math review – complex numbers (1)

2009 Fall ME451 - GGZ Week 1-2: Math Review and Laplace Transformation

Math Review Math Review –– Complex Numbers (1)Complex Numbers (1)

Complex number: ordered pair of two real numbers

Multiplication

1 and , where, −≅∈∈+= jRyxCjyxs

Conjugate: jyxss −== *

Addition:

)()( , ,212121222111

yyjxxssjyxsjyxs +++=++=+=

)()(1221212121

yxyxjyyxxss ++−=

22*yxss +=


Math Review Math Review –– Complex Numbers (2)Complex Numbers (2)

Euler’s identity

Phase:

j

eeeeje

jjjj

j

2sin ,

2cos where,sincos

θθθθθ θθθθ

−− −=

+=+=

Polar form:

θθθ sin,cos where, ryrxrejyxsj ===+=

Magnitude:22

yxr +=

)(

2

1

2

1

)(

21212211

21

2121

,

θθ

θθθθ

−

+

=

===

j

jjj

er

r

s

s

errssersers

x

y1tan

−=θ

x

y

Re

Im

rθ


Math Review Math Review –– LogarithmLogarithm

The logarithm of x to the base b is denoted by xb

log

The logarithm of 1000 to the base 10 is 3, i.e.,

Properties:

100010 that note ,31000log 3

10 ==

01log and 110log1010

==

yxy

x

yxxy

xyxxb

bbb

bbb

b

y

b

xb

loglog)(log

loglog)(log

)(log)(log and )(log

−=

+=

==


Math Review Math Review –– Matrix OperationsMatrix Operations

=

=

2221

1211

2221

1211 and

bb

bbB

aa

aaA

Determinant:12212211

det aaaaA −=

Multiplication: 2222122121221121

2212121121121111

++

++=

babababa

babababaAB

Inverse:

det

1121

1222

1

A

aa

aa

A

−

−

=−


Laplace TransformLaplace Transform

Definition:

∫∞

−==0

)()()]([ dtetfsFtfLst

The Laplace transform (LT) of f(t) is

LT: replace ODEs as linear input-output maps

solve ODEs w/ constant coefficients

t - domain

)(tf

s - domain

)(sF

LRe

Im

ts


Laplace Transform Laplace Transform –– Example (1)Example (1)

Example (1)

0 ,1)( ≥= ttf

ss

es

dtedtetfsFtfL

st

stst

110

1

)()()]([

0

0 0

=+=

−=

===

∞

−

∞ ∞

−−

∫ ∫

sLtfL

1]1[)]([ ==Q

∆≤

== ∆

→∆ 0;0

;lim)()(

1

0

tttf δ

11

lim

1lim

)()()()]([

0

00

0 0

=∆⋅∆

=

∆=

===

→∆

∆

−

→∆

∞ ∞

−−

∫

∫ ∫

dte

dtetdtetfsFtfL

st

stst δ

1)]([)]([ == tLtfL δQ

∆

1

∆ t

)(tδ

0→∆



Example (2)

0 ,)( ≥= tttf

22

00

0 0 0 0 0

0 0

1)

10()00(

11

)(

)(

1 , )()()]([

sse

sse

s

t

dtt

vuvdt

t

udt

t

uvvdt

t

udt

t

uv

es

vtudttedtetfsFtfL

stst

ststst

=−−+−=−

⋅−

−−

=

∂

∂+

∂

∂=

∂

∂

∂

∂−

∂

∂=

−=====

∞

−

∞

−

∞ ∞ ∞ ∞ ∞

∞ ∞

−−−

∫ ∫ ∫ ∫ ∫

∫ ∫

2

1][)]([

stLtfL ==Q


Laplace Transform Laplace Transform –– Property (1)Property (1)

Linearity

)()()]([)]([)]()([ sbGsaFtgbLtfaLtbgtafL +=+=+

)()(

)()(

)]()([)]()([

)(

0

)(

0

0

sbGsaF

dtetgbdtetfa

dtetbgtaftbgtafL

sG

st

sF

st

st

+=

+=

+=+

∫∫

∫∞

−

∞

−

∞

−

4342143421

Proof:



“s” Shifting property

)(lim)( where),()]([ then ),()]([ If sFasFasFtfeLsFtfLass

at

−→=−−==

)(

)(

)()]([

0

)(

0

asF

dtetf

dtetfetfeL

tas

statat

−=

=

=

∫

∫∞

−−

∞

−

Proof:



“t” Shifting property

)()]()([

then ,function stepunit is )( and )()]([ If

sFeatuatfL

tusFtfL

as−=−−

=

( )

)(

)()()(

)()()()()]()([

00

)(

0

sFe

atdefedeuf

dteatuatfdteatuatfatuatfL

as

sasas

a

stst

−

∞

−−

∞

+−

∞

−

∞

−

=

−===

−−=−−=−−

∫∫

∫∫

ττττττ ττ

Proof:

<

≥=−

at

atatu

,0

,1)(

a

1

)( atu −

t



Transforms of Derivatives

∂

∂=−==

t

tftffssFtfLsFtfL

)()( )0()()]([ then ,)()]([ If &&

Proof:

( )

)0()()()(

)(

)(

)( , )()]([

)(

00

0 0 0 0 0

00

fssFdtetfsetf

dtt

vuvdt

t

udt

t

uvvdt

t

udt

t

uv

tfveudtt

vudtetftfL

sF

stst

stst

−=

−−=

∂

∂+

∂

∂=

∂

∂

∂

∂−

∂

∂=

==∂

∂==

∫

∫ ∫ ∫ ∫ ∫

∫∫

∞

−∞

−

∞ ∞ ∞ ∞ ∞

−

∞∞

−

43421

&&

)0()0()()]([ 2fsfsFstfL &&& −−=



Transforms of Integrals

)(1

])([ then ,)()]([ If0

sFs

dfLsFtfL

t

== ∫ ττ

Proof:

.0)0( and )()( then,,)()(Let 0

=== ∫ gtftgdftg

t

&ττ

)()0()]([)]([)]([)(

sderivative of s transformUsing

)(

ssGgtgLstgLtfLsF

sG

=−===321

&

)(1

])([)(

Therefore

t

0

sFs

dfLs G == ∫ ττ



Linearity ttf 21)( +=

)].([G(s) find ,)( If 2tgLetg

t ==

2

11

]1[)]1([)(

2

2

2

−==

==

−→

−→

ss

LeLsG

ss

ss

t

2

1][

Therefore

2t

−=

se L

22

212

1][2]1[]21[)]([

s

s

sstLLtLtfL

+=+=+=+=

“s” Shifting properties:



Then, .3)( and ,)( ,3Let −=−== tatfttfa

).3()62()( where

)]([)( Find

−−=

=

tuttg

tgLsG

2

3 12

)]3()3[(2)]3()3(2[)]([)(

se

tutLtutLtgLsG

s−=

−−=−−==

3

3s-

2

2G(s)

Therefore,

es

=

“t” Shifting properties:

6 93−

6−

3

6)(tg

t



)0()0()]([)]([ using ][sin 2fsftfLstfLwtL &&& −−=

wtwtf

wfwtwtf

fwttf

sin)(

)0( ,cos)(

0)0( ,sin)(Let

2−=

==

==

&&

&&

{ {wtftf

ffswtLswtwL )0()0(]sin[]sin[ Thus,0)(

2

)(

2 &32143421

&&

−−=−

22][sin

Therefore,

ws

wwtL

+=

Calculating

wwtLswtLw −=− ][sin][sin 22

wwtLws =+ ][sin)( 22


Cosine

Sine

Exponential

nth order ramp

Unit ramp

Unit step

Unit impulse

Description

Laplace Transform Laplace Transform –– LT Table to rememberLT Table to remember

)(tf

t

s

1)(tu

)(tδ

)]([)( tfLsF =

1

2

1

s

nt 1

!+n

s

n

ateas −

1

wtsin22

ws

w

+

wtcos 22ws

s

+


Convolution

Transforms of integrals

Transforms of

derivatives

“s” shifting properties

Linearity

Laplace Transform Laplace Transform –– LT PropertiesLT Properties

)]([)]([)]()([ tgbLtfaLtbgtafL +=+

)0()()]([ fssFtfL −=&

)]([)()]()([ tfLesFeatuatfLasas −− ==−−

ass

attfLasFtfeL

−==−= )]([)()]([

)0()0()()]([2

fsfsFstfL &&& −−=

M

∫ ==

t

tfLs

sFs

dfL0

)]([1

)(1

])([ ττ

∫∫ −=−=−

tt

dtfgdtgfsGsFL00

1 )()()()()]()([ ττττττ

“t” shifting properties


Inverse Laplace TransformInverse Laplace Transform

)]([)( Find )],([)(Given 1sFLtftfLsF

−==

t - domain

)(tf

s - domain

)(sF

1−L

Approach:

etc. ,properties tables,using term-by- term)]([Obtain (2)

functionscommon of s transform theselike"look " )( Make (1)

1sFL

sF

−


Inverse Laplace Transform Inverse Laplace Transform ––Example (1)Example (1)

)]([)( Find ,6

1)(Given 1

23sYLty

sss

ssY −=

−+

+=

3

152

2

103

61-

6

1)(

nextit discuss willWe

23

444 3444 21 +

−

+−

+=−+

+=

ssssss

ssY

ttee

sL

sL

sLty

32

111

15

2

10

3

6

1

]3

1[

15

2]

2

1[

10

3]

1[

6

1)(

−

−−−

−+−

=

+−

−+

−=


Inverse Laplace Transform Inverse Laplace Transform –– PFE Case (1)PFE Case (1)

CA, Bs

C

s

B

s

A

sss

ssY and , Find ,

32)3)(2(

1)(

++

−+=

+−

+=

)2()3()3)(2(1 −++++−=+⇒ sCssBsssAs

PFE: Partial Fraction ExpansionsA way to expand general rational functions into forms that

appears in the LT table

Case 1: real and distinct roots

15

2 )5)(3(2;3

10

3 )5)(2(3;2

6

1 )3)(2(1;0

−=⇒−−=−−=

=⇒==

−=⇒−==

CCs

BBs

AAs



tt eetysss

sY 32

15

2

10

3

6

1)(

3

15/2

2

10/36/1)( −−+−=⇒

+

−+

−+

−=

AsCBAsCBA

CsCsBsBsAAsAs

sCssBsssAs

6)23()(

236

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

2

222

−−++++=

−+++−+=

−++++−=+

Case 1: real and distinct roots (cont’d)

unknowns 3 with equations 3

61:

231:

0:

0

1

2

−=

−+=

++=

As

CBAs

CBAs

Alternative approach:

We have



323 )2()2(2)2(

3)(

++

++

+=

+

+=

s

C

s

B

s

A

s

ssY

0 20

1 2Let

)2(21

1 2Let

)2()2(3 2

=⇒=⇒∂

∂

=⇒−=

++=⇒∂

∂

=⇒−=

++++=+⇒

AA s

Bs

BsA s

Cs

CsBsAs

Case 2: real and repeated roots

32 )2(

1

)2(

1)(

++

+=

sssY



tt

ssss

ssss

et

tety

tLtL

sssssY

2

2

2

2

2

2

2

3

2

232

2)(

]2

[][

11

)2(

1

)2(

1)(

−−

+=+=

+=+=

+=

+=

+=+

++

=

Case 2: real and repeated roots (cont’d)



)( bias ±=

]2cos[]2[cos2

)(1

1

22teLtL

s

ssY

t

ss

ss

−

+=

+=

==+

=

Case 3: complex conjugate roots

iss

s

ss

ssY 21 :Roots

2)1(

1

52

1)(

222±−=⇒

++

+=

++

+=

tetyt 2cos)(

Therefore

−=


Inverse Laplace Transform Inverse Laplace Transform –– Example (2)Example (2)

22222

2

21)4)(1(

1)(

+

++

+++=

++

+=

s

EDs

s

C

s

B

s

A

sss

ssF

)1()()4()4)(1()4)(1(1222222 ++++++++++=+ ssEDssCsssBssAss

20/3,4163

)](24[4

)12)(4)(2(32

5/2521

4/1410

−=−=⇒−=−=−⇒

++−−=

+−+=−⇒=

=⇒=⇒−=

=⇒=⇒=

EDEDED

EDiDE

iEiDis

CCs

BBs

4/1 440 0

])1(2)[()1(]2)4(2[

)]1(2)4[()]1(2)4()4)(1[(2

2232

2222

−=−=⇒+=⇒=

+++++++++

++++++++++=⇒∂

∂

BABAs

sssEDssDssssC

sssBssssssAss

)2sin2

12cos(

20

3

5

2

4

1

4

1)(

2

1

20

3

1

1

5

21

4

11

4

1)(

222ttettf

s

s

ssssF

t +−+++−=⇒+

+−+

+++−= −


Laplace Transform Laplace Transform –– ConvolutionConvolution

∫∫ −=−=

==

−

−−

tt

dttfgdttgfsGsFL

tgsGLtfsFL

00

1

11

)()()()()]()([

),()]([ and )()]([ If

ττττ

Example:

44 344 214342143421tttysGsF

sssssssY

3sin8

1sin

8

3)(

22

)(

2

)(

222 9

8/3

1

8/3

9

3

1

1

)9)(1(

3)(

−=

+−

+=

+

+=

++=

tsGLttsFLtf 3sin)]([)(g and sin)]([)( 11 ==== −−

tt

dtdtgfty

tt

3sin8

1sin

8

3

)(3sinsin)()()(00

−==

−=−= ∫∫

L

ττττττ


Laplace Transform Laplace Transform –– Solving Solving ODEsODEs

ODE in

domain t

AE in

domain s

Solution

)(ty

Direct solution

Solve for

)(sY

L

1−L)(sY


Laplace Transform Laplace Transform –– Solving Solving ODEsODEs ExampleExample

0)0()0( ,21)(4)( ==++−=+ −yyettyty

t &&&

44 344 214434421

&

&

&

Solution Particular

22

2

Solution sHomogeneou

2

2

2

2

2

2

)4)(1(

1

4

)0()0()(

)1(

1)0()0()()4(

1

211)(4)0()0()( :

++

++

+

+=

+

+++=+

+++−=+−−

sss

s

s

ysysY

ss

sysysYs

ssssYysysYsL

0

)2sin2

12cos(

20

3

5

2

4

1

4

1)( ttetty

t +−+++−= −

222 2

1

20

3

1

1

5

21

4

11

4

1)(

+

+−+

+++−=

s

s

ssssY

ILT Example (2)

math review – complex numbers (1)

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