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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.