taumulurtulkyoy
2022-10-14
Answered

Inverse Laplace transform of $\frac{1}{1+2{e}^{-s}+{e}^{-2s}}$

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honejata1

Answered 2022-10-15
Author has **10** answers

I guess that you would like to find a function g(t) such that $h(s)\ast g(s)=\delta (t)$ (where ∗ means convolution), where $h(t)=\delta (t)+2\delta (t-1)+\delta (t-2)$

After taking the Laplace Transform of both parts we get $H\cdot G=\mathcal{L}\{\delta \}=1$ (where ⋅ means multiplication) . Hence, we need to compute the inverse Laplace transform of

$$G(s)=\frac{1}{H(s)}=\frac{1}{1+2{e}^{-s}+{e}^{-2s}}=\frac{1}{(1+{e}^{-s}{)}^{2}}=\sum _{n=0}^{\mathrm{\infty}}(n+1)(-1{)}^{n}{e}^{-ns}.$$

By recalling that ${\mathcal{L}}^{-1}({e}^{-ns})=\delta (t-n)$, we obtain

$$g(t)=\sum _{n=0}^{\mathrm{\infty}}(n+1)(-1{)}^{n}\delta (t-n).$$

After taking the Laplace Transform of both parts we get $H\cdot G=\mathcal{L}\{\delta \}=1$ (where ⋅ means multiplication) . Hence, we need to compute the inverse Laplace transform of

$$G(s)=\frac{1}{H(s)}=\frac{1}{1+2{e}^{-s}+{e}^{-2s}}=\frac{1}{(1+{e}^{-s}{)}^{2}}=\sum _{n=0}^{\mathrm{\infty}}(n+1)(-1{)}^{n}{e}^{-ns}.$$

By recalling that ${\mathcal{L}}^{-1}({e}^{-ns})=\delta (t-n)$, we obtain

$$g(t)=\sum _{n=0}^{\mathrm{\infty}}(n+1)(-1{)}^{n}\delta (t-n).$$

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