System analysis in the time domain involves ( finding y(t)):
C
R
( )x t
( )y t
( ) ( ) ( )dy tRC y t x t
dt Solving the differential equation
Using the convolution integral
OR
Both Techniques can results in tedious (ممل )mathematical operation
y(t) = ( ( ) )x t h t
Chapter 7 The Laplace Transform
Consider the following RC circuit ( System)
Fourier Transform provided an alternative approach
C
R
( )x t
( )y t
( ) ( ) ( )dy tRC y t x t
dt
Differential Equation
( ) ( ) ( ) ( )RC j Y Y X
Algebraic Equation
( )( ) ( ) 1
XYj RC
solve for ( )Y
Inverse Back( )y t
( ) ( ) ( )dy tRC y t x t
dt
C
R
( )x t
( )y t
( ) ( ) ( )dy tRC y t x t
dt
both side
( ) ( ) ( ) ( )RC j Y Y X
Fourier Transform
Solve for ( )
( )( ) ( ) 1
Y
XYj RC
( )y t
Unfortunately , there are many signals of interest that arise in systemanalysis for whi(not
ch theabsolutley integrable)
( ) x t dt
Example
Fourier Transform does not exists
2 ( ) , ( ) , ( ) tx t t x t t x t e
Inverse Back
( ) 1 2 ( )
1( ) ( ) ( ) +
( ) ( ) [ ( )+ ( )]
In fact signals such as
are not strictly integrable and their Fourier transforms
all contain some non conventional fu io
nct
o o o
x t
x t u tj
x t cos t
n such as .( )
A more general transform is needed
Suppose we have a function or signal that is not absolutely integrable as shown x t
Now multiply ( ) by tx et
The signal ( ) is absolutley integrable tex t The signal ( ) has Fourier Transform tex t
1( ) ( ) ( ) ( )2
j t j tx t X e d X x t e dt
Fourier Transform pairs was defined as
. ( ) ( )t t j te eF T x t x t e dt
( )( ) j tx t e dt
( )X j
Let us deined this ( ) as Complex Frequency jsj
0
. ( ) ( ) tt sF T x t x te e dt
( )X s L ( )b x t
were Denotes the operation of o Bilateral Laplace Transfobtaining the rm L b
Now multiply ( ) by and takes the Fourier Transform tx t e
Fourier Transform of ( )
The domain is not but
te x t
j
7.1 DEFINITIONS OF LAPLACE TRANSFORMS
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 te
The Laplace transform of a time function is given by the integral
This definition is called the , or , Laplace transform — hence, the subscript bbilateral two-sided
Notice that the bilateral Laplace transform integral becomes the Fourier transform integral if is replaced by (i.e, 0)s j
The inverse Laplace transform
where
This transform is usually called, simply, the Laplace transform , the subscript b dropped
Laplace transform pairs
For most application , ( ) is zero for 0f t t
Unilateral (single sided) Laplace transform pairs
Forward transform
Inverse transform
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
Complex domain
Using Properties and table of known transform ( similar to Fourier)
0
L ( ) ([ ] ) tst t e dt
0
L ( ) ([ ] ) tst t e dt
0
ts
t
e
1
The value of s make no difference
Assuming the lower limit is 0
Region of Convergence
j domains
0
Region of Convergence
We also can drive it from the shifting in time property that will be discussed later
Laplace Transform for an impulse (t)
( )1lim 1j t
te
s
1
lim 1t j t
te e
s
For this to converge then > 0lim 0t
te
region of convergence (ROC)
Re(s) > 0
is were the Laplace Transform Function ( )F s Pole
We derive the Laplace transform of the exponential function
A short table of Laplace transforms is constructed from Examples 7.1 and 7.2and is given as Table 7.2
We know derive the Laplace Transform (LT) for a cosine function
Since
Proofat at st
0
[ ] [ ( )]e ( ) e eL f t f t dt
(s+a)t
0
( )ef t dt
Let s*= s+a at s*t
0
e ( ) ( )[ ] eL f t f t dt
( *) ( )F s F s a
0 20
2[ ] L cos t s
s
0 220
0
[ ] L sin ts
Example
0 L cos[ ] t 0 01 1
2 2 =L j t j te e
0 01 1
2 2[ ] = L + L[ ]j t j te e
1 L ( )[ ] = ( )
t
se u t
Since
Then 0
0
1L =(
[ ] )
j tej s
0
0
1L =( )
[ ] j tesj
00 0
1 11 1 L cos ( ) (2
2
])
[ s s
tj j
2 2
0
ss
0 220
0
[ ] L sin ts
Similarly
Real Shifting
Let
0.3( ) 5 tf t e 50.3s
0.3( ) 5 tf t e
2 0 0we need to put the function ( ) in a delay form ( ) ( ) in order to use
the Laplace Transform for time shift
f t f t t u t t
It is sometimes necessary to construct complex waveforms from simpler waveforms
(See example 7.3)
(Shift in time property)Since
Transform of Derivatives
( )L ( ) (0 )
df tf
dsF
ts
Proof
0
( ) ( )L
tsdf t df te dt
dt dt
Integrating by parts, ( ) ( )tsu e dv t df t
( ) ( )tsdu e v t ts f
b bb
aa a
udv uv vdu
0
( ) ( )L
tsdf t df te dt
dt dt
( ) ( )tsu e dv t df t ( ) ( )stdu e v t f ts
b bb
aa a
udv uv vdu
00
( ) + ( ) st stse f t f t e dt
( ) (0) ( ) (0 ) + ( ) s se f e f Fs s
0 1
( ) (0 ) F fs s
( ) ( ) (0 ) s sdf t F fdt
2( ) 2 ( )ti t e u t
=0
=0 =0 =0
2 1
2
1
2 3
1
( ) ( ) ( )
( ) ( ) ( )
t
t t t
n
n
n
n
nn n
n
s sd f t F f tdt
df t d f t d f tdt d t
sd
s
ts
2
2 =0
=0
2 ( )( ) ( ) ( ) t
t
s s sdf td f t F f t
dtdt
( ) ( ) (0 ) df t Fd
s fst
23
23 =0
=0 =0
3 2 ( ) ( )( ) ( ) ( ) t
tt
dx t d f td s s s sf t F f tdtdt dt
Proof
find the Laplace transforms of the following integrals:
Proof (See the book)
Since
Transfer Functions
Consider the following circuit
R( )x t ( ) i t
L
C
( )y t
input output
We want a relation (an equation) between the input x(t) and output y(t)
1( ) ' '( ) ( ) + ( )
t
di tx t L Ri t i t dtdt C
KVL
2
2
( ) ( ) ( )( ) + dx t di t i tdi tL Rdt dt Cdt
7.6 Response of LTI Systems
R( )x t ( ) i t
L
C
( )y t
input output
2
2
( ) ( ) ( )( ) + dx t di t i tdi tL Rdt dt Cdt
( )Since ( )
y ti t
R
2
2
( ) ( ) ( )( ) + dx t L dy t y tdy t Rdt R dt RCdt
Writing the differential equation as 2
2
( ) ( )( ) + ( ) dx t dy tdy tRC LC RC y tdt dtdt
R( )x t ( ) i t
L
C
( )y t
input output
2
2
( ) ( )( ) + { } { } { } { } ( ) 1RC dx t dy tdy t y tdt dtd
LCt
RC
Real coefficients, non negative which results from system components R, L, C
In general,
1 1
1 0 1 01 1
( ) ( ) ( ) ( ) + + + ( ) + + + ( )n n m m
n n m mn n m mdy t dy t dx t dx ty t y t
dt dt dt da a b b
ta b
' 'were , n ms ba s are real, non negative which results from system components R, L, C
Now if we take the Laplace Transform of both side (Assuming Zero initial Conditions)
1 01
11
0( ) + ( ) + + ( ) ( ) + ( ) + + ( )n nn m
mn m
ma a a b bs s s s s s s s s sY Y bY X X X
We now define the transfer function H(s) ,
all initial conditions are zero
( )( )
( )
YH
X
ss
s 1 0
1
11 0
+ + +
+ + + m m
n
m m
n nn
b b b
a a a
s s
s s
1 1
1 0 1 01 1
( ) ( ) ( ) ( ) + + + ( ) + + + ( )n n m m
n n m mn n m mdy t dy t dx t dx ty t x t
dt dt dt da a b b
ta b
1 0
11 0
1 + + +
+ + +
( )( )
( )m
m
n
m
n n
m
n
b b b
sa
s
as a
sYH
ssX
s
' 'Sin ce , n ms ba s are real, non negative
( )
( )N sD s
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) = ± ∞ )
The Degree of N(s) ( which is related to input) must be less than orEqual of D(s) ( which is related to output) for the system to be Bounded-input, bounded-output (BIBO)
( )
( )N sD s
1 0
11 0
1 + + +
+ + +
( )( )
( )m
m
n
m
n n
m
n
b b b
sa
s
as a
sYH
ssX
s
Example : 3 2
2
4 + 2 1 6 8
( ) s ss
ss
sH
Using polynomial division , we obtain 2
19 174 + 2 + 6 8
( ) sss s
sH
Now assume the input x(t) = u(t) (bounded input) 1 ( ) s sX
2
2 1 19 174 + + 6 8
( ) ( ) ( ) ss s s s
Y Xs s sH
1
2
19 174 ( ) + 2 + ( 6 8)
( ) sts s s
y t L
unbounded
)
(
We see that for finite boundedInput (i.e x(t) =u(t) )
We get an infinite (unbounded)output
for m n BIBO
( )
( )N sD s
1 0
11 0
1 + + +
+ + +
( )( )
( )m
m
n
m
n n
m
n
b b b
sa
s
as a
sYH
ssX
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) ?
The inverse transform, from Table 7.2
v(t) = 12u(t)
Now using the Laplace Table 7.2 to find i(t)
( ) ( )v t Ri t ( ) ( )V s RI s
( )( )L
di tv t Ldt
[ ]( ) ( ) (0 )L
V s L sI is s ( ) (0 )LI s Li
1( ) ( ) (s
0 ) L
I s V s isL
Source Transformation
R
Z R
L
Z sL
( )( ) Cdv ti t Cdt
( ) ([ ( ]0 ) )C
I s C sV vs
( ) (0 )C
sC s CvV
1( ) ( ) (0 ) C S
V s I s vsC
Source Transformation1
CZ
sC
10s
8 5s
1
2 s
2 2
0.5 2.5( )
1 1X
s ss
s s
( ) I s
1 R
( ) 0.5 cos 2.5 sin ( )x t t t u t
( ) i t
0 t 1 10
L H 5 8
C F
R1 Z 1 R
2 2
0.5 2.5( ) ( )
1 1
ssx t X
s
s s
L1 Z
10 10L H s
C 5 8 Z 58
Fs
C
(0 ) 2 Cv V
(0 ) 0 Li V
Example Using Laplace Transform Method find i(t)
10s
8 5s
1
2 s
2 2
0.5 2.5( )
1 1X
s ss
s s
( ) I s
KVL8 2( ) ( ) ( ) (1) ( ) 0
10 5X I
ss s s s
s sI I
2 2
8 20.5 2.5 1 ( )10 51 1
Iss s s
s ss s
2
2 2
15 25 20( )( 1)( 10 16)
s sss s s
I
10s
8 5s
1
2 s
2 2
0.5 2.5( )
1 1X
s ss
s s
( ) I s
2
2 2
15 25 20( )( 1)( 10 16)
s sss s s
I
( Imaginary Roots)
2
2 2
15 25 20( )
( 1)( 10 16)s s
ss s s
I
2(15 25 20)( )( )( 2)( 8)j
s ss js s s
1 2 3 4
( ) ( ) ( 2) ( 8)s s sA A A A
j sj
2 8( ) cos( ) sin( ) ( )t ti t t t e e 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 Rational Function
2
3 2
2 Example ( ) = 2
2 5 3
s sss s
Ys
When the degree of the numerator of rational function is less theDegree of the dominator
Proper Rational Function
2
3 2
2 Example ( ) = 2
2 5 3
s sss s
Ys
Highest Degree is 2
Highest Degree is 3
Examples of proper rational Functions
1 1( ) =1
Y ss
2
22 2
2 6 6( ) =( 2)(
2 2)
sss s
Y ss
Examples of not proper rational Functions
32( ) = 1
Y sss
However we can obtain a proper rational Function through long division
32( ) = 1
Y sss
1= 1 +1s
We will discuss different techniques of factoring Y(s) into simpleknown forms
Simple Factors
2
10Let ( ) =( 0 1
1 6)
ss
Ys
If we check the Table , we see there is no form similar to Y(s)
However if we expand Y(s) in partial fractions:
2
10( 10 16)s s
( 2) ( 8)s s
A B
and ( 2) ( 8)
A Bs s Are available on the Table
Next we develop Techniques of finding A and B
Heavisdie’s Expansion Theorem
2
10 10( ) ( 10 16) ( 2)( 8) ( 2) ( 8)
ss s s s s s
A BY
Multiply ( ) both side by (s+2) and set s =2
2 2 2
10 ( 2) ( 2) ( 2)( 2)( 8) ( 2) (
X X8)
Xss s
s s ss s s s
A B
10 ( 2) ( 8) 2(
282 )
BA
10 0(6)
A 5 3
A
X
Similarly Multiply both side by (s +8) and set s =8
8 8 8
10 ( 8) ( 8) ( 8)( 2)( 8) ( 2) (
X X8)
Xss s
s s ss s s s
A B
5 3
B
2 85 5(t) = ( ) ( )3 3
t ty e u t e u t 2 85= ( )3
t te e u t
2
2 2
(15 25 20)Let ( )
( 1)( 10 16)s s
ss s s
Y
Imaginary Roots
2(15 25 20)( )( )( 2)( 8)j
s ss js s s
1 2 3 4
( ) ( ) ( 2) ( 8)s s sA A A A
j sj
Using Heavisdie’s Expansions, by multiplying the left hand side andRight hand side by the factors
( ) , ( ) , ( 2) , ( 8)s sj s sj , , 2 , 8s s s sj j and substitute respectively
We obtain 1 2 3 4
1 1(1 ) , (1 ) , 1 , 22 2
A j A j A A
(1/2)(1 ) (1/2)(1 ) 1 2( )( ) ( ) ( 2) ( 8)
ss
j jYj sjs s
2 ( )te u t 8 ( )te u t
From Table
(1/2)(1 ) (1/2)(1 ) and ( ) ( )
j jj js s
Can be inverted in two methods:
(a)(1/2)(1 )
( )j
s j
(1/2)(1 ) ( )jtj e u t
(1/2)(1 ) ( )
js j
(1/2)(1 ) ( )jtj e u t
(1/2)(1 ) ( )
js j
(1/2)(1 ) ( )jtj e u t
(1/2)(1 ) ( )
js j
(1/2)(1 ) ( )jtj e u t
combine
(1/2)(1 ) (1/2)(1 )( ) ( )
js s
jj j
1 1 (1 ) ( ) (1 ) ( )2 2
jt jtj e u t j e u t
1( ) ( ) ( ) ( )2 2
jt jt jt jtje e u t e e u t
cos( ) ( ) sin( ) ( )t u t t u t
1 1( ) ( ) ( ) ( )2 2
jt jt jt jte e u t e e u tj
(1/2)(1 ) (1/2)(1 ) and ( ) ( )
j jj js s
Can be combined as
(1/2)(1 ) (1/2)(1 ) + ( ) ( )
j jj js s
(1/2)(1 )( ) (1/2)(1 )( )( )( )s ss
j j j jj js
2
11
ss
2 2
1 1 1
ss s
(b)
cos( ) ( ) sin( ) ( )t u t t u t
(1/2)(1 ) (1/2)(1 ) 1 2( )( ) ( ) ( 2) ( 8)
ss
j jYj sjs s
2 ( )te u t 8 ( )te u tcos( ) ( ) sin( ) ( )t u t t u t
2 8( ) cos( ) ( ) sin( ) ( ) ( ) ( )t ty t t u t t u t e u t e u t
2 8( ) cos( ) sin( ) ( )t ty t t t e e u t
Repeated Linear Factor
( )( ) ( ) ( )n
sss
Ps
YQ
If
Then its partial fraction
1 2 2 n2 3
( )( ) + + + ( ) ( )( ) ( ) ( )n
sss ss s sA A A A RY
Q
Were( )
m ( )
1 ( ) ( )( )!
n mn
n ms
s sdA Yn m ds
2
10 ( )
( 2) (exam
)e
8pl
ss
s sY
Repeated Linear Factor
2
10( )
( 2) ( 8)
ss
sY
s
Let
1 22
( )( 2) ( 8)( 2)
A A BY ss ss
Then
(2 1)2
1 2(2 1)
2
101 ( 2)(2 1)! ( 2) ( 8) s
ssd
s s sA
d
2
10
( 8)s
sdsd
s
2
2
10( 8) 10( 8)
s
s ss
2
2 210( 8) 10( )( 8)2
8036
209
( )
m ( )
1 ( ) ( )( )!
n mn
n ms
s sdA Yn m ds
2
1
n
m
2
2
n
m
Also A2 can be found using Heavisdie’s
by multiplying both sides by
B can be found using Heavisdie’s
by multiplying both sides by
(2 2)2
2 2(2 2)
2
101 ( 2)(2 2)! ( 2) ( 8) s
ssd
s s sA
d
2
10
( 8)s
ss
22
10( )( 8)
103
206
1 22
( )( 2) ( 8)( 2)
A A BY ss ss
209
103
1 22
( )( 2) ( 8)( 2)
A A BY ss ss
209
103
To find B , we use Heaviside
1 22
2
888 8
10( 8) ( 8) ( 8) ( 8)( 2) ( 8)( 2) ( 8) (
X2)
X Xsss s
ss s s ssA A
s sB
ss
0 0
20 9
B
209
2
(20/9) (10/3) (20/9)( )
( 2) ( 8)( 2)s
s ssY
2
20 1 1 (3/2)( ) ( 8) ( 2)9 ( 2)
ss
Ys s
8 2 220 3( ) ( )9 2
t t ty t e e te u t
Repeated Linear Factor
3
10( )
( 2) ( 8)
ss
sY
s
Let
1 2 32 3
( )( 2) ( 8)( 2) ( 2)
ss ss sA A A BY
Then
Can be found using Heaviside’s expansion
8( 8) ( )
ssB s Y
10
27
3 3
2( 2) ( )
ssA Ys
10
3
(3 2)3
2 (3 2) 2
1 ( 2) ( )(3 2)! s
ss
dA Yd
s
2
10
( 8)s
sdsd
s
209
1 2 32 3
( )( 2) ( 8)( 2) ( 2)
ss ss sA A A BY
209
103
1027
A1 Can be found using Heaviside differentiation techniques
(3 1)3
1 (3 1) 2
1 ( 2) ( )(3 1)! s
ss
dA Yd
s
2
2
2
101 ( 8)2
s
ssds
d
3
2
1601 ( 8)2
ss
2027
2027
2 3
1 110 20 101 1( )( 8) ( 2)27 9 3( 2) ( 2)
ss s
Ys s
8 2 210 5 4
27 3 3( ) ( )t t tt ty t e e e u t