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The question I had was

Find the Laplace transform of $$f(t)=10e^{-200t}u(t).$$

Would it be correct to take out the 10 because it is a constant, find the Laplace transform of $e^{-200t}$ and then multiply it by the Laplace transform of $u(t)$ to obtain a final answer of : $$10\left(\frac{1}{s+200}\right)\left(\frac{1}{s}\right)?$$ The $u(t)$ is what is really confusing me in this problem.

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3 Answers 3

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Yes, you can move the 10 out, but no, you cannot proceed as you do. Independent of your definition of Laplace transform (whole $\mathbb R$ or just $\mathbb R^+$ as domain of integration in the first step), you will find $$ (\mathcal L f)(s)=\int 10e^{-200t}u(t)e^{-st}\,dt=10\int_0^{+\infty}e^{-(200+s)t}\,dt=\frac{10}{s+200}. $$ As an alternative, you could also use rules for the Laplace transform: first that $\mathcal L u=1/s$ and that multiplication with the exponential shifts it, $1/(s+200)$ and that the multiplication of constant 10 just multiplies (since the Laplace transform is a linear operator), $10/(s+200)$.

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Another way to solve this problem is that the u(t) function means 1 for t>0,

0 for t <0 Since the laplace transform is restricted to the functions with t>0,i. e, for applying various properties of laplace transforms , functions must have t>0 condition, , So from earlier discussions our function becomes 10e^(-200t ), t>0 Now the laplace transform of 1 is 1/s and it is multiplied with 10e^(-200t ) so by shifting rule the laplace transform becomes 10/s+200. Just be careful while applying the properties or shortcut formulas of laplace transform for example if we have a function that is multiplied with Ua(t) i. e, u (t-a), then we can't directly just use the general shortcut formulas. Two methods can be used- 1. Either integrate from a to infinity.(laplace integral)

2 . or use the formula for u (t-a), Which is e^(-as)/s.

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Laplace Transform of $$e^{-200t}$$ does not exist

But Laplace transform of $$e^{-200t} u(t)$$ exist

$$ e^{-200t} u(t) \rightleftharpoons \frac{1}{s+200}$$

$$ 10e^{-200t} u(t) \rightleftharpoons \frac{10}{s+200}$$

And you can not multiply separately calculated Laplace transform

Multiplication in time domain gets converted into convolution in Laplace Domain

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