1

Inverse Laplace Transform

Note that most of the transforms are written as fractions

2

nnnn

mmmm

asasasa

bsbsbsb

sQ

sPsF

11

10

11

10

...

...

)(

)()(

where P(s) and Q(s) are polynomials in the complex frequency variables s and the coefficients ao, a1, . . . , an, bo, b1, . . . , bm are real numbers. A rational function is specified completely by the two sets of real coefficients which define the numerator and denominator polynomials. On the other hand, the polynomials can also be expressed in the factored form in terms of their zeros. Thus, an alternate representation of F(s) is given

n

jj

m

ii

ps

zsKsF

1

1)(

Partial-fraction Expansion

3

The first step in the partial-fraction expansion is to put the rationalfunction into a proper form. We say that a rational function is proper if the degree of the numerator polynomial is less than the degree of the denominator polynomial. If the given rational function F(s) is not proper, i.e.., if the degree of P(s) is greater than or equal to that of Q(s), we divide (long division) P(s) by Q(s) and obtain

)(

)()(ˆ

)(

)()(

sQ

sRsP

sQ

sPsF

The quotient, is a polynomial and R(s) is the remainder; therefore, R(s) has a degree less than that of Q(s), and the new rational function R(s)/Q(s) is proper. Since is a polynomial, the corresponding time function is a linear combination of ,(1),(2), etc., and can be determined directly using Tables. We therefore go ahead with the new rational function R(s)/Q(s) which is proper. In the remaining part of this section we assume that all rational functions are proper.

)(ˆ sP

)(ˆ sP

4

Second Order Polynomial • Poles: roots of are called and • Zeros: roots of are called. • Pole-Zero Plot: plot in the complex frequency

domain zeros as ‘o’ and poles as ‘x’)

• In general we may write a second order Q(s) as

• The roots are

2 22 n ns s 2

1,2 1 ns :

:n

damping ratio

naturalfrequency

j

• If the damping ratio >1 (roots are real & distinct ‘simple’)• If the damping ratio <1 (roots are complex conjugates• If the damping ratio =1 (roots are real & equal ‘repeated’)

5

We start with a simple example as follows:

)3)(2)(1(

53

)(

)()(

2

sss

ss

sQ

sPsF

We claim that there are constants K1, K2, and K3 such that

12

323 5

( 1)( 2)( 3) 1 2 3

Ks s

s s s s s s

K K

1

1

2 3 51.5

( 2)( 3)s

s s

sK

s

2

2

2

3 53

( 1)( 3)s

s s

s sK

3

3

2 3 52.5

( 1)( 2)s

s s

sK

s

Heaviside’s Expansion

Heaviside’s Expansion

Compare Heaviside’s Expansion with solving for K or substituting convenient values for s.Compare Heaviside’s Expansion with solving for K or substituting convenient values for s.

Case 1 Simple poles

6

21

1

2 3

3 5( )

( 1)( 2)( 3)

1.5 3 2.5

1 2 3

1.5 3 2.5 for 0

-

-

t t t

s sf t

s s s

s s s

te e e

L

L

n

j j

j

ps

KsF

1

)(

and the residue Kj of the pole pj is given by

jpsjj sFpsK

)(

7

Find the partial-fraction expansion of

23)1(

1)(

sssF

The function has two multiple poles at s =p1=-1 (third order, n1=3) and at s = p2 = 0 (second order,n2= 2). Thus, the partial fraction expansion is of the form

22221

313

21211

111)(

s

K

s

K

s

K

s

K

s

KsF

To calculate K11, K12, and K13, we first multiply F(s) by (s +1)3 to obtain

23 1

)()1(s

sFs we find

11

1213

ssK 2

21

13

1212

ss ssds

dK

( )

( )

1( ) ( )

( )!

n mn

m n m s

dK s p F s

n m ds

Case 2 Multiple poles

8

32

)1(

1)(

ssFs

21 1 6

(1 / 2) 311 2 2 42 11

dK

ds s s ss

Similarly, to calculate K21 and K22, we first multiply F(s) by s2 to obtain

Using, we find

1

1

1

0

322

s

sK

3

1

1

0

321

s

sds

dK

Therefore, the partial-fraction expansion is

23223

13

1

1

1

2

1

3

)1(

1)(

ssssssssF

The corresponding time function is

0for t 3-23 )1(

1)( 2

21

23

1

tetteess

tf ttt-L

9

Case 3Complex poles

The two cases presented above are valid for poles which are either real or complex. However, if complex poles are present, the coefficients in the partial-fraction expansion are, in general, complex, and further simplification is possible.

First, we observe that F(s) is a ratio of polynomials with real coefficients; hence zeros and poles, if complex, must occur in complex conjugate pairs. More precisely, if p1 = 1 + j1 is a pole, that is, Q1(p1)=0, then is also a pole; that is, . This is due to the fact that any polynomial Q(s) with real coefficients has the property that for all s.Let us assume that the rational function has a simple pole at s = p1 = -+ j; then it must have another pole at s = p2 = =- - j. The partial-fraction expansion of F(s) must contain the following two terms:

*1 1 1p j *

1 1 0Q (p )

*1p

1 2K K

s j s j

Using formula for simple poles, we obtain

10

1 ( )s j

K s j F s

2 ( )s j

K s j F s

Since F(s) is a rational function of s with real coefficients, it follows that K2 is the complex conjugate of K1.

Once K and K* are determined, the corresponding two complex terms cn be combined as follows: Suppose K=a+jb. Then K*=a-jb, abd

*

2 2 2 2

( )( ) ( )( )

( )( )

2 ( ) 2

( ) ( )

K K a jb s j a jb s j

s j s j s j s j

a s b

s s

Once, we find a and b, given, , using the Inverse Laplace transform, we get

2 cos 2 sin ( )t tae t be t u t

,

11

This formula, which gives the corresponding time function for a pair of terms due to the complex conjugate poles, is extremely useful.

There are good Matlab examples utilizing the command residueThere are good Matlab examples utilizing the command residue

Practice the examples