0
$\begingroup$

I am working on a problem in a signals and systems class where I am being asked to find the inverse Fourier transform of a function. I am given the equation: $$X(j \omega)= \begin{cases} -\omega e^{-3 \omega}, & {-1 \lt \omega \le 0} \\ \omega e^{-3 \omega}, & { 0 \lt \omega \lt 1} \\ 0 , & { otherwise } \end{cases} $$ I need to find $x(t)$. So I computed the integral for the inverse fourier transform: $$ x(t) = \int\limits_{-1}^1 X(j \omega) e^{j \omega t} d\omega $$ At the end I got: $$ x(t) = \frac{e^{jt-3}+e^{-(jt-3)}}{2 \pi ((jt-3)^2)}+\frac{e^{-(jt-3)}-e^{jt-3}}{2\pi (jt-3)}-\frac{1}{\pi (jt-3)^2} $$ but the answer should look like:

$$ x(t) = \frac{1}{\pi}(\frac{sin(t-3)}{t-3} + \frac{cos(t-3)-1}{(t-3)^2}) $$

I know that I will have the use euler's identity, but I'm not sure how to use it given that the numerator and denominator both have that -3 in them. Any help would be greatly appreciated.

$\endgroup$
  • $\begingroup$ you'll find that you probably made a mistake as the poles of $x(t)$ are not a the same $t$. and in general, expand everything : the expression they gave you for $x(t)$ using $2i\sin(x) = e^{ix}-e^{-ix}$ and $\cos(x) = \ldots$, the $(it-3)^2$ etc. $\endgroup$ – reuns Mar 7 '16 at 23:56
  • $\begingroup$ What you do you mean exactly by the poles of x(t) aren't the same as t? Are you suggesting that I factor out the $e^{-3}$? Or are you saying that in general, $e^{jt+b} - e^{-(jt+b)} = 2jsin(t+b) $ $\endgroup$ – user1505399 Mar 8 '16 at 0:02
  • $\begingroup$ I suggested to expand everything. I agree it is boring, but you should be able to prove or disprove that the functions are the same $\endgroup$ – reuns Mar 8 '16 at 0:05
  • $\begingroup$ Haven't looked closely, but $e^{jt-3}$ is going to give you something like $e^{-3}\sin t$. I can't see how it would ever give you $\sin(t-3)$. Are you sure it shouldn't be $e^{j(t-3)}$? $\endgroup$ – David Mar 8 '16 at 0:12
  • $\begingroup$ I'm pretty sure. Or at the very least I can't see where another t term would come from. I derived the equation for $X(j \omega)$ from graphs provided. The magnitude was the same as $X(j \omega)$ if you ignore the e part. Angle for the magnitude was given as a straight line with slope $-3 \omega$ I don't think any of that would introduce an extra t term $\endgroup$ – user1505399 Mar 8 '16 at 0:36

Your Answer

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

Browse other questions tagged or ask your own question.