Answer to Question #269618 in Electrical Engineering for Abubakar Tanveer

Question

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)

Answer 1

1

$g=u\left(t\right)+2e^{-2t}u\left(t\right)+2\cos \left(3000t\right)u\left(t\right)+e^{-5t}\sin \left(5000t\right)u\left(t\right)\\ ut+2e^{-2t}ut+2\cos \left(3000t\right)ut+e^{-5t}\sin \left(5000t\right)ut=g\\ ut+2e^{-2t}ut+2ut\cos \left(3000t\right)+e^{-5t}ut\sin \left(5000t\right)=g\\ ut\left(1+2e^{-2t}+2\cos \left(3000t\right)+e^{-5t}\sin \left(5000t\right)\right)=g\\ \mathrm{Divide\:both\:sides\:by\:}t\left(1+2e^{-2t}+2\cos \left(3000t\right)+e^{-5t}\sin \left(5000t\right)\right)\\ \frac{ut\left(1+2e^{-2t}+2\cos \left(3000t\right)+e^{-5t}\sin \left(5000t\right)\right)}{t\left(1+2e^{-2t}+2\cos \left(3000t\right)+e^{-5t}\sin \left(5000t\right)\right)}=\frac{g}{t\left(1+2e^{-2t}+2\cos \left(3000t\right)+e^{-5t}\sin \left(5000t\right)\right)}\\ u=\frac{g}{t\left(1+2e^{-2t}+2\cos \left(3000t\right)+e^{-5t}\sin \left(5000t\right)\right)}$

2

$L^{-1}\left\{\frac{1}{\left(s^2+6s+9\right)}\right\}\\ \frac{1}{s^2+6s+9}=\frac{1}{\left(s+3\right)^2}\\ =L^{-1}\left\{\frac{1}{\left(s+3\right)^2}\right\}\\ \mathrm{Apply\:inverse\:transform\:rule:\quad if\:}L^{-1}\left\{F\left(s\right)\right\}=f\left(t\right)\mathrm{\:then}\:L^{-1}\left\{F\left(s-a\right)\right\}=e^{at}f\left(t\right)\\ \mathrm{For\:}\frac{1}{\left(s+3\right)^2}:\quad a=-3,\:\quad F\left(s\right)=\frac{1}{s^2}\\ =e^{-3t}L^{-1}\left\{\frac{1}{s^2}\right\}\\ =e^{-3t}t$