# Laplace Transform

#### Description

ME 467 Systems Dynamics & Automatic Control

Chapter 2: LAPLACE Transforms

1

Transformation Definition 

Transforms -- a mathematical conversion from one domain (form) to another to make a problem easier to solve

problem in srcinal form

solution in srcinal form transform

solution in transform way of thinking

inverse transform

Laplace Transform problem in time domain

solution in time domain

Laplace transform

inverse solution in s domain

Other transforms • Fourier • z-transform • wavelets

Laplace transform

Nature of the Laplace Transform 

Why study Laplace Transforms 

The Laplace Transform provides a tool for the rapid solution of ordinary constant coefficient linear differential equations It gives us a straight forward procedure to solve and analyze dynamic systems.

Nature of the Laplace Transform

L(f(t)) Laplace Domain Differential Equation f(t)

Algebraic Equation f(s)

Time Domain Non Differential Equation f(t)

L-1(f(s))

Algebraic Manipulation

Algebraic Equation f(s)

Laplace Transform L[

f (t )]  F ( s) 

0

f (t )e  st dt Complex Number

Convert time-domain functions and operations into frequency-domain  

f(t)  F(s) (tR, sC) Linear differential equations (LDE)  algebraic expression in Complex plane

Graphical solution for LDE characteristics

The Complex Plane (review) Imaginary axis (j)

u  x  jy

y

 

y

u    tan 1 y

r

x

x

Real axis

r

| u | r | u |

u  x  jy (complex) conjugate

x2  y2

Basic Theorems of Linearity L[ Kf()]t  KL [ ()]f t ( )KF s

L[ f1( )t 2  f( )]t1 [ L(2)]f t [ (L)] f t

 F1 ( s)  F(2) s 

The Laplace transform of a product is not the product of the transforms.

L[ f1( )t2 f( )]t 1 2(F) s( )F s

Unit step function. F (s) 

 0

()f t e st dt

f(t) = u(t) = 1

F ( s)  (1) e st dt 

  e0  F ( s)   0    s  0  s  e

 st

0

1

s

Exponential Function (t )  e at  F (s )

 0

e

 at st

e

e

 ( s a )t

 dt

 

 0

   0  (s ) a  0 (  ) 

1

s a

ast (

e

)

dt e 0 s

a

Exponential function multiplied by another function

g ()t  e at()f t

at

G )s(  e f)( t e dt 0

 F )(s Example

s s a

 at

st

)(f t e

 

( F) s  a

L [1*e ] 

1

s

 s  s a

dtt sa

0

1

s a

(

)

Derive the Laplace transform of the sinusoidal function 1 L [ cos  t )]  L  e j 2

t

  e  j  t   

 1 L [e j ]t [ 1 L] e  j t  

2 1

1

2 1 1

2 sj 

s s 2 2

2 s  j

Complex Differential Theorem L [t f()]t

In general

d ds

  ( )F s

L [t f ( )]t (1)  n

n

dn (n) F s ds

example

L [t *1]  1

d 1 1  2 ds s s

Laplace Transforms of Common Functions Name Impulse

f(t)

1 f (t )   0

F(s) t0 t0

1

Step

f (t )  1

1

Ramp

f (t )  t

1

Exponential

f (t )  e at

Sine

f (t )  sin(t )

s

s2 1

sa 1

 2  s2

a is +ve

Example

(t ) 50 u( )t

F ( s) 

50

s

Example

vt ( ) e5

2t

V (s ) L[v (t )] 5 

20

sin t 4 4

( s 2) 2 (4) 20

2

 2 s  4s 4 16   s 4 20 s 2

Example

p(t) 5cos2 t 3 e4t s   3 s 2  (2) 2

P()s [ (L)] p5 t

5s

s 4 2

3

s4

1 s 4

Most mathematical handbooks have tables of Laplace transforms

Inverse Laplace Transforms by Identification When a differential equation is solved by

Laplace transforms, the solution is obtained as a function of the variable s. The inverse transform must be formed in order to determine the time response. The simplest forms are those that can be recognized within the tables and a few of those will now be considered.

Example F ( s) 

5 12

s

2

s

8

s3

f (t) 5 12 t 8 e3t

Example V ( s) 

200

s 2  100

 2 2  s  (10)  

V ( s) 20 

10

v(t)  20sin10 t

Determine the inverse transform of the function below. 8s  4 V ( s)  2 s  6s  13 

When the denominator contains a quadratic, check the roots. If they are real, a partial fraction expansion will be required. If they are complex, the table may be used. In this case, the roots are

s1,2  3  2i

Example 8. Continuation.

s  6s  13 2

   2

  2

s 2 6 s (3) 13 ( 3)  s  6s  9 4

 ( s 3) 2 (2)

2

2

Example, Cont. V ( s) 

8( s 3)

4 24 

2 2 ( s 3) (2) ( s3) 2

2

(2)

8( s 3)  2 10(2) 2 2 ( s 3) (2) ( s3)2 (2)

tv( )e 8

3t

tcose2 10 t

3t

sin 2

LAPLACE TRANSFORMS PARTIAL FRACTION EXPANSION

Definition 

Definition -- Partial fractions are several fractions whose sum equals a given fraction

Purpose -- Working with transforms requires breaking

complex

fractions

into

simpler

fractions to allow use of tables of transforms

Partial Fraction Expansions s 1 A B   ( s  2) ( s  3) s2 s3 s 1 A( s  3)  Bs  2) ( s  2) ( s  3)  ( s  2) ( s  3)

A  B 1

3 A  2 B 1

s 1 1 2   ( s  2) ( s  3) s  2 s3

Expand into a term for each factor in the denominator. Recombine RHS Equate terms in s and constant terms. Solve. Each term is in a form so that inverse Laplace transforms can be applied.

Example F ( s) 

s6 s6  s 2  3s 2 ( s 1)( s2)

F (s) 

s6 A A  1  2 ( s 1)( s 2)  1s 2s

A1 ()1) s(

F s

s 1

1  6 s  6  5 s  2  s 1 1  2

Example, cont. A2 ()s2) (

F s s 2

F ( s) 

5

s 1

s  6 2  6   4 s  1  s 2 2  1

4

s2

f ()t 5 e4t  e2t

Example 50( s  3)

F ( s) 

( s 1)( s 2)( s 22 s5)

F1 ( s) 

A1 s 1

A2 s2

Example, Cont. A1 

   ( s 2 )( s 2 2 s 5) 50( s 3)

(50)(2)

(1)(4)

 25

s 1

A2 

 (50)(1)     2 ( s 1)( s  2 s 5)  s 2 ( 1)(5) 50( s 3)

f1 (t) 25 et 10  e2t

10

Example, Cont. To get A and B, substitute by any two convenient numbers, (i.e. s=0 and s=1)

50( s 3)

25

( s 1)( s2)(  s22 5)s

50(3) (1)(2)(5) 50(4) (2)(3)(8)

25 1 0

25 10

1 2

F ( s) 

B 5

2

3

10

   2 1 s 2  s 2 5s

A B 8

25

10

s 1

s2

  

B  25

A  15 15  s25  2  2s 5

s

As  B s

Example, cont. F2 ( s) 

15s 25 s 2  2s  5

s 2  2s25 s 2 s15122 ( 1)  s(2) F2 ( s) 

s1)2 2  5(2) 15  s2 25 2 15(  2   2 ( s 1) (2) (  s1) (2)   ( 1)s (2)

f (t)  (f1) t ()f 2 t t  25et 10 e

t2

t e 15 t ecos 2t 5

sin 2

Example, Repeated Roots F (s) 

F (s) 

60

2

s( s 2)

A  )sF ( ss 0

C1 s( 2) )F ( s2

60

s( s  2)2

C1 A  s ( 2) s(

2

C2 2) s 

60  60  2  15  ( s 2)  s 0 (0 2)

 s 2

2

60  60   30 s  s 2 2

Example, Cont. F ( s) 

60

s( s  2)

15

 2

C2 2 2 s

30

s( 2)s 

Let s=1, and solve for C2

60

15

30

(1)(1  2) 2  1 (1 2)

F (s) 

60

s(s  2)

 2

2

15

C 2 (1 2) 

C2  15

30

s( 2)s 

f (t) 15 30 te2t15   e2t1515

15 2 s

2

(1 e22t )

t

Repeated Roots –General Form Given

) s zm B( s) K  s  z)1 s ) z 2  ,for n A( s)  s  p) It can be expressed as )F ( s 

)F ( s  B(s) A( s)

b2 2 b1 1  ...  s) p  )s p   )s p

 n B( s )  s  p  bn  )   A ( s )   s  p

1

 n !1ds )

mn

bn

n

d  n B( s )  s p   bn1  ) ds  As ( )  s  p

d n 1 

s  p) n 1 

)A s(

n

B (s ) 

 b1   s  p

Laplace of Differentiated Functions

L[ f '( )] t  sF ( ) s (0) f L[ f "( )] t  s2 F ( ) s (0) sf n 1 L [f n( )] t n  s n(F) s  s (0) f

n

f '(0)

2  s '(0).... f

(0) f 1 d f (t )

Note: s is called Differentiation Operator

(t )

dt

s

Laplace of Integrated Functions t F ( s)   L  f (t )dt   0  s

Note: 1/s is called Integrator operator (t )

1

s

 f (t ) dt

Solve This Differential Equation dy y (0)  10  2 y  12 dt  dy  L    2 L  y  L 12  dt  sY ( s) 10 2 Y ( ) s  12 s

 s  2 ) Y( s) 10  Y ( s) 

10

s2

12

12

s( s 2) 

s

Example Solve the following differential equation

dy  2 y 12sin4 t dt

y (0)  10 12(4)

sY ( s) 10 2 Y ( ) s  s 2  16

Y ( s) 

10

48

s  2 ( s 2)( s 2 16) 

48 ( s 2)( s 2 16)

A s2

B1s  B2 2 s16

Example, cont. 48    2.4 2  s  16  s 2 20 48

A

48

( s 2)( s 2 16) 48 (2)(16) 48 (1)(17)

2.4 2

2.4 1

2.4

s2

B2 16

B1s  B2 2 s16

B2  4.8

 B1  B2

B1  2.4

17

Example, Cont.

Y ( s) 

10

2.4

s  2  s2

2.4 s 2

s16

4.8 2

16 s

y(t) 12.4 e2t 2.4cos4 t 1.2sin 4 t

Example d 2y dy  3 2 2y 4 2 dt dt y(0)  10 and y'(0) 0

(sY) 1s0 2 ( )

s 2Y ( s) 1 0 s0 3

Y s

10 s 30   3s2 2 s s ( s 2 3 s2) 24 10 s30    s ( s 1)(s 2) (  s1)( s2)

Y ( s) 

24

24 s

Example, Cont. 24

12

s(s  1)( s2) 10s  30

24

 s 1s 20

12

s

4

s 1

2 s

10

 ( s 1)( s 2)  1s F ( s) 

12

2s 2

s2

f (t)  12 4 et2 e2t

Example d 2y dy  2 5 2y 0 2 dt dt y(0)  0 and y'(0) 10

s 2Y ( s) 010 2 Y ( s) 

(sY) 0s 20

s( s 2 2 s5)

5( ) Y s 10

 2 2s 5 s

20

s

Example, Cont. 20

s( s 2 2s5)

As  B  2 (s  2s 5) s 4

A B (1)(1 2 5) 1  (1 2 5) 20

20 (1)(1 2 5)

A  4

4

A  B  1 (1 2 5) 4

B  8

Example, Cont. Y ( s)  2

4  8s

4

 s 2  2s5

s 2

10

4 4 2

  2s 25 s

s

 2s5 s 2

22

s  2s 5 s 2 s151 ( 1)  s(2) Y ( s) 

4

s

4( s1)  3(2)  2 ( s 1)2 (2) ( 2s1)2 (2)

t t y (t) 4 4 ecos 2 t3 esin 2

t

s

Initial- and Final-Value Theorems

For the given differential equations find the initial and final value of y(t)

y (t) 6 (y )t 8 ( y) t 2( )u t

* Apply Initial- and Final-Value Theorems

y (t) 6 (y )t 8 ( y) t 2( ) u t Y ( s) 

lim t   f (t ) 

lim t 0  f (t ) 

2

s ( s  2) ( s  4) 2 (0)

(0) (0  2) (0  4) 2 () () (  2) (  4)

1

4

0

Laplace transform of the function. Apply final-value theorem Apply initialvalue theorem

Step functions Steps functions are used to form piecewise continuous functions.

Unit step function(Heaviside function):

A plot of uc(t) is :

Step functions For a given function f(t), if it is multiplied with uc(t), then

uc picks up the interval [c;). 

Rectangular pulse. The plot of the function looks like

for 0 · a < b < 1. We see it can be expressed as

and it picks up the interval [a; b).

Example For the function

In terms of the unit step function, f(t) can rewritten as:

Laplace transform of Step functions uc(t)

Shift of a function Given f(t), t > 0, then

is the shift of f by c units.

Shift of a function Let y = t - c, so t = y + c, and dt = dy,

which is equivalent to

Note now we are only considering the domain t  0. So u0(t) = 1 for all t  0.

Example For the function

In terms of the unit step function, f(t) can rewritten as:

The Laplace transform is

Example For the function

In terms of the unit step function, f(t) can rewritten as:

The Laplace transform is

Summary Steps for Solving an ODE d2y dy 6  8y  2 2 dt dt

y (0)  y ' (0)  0  ODE w/initial conditions

s 2 Y ( s)  6s Y ( s)  8 Y ( s)  2 / s

Y ( s) 

Y ( s) 

1 4s

2

s ( s  2) ( s  4)

1 1  2 ( s  2) 4 ( s  4)

y (t ) 

1 4

e 2t e 4t  2 4

Apply Laplace transform

to each term Solve for Y(s)

Apply partial fraction expansion Apply inverse Laplace transform to each term