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

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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Mary Herrera
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
$$\displaystyle={a}{\int_{{0}}^{{\infty}}}{e}^{{-{\left({b}+{s}\right)}{t}}}{\left.{d}{t}\right.}={\frac{{{a}}}{{{s}+{b}}}}$$
The region of convergence is $$\displaystyle{s}\in{C}:{R}{e}{\left({s}\right)}\succ{b}$$
###### Not exactly what youâ€™re looking for?
Jenny Sheppard
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: $$\displaystyle{L}{10}{e}^{{-{200}{t}}}{u}{\left({t}\right)}={10}{L}{e}^{{-{200}{t}}}={10}{\frac{{{1}}}{{{s}+{200}}}}$$