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I have to convolve the following signal and its impulse response, and I thought taking the Fourier Transform would be the best approach:

$$x(t) = te^{-2t}*u(t)$$

$$ h(t) = e^{-4t}*u(t)$$

Where $u(t)$ is the unit step/Heaviside function.

I know that convolution in the time domain is equivalent to a multiplication in the frequency domain, I performed the following conversions from a table intending to multiply:

$$X(w) = \frac{1}{(2 + jw)^2}$$

$$ H(w) = \frac{1}{(4 + jw)} $$

I got a little stuck here. When I multiply out, I'm not sure how to simply into something I can read off a table for an inverse transform, if at all, to get the time domain convolution.

Can anyone offer any assistance? Greatly appreciated! My apologies for my lack of LaTEX ability.

EDIT My solution attempt with partial fractions (and having a go at LaTEX as well!) Can anyone confirm?

$$Y(w) = \frac{1}{4(4 + jw)} - \frac{1}{(4(2 + jw)} + \frac{1} {2(2 + jw)^2}$$

Rome_Leader
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  • Multiply through (don't expand the denominator) and do a partial fractions decomposition. – mrf Jul 30 '13 at 23:29
  • OK, it wasn't straightforward due to the squared term as expected, but my result was Y(w) = 1/4(4 + jw) - 1/(4(2 + jw) + 1/ 2(2 + jw)^2. That certainly seems reasonable to me! – Rome_Leader Jul 30 '13 at 23:41

1 Answers1

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$$\frac{1}{(2+j\omega)^2(4+j\omega)}=\frac{1/j^3}{1/j^3(2+j\omega)^2(4+j\omega)}=\frac{j}{(\omega-j2)^2(\omega-4j)}$$

Now use partial fraction decomposition.

rurouniwallace
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