There are techniques for finding the system response of a system described by a differential equation, based on the replacement of functions of a real variable (usually time or distance) by certain frequency-dependent representations, or by functions of a complex variable dependent upon frequency. The equations are converted from the time or space domain to the frequency domain through the use of mathematical transforms.

Differentialequations

Inputexcitation e(t)

Outputresponse r(t)

Time Domain Frequency Domain

Algebraicequations

Inputexcitation E(s)

Outputresponse R(s)

Laplace Transform

Inverse Laplace Transform

Let f(t) be a real function of a real variable t (time) defined for t>0. Then

is called the Laplace transform of f(t). The Laplace transform is a function of a complex variable s. Often s is separated into its real and imaginary parts: s=+j , where and are real variables.

After a solution of the transformed problem has been obtained in terms of s, it is necessary to "invert" this transform to obtain the solution in terms of the time variable, t. This transformation from the s-domain into the t-domain is called the inverse Laplace transform.

Let F(s) be the Laplace transform of a function f(t), t>0. The contour integral

where c>0 (0 as above) is called the inverse Laplace transform of F(s).

It is seldom necessary to perform the integration in the Laplace transform or the contour integration in the inverse Laplace transform. Most often, Laplace transforms and inverse Laplace transforms are found using tables of Laplace transform pairs.

Time Domainf(t), t>0

Frequency DomainF(s)

1. 12. K K/s3. Kt K/s2

4. Ke-at K/(s+a)5. Kte-at K/(s+a)2

6. Ksint K/(s2+2)

7. Kcost Ks/(s2+2)8. Ke-atsint K/((s+a)2+2))

9. Ke-atcost K(s+a)/((s+a)2+2))

Time Domain f(t), t>0

Frequency Domain F(s)

10. t s 11. f(t) F(s) 12. L-1{F(s)}=f(t) L{f(t)}=F(s) 13. Af1(t) + Bf2(t) AF1(s)+BF2(s)

14. t

df0

)(

F s

s

( )

15. df t

dt

( )

sF s f( ) ( ) 0

The inverse Laplace transform is usually more difficult than a simple table conversion.

X ss s

s s s( )

( )( )

( )( )

8 3 8

2 4

If we can break the right-hand side of the equation into a sum of terms and each term is in a table of Laplace transforms, we can get the inverse transform of the equation (partial fraction expansion).

X ss s

s s s

K

s

K

s

K

s( )

( )( )

( )( )

8 3 8

2 4 2 41 2 3

In general, there will be a term on the right-hand side for each root of the polynomial in the denominator of the left-hand side. Multiple roots for factors such as (s+2)n will have a term for each power of the factor from 1 to n.

Y ss

s

K

s

K

s( )

( )

( ) ( )

8 1

2 2 221 2

2

Complex roots are common, and they always occur in conjugate pairs. The two constants in the numerator of the complex conjugate terms are also complex conjugates.

Z ss s

K

s j

K

s j( )

.

( ) ( )

*

52

2 5 1 2 1 22

where K* is the complex conjugate of K.

The solution of each distinct (non-multiple) root, real or complex uses a two step process.

The first step in evaluating the constant is to multiply both sides of the equation by the factor in the denominator of the constant you wish to find.The second step is to replace s on both sides of the equation by the root of the factor by which you multiplied in step 1

X ss s

s s s

K

s

K

s

K

s( )

( )( )

( )( )

8 3 8

2 4 2 41 2 3

Ks s

s s s

1

0

8 3 8

2 4

8 0 3 0 8

0 2 0 424

( )( )

( )( )

( )( )

( )( )

Ks s

s s s

2

2

8 3 8

4

8 2 3 2 8

2 2 412

( )( )

( )

( )( )

( )

Ks s

s s s

3

4

8 3 8

2

8 4 3 4 8

4 4 44

( )( )

( )

( )( )

( )

The partial fraction expansion is:

X ss s s

( )

24 12

2

4

4

The inverse Laplace transform is found from the functional table pairs to be:

x t e et t( ) 24 12 42 4

Any unrepeated roots are found as before.The constants of the repeated roots (s-a)m are found by first breaking the quotient into a partial fraction expansion with descending powers from m to 0:

B

s a

B

s a

B

s am

m( ) ( ) ( )

2

21

The constants are found using one of the following:

BP a

Q s s am m

s a

( )

( ) / ( )

1)/()(

)(

)!(

1

1 as

mim

im

i assQ

sP

ds

d

imB

Y ss

s

K

s

K

s( )

( )

( ) ( )

8 1

2 2 221 2

2

Ks s

ss

ss2

2

22

2

8 1 2

28 1 8

( )( )

( )( )

The partial fraction expansion yields:

Y ss s

( )( )

8

2

8

2 2

8)2/()2(

)1(8

)!12(

1

222

s

i ss

s

ds

dB

The inverse Laplace transform derived from the functionaltable pairs yields:

y t e tet t( ) 8 82 2

Y ss

s

K

s

K

s( )

( )

( ) ( )

8 1

2 2 221 2

2

21 )2()1(8 KsKs

211 288 KKsKs

Equating like terms:

211 288 KKandK

211 288 KKandK

2828 K

28168 K

Thus

22

8

2

8)(

sssY

tt teety 22 88)(

221

2 22)2(

)1(8)(

s

K

s

K

s

ssY

Ks s

ss

ss2

2

22

2

8 1 2

28 1 8

( )( )

( )( )

As before, we can solve for K2 in the usual manner.

2212

22

2

8)2(

2)2(

)2(

)1(8)2(

s

ss

Ks

s

ss

ds

Ksd

ds

sd 82)1(8 1

18 K

22 2

8

2

8

)2(

)1(8)(

sss

ssY

tt teety 22 88)(

Unrepeated complex roots are solved similar to the process for unrepeated real roots. That is you multiply by one of the denominator terms in the partial fraction and solve for the appropriate constant.Once you have found one of the constants, the other constant is simply the complex conjugate.

Z ss s

K

s j

K

s j( )

.

( ) ( )

*

52

2 5 1 2 1 22

Ks j

s j s jj

s j

5 2 1 2

1 2 1 213

1 2

. ( )

( )( ).

K j* . 13

)21(

3.1

)21(

3.1

52

2.5)(

2 js

j

js

j

sssZ

)21()21()(

3.13.1

js

e

js

esZ

jj