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10120. Laplacetransformtechnique: coverup

I

want

to

show

you

some

practical

tricks

which

will

help

you

to

find the inverse Laplace transform of a rational function. These arerefinementsonthemethodofpartialfractionswhichyoustudiedwhenyou learnedhowto integraterationalfunctions. Someofthiswillusecomplexnumbers.20.1. Simplecase. First,letsdoaneasycase:

1F(s) =

2 .s 4Tobegin,factorthedenominator,andwrite

1 a b(1) F(s) = = +(s2)(s+2) s2 s+ 2forasyetunknownconstantsaandb. Onewaytoproceed istocrossmultiply and collect terms in the numerator. That is fine but thefollowingismorefun.

To find a,firstmultiply through by the correspondingdenominator,(s2)inthiscase. Youget

1=a+ (s2)(otherterms),

s+ 2b

inwhichtheotherterms(namely, )donthaveafactorof(s2)s+ 2inthedenominator. Thensets=2:

1=a+(22)(otherterms)=a

2 + 2sincethesecondtermvanishes. Soa= 1/4. Inexactlythesameway,youcanmultiply(1)throughbys+2andthensets=2,tofind

1=b

22orb=1/4. Thus

1/4 1/4F(s) = .

s+ 2s2

Thetablesthenshowthatf(t)=(1/4)(e2te2t).

This approach to partial fractions has been called the cover-upmethod; you cover up the denominators of the terms you wish tocompute. You can do it without writing anything down;just cover

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102up the denominators and write down the answer. It works well butdoes not completely handle the case in which the denominator has arepeatedroot.20.2. Repeatedroots. Forexample lookat

sF(s) = .

s2 + 2s+ 1Thedenominatorfactorsas(s + 1)2. Wewanttofindaandbsuchthat

s a bF(s) = = + .

(s+1)2 s+ 1 (s+1)2Wecanbegintousethecoverupmethod: multiply throughby(s+1)2and

set

s=

1:

The

left

hand

side

is

just

1;

the

first

term

vanishes;

andthesecondtermisb: sob=1. Wecantgetathisway,though.Onewaytofindaistosetstosomeothervalue. Anyothervaluewilldo, and we might as well make our arithmetic as simple as possible.Letstakes=0: thenwehave

a+10 =

1 1soa=1:

1 1F(s) =

s+ 1(s+1)2.Nowthetablesshow

f(t) =ettet.20.3. Completing thesquare. Suppose

1F(s) = .

s2 + 2s+ 2The first part of the method here is to complete the square in thedenominator,andrewrite thenumerator inthesameterms:

s a(s+1)+b= .

s2 + 2s+ 2 (s+1)2 + 1Thisworkswitha=1andb=1:

F(s) = (s+1)1.(s+1)2 + 1

Now the s-shift rule applies, since F(s) is written in terms of sa(whereherea=1). Thesecondpartofthismethodgivesyouawaytousethes-shiftrulewithoutgettingtooconfused. Youshouldinvent

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103a newfunction namesay G(s)and use it to denote the functionsuchthatthatF(s) =G(sa). Thus,here,

G(s) = s1.s2 + 1

Nowthes-shiftrulesaysthatifg(t)G(s)thenetg(t)G(s+1)=F(s),which istosaythatf(t) =etg(t). Thetablesgive

g(t)=costsintso

f(t) =et(costsint).20.4. Complex coverup. Now lets take an example in which thequadratic

factor

does

not

occur

alone

in

the

denominator:

say

1

F(s) = .s3 +s22

Thedenominator factorsass3 +s22 = (s1)(s2 + 2s+2). Intheexampleabovewelearnedthatthefactors2+ 2s+ 2shouldbehandledbycompletingthesquareandgroupingthe(s+1)inthenumerator:

1 a b(s+1)+cF(s) = = + .

(s1)((s+1)2 +1) s1 (s+1)2 + 1Findajust asbefore: multiply throughby s1andthen set s=1,to get a = 1/5. To find b and c, multiply through by the quadraticfactor(s+1)2+ 1andthensetsequaltoarootofthatfactor. Havingalreadycompletedthesquare, itseasytofindaroot: (s+1)2 =1,sos+ 1 =iforexample,sos=1 +i. Weget:

1(1 +i)1 =b((1 +i)+1)+c

or,rationalizingthedenominator,2i

=c+bi5

Sincewewantbandcreal,wemusthavec=2/5andb=1/5:

1 1 (s+1)+2F(s) = .5 s1(s+1)2 + 1

Wereinpositiontoappealtothes-shiftrule,usingthetricksdescribedin20.3,andfind

1 f(t) = etet(sint+2cost) .

5

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10420.5. Completepartialfractions. Thereisanotherwaytodealwithquadraticfactors:justfactorthemoverthecomplexnumbersandusethe coverup method in its original form as in Section 20.1. I dontrecommend using this in practice, but its interesting to see how itworksout,andwewillusetheseideasinSection22. Usingtheexample

1F(s) =

s3 +s22again,wecanfindcomplexconstantsa,b,csuchthat

a b c(2) F(s) = + +

s1 s(1 +i) s(1i)Expectthatawillbereal,andthatbandcwillbecomplexconjugatesofeachother.

Find ajust as before; a = 1/5. To find b, do the same: multiplythroughbys(1 +i)toget

1(s1)(s(1i)) =b+ (s(1 +i))(otherterms)

andthensets=1 +itosee1

=b(2 +i)(2i)

orb= 1/(24i) = (1 + 2i)/10. Thecoefficientccanbecomputedsimilarly. Alternatively, you can use the fact that the two last termsin(2)mustbecomplexconjugatetoeachother(inorderforthewhole

expressiontocomeoutreal)andsodiscoverthatc=b= (12i)/10:F(s) = 1/5 + (1 + 2i)/10+(12i)/10.

s1 s(1 +i) s(1i)Thenumeratorsa,b,andc,inthisexpressionarecalledtheresidues

ofthepolesofF(s);see22.1below.Its perfectly simple to find the inverse Laplace transforms of the

termshere:f(t) = 1 t +1 + 2i (1+i)t +12i (1i)te e e .

5 10 10Thelasttwotermsarecomplexconjugatesofeachother,sotheirsumistwicetherealpartofeach,namely,

t te e2 Re((1 + 2i)(cost+isint)) = (cost2sint).

10 5Wewindupwiththesamefunctionf(t)asbefore.

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105ListofpropertiesoftheLaplace transform1.L

islinear:

af(t) +

bg(t)

aF

(s) +

bG(s).

2. F(s)essentiallydeterminesf(t).3. s-shifttheorem: Iff(t)F(s),theneatf(t)F(sa).4. t-shifttheorem: Iff(t)F(s),thenfa(t)easF(s),where

fa(t) = f(ta) ift > a .0 ift < a5. s-derivativetheorem: Iff(t)F(s),thentf(t)F(s).6. t-derivativetheorem: Iff(t)F(s),thenf(t)sF(s)f(0+)wheref(t)iscontinuousfort >0andthenotationf(t)indicatestheordinaryderivativeoff(t).7. Iff(t)F(s)andg(t)G(s),thenf(t) g(t)F(s)G(s).8. (t)1.

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