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-Everything still holds in the complex domain

-Eg L }{sin t

-Now, not all )(tf have a Laplace transform. The integral, intuitively, needs to have the

integrand, )(tfe st , go to zero sufficiently fast. From the examples, we seemed to

need the overall function to decay faster than kte , k>0, in order for the integral to exist

eg/nst

te

,atst

ee

can all be made to decay sufficiently fast but

2tst

ee

wont converge forany choice ofs

-Now, )(tf doesnt need to be continuous (eg. turning on a switch, pulse/square wave

generator)

-Lets assume )(tf is piecewise continuous. In other words, it is continuous on any

finite interval. If it is discontinuous, it will have finite right and left hand limits (ie canonly have finite jumps)

-Theorem. A function )(tf is piecewise continuous on every finite interval for 0t

and atMetf |)(| for some a (can be positive!). Then the Laplace transform exists for

asre }{

-When it exists, the Laplace transform is unique except at the discontinuities

-These are sufficient conditions

-eg. L }1

{t

eg. L }{2t

e

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eg. What is the Laplace transform of10,)( tttf

1,0 t

Soln/

-Inverse Laplace

-can apply L-1

i

i

stdsesF

isF

)(2

1)}({

(This is an integration in the complex plane. The depends on the region of

convergence

-We can, ifF(s) is rational, use tables and partial fraction expansions

eg/23

2

7

73)(

ss

sssF

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B. Application to ODEs

-We need to figure out what happens when we differentiate a function

-Theorem- suppose )(tf is continuous for 0t and satisfies the conditions to have a

Laplace transform. As well, suppose )(tdtdf is piecewise continuous on every finite

interval in 0t . Then, there exists an such that L )}({ tdt

dfexists for }Re{s and

L stdt

df)}({ L )0()}({ ftf

where )0(f is the initial value.

Proof/

-What do the conditions mean?

-Apply this twice to get

L stdt

fd)}({

2

2

L )0()}({ ftdt

df

-Similarly, L )}({ tdt

fdn

n

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-eg/ L }{cos t

-We can now solve ODEs with initial values using algebraic techniques

-Eg/ 1)0(, xxx

Eg/ 1)0(,0)0(,1 yyy

Eg/ 1)0(,0)0(,19 yyyy

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C. Shifting theorems and the Heaviside Function- s-Shifting

-Theorem: If )(tf has a transform )(sF (where s ), then

)(tfeat has the transform )( asF (where as )

ie L )()}({ asFtfeat

-Equivalently, L-1 )}({ asF )(tfeat

Pf/

-This now allows us to use PFE of all possible types of rational functions in s using ourprevious table

If ,)( nattetf then

If ,cos)( tetf at then

If ,sin)( tetf at then

Eg/)404(

1)(

22

ssssF

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Or solve using complex domain

Eg./ 4)0(,2)0(,052 yyyyy

Eg/ 0)0(,1)0(,2 yyteyyy t

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Eg/ 0)0(,1)0(,1136 yyyyy

Extra problem:

Look at16

14 s

Solution:

))()(2)(2(

1

16

113513545454 jjjj

esesesess

})2

2

2

2{2})(

2

2

2

2{2})(

2

2

2

2{2})(

2

2

2

2{2(

1

jsjsjsjs

})2}2})({2}2})({2}2})({2}2({

1

jsjsjsjs

)2}2)({2}2({

122

ss

This can be checked by multiplying the two terms in the denominator

Thus, these are these are terms involving quadratics which will lead to terms such as

),2cos(),2sin(),2cos(),2sin( 2222 tetetete tttt

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-Terms to use in general

-Now look at time-shifting

Thm- Suppose L )()}({ sFtf . Then the function

atatf

attf

),(

,0{)(

~

has the Laplace Transform )(sFe as

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-Heaviside Step Function )( atu

at

atatu

,1

,0{)(

- )(tu is the step function. Thus, )( atu is a delayed step

A delayed function is )()()(~

atuatftf

-Theorem (restated)

L )()}()({ sFeatuatf as

L1 )()()}({ atuatfsFeas

Pf/

-Show that Ls

eatu

as

)}({

Note: )()()()( atuatfatutf

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Eg/ FindL }/{ 331 se s and sketch

Eg/Find the transform of

2sin

20

02

)(

tt

t

t

tf

Eg/1

422

)(2

2

2

2

2

s

es

s

e

s

e

ssF

sss

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Eg. 0)0(,1)0(),3(244 yytutyyy

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Eg/ L })1({ ttu

-Handling periodic signals

-Theorem- iff is periodic with periodTand piece-wise continuous over this period, then

L dtetfe

tfst

T

sT

)(

1

1)}({

0

The integral is the Laplace transform over one intervalPf/

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Eg/ Pulse generator

Eg/ Sawtooth generator

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-Other properties:

Integration off(t)

Theorem: Iff(t) is piecewise-continuous and satisfiest

Metf|)(| , then

L

sdf

t 1})({

0 L )}({ tf

Pf/

Let t

dftg0

)()(

Now, one can show that

t

Metg |)(| which means there is a Laplace transform

L )}({ tf = L )}({ tg =sL )}({ tg -g(0)

The last term is zero

QED