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A voltage v(t)=2t^(2)e^(-3t) is passed through a delay line having a delay of 4 sec.Detrmine Laplace transform of the output voltage
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I think your question was: \[v(t) = 2t ^{2} e ^{-3t}\] You need to apply the following rules: \[L[u(t)] = 1/s\] \[L[e ^{at} f(t)] = F(s - a)\] \[L[t ^{n} f(t)] = (-1)^{n}F ^{n}(s)\] The Laplace transform is:\[ V(s) = 4/(s + 3)^{3}\] Now to account for the delay. The time shift manifests itself as a phase shift in s domain. So, an additional e^(-as) term joins the result. So, the final result after 4s delay is: \[ V(s) = (4 e^{-4s})/(s + 3)^{3}\]
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