y ( s) 


s s  2  j2 s  2  j2
K1
K1*

 2 | K1 | e t cos( t  K1 )u(t )
( s    j ) ( s    j )
10
K1  sY ( s) |s 0 
8
8  j2
K 2  ( s  2  j 2)Vo ( s ) |s  2 j 2 
(2  j 2)( j 4)
K2 
8.2514
 0.73  211
2.83135  490
 10

vo ( t )    1.46 cos( 2t  211) 
8

Second order networks: variation of poles with damping ratio
Normalized second order system
02
H ( s)  2
s  20 s  02
poles : s1, 2   0   0  2  1
Case 2 :   1 : Underdamped network
poles : s1, 2   0  j 0 1   2
  cos
LEARNING EXAMPLE
o2 
1
,
LC
2o 
R
L
1
V ( s)
LC

Gv ( s)  o

1
 R
Vin ( s) 1  Ls  R
s2   s 
Cs
LC
 L
1
Cs
Variation of poles.
Use  o  2000
LEARNING EXAMPLE
The Tacoma Narrows Bridge Revisited
Previously the event
was modeled as a
resonance problem.
More detailed studies
show that a model
with a wind-dependent
damping ratio provides
a better explanation
d 2
d
2

2



o
o  0
2
dt
dt
  0.0046  0.00013U
U  wind speed (mph)
Torsional Resonance
Model
Conditions at failure
wind speed
twist
 42mph
 12
time to collapse  45min
Problem: Develop a circuit that models this event
integrator
model d 2
d
2

2



 0
o
o
2
dt
dt
..
.
  2o   o2

..
adder
.
 0    (2o   o2 )

  (0.001156  00013U )  1.579
Using numerical values
vi
Vi ( s ) V ( s )

0
1
R
Cs
1
V ( s )  
Vi ( s )
R C s
 
v1
Simulation
building
blocks
v2
V1 V2 V


0
R1 R2 R f
Rf 
 Rf
V  
V1 
V2 
R
R
 1

2
d 2
 2
dt
Simulation using
dependent sources

d
dt

Simulation results
Wind speed=20mph
initial torsion=1 degree
Wind speed=35mph
initial torsion =1degree
Wind speed=42mph
initial torsion=1 degree
POLE-ZERO PLOT/BODE PLOT CONNECTION
Bode plots display magnitude and phase information of G ( s) |s  j
They show a cross section of G(s)
s2
G ( s)  2
s  2s  5
1
V ( s)
LC
G ( s)  o

1
Vin ( s )
 R
s2   s 
 L  LC
If the poles get closer to
imaginary axis the peaks
and valleys are more
pronounced
Cross section
shown by Bode
Cross section
Due to symmetry
show only positive
frequencies
Front view
Amplitude Bode plot
Uses log scales
STEADY STATE RESPONSE
Y ( s)  H ( s) X ( s)
Response when all initial conditions are zero
Laplace uses positive time functions. Even for sinusoids the response contains
transitory terms
EXAMPLE H ( s)  1 , X ( s) 
s 1
s
( x (t )  [cos  t ]u(t ))
s2   2
s
K1
K2
K 2*
Y ( s) 



( s  1)( s  j )( s  j ) s  1 s  j s  j


y(t )  Ket  2 | K 2 | cos( t  K 2 ) u(t )
transient
Steady state response
For the general case
X M cos t u(t ) 

If interested in the steady state response
only, then don’t determine residues
associated with transient terms
If x (t )  X M cos(o t   )u(t )
yss (t ) | X M | H ( jo ) | cos(o t  H ( jo )   )

X M jt
1 XM
XM 

e  e  jt  X ( s)  

2
2  s  jo s  jo 
K *x
1  X M
X M 
Kx
 
Y ( s )  H ( s )  


 transient terms
2
s

j

s

j

s

j

s

j

o
o 
o
o
 
y (t )  2 | K x | cos(o t  K 2 )  transient terms
1
K x  ( s  jo )Y ( s) |s  jo  X M H ( jo )
2
yss (t )  X M | H ( jo ) | cos( o t  H ( jo ))
LEARNING EXAMPLE
Determine the steady state response
If x (t )  X M cos(o t   )u(t )
yss (t ) | X M | H ( jo ) | cos(o t  H ( jo )   )
 o  2, X M  10
s2
s2
Vo ( s )  2
Vi ( s )  H ( s )  2
3s  4 s  4
3s  4 s  4
Transform the circuit to the Laplace domain.
Assume all initial conditions are zero
( j 2)2
H ( j 2) 
 0.35445
3( j 2)2  4( j 2)  4
 ys (t )  3.54 cos( 2t  45)V
V1  Vi V1 V1
 
0
2
2
2
1
s
1
Voltage divider : Vo 
V1
2
1
s
KCL@V1 :
LEARNING EXTENSION
Determine voss (t ), t  0
If x (t )  X M cos(o t   )u(t )
yss (t ) | X M | H ( jo ) | cos(o t  H ( jo )   )
 o  2, X M  12
Transform circuit to Laplace domain.
Assume all initial conditions are zero
Thevenin
Vi (s )
s
1
s
1
1
VOC ( s )  s Vi ( s ) 
Vi ( s )
1
s

1
1
s
1
1
s2  s  1
ZTh ( s )  s  || 1, || s 

s
s 1
s 1
Vo ( s ) 
2
VOC ( s )
2  ZTh ( s )
Vo ( s) 
2
1

Vi ( s)
s2  s  1 s  1
2
s 1
2
Vi ( s)
2
s  3s  3
H (s)
2
2
2
H ( j 2) 


 4  6 j  3  1  6 j 6.0899.46
Vo ( s) 
voss (t )  12 
2
cos(2t  99.46)
6.08
LEARNING BY APPLICATION
De-emphasize bass
Pole-zero map for GvR (s )
Enhances bass to original level
RIAA recording filter
K (1  s z1 )(1  s z 2 )
GvR ( s) 
(1  s p )
 z1  75s
| KG( s) |s  j 2 1000 1
 z 2  3180 s
 p  318s
zeros :  z1  13.33kr / s [2.12kHz ],
 z1  313.46r / s[50 Hz ]
pole : p  3,1346r / s[500 Hz ]
The playback filter is the reciprocal
Gvp ( s) 
1
Grp ( s)
Pole/zero of playback filter cancels
pole/zero of recording filter
LEARNING BY DESIGN
Filtering noise in a data transmission line
Data bits at 1000bps
Noise source is 100kHz
SOLUTION: Insert a second order low-pass
filter in the path. Should not affect data
signal and should attenuate noise

 R 
1

  3 
R
R
R
C
C
Vo ( s)
 1  2 3 1 2 

Vdata ( s)
 1
1
1 
1
 
s 2  s


 R1C1 R2C1 R3C1  R2 R3C1C 2
Proposed filter
V1
s 2  2o s  o2
V  0
R1  R2  R3  o 
V1  Vdata
V1
V V V

 1  1 o 0
R1
(1 / C1s ) R2
R3
V1
Vo

0
R2 (1 / C 2 s )
Design equations
1
3 C2
, 
R C1C2
2 C1
Well below 100k
above 1k
Filter design
criteria
 1
Below 100k
Selected pole location or filter
  2 105
Select R  40k,  0  25,000,   2.
Use design equations and determine
C1  0.75nF , C 2  1.33nF
The filter eliminates noise
but smooths data pulse
noise
25kbps
data transfer
rate
Filtered signal
is useless
Circuit simulating the filter
REDESIGN!
New pole-zero selection
 1
Design equations
 0  150,000 
 1
Simulation for 25kbps
3 C2
2 C1
1
40,000 C1C 2
LEARNING EXAMPLE
THIS FILTER EXHIBITS TOO MUCH OSCILLATION TO PULSE INPUTS
Must be made critically damped
R ||
1
Cs
Analysis of existing circuit
  0.25   R  (for no change in  )
O
Decrease resistance by a factor of four
Proposed Solution
Design equation
DESIGN EXAMPLE
A 10kHz Wein Bridge Oscillator

1
3

Ideal OpAmp
Output should be a sinusoid of a frequency such that
Select C=100nF then
APPLICATION
R=1.59k
LAPLACE
 R  2R
2
1