Find the Laplace transform of NSK PSKf(t)=10e^{-200t}u(t)ZSK NSK Would it be correct to take

Patricia Crane 2021-12-11 Answered
Find the Laplace transform of
\(\displaystyle{f{{\left({t}\right)}}}={10}{e}^{{-{200}{t}}}{u}{\left({t}\right)}\)
Would it be correct to take out the 10 because it is a constant, find the Laplace transform of \(\displaystyle{e}^{{-{200}{t}}}\) and then multiply it by the Laplace transform of u(t) to obtain a final answer of
\(\displaystyle{10}{\left({\frac{{{1}}}{{{s}+{200}}}}\right)}{\left({\frac{{{1}}}{{{s}}}}\right)}\)
The u(t) is what is really confusing me in this problem.

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Expert Answer

Mary Herrera
Answered 2021-12-12 Author has 4127 answers
There is an extra 1/s in your final formula. Let \(\displaystyle{f{{\left({t}\right)}}}={a}{e}^{{-{b}{t}}}{u}{\left({t}\right)}\) with \(\displaystyle{a},{b}\in{R}\) (I'm assuming your u(t) is the Heaviside step function). Then
PSK(Lf)(s)=\int_0^{\infty} f(t)e^{-st} dt=\int_0^{\int} ae^{-bt}e^{-st}u(t)dtask
\(\displaystyle={a}{\int_{{0}}^{{\infty}}}{e}^{{-{\left({b}+{s}\right)}{t}}}{\left.{d}{t}\right.}={\frac{{{a}}}{{{s}+{b}}}}\)
The region of convergence is \(\displaystyle{s}\in{C}:{R}{e}{\left({s}\right)}\succ{b}\)
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Jenny Sheppard
Answered 2021-12-13 Author has 2393 answers
It's OK to break out the 10 since the Laplace transform is linear, that is fox \(\displaystyle{L}\alpha{f}=\alpha\) Lf for any constant \(\displaystyle\alpha\)
But to separately transform \(\displaystyle{e}^{{-{200}{t}}}\) and u(t) and multiply the results is wrong. Multiplication under the L corresponds to convolution outside it.
However if you're talking about the one-sided laplace transform then the transform of \(\displaystyle{e}^{{-{200}{t}}}{u}{\left({t}\right)}\) is the same as for \(\displaystyle{e}^{{-{200}{t}}}\) since we only integrate over positive numbers anyway. The result would consequently be: \(\displaystyle{L}{10}{e}^{{-{200}{t}}}{u}{\left({t}\right)}={10}{L}{e}^{{-{200}{t}}}={10}{\frac{{{1}}}{{{s}+{200}}}}\)
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