Transcript Chapter 7-Laplace

Chapter 7 The Laplace Transform
Consider the following RC circuit ( System)
R


x (t )

C
y (t )

System analysis in the time domain involves ( finding y(t)):
Solving the differential equation
RC dy (t )  y (t )  x (t )
dt
OR
Using the convolution integral
y(t) = x (t )h (t )
Both Techniques can results in tedious ( ‫) ممل‬mathematical operation
R


C
x (t )
y (t )


dy
(
t
)
RC
 y (t )  x (t )
dt
Fourier Transform provided an alternative approach
Differential Equation
RC dy (t )  y (t )  x (t )
dt
Algebraic Equation
RC( j) Y ()  Y ()  X ()
solve for Y ( )
y (t )
Inverse Back
Y () 
X ()
( j RC)  1
RC dy (t )  y (t )  x (t )
dt
R

C
x (t )

Fourier Transform both side
y (t )
RC ( j ) Y ( )  Y ()  X ()


Solve for Y ( )
y (t )
Inverse Back
Y ( ) 
X ( )
( j RC)  1
Unfortunately , there are many signals of interest that arise in system
analysis for which the Fourier Transform does not exists
(not absolutley integrable)



x(t ) dt  
Example x(t )  t , x(t )  t 2 , x(t )  et
In fact signals such as
x(t )  1  2 ( )
1
j
x(t )  cos (ot )   [ (  o )+ (  o )]
x(t )  u (t )   ( ) +
are not strictly integrable and their Fourier transforms
all contain some non  conventional function such as  ( ).
A more general transform is needed
Suppose we have a function or signal x  t  that is not absolutely integrable as shown
Now multiply x(t) by e t
The signal x(t)e t is absolutley integrable
The signal x(t)e t has Fourier Transform
Fourier Transform pairs was defined as


x(t)  1 X ()e jt d  X ()  x(t)e  jt dt
2 



Now multiply x (t ) by e t and takes the Fourier Transform



F.T e x(t)   x(t)e e


 t
 t
 jt






Let us deined this (  j) as
(  j )t

x
(
t
)
e
dt  X (  j )
dt
s
Fourier Transform of e t x(t )
The domain is not  but   j
  j Complex Frequency
 t
 st
F.T e x(t)   x(t)e dt  X (s )


Lb  x(t) 
0
were
Lb  
Denotes the operation of obtaining the Bilateral Laplace Transform
7.1 DEFINITIONS OF LAPLACE TRANSFORMS
The Laplace transform of a time function is given by the integral
This definition is called the bilateral, or two-sided, Laplace transform —hence, the subscript b
Notice that the bilateral Laplace transform integral becomes the Fourier transform integral
if s is replaced by j (i.e,   0)
The Laplace transform variable is complex
This show that the bilateral Laplace transform of a signal can be interpreted as the
Fourier transform of that signal multiplied by an exponential function e  t
The inverse Laplace transform
where
For most application , f (t) is zero for t  0
This transform is usually called, simply, the Laplace transform , the subscript b dropped
Laplace transform pairs
Unilateral (single sided) Laplace transform pairs
Forward transform
Inverse transform
Complex domain
Because of the difficulty in evaluating the complex inversion integral
Simpler procedure to find the inverse of Laplace Transform (i.e f(t) ) will be
presented later
Using Properties and table of known transform ( similar to Fourier)
Laplace Transform for an impulse (t)


L[  (t ) ]   (t )e dt
st
0
Assuming the lower limit is 0

L[  (t ) ]   (t )e dt  e

0
The value of s make no difference
st
s t
1
t 0

j
s domain
Region of Convergence
Region of Convergence
0

We also can drive it from the shifting in time property that will be discussed later

1 
1
lim e  (  j  )t  1  lim e  t e  j t  1


s  t 
s  t 
For this to converge then
lim e  t  0
t 
  >0
 Re(s) > 0
region of convergence (ROC)
Pole is were the Laplace Transform Function F (s)  
We derive the Laplace transform of the exponential function
A short table of Laplace transforms is constructed from Examples 7.1 and 7.2
and is given as Table 7.2
We know derive the Laplace Transform (LT) for a cosine function
Since
Proof


 (s+a)t

f
(
t
)e
dt
L [e f (t )]   [e f (t )]e dt

 at
 at
 st
0
0

L [e f (t )]   f (t )e
 at
Let s*= s+a
0
L[ cos0t ] 
L[ sin0t ] 
s
s 2 02
0
s 2 02
 s*t
dt
 F (s *)  F (s  a )
Example
L[ cos0t ] =L  e j 0t  e  j 0t  = 1 L[e j  t ] + 1 L[e  j  t ]
2

2
2
2


L[ e tu (t ) ] = 1
Since
(  s )
1
Then
L[e j  t ] =
0
1
1
( j 0  s )
0
L[e  j  t ] =
0
0
1
( j 0  s )
1
1
1
1
 2s 2
L[ cos0t ] 

s 0
2 (s  j 0 ) 2 (s  j 0 )
Similarly
0
L[ sin0t ]  2 2
s 0
Real Shifting
Let
f (t )  5e 0.3t
5
s  0.3
f (t )  5e 0.3t
we need to put the function f (t ) in a delay form f (t  t )u (t  t ) in order to use
2
the Laplace Transform for time shift
0
0
It is sometimes necessary to construct complex waveforms from simpler waveforms
(See example 7.3)
Since
(Shift in time property)
Transform of Derivatives
df (t ) 


L
 sF (s )  f (0 )

d
t


Proof
df (t ) 

L


dt


Integrating by parts,

b

a
udv  uv
u e


0
st
du se
b
a
b

 vdu
a
st
 df (t ) e dt
 dt 
dv (t ) df (t )
st
v (t )  f (t )
b

udv  uv

 vdu
a
du se
dv (t ) df (t )
st
df (t ) 

L


dt


0
 s ()


 e
b
a
a
u e
b


0
f () e
st
st
v (t )  f (t )


 df (t ) e dt  e st f (t ) + s f (t )e st dt

 dt 
0

0
1
s (0)
 



f (0 ) + sF (s )  sF (s )  f (0)

df
(
t
)

 sF (s )  f (0 )
dt
i (t )  2e u (t )
2t
df (t )  sF (s )  f (0)
dt
2
d f (t )  s 2F (s )  s f (t )  df (t )
2
dt
t =0
dt
t =0
2
d 3f (t )  s 3F (s )  s 2 f (t )  s dx (t )
dt
t =0
dt 3

t =0
d f (t )
2
dt
t =0
n
d f (t )  s n F (s )  s n 1 f (t ) 
n
t =0
dt
2
n  2 df (t )
n 3 d f (t )
s
s

2
dt t =0
dt t =0
n 1
 d fn (1t )
dt
t =0
Proof
find the Laplace transforms of the following integrals:
Proof
(See the book)
Since
7.6 Response of LTI Systems
Transfer Functions
Consider the following circuit
L

input
x (t )


i (t )
R
y (t )
output

C
We want a relation (an equation) between the input x(t) and output y(t)
t
KVL
1
di
(
t
)
x (t )  L
 Ri (t ) +
dt
C

i (t ')dt '

2
dx (t )
di (t ) i (t )
di
(
t
)
L
R
+
2
dt
dt
C
dt
L

input
x (t )


R
i (t )
y (t )
output

C
2
dx (t )
di
(t )  R di (t ) + i (t )
L
dt
dt
C
dt 2
Since i (t ) 
y (t )
R
dx (t ) L dy 2 (t )
dy (t ) y (t )



R
+
2
dt
R dt
dt
RC
Writing the differential equation as
2
dx (t )
dy
(t )  RC dy (t ) + y (t )
RC
 LC
dt
dt
dt 2
L

input
x (t )


i (t )
R
y (t )
output

C
2
dx (t )
dy
(t )  {RC }dy (t ) + {1}y (t )
{RC }
 {LC }
dt
dt
dt 2
Real coefficients, non negative which results from system components
R, L, C
In general,
dy n (t )
dy n 1(t )
an
+ a n 1
+
n
n 1
dt
dt
dx m (t )
dx m 1(t )
+ a0 y (t )  b m
+ b m 1
+
m
m 1
dt
dt
were a n 's , b m 's are real, non negative which results from system components
+ b 0 y (t )
R, L, C
Now if we take the Laplace Transform of both side (Assuming Zero initial Conditions)
a n s nY (s ) + a n 1s n 1Y (s ) +
+ a0Y (s )  b m s m X (s ) + b m 1s m 1 X (s ) +
We now define the transfer function H(s) ,
H (s )
Y (s )
X (s )

b m s m + b m 1s m 1 +
a n s n + a n 1s n 1 +
all initial conditions are zero
+ b0
+ a0
+ b 0 X (s )
dy n (t )
dy n 1(t )
an
+ a n 1
+
n
n 1
dt
dt
H (s )
Since a n 's , b m 's
dx m (t )
dx m 1(t )
+ a0 y (t )  b m
+ b m 1
+
m
m 1
dt
dt
Y (s )

X (s )
b m s m + b m 1s m 1 +
a n s n + a n 1s n 1 +
+ b0
+ a0
+ b 0 x (t )
N (s )
D (s )
are real, non negative
The roots of the polynomials N(s) , D(s) are either real or
occur in complex conjugate
The roots of N(s) are referred to as the zero of H(s) ( H(s) = 0 )
The roots of D(s) are referred to as the pole of H(s) ( H(s) = ± ∞ )
H (s )
b m s m + b m 1s m 1 +
Y (s )

X (s )
a n s n + a n 1s n 1 +
+ b0
+ a0
N (s )
D (s )
The Degree of N(s) ( which is related to input) must be less than or
Equal of D(s) ( which is related to output) for the system to be
Bounded-input, bounded-output (BIBO)
Example :
H (s ) 
4s 3 + 2s 2  s 1
s 2  6s  8
Using polynomial division , we obtain
H (s ) 
4s + 2 +
19s 17
s 2  6s  8
Now assume the input x(t) = u(t) (bounded input)  X (s )  1s
Y (s )  X (s )H (s ) 
y (t ) 
4+
4 (t ) + 2 +
unbounded ( )
2 1
+
s s
L
1 







19s 17 
s 2  6s  8 
19s 17 
s (s 2  6s  8) 
We see that for finite bounded
Input (i.e x(t) =u(t) )
We get an infinite (unbounded)
output
m  n for BIBO
H (s )
Y (s )

X (s )
b m s m + b m 1s m 1 +
a n s n + a n 1s n 1 +
+ b0
+ a0
N (s )
D (s )
The poles of H(s) must have real parts which are negative
The poles must lie in the left half of the s-plan
Proof (See the book)
( It is very similar to the Fourier Transform Property )
The loop equation (KVL) for this circuit is given by
If v(t) = 12u(t) ,find i(t) ?
v(t) = 12u(t)
Now using the Laplace Table 7.2 to find i(t)
The inverse transform, from Table 7.2
v (t )  Ri (t )  V (s )  RI (s )
Z R R 
v L (t )  L di (t )  V L (s )  L[sI (s ) i (0)]
dt
 sLI (s )  Li (0)
Z L  sL 
)
i
(
0
I (s )  V L (s )  s
sL
1
Source Transformation
i (t ) C dv C (t )  I (s )  C [sV C (s ) v (0)]
dt
 sCV C (s ) Cv (0)
ZC 
1
sC

)
v
(0
1
V C (s ) 
I (s ) 
sC
S
Source Transformation
Example Using Laplace Transform Method find i(t)
t 0


L 1 H
10


C 5 F
8
R 1 
i (t )
x (t )  0.5 cos t  2.5 sin t u (t )
x (t )  X (s ) 
i L (0 )  0 V
0.5s 2.5s
 2
2
s 1 s 1
v C (0 )   2 V
L  1 H  ZL  s 
10
10
C  5 F  ZC  8 
5s
8
s
10
R 1   ZR  1 
0.5s 2.5s
X (s )  2  2
s 1 s 1


8
5s
I (s )
2
s
 
1
s
10
X (s ) 
KVL
0.5s 2.5s
 2
2
s 1 s 1


8
5s
I (s )
s 
8


X (s )    I (s )  
 10 
 5s
2
s
 
1
 I (s )  2  (1)I (s )  0

s

s 8 
2
0.5
s
2.5
s
 2  2    1 I (s ) 
s
s 1 s 1  10 5s 
2
15
s
 25s  20
I (s )  2
(s 1)(s 2 10s 16)
s
10
X (s ) 
0.5s 2.5s
 2
2
s 1 s 1


8
5s
I (s )
2
s
 
1
2
15
s
 25s  20
I (s )  2
(s 1)(s 2 10s 16)
15s 2  25s  20
(15s 2  25s  20)
I (s )  2

2
(s 1)(s 10s 16) (s  j )(s  j )(s  2)(s 8)
( Imaginary Roots)
A1
A2
A3
A4




(s  j ) (s  j ) (s  2) (s 8)
i (t )   cos(t ) sin(t ) e 2t e 8t u (t )
Inversion of Rational Function ( Inverse Laplace Transform)
Let Y(s) be Laplace Transform of some function y(t) .
We want to find y(t) without using the inversion formula .
We want to find y(t) using the Laplace Transform known table
and properties
Objective : Put Y(s) in a form or a sum of forms that we know it is
in the Laplace Transform Table
Y(s) in general is a ratio of two polynomials
2  2s
s
Example Y (s ) = 3
2s 5s 2  3s  2
Rational Function
When the degree of the numerator of rational function is less the
Degree of the dominator
Proper Rational Function
Highest Degree is 2
2  2s
s
Example Y (s ) = 3
2s 5s 2  3s  2
Highest Degree is 3
Examples of proper rational Functions
1
Y 1(s ) =
s 1
2  6s  6
2
s
Y 2(s ) =
(s  2)(s 2  2s 2  2)
Examples of not proper rational Functions
s

2
Y 3(s ) =
s 1
However we can obtain a proper rational Function through long division
Y 3(s ) = s  2 = 1 + 1
s 1
s 1
We will discuss different techniques of factoring Y(s) into simple
known forms
Simple Factors
10
Let Y (s ) = 2
(s 10s 16)
If we check the Table , we see there is no form similar to Y(s)
However if we expand Y(s) in partial fractions:
10
A  B

(s 2 10s 16) (s  2) (s 8)
A and B
Are available on the Table
(s  2)
(s 8)
Next we develop Techniques of finding A and B
Heavisdie’s Expansion Theorem
Y (s ) 
10
10
A  B


(s 2 10s 16) (s  2)(s 8) (s  2) (s 8)
Multiply ( X) both side by (s+2) and set s = 2
10
X(s  2)
 A X(s  2)
 B X(s  2)
(s  2)(s 8)
(s  2)
(s 8)
s  2
s  2
s  2
10  A  B (2  2)
(2 8)
(2 8)
10  A  0
(6)
A 5
3
Similarly Multiply both side by (s +8) and set s = 8
10
X(s 8)
 A X(s 8)
 B X(s 8)
(s  2)(s 8)
(s  2)
(s 8)
s  8
s  8
s  8
5
B 
3
y (t) = 5e 2tu (t )  5e 8tu (t ) = 5 e 2t e 8t u (t )
3
3
3
Imaginary Roots
(15s 2  25s  20)
(15s 2  25s  20) 
Let Y (s )  2
(s 1)(s 2 10s 16) (s  j )(s  j )(s  2)(s 8)
A1
A2
A3
A4




(s  j ) (s  j ) (s  2) (s 8)
Using Heavisdie’s Expansions, by multiplying the left hand side and
Right hand side by the factors
(s  j ) , (s  j ) , (s  2) , (s 8)
s  j , s  j , s 2 , s 8
and substitute
We obtain
respectively
1
1
A 1  (1 j ) , A 2  (1 j ) , A 3 1 , A 4 2
2
2
From Table
e 2tu (t ) e 8tu (t )
Y (s ) 
(1/2)(1 j ) (1/2)(1 j )
1
2



(s  j )
(s  j )
(s  2) (s 8)
(1/2)(1 j )
(1/2)(1 j )
and
(s  j )
(s  j )
(a)
Can be inverted in two methods:
(1/2)(1 j )  (1/2)(1 j ) e  jtu (t )
(s  j )
(1/2)(1 j )  (1/2)(1 j ) e jtu (t )
(s  j )
combine
(1/2)(1 j )  (1/2)(1 j ) e  jtu (t )
(s  j )
(1/2)(1 j )  (1/2)(1 j ) e jtu (t )
(s  j )
(1/2)(1 j ) (1/2)(1 j )

(s  j )
(s  j )








1
1
 jt
 (1 j ) e u (t )  (1 j ) e jtu (t )
2
2
1
j
jt
 jt
 (e e ) u (t )  (e  jt e jt )u (t )
2
2
1
1
jt
 jt
 (e e ) u (t )  (e jt e  jt )u (t )
2
2j
 cos(t ) u (t ) sin(t ) u (t )
(b)
(1/2)(1 j )
(1/2)(1 j )
and
(s  j )
(s  j )
Can be combined as
(1/2)(1 j )
(1/2)(1 j )
+
(s  j )
(s  j )
(1/2)(1 j )(s  j )  (1/2)(1 j )(s  j )

(s  j )(s  j )
s 1  s  1
 cos(t ) u (t ) sin(t ) u (t )
 2
2
2
s 1 s 1 s 1
cos(t ) u (t ) sin(t ) u (t )
Y (s ) 
e 2tu (t ) e 8tu (t )
(1/2)(1 j ) (1/2)(1 j )
1
2



(s  j )
(s  j )
(s  2) (s 8)
y (t )  cos(t ) u (t ) sin(t ) u (t ) e 2tu (t ) e 8tu (t )
y (t )   cos(t ) sin(t ) e 2t e 8t u (t )
Repeated Linear Factor
P
(
s
)
If Y (s ) 
n
(s  ) Q (s )





10s
example Y (s ) 
2
(s  2) (s 8)
Then its partial fraction
A
A
A
1
2
2
Y (s ) 


+
2
3
(s  ) (s  ) (s  )
Were
(n  m )
1
d
Am
(n  m )! ds (n m )
A
R
(
s
)
n
+
+
n
Q (s )
(s  )
(s  ) Y (s ) 

 s 
n





Repeated Linear Factor
Let
10s
Y (s ) 
(s  2) 2 (s 8)
Also A2 can be found using Heavisdie’s
by multiplying both sides by
B can be found using Heavisdie’s
n 2
m 1
Then
n 2
m 2
Y (s )  A 1  A 2 2  B
(s  2) (s  2) (s 8)
1 d (21)
A 1
(21)! ds (21)
by multiplying both sides by
1
d (n  m )
Am
(n  m )! ds (n m )
(s  ) Y (s ) 

 s 
n
2
d  10s 
10s



(
s

2)
2
 (s 8) 


d
s

 s 2
(s  2) (s 8)  s 2

10(s 8) 10s 
10(28) 10(2)    80   20


2
2


9
36


(
s

8)
(

2

8)

 s 2 
  
20
9
10
3
Y (s )  A 1  A 2 2  B
(s  2) (s  2) (s 8)
1
d (22)
A2
(2 2)! ds (22)
 10(2) 


(

2

8)


2
10s



(
s

2)
2


(
s

2)
(
s

8)

 s 2
 20   10


3
6


 10s 
 (s 8) 

 s 2
20
9

10
3
20
9
Y (s )  A 1  A 2 2  B
(s  2) (s  2) (s 8)
To find B , we use Heaviside
(s 8)
0
0
10s
A 1 X(s 8)
A 2 X(s 8)
B X(s 8)



2
(s 8)
(s  2) 2 (s 8) s 8 (s  2)
s 8
s 8 (s  2)
s 8
(20/9)
(10/3)
(20/9)
20
Y (s ) 


 B 
2
(s  2)
(s 8)
9
(s  2)

20 
1
1
(3/2)

Y (s )  


2 

(
s

8)
(
s

2)
9
(s  2) 


20
3
y (t )   e 8t e 2t  te 2t u (t )

9 
2

Repeated Linear Factor
Let
Then
10s
Y (s ) 
(s  2)3 (s 8)
Can be found using Heaviside’s expansion
Y (s )  A 1  A 2 2  A 3 3  B
(s  2) (s  2) (s  2) (s 8)
B  (s 8) Y (s ) s  8  10
27
(3 2)
1
d
A2
(3 2)! ds (32)
A 3  (s  2)3 Y (s )
(s  2) Y (s ) 

 s 2
3
s  2
10

3
d  10s 
20


ds  (s 8)  s 2 9
 20
27
20
9
10
3
10
27
Y (s )  A 1  A 2 2  A 3 3  B
(s  2) (s  2) (s  2) (s 8)
A1 Can be found using Heaviside differentiation techniques
1 d (31)
A 1
(31)! ds (31)
2
1
d
 10s 
(s  2) Y (s ) 


 s 2 2 ds 2  (s 8) 

 s 2
20

27
3
 160 
 (s 8)3 

 s 2
1  20 1
10  1
10 1
Y (s )  



27  (s 8) (s  2)  9 (s  2) 2 3 (s  2)3
1

2
 y (t ) 

 10 


 27 

e 8t e 2t  
5
3
t


t


4 
 e
3 

2t 



u (t )