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Chapter 6

The Laplace

Transform and Its

Applications

.

Which Transform to Use?

Application

Continuous

Domain

Discrete

Domain

Signal

Processing

Fourier T.

Discrete F.T.

(DFT/FFT)

Control Theory Laplace T. z-Transform

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The Laplace transformation is a technique

employed primarily to solve ordinary

differential equations. It is also used in

modelling engineering systems.

GENERAL OBJECTIVE

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Math behind the s-plane

• The actual Laplace Transform*:
  • h(t) could be a signal, or a system
    • As a signal:
    • As a system:
  • H(s): Laplace Transform of h(t)
  • s = σ + jω: Complex Frequency Variable

∫

∞

−

=

0

H ( s) h(t)e dt

st

h(t )= sin( 15 t+ 0. 16 )

( )

( )

( )

v t

v t

h t

in

out

=

The Laplace Transform of a time-domain function gives us

a function of the complex frequency variable ‘s’

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In summary, The Laplace Transform

The Laplace Transform of a function, f(t), is defined as;





0

L[ f(t)] F(s) f (t )e dt

st

The Inverse Laplace Transform is defined by

 

 



j

j

ts

F s e ds

j

L F s f t





[ ( )] ( )

1

Eq A

Eq B

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• 12/03/2018 CnS2053

Examples

• 3. Let f (t) = e

(a+bi)t

, t ≥ 0. Then

Examples

• 1. Find the Laplace transform of

f (t) = sin at, t ≥ 0.

• 2. Find the Laplace transform of

f(t)=2 + 5e

−2t

− 3 sin 4t, t ≥ 0.

5

1 4

s

F s

s s





 

Perform a partial fraction expansion (PFE)

  

1 2

5

( )

1 4 1 4

s

F s

s s s s

  

  

   

where coefficients 1 and have to be determined.

2

PARTIAL FRACTION

To find : Multiply both sides by s + 1 and let s = - 1

1 2

1 4

5 4 5 1

4 3 1 3 s s

s s

s s

 

 

 

     

 

To find : Multiply both sides by s + 4 and let s = - 2

1 1

1 1 4

4 1 1 1

( ) { ( )} { }

3 1 3 4

4 1 1 1 4 1

{ } { }

3 1 3 4 3 3

t t

f t L F s L

s s

L L e e

s s

 

   

  

 

   

 

Therefore,

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CnS

13

Solution of Ordinary Differential

Equations (ODEs) by Laplace Transforms

Procedure:

1. Take the L of both sides of the ODE.
2. Rearrange the resulting algebraic equation in the

s domain to solve for the L of the output variable,

e.g., F(s).

3. Perform a partial fraction expansion.

4. Use the L

to find f(t) from the expression for F(s).

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EXAMPLE 2.2.

x

f f e f



Taking Laplace transforms of both sides of this equation gives:

3[ ( ) (0)] 2 ( )

s

sF s f F s

s s s s

s

F s

s s s s s s

s s s

2 /

x x

f t e e



K.A. Stroud. Engineering

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EXAMPLE 2.3.

CnS

Hence, we have

The Laplace-transformed differential equation is

Recall the inverse transforms: ?????

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C

C A

( )

( ) ( )

dv t

RC v t V u t

dt

 

0

0

0

0

( )

( ) ( )

[ ( ) ] ( )

( )[ 1]

( )

( )

1

( )

1 1

c

c A

A

C C

A C C A C A C

dv t

L RC v t V u t

dt

V

RC sV s V V s

s

V

V s RCs V RC

s

Solving V s

V

V RC

s

V s

RCs

Rooting s

V

V RC

V s

s s s

RC RC

 

   

 

  

  







 

 

   

 

Convert the differential equation into an algebraic one

SOLUTION

P.T.O

0

: ( ) ( ) ( ) 0

: ( ) ( )

Re : ( ) ( )

( )

: ( )

(0 ) V

S R C

S A

R

C

C

KVL v t v t v t

Source v t V u t

sistor v t i t R

dv t

Capacitor i t C

dt

v V

   







 

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1 2

1 2

1 2

1 0

0

1

0

1 1

, sin cov lg

1

( ) V

1 1

invert ( )

A

A A

A A

s s RC

A A

C

t t

RC RC

C A A

V

k k RC

s

s s s

RC RC

residue k k are found u g er up a orithm

V V

RC RC

k V and k V

s s

RC

V V V

V s

s s s

RC RC

L v t V V e V e

 

  

 

   

     

   



    



   

 

    