Answer to Question #269618 in Electrical Engineering for Abubakar Tanveer

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\\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)\\\\\nut+2e^{-2t}ut+2\\cos \\left(3000t\\right)ut+e^{-5t}\\sin \\left(5000t\\right)ut=g\\\\\nut+2e^{-2t}ut+2ut\\cos \\left(3000t\\right)+e^{-5t}ut\\sin \\left(5000t\\right)=g\\\\\nut\\left(1+2e^{-2t}+2\\cos \\left(3000t\\right)+e^{-5t}\\sin \\left(5000t\\right)\\right)=g\\\\\n\\mathrm{Divide\\:both\\:sides\\:by\\:}t\\left(1+2e^{-2t}+2\\cos \\left(3000t\\right)+e^{-5t}\\sin \\left(5000t\\right)\\right)\\\\\n\\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)}\\\\\nu=\\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\\}\\\\\n\\frac{1}{s^2+6s+9}=\\frac{1}{\\left(s+3\\right)^2}\\\\\n=L^{-1}\\left\\{\\frac{1}{\\left(s+3\\right)^2}\\right\\}\\\\\n\\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)\\\\\n\\mathrm{For\\:}\\frac{1}{\\left(s+3\\right)^2}:\\quad a=-3,\\:\\quad F\\left(s\\right)=\\frac{1}{s^2}\\\\\n=e^{-3t}L^{-1}\\left\\{\\frac{1}{s^2}\\right\\}\\\\\n=e^{-3t}t"


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