Question #269618

1.Find the Laplace Transform of the function g(t)=u(t)+2exp(-2t)u(t)+2cos(3000t)u(t)+

exp(-5t)sin(5000t)u(t) . Indicate the positions of zeros and poles .

2.) Find the Inverse Transform of the function 1/(s^2+6s+9)


1
Expert's answer
2021-11-22T17:11:03-0500

1

g=u(t)+2e2tu(t)+2cos(3000t)u(t)+e5tsin(5000t)u(t)ut+2e2tut+2cos(3000t)ut+e5tsin(5000t)ut=gut+2e2tut+2utcos(3000t)+e5tutsin(5000t)=gut(1+2e2t+2cos(3000t)+e5tsin(5000t))=gDividebothsidesbyt(1+2e2t+2cos(3000t)+e5tsin(5000t))ut(1+2e2t+2cos(3000t)+e5tsin(5000t))t(1+2e2t+2cos(3000t)+e5tsin(5000t))=gt(1+2e2t+2cos(3000t)+e5tsin(5000t))u=gt(1+2e2t+2cos(3000t)+e5tsin(5000t))g=u\left(t\right)+2e^{-2t}u\left(t\right)+2\cos \left(3000t\right)u\left(t\right)+e^{-5t}\sin \left(5000t\right)u\left(t\right)\\ ut+2e^{-2t}ut+2\cos \left(3000t\right)ut+e^{-5t}\sin \left(5000t\right)ut=g\\ ut+2e^{-2t}ut+2ut\cos \left(3000t\right)+e^{-5t}ut\sin \left(5000t\right)=g\\ ut\left(1+2e^{-2t}+2\cos \left(3000t\right)+e^{-5t}\sin \left(5000t\right)\right)=g\\ \mathrm{Divide\:both\:sides\:by\:}t\left(1+2e^{-2t}+2\cos \left(3000t\right)+e^{-5t}\sin \left(5000t\right)\right)\\ \frac{ut\left(1+2e^{-2t}+2\cos \left(3000t\right)+e^{-5t}\sin \left(5000t\right)\right)}{t\left(1+2e^{-2t}+2\cos \left(3000t\right)+e^{-5t}\sin \left(5000t\right)\right)}=\frac{g}{t\left(1+2e^{-2t}+2\cos \left(3000t\right)+e^{-5t}\sin \left(5000t\right)\right)}\\ u=\frac{g}{t\left(1+2e^{-2t}+2\cos \left(3000t\right)+e^{-5t}\sin \left(5000t\right)\right)}


2

L1{1(s2+6s+9)}1s2+6s+9=1(s+3)2=L1{1(s+3)2}Applyinversetransformrule:ifL1{F(s)}=f(t)thenL1{F(sa)}=eatf(t)For1(s+3)2:a=3,F(s)=1s2=e3tL1{1s2}=e3ttL^{-1}\left\{\frac{1}{\left(s^2+6s+9\right)}\right\}\\ \frac{1}{s^2+6s+9}=\frac{1}{\left(s+3\right)^2}\\ =L^{-1}\left\{\frac{1}{\left(s+3\right)^2}\right\}\\ \mathrm{Apply\:inverse\:transform\:rule:\quad if\:}L^{-1}\left\{F\left(s\right)\right\}=f\left(t\right)\mathrm{\:then}\:L^{-1}\left\{F\left(s-a\right)\right\}=e^{at}f\left(t\right)\\ \mathrm{For\:}\frac{1}{\left(s+3\right)^2}:\quad a=-3,\:\quad F\left(s\right)=\frac{1}{s^2}\\ =e^{-3t}L^{-1}\left\{\frac{1}{s^2}\right\}\\ =e^{-3t}t


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