Question

L^{-1}bigg(frac{5}{s^{2}(s^{2}-4s+5)}bigg)=5int_{0}^{infty}(t-T)e^{2T}sin 2T dT Select one: True or False

Laplace transform
ANSWERED
asked 2021-03-09
\(L^{-1}\bigg(\frac{5}{s^{2}(s^{2}-4s+5)}\bigg)=5\int_{0}^{\infty}(t-T)e^{2T}\sin 2T\ dT\)
Select one: True or False

Answers (1)

2021-03-10

Step 1
Given:
Let us consider,
\(L^{-1}\bigg(\frac{5}{s^{2}(s^{2}-4s+5)}\bigg)\)
Step 2
\(L^{-1}\bigg(\frac{5}{s^{2}(s^{2}-4s+5)}\bigg)\)
\(=L^{-1}\bigg(\frac{5}{s^{2}}\cdot\frac{1}{s^2-4s+5}\bigg)\)
\(\text{Let us consider,}\)
\(H(s)=\frac{5}{s^{2}}\)
\(\text{By Applying the inverse laplace transform as below,}\)
\(L^{-1}(H(s))=5L^{-1}\bigg(\frac{1}{s^{2}}\bigg)\)
\(h(t)=5t\)
\(\text{Step 3}\)
\(\text{Let us consider,}\)
\(G(s)=\frac{1}{s^{2}-4s+5}\)
\(G(s)=\frac{1}{s^{2}-4s+4+1}\)
\(G(s)=\frac{1}{(s-2)^{2}+1}\)
\(\text{By Applying the inverse Laplace Transform, }\)
\(L^{-1}(G(s))=L^{-1}\bigg(\frac{1}{(s-2)^{2}+1}\bigg)\)
\(=e^{2t}\sin t\)
\(\text{Step 4}\)
\(L^{-1}(H(s)G(s))=\int_0^\infty h(t-T)\cdot g(T)\)
\(\text{Therefore, }\)
\(L^{-1}\bigg(\frac{5}{s^{2}(s^{2}-4s+5)}\bigg)=5\int_{0}^{\infty}(t-T)e^{2T}\sin T\)
Hence ,
The given Equation is False.

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