1. LOGICS OF The Laplace Transform BY TARUN GEHLOT
Introduction
x(t) input y(t) output
(1) System analysis Dynamic
System
A processor which
processes the input signal
to produce the output
static system: y(t) = ax(t) => easy (simple processing)
dynamic system:
dy ( n ) (t ) dy ( n 1) (t ) dx ( m ) (t )
a1 ... a n y (t ) b0 ... bm x(t )
dt n dt n 1 dt m
Can we determine y(t) for given u(t) easily?
Easier solution method
X(f) Y(f )
Dynamic
System
H(f )
Algebraic equation, not
Y( f ) H( f )X ( f ) differential equation
y(t ) F 1 (Y ( f ))
(1) we have systematic way to obtain H(f) based on the differential equation
(2) we can obtain X(f)
 Fourier transform: an easier way
(2) Problem
Fourier transform of the input signal:
Page 5-1

2. j 2 ft
X( f ) x (t )e dt
j 2 ft
|e | 1
if x(t) does not go to zero when t and t
 X(f) typically does not exist! (the existence of X(f) if not guaranteed)
A very strong condition, can not be satisfied by many signals!
(3) Solution:
Why should we care about t<0 for system analysis? We do not care!
x(t) does not go to zero (when t ), but x(t )e t may!
(Will be much easier. )
 use “single-sided” Fourier transform of x(t )e t , instead of “double-
sided” Fourier transform of x(t).
t j t ( j )t
 ( x (t )e )e dt x (t )e dt
0 0
very useful, let’s use a new name for it: Laplace transform.
(4) Laplace Transform
Definition: Laplace transform of x(t)
L[ x(t )] X ( s) x(t )e st dt (s j )
0
What is a Laplace transform of x(t)?
A time function? No, t has been eliminated by the integral with respect to t!
A function of s ( s is complex variable)
(5) System analysis using Laplace transform
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3. X(s) Y(s )
Dynamic
System
G(s )
Inverse Laplace transform
Y ( s) G( s) X ( s)
y(t ) L 1 (Y ( s))
(6) How is the inverse transform defined?
j
1
x (t ) X ( s )e st ds
2 j j
(7) Will we often use the definition of the inverse transform to find time
function?
No!
What will we do?
Express X(s) as sum of terms for which we know the inverse transforms!
5-2 Examples of Evaluating Laplace Transforms using the definition
(1) x(t)=1 and step function x(t)=u(t)
t
st st 1
L[ x(t ) u (t )] x(t )e dt e dt e st d ( st )
0 0
st 0
st t 0
e t e j t
t e j e j 0
e e e
s t 0 s t 0 s s
j
(| e | 1 e 0 , (if 0) e , (if 0)
1 1
(cos0 j sin 0)
s s
1
L(1) L[u (t )] (Re(s)) 0
s
t
(2) x(t ) e u(t )
t t
L[e u(t )] e e st dt e ( s )t
dt
0 0
~
Define a new complex variable s s
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4. ~
st
 e dt
0
st 1
we know e dt Re( s ) 0
0
s
~ ~
st 1
e dt ~
Re( s ) 0
0 s
( s )t 1
e dt Re( s ) 0
0
s
t 1
L[e u(t )] Re( s ) 0 or Re( s ) Re( )
s
t 1
L[e ]
s
(3) x (t ) (t )
L[ (t )] (t )e st dt
0
st t j t
e e e
t 0 t 0
t
e (cos t j sin t )
t 0
1
No constraint on s.
5-2B Discussion: Convergence of the Laplace Transform
(1) To assure x(t )e st dt x (t )e t
e j t
dt converge, Re(s) must be psotive
0 0
t
enough such that x(t )e goes to zero when t goes to positive infinite
(2) Region of absolute convergence and pole
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5. (3) How to obtain Fourier transform form Laplace transform:
s j
L[ x(t )] X ( s) X( j ) F ( x(t ))
Important: why introduce Laplace transform; definition of Laplace transform as a
modification of Fourier transform; find the Laplace transforms of the three basic
functions based on the (mathematical) definition of Laplace transform.
Chapter structure
Part one: Definition (5.1)
Part two: Direct evaluation of Laplace transforms of simple time-
domain function.
Easy? Not at all!
 complex (not simple) functions: even harder
 tools for application of Laplace transform in system analysis
Part three: Rest of the chapter
What tools: tools for easier Laplace transform evaluation
tools for easy inverse transform
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6. 5-3 Some Laplace Transform theorems
(Tools for evaluating Laplace transform based on the Laplace
transforms of the basic functions)
5.3.1 Linearity
Assume x(t ) a1x1(t ) a2 x2 (t ) ( a1 and a 2 are time independent)
X1( s) L[ x1(t )], X 2 ( s) L[ x2 (t )]
then X ( s) L[ x(t )] a1 X1( s) a2 X 2 ( s)
HW#2-1: Assume x(t ) a1 (t ) x1 (t ) a2 (t ) x2 (t ) , X1( s) L[ x1(t )] and X 2 ( s) L[ x2 (t )] .
(a) Is X ( s ) L[ x (t )] equal to a1(t ) X1( s) a2 (t ) X 2 ( s) ?
(Answer: Yes or No)
(b) If A ( s) L[a1(t )] , is it true
1
L[ x(t )] A1( s) X1( s) A2 ( s) X 2 ( s) ?
(Answer: Yes or No)
Example 5.1
(1) Find L(cos 0t )
Key to solution : express (cos 0t ) as linear combination of (t ) , u(t ) ,
t
and/or e :
L[ (t )] 1
1
L[u(t )]
s
t 1
L[e ]
s
let j 0
j 0t
1
L[e ]
s j 0
let j 0
1
L[e j 0t
] L[e ( j 0 )t
]
s j 0
Page 5-6

7. j 0t
Can we use e and e j 0t
to express cos( 0t ) ?
j 0t
e cos( 0t ) j sin( 0t )
cos( 0t ) j sin( 0t )
ej 0t
cos( 0t ) j sin( 0t )
j 0t
e ej 0t
2 cos( 0 t )
j 0t
e ej 0t
cos( 0 t )
2
1
L[cos( 0 t )] [ L(e j 0t ) L(e j 0t )]
2
1 1 1
2 s j 0 s j 0
1 (s j 0 ) (s j 0 )
2 ( s j 0 )(s j 0 )
s
2
s2 0
(2) Find L[sin 0t ]
5-3-2 Transforms of Derivatives
Assume X ( s) L[ x (t )]
dx(t )
Then L sX ( s ) x (0 )
dt
Proof:
dx(t ) dx(t ) st
(1) Definition L e dt e st dx(t )
d (t ) 0
dt 0
(2) Integration by parts:
General equation:
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8. b b
t b
u (t )dv(t ) u (t )v(t ) t a
v(t )du(t )
a a
(3) Use the above equation
st
Why? If we assume v(t ) x(t ) , u (t ) e
st
v(t )du(t ) x(t )de s x(t )e st dt
0 0 0
X ( s) L x(t )
st
u (t ) e , v(t ) x(t )
dx(t )
u (t )dv(t ) e st dx(t ) [ L from (1) ]
0 0
dt
b
t
u (t )v(t ) t 0
v(t )du(t )
a
t
e st x(t ) t 0
s x(t )e st dt
0
t
e st x(t ) t 0
sL[ x(t )]
st
lim e x(t ) must go to zero. Otherwise,
t
L[ x(t )] x(t )e st dt does not exist !
0
st t s0
e x (t ) t 0
e x(0) x(0)
use 0 as lower limit => x (0 )
t
u (t )v(t ) t 0
v(t )du (t )
0
x(0 ) sX ( s ) sX ( s ) x(0 )
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9. HW#2-2: Assume X ( s ) L[ x (t )] . Prove
d 2 x(t )
L s 2 X ( s) sx (0 ) x (1) (0 )
dt
d ( n ) x (t )
HW#2-3: Express L using L[ x ( t )]
dt n
Example 5-2
Find i(t) using Laplace transform method
for t>0
Solution:
(1) Before switched from 1 to 2 at t=0
4
i 2A i (0 ) 2A
2
(2) System equation (t>0)
di(t )
L Ri (t ) 0 ( L 1H ) ( R 2ohm)
dt
KVL:
di(t )
2i (t ) 0
dt
(3) Solve system equation using Laplace transform
di(t ) di(t )
L 2i (t ) L 2 L[i (t )]
dt dt
sI ( s ) i (0 ) 2 I ( s )
( s 2) I ( s ) 2 0
2
I (s)
s 2
i (t ) 2e 2t u (t ) A
Page 5-9

10. 5-3-3 Laplace Transform of an integral
t
Assume y(t ) x( ) d , X ( s) L[ x(t )]
t
X ( s) y (0 )
Then L x ( )d
s s
0
where y (0 ) x ( )d
Proof :
t 0
st
L x ( )d x ( )d e dt
0
t
udv uv t 0
vdu
0 0
(1)
t st t
t e
uv t 0
x( )d
s t 0 1
st t s0 0 y (0 )
e e
lim x( )d x( )d
t s s
y (0 ) 0
s
(2)
st
e 1 1
vdu x (t )dt x (t )e st dt X ( s)
0 0
s s0 s
(3)
t
y (0 ) X ( s) X ( s) y (0 )
L x ( )d Proved!
s s s s
Page 5-10

11. Example 5.3:
Find I(s) = L(i(t))
Solution:
(1) Differential equation
KVL :
x(t ) vL (t ) vC (t ) vR (t )
dvC (t )
i (t ) C
dt di(t )
1 v L (t ) L
dvC (t ) i (t )dt dt
C t
t t 1
1 vC (t ) i ( )d
dvC ( ) i ( )d C
C
t v R (t ) Ri (t )
1
vC (t ) vC ( ) i ( )d
C
t
1
vC ( ) 0 vC (t ) i ( )d
C
(2) Laplace transform
X ( s ) VL ( s ) VC ( s ) VR ( s )
VL ( s ) L[ sI ( s ) i (0 )] LsI ( s ) i (0 ) zero
0
1 I ( s) 1
VC ( s ) i ( )d
C s s
0
1 11
I ( s) i ( )d
Cs sc
1 1
I ( s) vc ( 0 )
Cs s
VR ( s ) RI ( s )
Page 5-11

12. 1 1
X ( s) LsI ( s ) I ( s) vC (0 ) RI ( s )
Cs s
1
X ( s) vC (0 )
I ( s) s
1
Ls R
Cs
sX ( s ) vC (0 )
1
Ls 2 sR
C
sX ( s ) vC (0 )
R 1
L s2 s
L LC
5-3-4. Complex Frequency shift (s-shift) Theorem
Assume y(t ) x(t )e t
X (s) L[ x(t )] Y ( s) L[ y(t )]
Then Y (s) X (s )
1 t 1
L[u (t )] , L[u (t )e ]
s s
s 0
L[cos 0t ] 2
L[sin 0t ] 2
s2 0 s 2
0
t s t 0
=> L[cos 0t e ] 2
L[sin 0t e ] 2
2 2
(s ) 0 (s ) 0
s 8
Example 5-4 Find x(t ) L 1[ X ( s)] L 1
s 2 6s 13
Solution:
s 8 ( s 3) 5
X (s)
s 2 6s 13 s 2 6s 9 4
s 3 (5 / 2) 2
( s 3) 2 2 2 ( s 3) 2 2 2
Page 5-12

13. 5
x(t ) L 1[ X ( s)] e 3t
cos 2t e 3t
sin 2t (t 0)
2
5-3-4 Delay Theorem
question: How to express delayed function?
Assume L[ x(t )] L[ x(t )u(t )] X ( s)
Then L[ x(t t0 )u (t t0 )] e st 0 X (s) (t0 0)
(If (t0 0) , it will not be a delay!)
Proof :
L[ x(t t0 )u (t t0 )]
x(t t0 )u (t t0 )e st dt
0
t0
x(t t0 )u (t t0 )e st dt x(t t0 )u (t t0 )e st dt
0 t0
s ( t t0 ) st0
x(t t0 )e st dt x(t t0 )e dt
t0 t0
st0 s ( t t0 )
e x(t t0 )e d (t t0 )
t0
t t0
st0 s st0 st0
e x ( )e d e x(t )e st dt e X (s)
0 0
Page 5-13

14. st 0
Question: will L[ x(t t0 )u(t t0 )] e X (s) be true if t0 0?
No! (it will not be a delay)
Example 5-5: Square wave beginning at t = 0
T0 3T0
1 1 2
s 1 T0 s 1 2
s 1 2T0 s
L[ xsq (t )] 2 e 2 e 2 e 2 e
s s s s s
1
T0 s
e 2
1 1 1 2 1 3 1 4
2 2 2 2 ...
s s s s s
1 2 2 3
( ...)
s s
T0
s
2
1 2 1 1 1 1 e
T0
s s (1 ) s 1 s 2
s
1 e
5-3-5 Convolution
Signal 1: x1 (t ) Signal 2 : x2 (t )
y(t ) x1 (t ) * x2 (t ) x1 ( ) x2 (t )d
if x1 (t ) 0 t 0
y (t ) x1 ( ) x2 (t )d
0
Page 5-14

15. if x2 (t ) 0 t 0 ( x2 (t ) 0 t)
t
y (t ) x1 ( ) x2 (t )d
0
Therefore, if x1 (t ) 0, x2 (t ) 0 t 0
t
x1 ( ) x2 (t )d x1 ( ) x2 (t )d x1 ( ) x2 (t )d
0 0
t
L[ x1 ( ) x2 (t )d ] L[ x1 ( ) x2 (t )d ] L[ x1 ( ) x2 (t )d ]
0 0
[ x1 ( ) x2 (t )d ]e st dt x1 ( )[ x2 (t )e st dt ]d
0 0 0 0
t
Look at
x 2 (t )e st dt x 2 (t )e s (t )
e s
dt
0 0
t
s s
e x 2 ( )e d
d dt
t 0
t
x2 ( ) 0 0
s s s
e x 2 ( )e d e X 2 ( s)
0
Then
s
Y (s) x1 ( )e X 2 ( s)d
0
s
X 2 ( s ) x1 ( )e d
0
X 1 (s) X 2 (s)
Page 5-15

16. 5-3-7 Product
5-3-8 Initial Value Theorem
x(t ) L 1[ X ( s )]
x (0 ) lim sX ( s )
s
Example: A demonstration where x(0) is obvious
t
x(t ) e cos 0tu(t )
It is evident: x(0) e 0 cos 0 0 1
Using Laplace transform
s
X ( s) L[ x(t )] 2
(s )2 0
s(s )
x(0) lim sX ( s ) lim 2
s s (s )2 0
s2 s
lim 2
s s2 2 s 2
0
d (s 2 s ) / ds
lim 2
s d (s 2 2 s 2
0 ) / ds
2s d (2s ) / ds 2
lim lim lim 1
s 2s 2 s d (2s 2 ) / ds s 2
5-3-9 Final Value Theorem: if x(t ) and dx(t ) / dt are Laplace transformable, then
lim x(t ) lim sX (s)
t s 0
(condition: sX (s) has no poles on j axis or in the right-half s-plan
or lim t x(t ) exists)
5-3-10 Scaling
a>0: x(at) a times fast (if a>1)
or slow (if a<1) as x(t)
X ( s) L[ x(t )]
What do we expect on L[ x(at )] ?
Page 5-16

17. s
L[ x(at)] X ( )?
a
a 0 s
st 1 a
( at )
L[ x(at )] x(at )e dt x(at )e d (at )
0
a0
s
at 1 1 s
a
x ( )e d X
a 0 a0 a a
t
5-4 Inversion of Rational Functions
(1) Ways to find x(t ) from X (s )
1
(1) x(t ) X ( s)e st dt (Contour Integral)
2 j
(2) Transform pair
1
u (t )
s
1 t
e u (t )
s
1
Therefore X ( s) x(t ) e t u (t )
s
(2) All kinds of Laplace Transform ? No!
We almost only see
b0 s m b1s m 1
... bm 1s bm s
e
s n a1s n 1
...an 1s an
rational function X(s) Delay
x(t ) L 1[ X ( s)]
y (t ) L 1[ X ( s)e s
] x(t )u (t )
Page 5-17

18.  Consider Rational Functions only!
(3) Non proper Rational Function
proper Rational Function
Non proper m>=n (b0 0)
proper m<n
Non proper => proper + Polynomial (using long division)
s3 4s 2 6s 7 s 5
s 1
s 2 3s 2 s 2 3s 2
s 1
s 2 3s 2
s3 4s 2 6s 7
s2 4s 7
3s s s2
s 5
How to find inverse Laplace transform for polynomials?
sn (n)
(t )
L 1 ( s 1) L 1 ( s) L 1 (1) (1)
(t ) (t )
consider proper rational functions only!
(4) Proper Rational Functions: Partial Fraction Expansion
1 t ne t
u (t )
(s )n 1
n!
s t
2
e cos 0 tu (t )
(s )2 0
 sum of 0
e t
sin tu (t )
2 2 0
(s ) 0
1 (t )
1
u (t )
s
1 t
e
s
Page 5-18

19. Let’s look at examples, and then summarize!
Techniques:
Common Denominator Factorize first!
Specific value of s Expand second!
Heaviside’s Expansion Find coefficients third!
Matlab
Example 5-9: Simple Factors
10
Y ( s)
s2 10s 16
Solution:
10 A B
(1) Factorize and expand Y ( s )
( s 8)(s 2) s 8 s 2
(2) Common Denominator Methods
10 A( s 2) B( s 8)
( s 8)(s 2) ( s 8)(s 2)
A( s 2) B( s 8) 10
A B 0 A B
2 A 8B 10 2 B 8B 10
B 10 / 6 5/3 A 5/3
5 1 5 1
Y (s)
3 s 8 3 s 2
5 5 2t
y (t ) ( e 8t e )u (t )
3 3
specific values of s
10 A B
( s 8)(s 2) s 8 s 2
s 0 10 A B
8 2 8 2
s 2 10 A B
10 4 10 4
Can you solve for A and B?
Page 5-19

20. Heaviside Expansion
10 A B
( s 8)( s 2) s 8 s 2
10 B ( s 8) s 8 10
A A 5/3
s 2 s 2 8 2
10 A( s 2) s 2 10
and B B 5/3
s 8 s 8 2 8
Example 5-10 Imaginary Roots
15s 2 25s 20
Y ( s)
( s 2 1)(s 2)(s 8)
Same as real roots!
Solution: what do we have:
A1 A2 A3 A4
Y (s)
s j s j s 2 s 8
A1 ( s j ) A2 ( s j ) A3 A4
Y (s) 2
s 1 s 2 s 8
( A1 A2 ) s ( A2 A1 ) j A3 A4
2
s 1 s 2 s 8
A1+A2 must be real number
(-A1+A2)j must be real number
c1s c2 A3 A4
Y ( s)
s2 1 s 2 s 8
Heaviside Expansion => A3=1 and A4= - 2.
15s 2 25s 20 c1 s c2 1 2
Y ( s)
(s 2 1)(s 2)(s 8) s2 1 s 2 s 8
15 25 20 c1 c2 1 2
s=1 =>
2 3 9 2 3 9
15 4 25 2 20 2c1 c2 1 2
s=2 =>
5 4 10 5 4 10
Page 5-20

21. Can we solve for c1 and c2?
c1=1 c2=1
s 1 1 2
Y (s)
s2 1 s 2 s 8
s 1 1 2
s2 1 s2 1 s 2 s 8
=> [cos t sin t e 2t 2e 8t ]u (t )
Too complex: use MATLAB
Example 5-11 Repeated linear Factors
10s
Y (s)
( s 2) 2 ( s 8)
A1 A2 A3
s 8 s 2 ( s 2) 2
Example 5-12
10s
Y ( s)
( s 2) 3 ( s 8)
A1 A2 A3 A4
s 8 s 2 ( s 2) 2 ( s 2) 3
Example 5-13 Complex - Conjugate Factors
2s 2 6s 6
Y ( s)
( s 2)(s 2 2s 2)
2s 2 6s 6 A1 A2 A3
( s 2)[(s 1) 2 1] s 2 s 1 j s 1 j
Example 5-14 Repeated Quadratic Factors
Page 5-21

22. s4 5s 3 12s 2 7 s 15
Y (s)
( s 2)(s 2 1) 2
A1 B1s c1 B2 s c2
s 2 s 2 1 ( s 2 1) 2
* Summary of Partial–Fraction Expansion
(1) Expansion Structure:
Simple Roots (including complex conjugate)
Aj
=> could be complex.
s j
Repeated Roots: m multiplicity
B1 B2 Bm
=> 2
... m
s j (s j) (s j)
real number or complex number
(2) Avoid complex number
For complex conjugates: j a jb
Aj Aj* cs D
*
s j s j ( s a) 2 b 2
Bk Bk * cs D
k * k
(s j) s [( s a) 2 b 2 ]k
j
(3) Inverse Laplace transform
Aj jt
Aje u (t )
s j
t
Aj t k 1e j
k
Ak u (t ) k 2
(s j) (k 1)!
Page 5-22

23. Aj Bj tk 1 jt j
*
t
k * k
[ Aj e Bje ]u (t )
(s j) (s j ) (k 1)!
j a jb
t k 1 at
*
e [ A j (cosbt j sin bt ) B j (cosbt j sin bt ]u (t )
j a jb ( k 1)!
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