PARTIAL FRACTIONS IN SHOCK AND VIBRATION ANALYSIS Revision K

By Tom IrvineEmail: tom@vibrationdata.com

January 11, 2013

Introduction

Transforming a fraction into the sum of partial fractions is an intermediate step in the solution of certain shock and vibration problems. The purpose of this tutorial is to summarize some common cases.

Term Appendix

{1s }{1 s2+2 ξ ωn s+ωn 2 } A

{1s2+ω2 }{αs+ β s2+2 ξ ωns+ωn 2 } B

αs+β ( s+ λ ) (s+σ ) , λ≠σ

C

[ 1ρ2− j 2 ξρ−1 ] [ 1

ρ2+ j 2ξρ−1 ] D

[ 1+ j 2ξρ( 1−ρ2 )+ j 2ξρ ] [ 1− j 2ξρ

( 1−ρ2)− j 2ξρ ] E

s2

(s+a ) ( s+b ), a≠b F

s4

(s+a ) ( s+b ) ( s+c ) ( s+d ), a≠b , a≠c , a≠d , b≠c , b≠d , c≠d G

1

Term Appendix

s(s+a )2

H

1(s2+α2 ) ( s+β )2

I

1s2 { 1

s2+ω2 }

J

Note that

s2+2 ξ ωn s+ωn 2=(s+ξ ωn)

2+ωn 2−ξ 2ω

n2 (1)

s2+2 ξ ωn s+ωn 2=(s+ξ ωn)2+ωn

2 [1−ξ 2 ] (2)

s2+2 ξ ωn s+ωn 2=(s+ξ ωn)

2+ωd 2

(3)

where

ωd =ωn

√1−ξ 2 (4)

Thus,

1s2+2 ξ ωns+ωn

2= 1

(s+ξ ωn)2+ωd 2

(5)

2

The inverse Laplace transform f(t) of equation (5) is

f ( t )= 1ωd

exp (−ξωn t )sin (ωd t ) (6)

Furthermore, consider the Laplace transform

F̂ (s )=s+ξωn

(s+ξ ωn)2+ωd 2

(7)

The inverse Laplace transform f̂ ( t ) is

f̂ ( t )=exp (−ξωn t )cos (ωd t ) (8)

3

APPENDIX A

Example 1

{1s }{1 s2+2 ξ ωn s+ωn 2 } ={ρs }+{σs+φ s2+2 ξ ωn s+ωn

2 } (A-1)

1=[ s2+2 ξ ωns+ωn 2] ρ + [σs+φ ] s (A-2)

1=[ ρs2+2 ξ ωn ρs+ρωn 2 ] + [σs2+φs ] (A-3)

1= ( ρ+σ ) s2+ (2 ξ ωn ρ+φ ) s+ρωn 2 (A-4)

Equation (A-4) implies three separate equations.

( ρ+σ )=0 (A-5)

(2 ξ ωn ρ+φ )=0 (A-6)

ρωn 2=1 (A-7)

Equation (A-7) yields

ρ= 1ωn 2

(A-8)

σ=−ρ (A-9)

σ = − 1ωn 2

(A-10)

φ=−2 ξ ωn ρ (A-11)

4

φ=−2 ξ ωn[ 1ωn 2 ]

(A-12)

φ= −2 ξωn

(A-13)

Equation (A-1) can thus be expressed as

{1s }{1 s2+2 ξ ωn s+ωn 2 } ={1ωn 2

s }+{−[1ωn 2 ] s−2 ξ

ωn

s2+2 ξ ωns+ωn 2 }

(A-14)

{1s }{1 s2+2 ξ ωn s+ωn 2 } =1

ωn 2 {1s }−1

ωn 2 {s+2 ξωn

s2+2 ξ ωns+ωn 2 }

(A-15)

{1s }{1 s2+2 ξ ωn s+ωn 2 } =1

ωn 2 {1s }−1

ωn 2 {s+2 ξωn

(s+ξ ωn)2+ωd 2 }

(A-16)

5

The following form is a more convenient format prior to taking the inverse Laplace transformation,

{1s }{1 s2+2 ξ ωn s+ωn 2 } =1

ωn 2 {1s }

−1ωn 2 {s+ξωn (s+ξ ωn)

2+ωd 2 }

−1ωn 2 {ξωn (s+ξ ωn)

2+ωd 2 }

(A-17)

{1s }{1 s2+2 ξ ωn s+ωn 2 } =1

ωn 2 {1s }

−(1ωn 2 ){s+ξωn (s+ξ ωn)

2+ωd 2 }

−(1ωn 2 )(ξωn

ωd ){ωd (s+ξ ωn)2+ωd

2 }

(A-18)

The inverse Laplace transform is

f ( t )= 1ωn 2 u( t )−

1ωn 2 exp (−ξ ωn t ) [cos (ωd t )+

ξωn

ωdsin (ωd t )] , t≥0

(A-19)

where u(t) is the unit step function.

6

APPENDIX B

Example 2

Equation (B-1) effectively represents a number of cases since and may each be set equal to zero.

{1s2+ω2 }{αs+ β s2+2 ξ ωns+ωn 2 } ={λs+ ρs2+ω2 }+{σs+φ s2+2 ξ ωn s+ωn

2 } (B-1)

Multiply through by the common denominator.

αs+β = {λs+ρ } {s2+2 ξ ωn s+ωn 2 }+ {σs+φ } {s2+ω2}

(B-2)

αs+β = λ s3+( ρ+2ξωn λ )s2+(2 ξ ωn ρ+λ ωn 2) s+( ρωn 2)

+σ s3+φ s2+σ ω2 s+φ ω2

(B-3)

αs+β =[ λ+σ ] s3

+ [ρ+2ξωn λ+φ ] s2

+ [2 ξ ωn ρ+ λωn 2+σω2] s+ [ρωn 2+φω2 ] (B-4)

Equation (B-4) implies four separate equations.

λ+σ=0 (B-5)

ρ+2 ξωn λ+φ=0 (B-6)

7

2 ξ ωn ρ+ λωn 2+σω2=α

(B-7)

ρ ωn 2+φ ω2=β

(B-8)

Equations (B-5) through (B-8) can be assembled into matrix form.

[ 1 0 1 02ξωn 1 0 1ωn 2 2 ξωn ω2 00 ωn

2 0 ω2 ][ λρσφ ]=[ 00αβ ]

(B-9)

Gaussian elimination is used to simplify the coefficient matrix.

[1 0 1 00 1 −2 ξωn 10 2ξωn ω2−ωn

2 00 ωn

2 0 ω2 ] [ λρσφ ]=[00αβ ] (B-10)

Equation (B-10) can be reduced to a 3 x 3 matrix.

[ 1 −2ξωn 12ξωn ω2−ωn

2 0ωn 2 0 ω2 ] [ ρσφ ]=[ 0

αβ ]

(B-11)

Complete the solution using Cramer’s rule.

det [ 1 −2ξωn 12ξωn ω2−ωn

2 0ωn 2 0 ω2 ]=ω2 (ω2−ωn

2)+ω2(2ξωn )2−ωn 2(ω2−ωn 2)

(B-12)

8

det [ 1 −2ξωn 12ξωn ω2−ωn

2 0ωn 2 0 ω2 ]=(ω2−ωn

2 )2+ (2ξ ω ωn )2

(B-13)

ρ= 1

(ω2−ωn 2)2+(2 ξ ω ωn )2

det [ 0 −2 ξωn 1α ω2−ωn

2 0β 0 ω2 ]

(B-14)

ρ=2ξ ω 2ωn α−(ω2−ωn

2) β(ω2−ωn

2)2+(2ξ ω ωn )2 (B-15)

σ= 1

(ω2−ωn 2)2+(2 ξ ω ωn)

2det [ 1 0 1

2ξωn α 0ωn 2 β ω2 ]

(B-16)

σ=ω2α+2 ξωn β−ωn

2α

(ω2−ωn 2)2+(2ξ ω ωn)2 (B-17a)

σ=(ω2−ωn

2) α+2 ξωn β

(ω2−ωn 2)2+(2ξ ω ωn)2 (B-17b)

Recall equation (B-5).

λ=−σ (B-18)

9

λ=−(ω2−ωn

2) α+2ξωn β

(ω2−ωn 2)2+(2 ξ ω ωn)2 (B-19)

φ= 1

(ω2−ωn 2)2+(2ξ ω ωn)2

det [ 1 −2ξωn 02ξωn ω2−ωn

2 αωn 2 0 β ]

(B-20)

φ=(ω2−ωn

2) β+(2ξωn )2 β−2ξωn 3 α

(ω2−ωn 2)2+(2 ξ ω ωn)2 (B-21a)

φ=[ (ω2−ωn

2)+(2ξωn )2 ] β−2ξωn 3α

(ω2−ωn 2 )2+ (2ξ ω ωn)2 (B-21b)

The coefficients are summarized in equation (B-22).

[ λρσφ ]= 1

(ω2−ωn 2)2+(2ξ ω ωn )2 [ −(ω2−ωn

2 ) α−2ξωn β

2ξ ω 2ωn α−(ω2−ωn 2) β

(ω2−ωn 2) α+2ξωn β

−2ξωn 3α+[(ω2−ωn

2)+(2ξωn )2 ] β ]

(B-22a)

10

[ λρσφ ]= 1

(ω2−ωn 2)2+(2ξ ω ωn )2 { [−(ω2−ωn

2)2 ξ ω 2ωn(ω2−ωn

2)−2 ξωn

3 ] α+[−2ξωn

−(ω2−ωn 2 )

2ξωn

[ (ω2−ωn 2)+(2 ξωn )2] ] β }

(B-22b)

Equation (B-1) can thus be rewritten as

{1s2+ω2 }{αs+ β s2+2 ξ ωns+ωn 2 } =

[−(ω2−ωn 2) α−2ξωn β ] s+[2ξ ω 2ωn α−(ω2−ωn

2) β ][ (ω2−ωn

2)2+(2 ξ ω ωn)2] [ s2+ω2 ]

+[ (ω2−ωn

2) α+2ξωn β ]s+[−2ξωn 3 α+[(ω2−ωn

2)+(2ξωn)2] β ] [ (ω2−ωn

2)2+(2ξ ω ωn )2 ] [ s2+2 ξ ωn s+ωn 2 ]

(B-23)

11

An alternate form is

{1s2+ω2 }{αs+ β s2+2 ξ ωns+ωn 2 } =

α[−(ω2−ωn

2) ] s+[ 2ξ ω 2ωn ][ (ω2−ωn

2)2+(2ξ ω ωn )2] [ s2+ω2 ]

+β[−2ξωn ] s+[−(ω2−ωn

2) ][ (ω2−ωn

2)2+(2ξ ω ωn )2] [ s2+ω2]

+α [(ω2−ωn 2) ] s+[−2ξωn

3 ]¿ [ (ω2−ωn

2)2+(2ξ ω ωn)2] [ s2+2 ξ ωns+ωn

2] ¿¿

¿ +β[2ξωn ] s+[ (ω2−ωn

2)+(2ξωn )2 ] [ (ω2−ωn

2)2+ (2 ξ ω ωn)2] [ s2+2 ξ ωns+ωn

2]¿¿

(B-24)

12

{1s2+ω2 }{αs+ β s2+2 ξ ωns+ωn 2 } =

α[−(ω2−ωn

2) ] s+[ 2ξ ω 2ωn ][ (ω2−ωn

2)2+(2ξ ω ωn )2] [ s2+ω2]

+β[−2ξωn ] s+[−(ω2−ωn

2) ][ (ω2−ωn

2)2+( 2ξ ω ωn )2] [ s2+ω2]

+α [(ω2−ωn 2) ] s+[−2ξωn

3]¿ [ (ω2−ωn

2)2+(2ξ ω ωn)2] [ (s+ξ ωn)

2+ωd 2] ¿

¿

¿ +β[2 ξωn ] s+[(ω2−ωn

2 )+(2ξωn)2] [ (ω2−ωn

2)2+(2ξ ω ωn )2] [ (s+ξ ωn)2+ωd

2 ]¿¿

(B-25)

13

The inverse Laplace transforms for the four terms on the left-hand-side are found as follows

F1(s )=α[−(ω2−ωn

2) ] s+[2ξ ω 2ωn ][ (ω2−ωn

2)2+(2ξ ω ωn)2] [ s2+ω2] (B-26)

f 1( t )=α

[ (ω2−ωn 2)2+(2ξ ω ωn )2] { [−(ω2−ωn

2) ] cos (ωt )+[2ξ ω ωn ] sin (ωt )}

(B-27)

F2(s )=+ β[−2ξωn ] s+[−(ω2−ωn

2) ][(ω2−ωn

2)2+(2ξ ω ωn)2 ] [ s2+ω2 ] (B-28)

f 2( t )=β

[ (ω2−ωn 2)2+(2ξ ω ωn )2] { [−2ξωn ] cos (ωt )+[−1

ω (ω2−ωn 2) ] sin (ωt )}

(B-29)

14

F3( s )=+α [ (ω2−ωn 2) ]s+[−2 ξωn

3 ]¿ [(ω2−ωn

2 )2+ (2ξ ω ωn)2 ] [(s+ξ ωn )2+ωd 2] ¿¿

¿ (B-30)

F3( s )=+α [ (ω2−ωn 2) ]s ¿

[(ω2−ωn 2 )2+ (2 ξ ω ωn )2 ] [(s+ξ ωn )2+ωd 2 ] ¿

¿¿ +α [−2 ξωn

3 ]¿ [(ω2−ωn

2)2+(2 ξ ω ωn)2 ] [ (s+ξ ωn )2+ωd 2] ¿¿

¿ ¿¿ (B-31)

F3( s )=+α [ (ω2−ωn 2) ]s ¿

[(ω2−ωn 2 )2+ (2 ξ ω ωn )2 ] [ (s+ξ ωn )2+ωd 2 ] ¿

¿¿ +α(−2ξωn

3

ωd )ωd ¿ [ (ω2−ωn 2)2+(2 ξ ω ωn)2] [ (s+ξ ωn)2+ωd

2] ¿¿

¿ ¿¿(B-32)

15

f 3( t )=α exp (−ξωn t )¿ [ (ω2−ωn

2)2+(2ξ ω ωn)2] ¿

¿ [(ω2−ωn 2 )cos (ωd t )+(−2ξωn

3

ωd )sin (ωd t )]¿(B-33)

F4 (s )=+β[2ξωn ] s+[(ω2−ωn

2 )+(2ξωn)2 ] [(ω2−ωn

2)2+(2ξ ω ωn)2 ] [(s+ξ ωn )2+ωd 2]

(B-34)

F4 (s )=+β [2ξωn ]s+ 1

[2ξωn ] [ (ω2−ωn

2)+(2ξωn )2] [(ω2−ωn

2 )2+ (2ξ ω ωn )2 ] [(s+ξ ωn )2+ωd 2 ] (B-35)

16

f 4 ( t )=β [2ξωn ]exp (−ξωn t )

[(ω2−ωn 2 )2+ (2ξ ω ωn )2 ] {cos (ωd t )+[ 1

[2ξωn ] [(ω2−ωn

2)+(2ξωn)2 ]−ξωnωd ]sin (ωd t )}

(B-36)

f 4 ( t )=β exp (−ξωn t )

[(ω2−ωn 2)2+(2ξ ω ωn )2 ] {2ξωncos (ωd t )+[ [(ω2−ωn

2 )+(2ξωn)2 ]−ξωn [2 ξωn ]

ωd ]sin (ωd t )}

(B-37)

f 4 ( t )=β exp (−ξωn t )

[(ω2−ωn 2 )2+(2ξ ω ωn )2 ] {2ξωncos (ωd t )+[ [(ω2−ωn

2 )+(2ξωn)2]−2ξ2ωn2

ωd ]sin (ωd t )}

(B-38)

f 4 ( t )=β exp (−ξωn t )

[(ω2−ωn 2 )2+(2ξ ω ωn )2 ] {2ξωncos (ωd t )+[ [(ω2−ωn

2 )+2 (ξωn)2]

ωd ]sin (ωd t )}

(B-39)

17

APPENDIX C

Example 3

Equation (C-1) effectively represents a number of cases since and may each be set equal to zero.

{αs+β ( s+λ ) ( s+σ ) } ={ps+λ }+{ms+σ } , λ≠σ (C-1)

{αs+β } ={ps+λ }( s+λ ) (s+σ )+{ms+σ } (s+λ ) ( s+σ ) (C-2)

(αs+β ) = (p ) ( s+σ )+(m ) ( s+ λ ) (C-3)

(αs+β ) =( p s+ pσ )+ ( m s+mλ ) (C-4)

(αs+β ) =(p+m ) s+( pσ+mλ ) (C-5)

Equation (C-5) implies equations (C-6) and (C-7).

α= (p+m) (C-6)

β=( pσ+mλ ) (C-7)

Equation (C-6) yields

m=α−p (C-8)

Substitute equation (C-8) into (C-7).

β=( pσ+(α−p ) λ ) (C-9)

β=(σ−λ ) p+αλ (C-10)

18

β−αλ=(σ−λ ) p (C-11)

(σ−λ ) p=β−αλ (C-12)

p= β−αλσ−λ (C-13)

Substitute equation (C-13) into (C-8).

m=α−( β−αλσ−λ ) (C-14)

m=α (σ− λ )−(β−αλ )

σ−λ (C-15)

m=(ασ−αλ )−(β−αλ )

σ−λ (C-16)

m=ασ−αλ−β+αλσ−λ (C-17)

m=ασ−βσ−λ (C-18)

Substitute equations (C-18) and (C-13) into (C-1).

{αs+β ( s+λ ) ( s+σ ) } ={1σ−λ }{ [ β−αλs+λ ]+[ασ−β s+σ ] } (C-19)

The inverse Laplace transform is

f ( t ) ={1σ−λ }{1λ

[ β−αλ ]sin ( λt )+1σ

[ασ−β ] sin (σ t ) } (C-20)

19

APPENDIX D

Example 4

The transfer function H ( ρ ) times its complex conjugate is

H ( ρ )H∗(ρ )=[ 1ρ2− j 2ξρ−1 ] [ 1

ρ2+ j 2 ξρ−1 ] (D-1)

Solve for the roots R1 and R2 of the first denominator.

R1 ,R2= j 2ξ±√ (− j 2 ξ )2−4(−1)2 (D-2)

R1 ,R2= j 2ξ±√ −4 ξ2+42 (D-3)

R1 ,R2= jξ±√ 1−ξ2 (D-4)

Solve for the roots R3 and R4 of the second denominator.

R3 ,R 4=− jξ±√ ( j2ξ )2−4 (−1 )

2 (D-5)

R3 , R 4=− j2ξ±√−4ξ2+42 (D-6)

R3 ,R 4=− jξ±√ 1−ξ2 (D-7)

(D-8)Summary,

20

R1=+ jξ+√ 1−ξ2 (D-9)

R2=+ jξ−√ 1−ξ2 (D-10)

R3=− jξ+√ 1−ξ2 (D-11)

R4=− jξ−√ 1−ξ2 (D-12)

Note

R2 = -R1* (D-13)

R3 = R1* (D-14)

R4 = -R1* (D-15)

Now substitute into the denominators.

H ( ρ )H∗(ρ )=[ 1( ρ− jξ−√ 1−ξ2) (ρ− jξ+√ 1−ξ2) ( ρ+ jξ−√ 1−ξ2) ( ρ+ jξ+√ 1−ξ2 ) ]

(D-16)

H ( ρ ) H∗( ρ )=[ 1(ρ−R1 ) ( ρ−R2 ) ( ρ−R3 ) (ρ−R 4 ) ] (D-17)

H ( ρ ) H∗( ρ )=¿¿ (D-18)

21

Expand into partial fractions.

¿¿¿

¿

(D-19)

Multiply through by the denominator on the left-hand side of equation (D-19).

1= +α( ρ−R1 )

( ρ−R1 ) ¿¿

¿¿

(D-20)

1= +α ¿¿¿

¿ (D-21)

22

1= +α ¿¿¿

¿

(D-22)

1= +α ¿¿¿

¿(D-23)

1= +α ¿¿¿

¿(D-24)

1= + [ α+ β+ λ+σ ] ρ3

+ [ R1 α+R 1∗β−R1∗λ−R 1 σ ] ρ 2

+[ −R1∗2α−R 12 β−R12 λ−R1∗2σ ] ρ

+[ −R1R1∗2α−R12R1∗β+R12R1∗λ+R 1R1∗2σ ]

(D-25)

23

1= + [α+β+λ+σ ] ρ3

+ [R1 α+R1∗β−R1∗λ−R1 σ ] ρ 2

+ [ −R1∗2α−R12 β−R12 λ−R1∗2σ ] ρ

+ [ −R1∗α−R1 β+R1 λ+R1∗σ ] R1R1∗ (D-26)

Equation (D-26) can be broken up into four separate equations,

α+β+ λ+σ=0 (D-27)

[ R1α+R1∗β−R1∗λ−R1σ ]=0 (D-28)

[ −R1∗2α−R12 β−R12 λ−R1∗2σ ] =0 (D-29)

[ −R1∗α−R1 β+R1 λ+R1∗σ ] R1R1∗ =1 (D-30)

The four equations are assembled into matrix form.

[ 1 1 1 1R1 R1∗ −R1∗ −R1

−R1∗2 −R12 −R12 −R1∗2

−R1∗ −R1 +R1 +R1∗ ] [αβλσ ]=¿¿

(D-31)Recall

R1=+ jξ+√1−ξ2 (D-32)

24

R1 R1∗¿ [+ jξ+√1−ξ2] [− jξ+√1−ξ2] (D-33)

R1 R1* =1 (D-34)

[ 1 1 1 1R1 R1∗ −R1∗ −R1

−R1∗2 −R12 −R12 −R1∗2

−R1∗ −R1 +R1 +R1∗ ] [αβλσ ]=[0001 ](D-35)

Multiply the first row by R1∗2¿ ¿ and add to the third row.

[ 1 1 1 1R1 R1∗ −R1∗ −R10 −R12+R1∗2 −R12+R1∗2 0

−R1∗ −R1 +R1 +R1∗] [αβλσ ]=[0001 ](D-36)

Scale the third row.

[ 1 1 1 1R1 R 1∗ −R 1∗ −R10 1 1 0

−R1∗ −R1 +R1 +R1∗] [αβλσ ]=[0001 ]

(D-37)Multiply the third row by -1 and add to the first row.

[ 1 0 0 1R1 R 1∗ −R 1∗ −R10 1 1 0

−R1∗ −R1 +R1 +R1∗] [αβλσ ]=[0001 ]25

(D-38)

Multiply the first row by -R1 and add to the second row. Also multiply the first row by R1* and add to the fourth row.

[1 0 0 10 R1∗ −R1∗ −2 R 10 1 1 00 −R1 +R1 +2 R1∗] [αβλσ ]=[0001 ]

(D-39)

Multiply the third row by -R1* and add to the second row. Also, multiply the third row by R1 and add to the fourth row.

[1 0 0 10 0 −2R1∗ −2 R10 1 1 00 0 +2R1 +2 R1∗ ] [αβλσ ]=[0001 ]

(D-40)

The first row equation yields

α=−σ (D-41)

The third row equation yields

β=− λ (D-42)

Equation (D-40) thus reduces to

[−2R1∗ −2R1+2 R1 +2 R1∗] [λσ ]=[01 ]

(D-43)

−2 R1∗λ−2R1 σ=0 (D-44)

−R1∗λ=R1 σ (D-45)

26

λ= −R1R1∗¿

σ ¿ (D-46)

2 R1 λ + 2R1 ∗σ=1 (D-47)

2 R1 ¿¿ (D-48)

¿¿ (D-49)

σ=12

1¿¿ ¿¿

(D-50)

σ=12R1 ∗¿

[ −R12 + R1 ∗2 ]¿ (D-51)

λ= −R1R1∗¿

σ ¿ (D-52)

λ=−12

R1[ −R12 + R1 ∗2 ] (D-53)

Recall,α=−σ (D-54)

β=− λ (D-55)

The complete solution set is thus

27

[αβλσ ]=1

21

[ −R12 + R1 ∗2 ]¿ ¿

(D-56)

28

Note that

−R12+R1∗2¿− j [ 4ξ √1−ξ2] (D-57)

[αβλσ ]=12 [ 1

− j [ 4ξ √1−ξ2] ] [−√1−ξ2+ jξ+√1−ξ2+ jξ−√1−ξ2− jξ+√1−ξ2− jξ

](D-58)

[αβλσ ]= j[8ξ √1−ξ2 ] [−√1−ξ2+ jξ

+√1−ξ2+ jξ−√1−ξ2− jξ+√1−ξ2− jξ

](D-59)

[αβλσ ]= 1[ 8 ξ √1−ξ2 ] [−ξ− j√1−ξ2

−ξ + j√1−ξ2

+ξ− j√1−ξ2

+ξ+ j√1−ξ2 ](D-60)

Let

ψ= ξ√1−ξ2

(D-61)

29

[αβλσ ]= 18 ξ [−ψ− j

−ψ + j+ψ− j+ψ+ j ]

(D-62)

Recall

R1=+ jξ+√1−ξ2 (D-63)

¿¿¿

¿(D-64)

30

APPENDIX E

Example 5

A transfer function times its complex conjugate is

H ( ρ )H∗( ρ )=[ 1+ j 2ξρ( 1−ρ2 )+ j 2ξρ ] [ 1− j 2ξρ

( 1−ρ2 )− j 2ξρ ] (E-1)

Rearrange equation (E-1) as

H ( ρ )H∗(ρ )=[ 1+ 4 ξ2 ρ2

ρ2− j 2ξρ−1 ] [ 1ρ2+ j 2 ξρ−1 ]

(E-2)

Solve for the roots R1 and R2 of the first denominator.

R1 ,R2= j 2ξ±√ (− j 2 ξ )2−4(−1)2 (E-3)

R1 ,R2= j 2ξ±√ −4 ξ2+42 (E-4)

R1 ,R2= jξ±√ 1−ξ2 (E-5)

Solve for the roots R3 and R4 of the second denominator.

R3 ,R 4=− j2 ξ±√ ( j2ξ )2−4 (−1 )

2 (E-6)

R 3 ,R 4=− j2ξ±√−4ξ2+42 (E-7)

31

R3 ,R 4=− jξ±√ 1−ξ2 (E-8)

Summary,

R1=+ jξ+√ 1−ξ2 (E-9)

R2=+ jξ−√ 1−ξ2 (E-10)

R3=− jξ+√ 1−ξ2 (E-11)

R4=− jξ−√ 1−ξ2 (E-12)

Note

R2 = -R1* (E-13)

R3 = R1* (E-14)

R4 = -R1* (E-15)

Now substitute into the denominators.

H ( ρ )H∗(ρ )=[ 1+ 4 ξ2 ρ2

( ρ− jξ−√ 1−ξ2) (ρ− jξ+√ 1−ξ2) ( ρ+ jξ−√ 1−ξ2) ( ρ+ jξ+√ 1−ξ2 ) ](E-16)

H ( ρ ) H∗( ρ )=[ 1+ 4 ξ2 ρ2

(ρ−R1 ) ( ρ−R2 ) ( ρ−R3 ) ( ρ−R 4 ) ] (E-17)

By substitution,

32

H ( ρ ) H∗( ρ )=¿¿ (E-18)

Expand into partial fractions.

¿¿¿

¿

(E-19)

Multiply through by the denominator on the left-hand side of equation (E-19).

[ 1+ 4 ξ2 ρ2 ]=

+α( ρ−R1 )

( ρ−R1 ) ¿

¿

¿

(E-20)

33

[ 1+ 4 ξ2 ρ2 ]=

+α ¿¿

¿

(E-21)

[ 1+ 4 ξ2 ρ2 ]=

+α ¿¿

¿

(E-22)

[ 1+ 4 ξ2 ρ2 ]=

+α ¿¿

¿

(E-23)

34

[ 1+ 4 ξ2 ρ2 ]=

+α ¿¿

¿

(E-24)[ 1+ 4 ξ2 ρ2 ]=

+ [ α+ β+λ+σ ] ρ3

+ [ R1 α+R1∗β−R1∗λ−R1 σ ] ρ 2

+ [ −R1∗2α−R12 β−R12 λ−R1∗2σ ] ρ

+ [ −R1 R1∗2α−R12R1∗β+R12R1∗λ+R1R1∗2σ ]

(E-25)

[ 1+ 4 ξ2 ρ2 ]=

+[α+β+λ+σ ] ρ3

+[R1 α+R1∗β−R1∗λ−R1 σ ] ρ2

+[ −R1∗2α−R12 β−R12 λ−R1∗2σ ] ρ

+[ −R1∗α−R1 β+R1 λ+R1∗σ ] R1 R1∗

(E-26)

35

Equation (E-26) can be broken up into four separate equations.

α+β+ λ+σ=0 (E-27)

[ R1α+R1∗β−R1∗λ−R1σ ]=4 ξ2 (E-28)

[ −R1∗2α−R12 β−R12 λ−R1∗2σ ]=0 (E-29)

[ −R1∗α−R1 β+R1 λ+R1∗σ ] R1R1∗¿1 (E-30)

The four equations are assembled into matrix form.

[ 1 1 1 1R1 R1∗ −R1∗ −R1

−R1∗2 −R12 −R12 −R1∗2

−R1∗ −R1 +R1 +R1∗ ] [αβλσ ]=¿¿

(E-31)Recall

R1=+ jξ+√1−ξ2 (E-32)

R1 R1∗¿ [+ jξ+√1−ξ2] [− jξ+√1−ξ2] (E-33)

R1 R1* =1 (E-34)

By substitution,

[ 1 1 1 1R1 R1∗ −R1∗ −R1

−R1∗2 −R12 −R12 −R1∗2

−R1∗ −R1 +R1 +R1∗ ] [αβλσ ]=[ 04 ξ2

01 ]

(E-35)

36

Multiply the first row by R1∗2¿ ¿ and add to the third row.

[ 1 1 1 1R1 R1∗ −R1∗ −R10 −R12+R1∗2 −R 12+R 1∗2 0

−R1∗ −R1 +R1 +R1∗] [αβλσ ]=[ 04 ξ2

01 ]

(E-36)Scale the third row.

[ 1 1 1 1R1 R1∗ −R1∗ −R 10 1 1 0

−R1∗ −R1 +R1 +R1∗] [αβλσ ]=[ 04 ξ2

01 ]

(E-37)Multiply the third row by -1 and add to the first row.

[ 1 0 0 1R1 R1∗ −R1∗ −R 10 1 1 0

−R1∗ −R1 +R1 +R1∗] [αβλσ ]=[ 04 ξ2

01 ]

(E-38)

Multiply the first row by -R1 and add to the second row. Also multiply the first row by R1* and add to the fourth row.

[1 0 0 10 R1∗ −R1∗ −2 R10 1 1 00 −R 1 +R1 +2 R1∗] [αβλσ ]=[ 0

4 ξ2

01 ]

(E-39)

Multiply the third row by -R1* and add to the second row. Also, multiply the third row by R1 and add to the fourth row.

37

[1 0 0 10 0 −2R 1∗ −2 R10 1 1 00 0 +2R1 +2 R 1∗ ] [αβλσ ]=[ 0

4 ξ2

01 ]

(E-40)

The first row equation yields

α=−σ (E-41)

The third row equation yields

λ=−β (E-42)

Equation (E-40) thus reduces to

[−2R1∗ −2R1+2 R1 +2 R1∗] [λσ ]=[4ξ2

1 ] (E-43)

Complete the solution using Cramer's rule.

det er minant [−2 R1∗ −2 R1+2R1 +2R1∗] [ λσ ]=4 [ R12−R1∗2 ]

(E-44)

Recall

R1=+ jξ+√1−ξ2 (E-45)

R12= [+ jξ+√1−ξ2 ] [+ jξ+√1−ξ2 ] (E-46)

R12=−ξ2+(1−ξ2 )+ j [2ξ √1−ξ2] (E-47)

R12= ( 1−2ξ2)+ j [2ξ √1−ξ2 ] (E-48)

R1∗¿− jξ+√1−ξ2 (E-49)

38

R1∗2¿ [− jξ+√1−ξ2 ] [− jξ+√1−ξ2 ] (E-50)

R1∗2¿−ξ2+( 1−ξ2)− j [2 ξ √1−ξ2] (E-51)

R1∗2¿ ( 1−2ξ2)− j [2ξ √1−ξ2 ] (E-52)Thus,

R12−R1∗2¿ j [4 ξ √1−ξ2 ] (E-53)

4 [ R12−R1∗2 ]= j [ 16 ξ √1−ξ2 ] (E-54)

determinant [−2 R1∗ −2R1+2R1 +2 R1∗] [ λσ ]= j [16 ξ √1−ξ2 ]

(E-55)

λ= 1j [16 ξ √1−ξ2 ]

determinant [4 ξ2 −2 R11 +2R1∗]

(E-56)

λ=( 4ξ2) ¿¿¿¿ (E-57)

λ=( 4ξ2) ¿¿¿¿ (E-58)

Recall,

39

R1=+ jξ+√1−ξ2 (E-59)

λ=(4 ξ2 ) (− jξ+√1−ξ2)+ jξ+√1−ξ2

j [8 ξ √1−ξ2] (E-60)

λ=( 1+4 ξ2) (√1−ξ2)+ jξ ( 1−4 ξ2 )

j [ 8 ξ √1−ξ2 ] (E-61)

λ=+ξ ( 1−4ξ2 )− j ( 1+4ξ2) (√1−ξ2 )

[8 ξ √1−ξ2] (E-62)

σ= 1j [16 ξ √1−ξ2 ]

det ermin ant [−2 R1∗ 4 ξ2

2R1 1 ] (E-63)

σ=[ −2R1∗−(4 ξ2 ) (2R1 ) ]

j [16 ξ √1−ξ2 ] (E-64)

σ=[ −R1∗−( 4 ξ2) (R1 ) ]

j [ 8 ξ √1−ξ2 ] (E-65)

R1=+ jξ+√1−ξ2 (E-66)

40

σ=[ −(− jξ+√1−ξ2)−( 4 ξ2) (+ jξ+√1−ξ2 )]

j [8 ξ √1−ξ2 ] (E-67)

σ=[ + jξ−√1−ξ2− j 4 ξ3−4ξ2√1−ξ2 ]

j [8 ξ √1−ξ2] (E-68)

σ=[ −( 1+4 ξ2)√1−ξ2+ jξ ( 1−4 ξ2) ]

j [8 ξ √1−ξ2 ] (E-69)

σ=[ +( 1−4 ξ2)+ j ( 1+4ξ2 ) √1−ξ2]

j [8 ξ √1−ξ2] (E-70)

Recall,α=−σ (E-71)

λ=−β (E-72)

The complete solution set is thus

[αβλσ ]= 18 ξ √1−ξ2 [−( 1−4 ξ2 )− j ( 1+4 ξ2 ) √1−ξ2

−( 1−4 ξ2)+ j ( 1+4ξ2 ) √1−ξ2

+( 1−4 ξ2 )− j ( 1+4 ξ2 ) √1−ξ2

+( 1−4 ξ2 )+ j ( 1+4ξ2) √1−ξ2 ]

(E-73)

41

The partial fraction expansion is thus

H ( ρ ) H∗( ρ )=

+α( ρ−R1 )

+β¿¿ ¿¿¿

¿

(E-74)

42

APPENDIX F

s2

(s+a ) ( s+b )=α+ β

s+a+ λs+b (F-1)

s2 =α (s+a ) ( s+b )+ β( s+a )

( s+a ) (s+b )+ λ(s+b )

(s+a ) ( s+b ) (F-2)

s2 =α (s2+(a+b) s+ab )+β (s+b )+ λ (s+a ) (F-3)

s2 =αs2+α (a+b ) s+α ab+βs+ βb+λs+ λa (F-4)

s2 =αs2+[α (a+b ) +β+λ ] s+α ab+βb+ λa (F-5)

Equation (F-5) yields three individual equations.

α=1 (F-6)

[α (a+b) +β+λ ] =0 (F-7)

α ab+βb+ λa=0 (F-8)

[(a+b ) +β+λ ] =0 (F-9)

ab+βb+λa=0 (F-10)

β+ λ=−(a+b) (F-11)

βb+ λa=−ab (F-12)

43

The equations are assembled into matrix form.

[1 1b a ] [ βλ ]=[−( a+b )

−ab ] (F-13)

Apply Cramer’s rule.

β= 1a−b

det [−(a+b ) 1−ab a ] [βλ ] (F-14)

β=−a2−ab+ab

a−b (F-15)

β= −a2

a−b (F-16)

λ= 1a−b

det [1 −(a+b )b −ab ]

(F-17)

λ=−ab+b (a+b )a−b (F-18)

λ=−ab+ab+b2

a−b (F-19)

λ= b2

a−b (F-20)

The partial fraction expansion is thus

44

s2

(s+a ) ( s+b )=1+[ −a2

a−b ][ 1s+a ]+[ b2

a−b ] [ λs+b ] , a≠b

(F-21)

APPENDIX G

s4

( s+a ) (s+b ) (s+c ) (s+d )

={1+[−a2

a−b ] [1s+a ]+[b2

a−b ] [1s+b ]}{1+[−c2

c−d ] [1s+c ]+[d2

c−d ] [1s+d ]}(G-1)

Let

A=[ −a2

a−b ] (G-2)

B=[b2

a−b ] (G-3)

C=[−c2

c−d ] (G-4)

D=[d2

c−d ] (G-5)

45

s4

( s+a ) (s+b ) (s+c ) (s+d )

={1+A [1s+a ]+B [1s+b ]}{1+C [1s+c ]+D [1s+d ]}(G-6)

s4

( s+a ) (s+b ) (s+c ) (s+d )

¿ {1+C [1s+c ]+D [1s+d ]}+A [1s+a ]{1+C [1s+c ]+D [1s+d ]}+B [1s+b ]{1+C [1s+c ]+D [1s+d ]}

(G-7)

s4

( s+a ) (s+b ) (s+c ) (s+d )

¿ {1+C [1s+c ]+D [1s+d ]}+A {[1s+a ]+C [1s+a ] [1s+c ]+D [1s+a ] [1s+d ]}+B {[1s+b ]+C [1s+b ] [1s+c ]+D [1s+b ] [1s+d ]}

(G-8)

Recall

1(s+a ) ( s+b )

= [ 1a−b ] [ −1

s+a+ 1s+b ] , a≠b

(G-9)

46

s4

( s+a ) (s+b ) (s+c ) (s+d )

¿ {1+C [1s+c ]+D [1s+d ]}+A {[1s+a ]+C [1a−c ] [−1

s+a+1s+c ]+D [1a−d ] [−1

s+a+1s+d ]}

+B {[1s+b ]+C [1b−c ] [−1s+b

+1s+c ]+D [1b−d ][−1

s+b+1s+d ]}

(G-10)

s4

( s+a ) (s+b ) (s+c ) (s+d )

¿ {1+C [1s+c ]+D [1s+d ]}+A {[1s+a ]+C [1a−c ] [−1

s+a+1s+c ]+D [1a−d ] [−1

s+a+1s+d ]}

+B {[1s+b ]+C [1b−c ] [−1s+b

+1s+c ]+D [1b−d ][−1

s+b+1s+d ]}

(G-11)

s4

( s+a ) (s+b ) (s+c ) (s+d )

¿{1+C [1s+c ]+D [1s+d ]}+ {[As+a ]+[ACa−c ] [−1

s+a ]+[ACa−c ] [1s+c ]+[ADa−d ] [−1s+a ]+[ADa−d ] [1s+d ]}

+ {[Bs+b ]+[BCb−c ] [−1s+b ]+[BCb−c ] [1s+c ]+[BDb−d ][−1

s+b ]+[BDb−d ] [1s+d ]}(G-12)

47

s4

( s+a ) (s+b ) (s+c ) (s+d )=

1+A [1−[Ca−c ]−[Da−d ]] [1s+a ]+B[1−[Cb−c ]−[Db−d ]][1s+b ]+C [1+[Aa−c ]+[Bb−c ]] [1s+c ]+D[1+[Aa−d ]+[Bb−d ]] [1s+d ]

(G-13)

s4

( s+a ) (s+b ) (s+c ) (s+d )=

1+[−a2

a−b ][1−[−c2

(a−c ) (c−d ) ]−[d2

(a−d ) (c−d ) ] ] [1s+a ]+[b2

a−b ][1−[−c2

(b−c ) ( c−d ) ]−[d2

(c−d ) (b−d ) ] ][1s+b ]+[−c2

c−d ] [1+[−a2

(a−b ) (a−c ) ]+[b2

(a−b ) (b−c ) ] ] [1s+c ]+[d2

c−d ] [1+[−a2

(a−b ) (a−d ) ]+[b2

(a−b ) (b−d ) ] ] [1s+d ](G-14)

48

s4

( s+a ) (s+b ) (s+c ) (s+d )=

1+[−a2

a−b ][1+[c2

(a−c ) (c−d ) ]−[d2

(a−d ) (c−d ) ]] [1s+a ]+[b2

a−b ][1+[c2

(b−c ) (c−d ) ]−[d2

(c−d ) (b−d ) ]] [1s+b ]+[−c2

c−d ] [1−[a2

(a−b ) (a−c ) ]+[b2

(a−b ) (b−c ) ]] [1s+c ]+[d2

c−d ] [1−[a2

(a−b ) (a−d ) ]+[b2

(a−b ) (b−d ) ]] [1s+d ] (G-15)

49

APPENDIX H

s

(s+a )2= As+B

(s+a )+ C

( s+a )2 (H-1)

s=( As+B ) (s+a )2

(s+a )+C

(H-2)

s =(As+B ) ( s+a )+C (H-3)

s=(As2+Bs )+a ( As+B )+C (H-4)

s=As2+ (aA+B ) s+aB+C (H-5)

A=0 (H-6)

s=Bs+aB+C (H-7)

B=1 (H-8)

C=−a (H-9)

s(s+a )2

= 1(s+a )

+ −a( s+a )2 (H-10)

50

APPENDIX I

1( s2+α2 ) ( s+β )2

= As+Bs2+α 2

+Cs+D( s+β )2 (I-1)

1( s2+α2 ) ( s+2β+β2)

= As+Bs2+α 2

+Cs+D(s+β )2 (I-2)

The expansion can be performed using the derivation in Appendix B.

{1s2+α2 }{1 s2+2 βs+β2 } =

+[−2 β ] s+[− (α 2−β2 ) ][ (α2−β2)2+(2 α β )2 ] [ s2+α 2 ]

+[ 2 β ] s+[ (α2−β2)+ (2β )2 ] [ (α 2−β2 )2+(2 α β )2] [ s2+2 βs+β2]

(I-3)

51

{1s2+α2 }{1 s2+2 βs+β2 } =

+[−2 β ] s+ [−α2+β2 ][α 4−β4 ] [ s2+α2 ]

+[ 2β ] s+ [α 2+3 β2 ] [α 4−β 4 ] [s+β ]2

(I-4)

{1s2+α2 }{1 s2+2 βs+β2 } =

2 β(α 4−β4 ) {− s+[α 2−β2 ] /[ 2β ]

[ s2+α2 ]+s+[α2+3β2 ] /[2 β ]

[ s+ β ]2 }(I-5)

The inverse Fourier transform is

f ( t )=1

[ (α2−β2)2+(2α β )2 ] { [−2 β ] cos (αt )+[−1α

(α 2−β2) ] sin (αt )}

+β exp (−βt )

[ (α2−β2)2+(2 α β )2 ] {2 β cos (βt )+[[ (α2−β2)+2 (β2 ) ]β ]sin (βt )}

(I-6)

52

f ( t )=1[α 4−β4 ] { [−2 β ] cos (αt )+[−1

α(α 2−β2 ) ] sin (αt )}

+β exp (−βt )[α 4−β4 ] {2 β cos ( βt )+[α 2+β2

β ]sin ( βt )} (I-7)

53

APPENDIX J

1s2 { 1

s2+ω2 }=αs+ ρs2 +

βs+λs2+ω2

(J-1)

1= (αs+ ρ ) ( s2+ω2)+βs3+ λs2 (J-2)

1=αs3+αω2 s+ρs2+ρω2+βs3+λs2 (J-3)

1= (α+β ) s3+( ρ+ λ ) s2+αω2s+ρω2 (J-4)

(α+β )=0 (J-5)

( ρ+λ )=0 (J-6)

α=0 (J-7)

β=0 (J-8)

ρ=1/ω2 (J-9)

λ=−1/ω2 (J-10)

1s2 { 1

s2+ω2 }=1 /ω2

s2 +−1/ω2

s2+ω2 (J-11)

The inverse Laplace transform is

f ( t )= 1ω2t− 1ω3

sin(ωt ) (J-12)

54