Question #127086
1. Find the inverse Laplace transform ℒ-^1 { e^-5s / s^2 - 25 }

2.Find the Laplace transform of the given function.

f(t) = {1, 0 ≤ t < 6
{ 0, t ≥ 6

Enclose numerators and denominators in parentheses. For example, (a−b)/(1+n).


L {f(t)} = __________ , s>0.
1
Expert's answer
2020-07-29T11:56:04-0400

L(sh(at))=ap2a2e5ss225=15e5s5p225L1(e5ss225)=15sh(5t)η(t5)Where η(t) is a Heavisidesfunction2.f(t)=H0,6(t)=η(t)η(t6)L(f(t))=1se6s1s=1e6ssL(\sh(at))=\dfrac{a}{p^2-a^2}\newline \dfrac{e^{-5s}}{s^2-25}=\dfrac{1}{5}e^{-5s}\dfrac{5}{p^2-25}\newline L^{-1}(\dfrac{e^{-5s}}{s^2-25})=\dfrac{1}{5}\sh(5t)\eta(t-5)\newline Where \space \eta(t) \space is \space a\space Heaviside's function\newline \newline 2. f(t)=H_{0,6}(t)=\eta(t)-\eta(t-6)\newline L(f(t))=\dfrac{1}{s}-e^{-6s}\dfrac{1}{s}=\dfrac{1-e^{-6s}}{s}


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Comments

Assignment Expert
10.05.21, 00:06

Dear B. Rama krishna. What should be done in your question. Please use the panel for submitting a new question.

B. Rama krishna
01.05.21, 08:39

(2s-5) /(9s*s-25)

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