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I have $x_1(t)$ here. To get $x_2(t)$, I need to differentiate $x_1(t)$. Express $x_2(t)$ as $2u(t+2)-4u(t)+2u(t-2)$.
From Fourier transform definition integral, I got $X_2(j\omega)=\frac{2e^{j\omega 2}}{j\omega}-\frac{4\pi \delta(\omega)}{j\omega}+\frac{2e^{-j\omega2}}{j\omega}$.
Is this correct? It seems weird and would be complicated to calculate the magnitude and phase spectrum. Thanks! enter image description here

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  • $\begingroup$ Simplifying it, I got $X_2(j\omega) = \frac{4\cos(2\omega)}{j\omega} - \frac{4\pi \delta(\omega)}{j\omega}$ $\endgroup$ – keanehui Apr 5 '20 at 8:42
  • $\begingroup$ You are on the right path. Since $u(t)$ is not a stable function, it's fourier transform will have discontinuity. $\endgroup$ – jithin Apr 5 '20 at 9:49
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    $\begingroup$ @jithin that's not right: the rect function (not stable) has the Fourier transform sinc, and that is as continuous as any function will ever get! $\endgroup$ – Marcus Müller Apr 5 '20 at 12:21
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    $\begingroup$ @MarcusMüller Pssssssst, not all functions have plugged discontinuities. $\endgroup$ – Cedron Dawg Apr 5 '20 at 12:27
  • $\begingroup$ @CedronDawg :) but really, the rect hint is all I'm willing to give here – I must have tried to derive the Fourier transform of the Heaviside function so many times that I forgot how to do it right, because because I don't get the (ugly) right result when I try to do it again. Keanehul, $u(t)$ is not the function you're looking for, honestly! $\endgroup$ – Marcus Müller Apr 5 '20 at 12:33
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It looks like the idea of that exercise is to compute the Fourier transform of $x_3(t)$, and then from that derive the Fourier transform of the original function $x_1(t)$.

Note that $x_3(t)$ is just a sum of scaled and shifted Dirac impulses, so computing its Fourier transform is trivial. The Fourier transform of $x_2(t)$ is then easily derived by multiplying $X_3(j\omega)$ by $\pi\delta(\omega)+1/j\omega$, and noting that the Dirac delta is cancelled because $X_3(0)=0$. So you just end up dividing by $j\omega$. In the same way, $X_1(j\omega)$ is obtained from $X_2(j\omega)$. Also note that none of the three Fourier transforms $X_1(j\omega)$, $X_2(j\omega)$, and $X_3(j\omega)$ have any Dirac impulses.

It's a good idea to cross-check your result by realizing that $x_1(t)$ can also be represented as the convolution of two rectangular functions, i.e., its Fourier transform must be a squared sinc-function.

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  • $\begingroup$ Thanks! And is it possible to find $X_2(j\omega)$ from only $x_1(t)$ or $x_2(t)$ using the Fourier transform table, which is a hint from the exercise? $\endgroup$ – keanehui Apr 6 '20 at 3:49
  • $\begingroup$ @keanehui: Sure, you can find $X_2(j\omega)$ from $x_2(t)$ if you know the transform of $u(t)$ and if you know the frequency domain consequence of shifting in the time domain. $\endgroup$ – Matt L. Apr 6 '20 at 7:53

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