Transcription of The Inverse Laplace Transform
1 The Inverse Laplace Transform1. IfL{f(t)}=F(s), then theinverse Laplace transformofF(s) isL 1{F(s)}=f(t).(1)The Inverse transformL 1is a linear operator:L 1{F(s) +G(s)}=L 1{F(s)}+L 1{G(s)},(2)andL 1{cF(s)}=cL 1{F(s)},(3)for any :The Inverse Laplace Transform ofU(s) =1s3+6s2+ 4,isu(t) =L 1{U(s)}=12L 1{2s3}+ 3L 1{2s2+ 4}=s22+ 3 sin 2t.(4) :Suppose you want to find the Inverse Laplace transformx(t) ofX(s) =1(s+ 1)4+s 3(s 3)2+ use the shift property (paragraph 11 from the previous set of notes):x(t) =L 1{1(s+ 1)4}+L 1{s 3(s 3)2+ 6}=e tt36+e3tcos :Lety(t) be the Inverse Laplace Transform ofY(s) =e 3sss2+ t worry about the exponential term. Since the Inverse Transform ofs/(s2+4) is cos 2t,we have by the switchig property (paragraph 12 from the previous notes):y(t) =L 1{e 3sss2+ 4}=H(t 3) cos 2(t 3).
2 :LetG(s) =s(s2+ 4s+ 5) 1. The Inverse Transform ofG(s) isg(t) =L 1{ss2+ 4s+ 5}=L 1{s(s+ 2)2+ 1}=L 1{s+ 2(s+ 2)2+ 1} L 1{2(s+ 2)2+ 1}=e 2tcost 2e 2tsint.(5)6. There is usually more than one way to invert the Laplace Transform . For example,letF(s) = (s2+ 4s) 1. You could compute the Inverse Transform of this function bycompleting the square:f(t) =L 1{1s2+ 4s}=L 1{1(s+ 2)2 4}=12L 1{2(s+ 2)2 4}=12e 2tsinh 2t.(6)You could also use the partial fraction decomposition (PFD) ofF(s):F(s) =1s(s+ 4)=14s 14(s+ 4).Therefore,f(t) =L 1{F(s)}=L 1{14s} L 1{14(s+ 4)}=14 14e 4t=12e 2tsinh 2t.(7) :Compute the Inverse Laplace transformq(t) ofQ(s) =3s(s2+ 1) could computeq(t) by partial fractions, but there s a less tedious way.
3 Note thatQ(s) = 32dds1s2+ ,q(t) =L 1{Q(s)}= 32L 1{dds1s2+ 1}=32tsint.(8) :Theconvolutionof functionsf(t) andg(t) is(f g)(t) = t0f(t v)g(v)dv.(9)As we showed in class, the convolution is commutative:(f g)(t) = t0f(t v)g(v)dv= t0g(t v)f(v)dv= (g f)(t).(10) :Letf(t) =tandg(t) =et. The convolution offandgis(f g)(t) = t0(t v)evdv=t t0evdv t0vevdv=et t 1.(11) :(The Convolution Theorem) If the Laplace transforms off(t) andg(t)areF(s) andG(s) respectively, thenL{(f g)(t)}=F(s)G(s),(12)that is,L 1{F(s)G(s)}= (f g)(t).(13)11. Suppose that you want to find the Inverse transformx(t) ofX(s). If you can writeX(s) as a productF(s)G(s) wheref(t) andg(t) are known, then by the above result,x(t) = (f g)(t). :Consider the previous example: Find the Inverse transformq(s) ofQ(s) =3s(s2+ 1) (s) =F(s)G(s), whereF(s) =3s2+ 1,andG(s) =ss2+ Inverse transforms are ofF(s) andG(s) aref(t) = 3 sintandg(t) = cost.
4 Thereforeq(s) =L 1{Q(s)}=L 1{F(s)G(s)}= (f g)(t)= 3 t0sin (t v) cosvdv.(14)Even if you stop here, you at least have a fairly simple, compact expression forq(s). Todo the integral (14), use the trigonometric identitysinAcosB=sin (A+B) + sin (A B) this, (14) becomesq(s) =32 t0sintdv+ t0sin (t 2v)dv=32tsint.(15) :Find the Inverse Laplace transformx(t) of the functionX(s) =1s(s2+ 4).If you want to use the convolution theorem, writeX(s) as a product:X(s) =1s1s2+ 1{1s}= 1,andL 1{1s2+ 4}=12sin 2t,we havex(t) =12 t0sin 2vdv=14(1 cos 2t).You could also use the PFD:X(s) =14s s4(s2+ 4).Therefore,x(t) =L 1{14s} L 1{s4(s2+ 4)}=14(1 cos 2t).
