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I am puzzled by what should be a simple problem. I am deriving the Laplace transform for

$$ (t-120) u(t-120) $$

where $u(t)$ is the unit step signal. Using the second shift theorem, ie

$$ \mathcal{L} (f(t-a) u(t-a) ) = e^{-as} F(s) $$

I would have thought that:

$$ \mathcal{L} ((t-120) u(t-120) = e^{-120 s} \left( \frac{1}{s^2} - \frac{120}{s} \right) $$

To check I ran the same problem through Mathematica 10 and it gave:

$$ \mathcal{L} ((t-120) u(t-120) = \frac{e^{-120 s}}{s^2} $$

which is different. Where have I gone wrong?

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The thing is that you have to identify $f(t)=t$. Then you have $F(s)=1/s^2$. If you consider $f(t)=t-120$, where is your $f(t-a)$? It would be $t-a-120$, and you can't apply the theorem. However if $f(t)=t$ then your problem ($(t-120)\,\mathcal U(t-120)$) can be expressed as $f(t-a)\,\mathcal U(t-a)$, where $a=120$, and now you can apply the theorem.

So what you are doing is to take the laplace transform of $120$; don't, because you aren't using the theorem anymore.

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  • $\begingroup$ Ah yes, I didn't read the theorem properly, the missing piece is: if f(t) has the transform F(s) then.... $\endgroup$
    – rhody
    Feb 7, 2015 at 0:45

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