# Simple Laplace Question

Discussion in 'Homework Help' started by jegues, Mar 2, 2011.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Hello all.

I designed myself a laplace question for an upcoming test and I want to make sure I've got the analysis down correctly.

Can someone check my work?

Also, for the inverse laplace of a constant over, $\frac{\gamma}{s}$ the inverse laplace is simply $\frac{\gamma}{}$ correct?

My table for laplace transforms says unit step function, why is that?

Thanks again!

File size:
283.5 KB
Views:
14
2. ### Vahe Member

Mar 3, 2011
75
9
In your final answer, you forgot the "t" after the 250. It should be $e^{-250t}$. Also you can just multiply your entire answer by the step function $u(t)$ instead of writing "for $t \ge 0$". So your final answer should be $(-4 + 4 e^{-250t}) u(t)$.

And yes the Laplace transform of $\gamma u(t)$ is $\gamma/s$. You can simply use the definition of the Laplace transform to show this. If $f(t)=\gamma u(t)$, its Laplace transform $F(s)$ is given by
$
F(s) = \int_{0}^{\infty} f(t) e^{-st} dt = \int_{0}^{\infty} \gamma u(t) e^{-st} dt = \gamma \int_{0}^{\infty} e^{-st} dt
$

where the last equality is due to the fact that $\gamma$ is a constant and is therefore pulled outside of the integral and the step function is simply unity (=1) over the range of the integral. Now if we perform the integral we get
$
F(s) = \gamma \int_{0}^{\infty} e^{-st} dt = \gamma \left[ -\frac{e^{-st}}{s}\right] |_{0}^{\infty}=-\frac{\gamma}{s} \left[ 0-1\right] = \frac{\gamma}{s}
$

Hopefully this clears up the matter for you. Please go over the details and show this to be true for yourself. Also, your book should contain some examples on how to derive the Laplace transforms for simple functions like the step function, delta function, etc.

Cheers,
Vahe