Download Laplace Transforms I..

Applications of Laplace
Transforms
Instructor: Chia-Ming Tsai
Electronics Engineering
National Chiao Tung University
Hsinchu, Taiwan, R.O.C.
Contents
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Introduction
Circuit Element Models
Circuit Analysis
Transfer Functions
State Variables
Network Stability
Summary
Introduction
• To learn how easy it is to work with circuits
in the s domain
• To learn the concept of modeling circuits in
the s domain
• To learn the concept of transfer function in
the s domain
• To learn how to apply the state variable
method for analyzing linear systems with
multiple inputs and multiple outputs
• To learn how the Laplace transform can be
used in stability analysis
Circuit Element Models
• Steps in applying the Laplace transform:
– Transform the circuit from the time domain to the s
domain (a new step to be discussed later)
– Solve the circuit using circuit analysis technique
(nodal/mesh analysis, source transformation, etc.)
– Take the inverse Laplace transform of the solution
and thus obtain the solution in the time domain
s-Domain Models for R and L
For a resistor,
 v(t )  Ri (t )
Lv(t )  LRi (t )
 V ( s )  RI ( s )
Time domain
For an inductor,
di (t )
 v(t )  L
dt
 V ( s)  L sI ( s)  i (0  )  sLI ( s)  Li(0  )


1
i (0  )
or I ( s )  V ( s) 
sL
s
s domain
s domain
s-Domain Model for C
For a capacitor,
dv(t )
 i (t )  C
dt
 I ( s)  C sV ( s)  v(0  )  sCV ( s)  Cv(0  )


1
v (0  )
or V ( s ) 
I ( s) 
sC
s
Time domain
s domain
s domain
Summary
For inductor:
For capacitor:
V ( s )  sLI ( s )  Li (0  )
I ( s )  sCV ( s )  Cv(0  )
1
i (0  )
I ( s)  V ( s) 
sL
s
1
v (0  )
V (s) 
I ( s) 
sC
s
Summary
• Impedance in the s domain
– Z(s)=V(s)/I(s)
• Admittance in the s domain
– Y(s)=1/Z(s)=V(s)/I(s)
Element
Z(s)
Resistor
R
Inductor
sL
Capacitor
1/sC
*Assuming zero initial conditions
Example 1
(1) Transforma tion to the s domain :
1

u (t )  s

 Z (1 H)  sL  s
 1
1 3
Z ( 3 F)  sC  s

(2) Applying mesh analysis :
 1  3
3
 s  1  s  I1  s I 2  0



 3 I1   s  5  3  I 2  0
 s
s

3
 I2  3
s  8s 2  18s
3
 Vo ( s )  sI 2  2
s  8s  18
3
2

2 ( s  4) 2  ( 2 ) 2
3  4t
 vo (t ) 
e sin 2t t  0
2
Example 2
vo (0)  5 V
Example 2 (Cont’d)
Applying nodal analysis at node a :
10 ( s  1)  Vo
Vo Vo
 2  0.5  
10
10 10 s
25s  35
A
B
 Vo 


( s  1)( s  2) s  1 s  2
Applying the residue method,
A  ( s  1)Vo ( s ) |
 10
B  ( s  2)Vo ( s ) |
 15
s  1
s  2
 vo(t)  (10e t  15e  2t )u (t )
Example 3
io (0)  I 0
V0
I ( s )( R  sL )  LI 0   0
s
LI 0
V0
 I (s) 

R  sL s ( R  sL )
V0 R I 0  V0 R


s
sR L
V0  t / V0
R

 i (t )   I 0  e 
, 
R
R
L

Circuit Analysis
• Operators (derivatives and integrals) into
simple multipliers of s and 1/s
• Use algebra to solve the circuit equations
• All of the circuit theorems and relationships
developed for dc circuits are perfectly valid in
the s domain
Example 1
10
Vs 
s
i ( 0)  1 A
v ( 0)  5 V
V1  Vs V1  0 i (0) V1  v(0) s 



0
10 3
5s
s
1 (0.1s )
40  5s
35
30
 V1 


( s  1)( s  2) s  1 s  2


 v1 (t )  35e t  30e  2t u (t )
Example 2
Solved example 1 by using superposit ion :
30
30
30
V1 


( s  1)( s  2) s  1 s  2


 v1 (t )  30e t  30e  2t u (t )
10
10
10
V2 


( s  1)( s  2) s  1 s  2


 v2 (t )  10e t  10e  2t u (t )
5s
5
10
V3 


( s  1)( s  2) s  1 s  2


 v3 (t )   5e t  10e  2t u (t )
v(t )  v1 (t )  v2 (t )  v3 (t )

 35e

 (30  10  5)e t  (30  10  10)e  2t u (t )
t

 30e  2t u (t )
Example 3
Assume that no initial energy is stored.
(a)Find Vo(s) using Thevenin’s theorem.
(b)Find vo(0+) and vo() by apply the
initial- and final-value theorems.
(c) Determine vo(t).
=10u(t)
Example 3: (a)
 10  50
Voc  VTh  5  
 s  s
To find Z Th , use Z Th  Voc I sc
Applying nodal analysis :
10 (V1  2 I x )  0 V1  0
 

0
s
5
2s
V
and I x  1
2s
100
V1
50
 V1 
, Ix 

2s  3
2 s s (2 s  3)
V
50 s
Z Th  oc 
 2s  3
I sc 50 s (2 s  3)
Vo 
5
5
125
 50 
VTh 

 
5  Z Th
5  2 s  3  s  s ( s  4)
Example 3: (b), (c)
Solution (b) :
Solution (c) :
125
s ( s  4)
The initial - value theorem gives
125
vo (0)  lim sVo ( s )  lim
0
s 
s  s  4
The final - value theorem gives
125
A
B
Vo ( s ) 
 
s ( s  4) s s  4
Applying the residue method,
Vo ( s ) 
125 125
vo ()  lim sVo ( s )  lim

s 0
s 0 s  4
4
125

 A  sVo ( s )|s 0  4  31.25

125
 B  ( s  4)Vo ( s )| 
 31.25
s


4
4

 vo (t )  31.25(1  e  4t )u (t )
Transfer Functions
• 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) 
, Y ( s)  X ( s) H ( s)
X ( s)
If x(t )   (t ) ,  X ( s )  1
Thus Y ( s )  H ( s ) or y (t )  h(t )
It implies
h(t ) : the unit impulse response of the network

 H ( s ) : the Laplace transform of h(t )
Transfer Functions (Cont’d)
• Two ways to find H(s)
– Assume an input and find the output
– Assume an output and find the input (the ladder
method: Ohm’s law + KCL)
• Four kinds of transfer functions
Vo ( s)
I o ( s)
H ( s)  Voltage gain 
, H ( s)  Current gain 
Vi ( s)
I i ( s)
V ( s)
H ( s)  Impedance 
I ( s)
I ( s)
, H ( s)  Admittance 
V ( s)
Example 1
If y (t )  10e t cos 4t u (t ) when x(t )  e t u (t ).
Find the transfer function and its impulse response.
Solution :
1
10( s  1)
X (s) 
, Y ( s) 
2
2
s 1
( s  1)  4
Y (s)
10( s  1)
4
 H ( s) 

 10  40
2
2
2
2
X ( s) ( s  1)  4
( s  1)  4
2
 h(t )  10 (t )  40e sin 4t u (t )
t
Example 2
Find
H(s)=V0(s)/I0(s).
Solution :
By current division,
( s  4) I 0
I2 
( s  4)  (2  1 2s)
2( s  4) I 0
V0  2 I 2 
s  6  1 2s
V0 ( s )
4 s ( s  4)
 H ( s) 
 2
I 0 ( s ) 2 s  12 s  1
Example 2 (The Ladder Method)
Let V0  1 V,
By Ohm' s law, I 2  V0 2  1 2
1 
1 4s  1

V1  I 2  2    1 

2s 
4s
4s

V1
4s  1
 I1 

s4
4 s ( s  4)
Applying KCL gives
2 s 2  12 s  1
I 0  I1  I 2 
4 s ( s  4)
V0 1
4s ( s  4)
 H (s) 
  2
I 0 I 0 2s  12 s  1
Example 3
Find
(a) H(s) = Vo/Vi,
(b) the impulse response,
(c) the response when vi(t) = u(t) V,
(d) the response when vi(t) = 8cos2t V.
Example 3: (a), (b)
Solution :
By voltage division,
1
Vo 
Vab
s 1
1 || ( s  1)
Vab 
Vi
1  1 || ( s  1)
( s  1) ( s  2)

Vi
1  ( s  1) ( s  2)
s 1

Vi
2s  3
Vi
1 s 1
 Vo 
Vi 
s  1 2s  3
2s  3
Vo
1
1 1
 H ( s)  

Vi 2s  3 2 s  3 2
 h(t )  0.5e 3t 2u (t )
Example 3: (c), (d)
Sol : (c)
Sol : (d)
1
vi (t )  u (t )  Vi ( s ) 
s
Vo ( s )  H ( s )Vi ( s )
8s
vi (t )  8 cos 2t  Vi ( s )  2
s 4
Vo ( s )  H ( s )Vi ( s )
4s
A
Bs  C
A
B


 2

 
3
3 2
s 4

3
3 s

s

s

s

4
s


2 s s  
2
2
2

2

24
24
64
1
1
A , B , C 
A , B
25
25
25
3
3
1
24   1
s
4 2 
 3t 2

 vo (t )  1  e
u (t ) V  Vo ( s )  
 2

2
3
25  s  3 2 s  4 3 s  4 
24  3t 2
4

 vo (t )    e
 cos 2t  sin 2t u (t ) V
25 
3

1




State Variables
• The state variables are those variables which,
if known, allow all other system parameters to
be determined by using only algebraic
equations.
• In an electric circuit, the state variables are the
inductor current and the capacitor voltage
since they collectively describe the energy state
of the system.
State Variable Method
 z1 (t ) 
 z (t ) 
z(t )   2 
  


 zm (t )
The state equation can be arranged as
x  Ax  Bz
where
 x1 (t ) 
 x (t ) the state vector
x (t )   2   representi ng n
  

 state variable s
 xn (t )
 y1 (t ) 
 y (t ) 
2

y(t )  
  


)
t
(
y
 p 
 dx1 (t ) dt   x1 (t ) 
dx (t ) dt   x (t )
 2 
x   2
     

 

dxn (t ) dt   xn (t )
x  Ax  Bz
y  Cx  Dz
State Variable Method (Cont’d)
x  Ax  Bz
y  Cx  Dz
Assuming zero initial conditions
Y (s)
 H (s) 
 C ( sI  A) 1 B  D
Z (s)
 A  system matrix
 B  input coupling matrix

and applying the Laplace transform , where 
C  output matrix

sX( s )  AX( s )  BZ( s )
 D  feedforwad matrix
( sI  A)X( s )  BZ( s )
 X( s )  ( sI  A) 1 BZ( s )
I : the identity matrix
Y( s )  CX( s )  DZ(s)
 C(sI  A) 1 BZ( s )  DZ(s)
In most cases, D  0 .
So the degree of the numerator of
H ( s ) is less than the degree of the
denominato r of H ( s ).
 H ( s )  C ( sI  A) 1 B
How to Apply State Variable Method
• Steps to apply the state variable method to
circuit analysis:
– Select the inductor current i and capacitor voltage v
as the state variables (define vector x, z)
– Apply KCL and KVL to obtain a set of first-order
differential equations (find matrix A, B)
– Obtain the output equation and put the final result
in state-space representaion (find matrix C)
– H(s)=C(sI-A)-1B
Network Stability
• A circuit is stable if its impulse response h(t) is
bounded as t approaches ; it is unstable if h(t)
grows without bound as t approaches .
• Two requirements for stability
– Degree of N(s) < Degree of D(s)
N (s)
R( s)
n
n 1
If H ( s) 
 k n s  k n 1s      k1s  k0 
D( s )
D( s )
 lim h(t )  
t 
if n  1
– All the poles must lie in the left half of the s plane
e  pi t  e  ( i  ji )t  e  it  0 only if  i  0
Network Stability (Cont’d)
• A circuit is stable when all the poles of its
transfer function H(s) lie in the left half of
the s plane.
• Circuits composed of passive elements (R, L,
and C) and independent sources either are
stable or have poles with zero real parts.
• Active circuits or passive circuits with
controlled sources can supply energy, and
can be unstable.
Example 1
Vo
1 sC
H ( s) 

Vs R  sL  1 sC
1L
 2
s  s R L  1 LC
 p1, 2     2  02
R

  2 L
where 
  1
 0 LC
For R, L, C  0 :
   0 (stable)


For R  0 :
   0 (unstable)
Example 2
Applying mesh analysis gives
Find k for a stable circuit.

1 
I2

 Vi   R  sC  I1  sC  0




 R  1  I  I1  kI  0
2
1

sC 
sC

1 
1 

R


  I1 
V
 i
sC
sC


 
 
 0    k  1   R  1   I 2 
 
sC  
sC 
The determinan t is
2
1 
1 
1 

  R

k




sC  sC 
sC 

sR 2C  2 R  k

sC
Let   0 ,
the single pole is given as
k  2R
R 2C
For stable operation,
p
k  2R
p 2 0
RC
 k  2R
Summary
• The methodology of circuit analysis using
Laplace transform
– Convert each element to its s-domain model
– Obtain the s-domin solution
– Apply the inverse Laplace transform to obtain the
t-domain solution
Summary
• The transfer function H(s) of a network is the
Laplace transform of the impulse response h(t)
Y ( s)
H (s) 
, Y ( s)  X ( s) H ( s)
X ( s)
• A circuit is stable when all the poles of its
transfer function H(s) lie in the left half of the
s plane.