# Laplace transform of triangle pulse. Step function confusion

Discussion in 'Math' started by RolfRomeo, Dec 7, 2009.

1. ### RolfRomeo Thread Starter Member

Apr 27, 2009
17
0
Exams are approaching, and I'm working through some old assignments. One of which is a cause of great confusion:

I'm supposed to find the time-domain description of a symmetric triangle pulse with halfperiod T=1, and maximum amplitude A=1, starting at t=0 and returning to 0 at t=2T, and then Laplace-transform it. In of itself this is not very difficult, and I think I know how to do it correctly, BUT:

Take a look at the attached plot. The blue curve correctly describes the signal, but the red curve is what gives the correct Laplace-transform. Can someone please explain to me why this is, as I can't seem to wrap my head around it.

The only thing I can imagine is if the delayed unit step function u(t-d) has some additional operator-like properties, so that u(t-d)*f(t) becomes u(t-d)*f(t-d). Or something...

Any thoughts are appreciated.

edit: a = A/T

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2. ### RolfRomeo Thread Starter Member

Apr 27, 2009
17
0
Replying to myself here, but since I'm partially answering my own question I figured it was called for.

After digging through the web some more I found this:
Time-displacement theorem
L[u(t-a)*g(t-a)] = exp(-as)*G(s)

I have only found it referenced in a table, with no proof or discussion, but that may be what I'm looking for.

Eureka!
By looking at the line segments as functions of t: f(t) = a*t I am able to arrive at the correct transform A/(T*s^2) * (1 - exp(-Ts))^2 from the time-description that looks right. What still puzzles me is that the shortcut from the not-right-looking time-description works. Or maybe rather the extra s-terms the correct looking one gives in the laplace domain, when not using the above theorem.

Last edited: Dec 7, 2009