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There are some similar questions around but they didn't help me much.

I've got a graph that I'm supposed to perform a laplace transform on. From $0<t<2$, it's a ramp function from $f(0)=0 to f(2) = 10$ (slope of 5) and then from 2 to 6 it's a -2.5 slope from $f(2) = 10$ to $f(6) = 0$.

Another way to solve this, I think, would be to take derivatives until it's in the form of impulse functions and then do the laplace using the transformation for a second derivative, but this is what I was doing first and I got stuck ...

I translated the intervals into $5tu(t) -7.5tu(t-2) + 2.5tu(t-6)$, where $u(t)$ is a heaviside step function.

I did this thinking I could use the time shift transform, but got stuck with the $t*u(t-2)$ part. The time shift transform is $f(t-a)u(t-a) = e^{-as}$, but in all but the first case, t is not shifted, only the step function is.

Would I have to use the integral definition of the transform from this point, or am I missing something?

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  • $\begingroup$ If anybody knows how I can draw a graph in these questions, that'd be helpful :) $\endgroup$ – Daniel B. May 6 '14 at 5:10
  • $\begingroup$ Also bear in mind I'm coming from a rudimentary crash course in these transforms from an engineering textbook, I haven't taken differential equations yet. $\endgroup$ – Daniel B. May 6 '14 at 5:13
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The cleanest way to handle such problems is to write $$ t u(t-2) = (t-2+2) u(t-2) = (t-2) u(t-2) + 2 u(t-2)$$ Now take Laplace transforms and use the time shift property you wrote.

You are almost there.

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  • $\begingroup$ I ... wow. Um what just happened? Oh, I see, distributive. Clever :D But ... what do I do with the 2u(t-2)? I can't use the transform for u(t) on that can I? $\endgroup$ – Daniel B. May 6 '14 at 5:19
  • $\begingroup$ Yes. $u(t) \to 1/s$ Hence $u(t-2) \to e^{-2s}/s$. $\endgroup$ – user44197 May 6 '14 at 5:21
  • $\begingroup$ It'd be ... 2*integral(e^(-st) from 2->infinity ... scribble scribble $\endgroup$ – Daniel B. May 6 '14 at 5:22
  • $\begingroup$ Use time shift property! see my comment above $\endgroup$ – user44197 May 6 '14 at 5:22
  • $\begingroup$ Oh hey. You responded ... Oh right, because u(t) is a special case of u(t-a) where a is zero, right? I don't see that on my little chart. I'm allowed a cheat sheet on my exam, I should put that on there. $\endgroup$ – Daniel B. May 6 '14 at 5:24

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