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I have problems finding the solution of this exercise:

Find the Fourier transfrom of the signal. $$ x(t) = e^{-t} \delta' $$

I have tried the following but i'm not sure if it's correct.

$$ X(w) = \int_{-\infty}^\infty e^{-t} \delta'e^{-iwt} dt$$

$$ X(w) = \int_{-\infty}^\infty \frac d {dt}\delta e^{-(1+iw)t} dt$$

Using integration by parts $$ u = e^{-(1+iw)t} \\du = \frac {-1}{1+iw}e^{-(1+iw)t} \\ v = \delta(t) \\ dv = \frac d{dt} \delta(t)$$

$$ X(w) = \delta e^{-(1+iw)t} + \int_{-\infty}^\infty \delta(t) \frac {1}{1+iw}e^{-(1+iw)t} dt $$

Evaluating the first parts becomes zero $$ X(w) = \int_{-\infty}^\infty \delta(t) \frac {1}{1+iw}e^{-(1+iw)t} dt $$

At this point i don't know how to transform this. I would like to know if my calculations are correct and how to continue.

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  • $\begingroup$ The Dirac delta distribution is designed to pull out the value of the function at zero. That is all that is left to do here. $\endgroup$ – Cameron Williams Jun 28 '17 at 16:44
  • $\begingroup$ The derivative of $e^{-(1+i\omega)t}$ is $-(1+i\omega)e^{-(1+i\omega)t}$. $\endgroup$ – Mark Viola Jun 28 '17 at 17:27

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