0
$\begingroup$

I'm seeking the Fourier transform of $t \cdot (u(t+1)-u(t-1))$ enter image description here

Given the fact that (The first minus here should be a plus) enter image description here

Using properties.

My initial approach was: enter image description here but I wasn't sure how I should go around nesting properties, so I decided that I can perhaps: enter image description here However, checking Wolfram, we have: enter image description here

I don't understand what's going on, my questions are:

1- What went wrong in my second attempt?

2- How would I carry on if I chose to go with my first attempt?

Sorry if these questions seem naive or anything, I've just been so recently introduced to the topic.

$\endgroup$
6
  • $\begingroup$ Hi! Your very beginning given fact... is plain wrong, from LHS to RHS? From which publication did you get that plain wrong image? Anyway, the required Fourier transform of the time-limited ramp is quite easy with the definition of the Fourier transform integral (you do not have to use the Fourier properties here). Indeed it will be more involved to use the properties to get the same result. $\endgroup$
    – Fat32
    Apr 22 at 13:44
  • $\begingroup$ Your math handwriting is very nice. But we need to get you to learn $\LaTeX$ markup so that your equations can be copied and edited. $\endgroup$ Apr 22 at 13:51
  • $\begingroup$ Your second equation is plain non-sense. $u(t-T_1)-u(t-T_1)$ is just plain zero, I think there should be a $+T_1$ somewhere in there. $\endgroup$
    – Hilmar
    Apr 22 at 14:29
  • $\begingroup$ @Fat32 Yes, indeed. It was a typo that I forgot to mention. I agree with your second point, however, the exercise is there to practice using properties. $\endgroup$
    – Essam
    Apr 22 at 17:57
  • $\begingroup$ @robertbristow-johnson Thank you, it takes far more effort if I choose to use MathJax, and also because the equations there were related to what I drew on the side, thank you for the advice whatsoever. $\endgroup$
    – Essam
    Apr 22 at 18:01
1
$\begingroup$

Your first approach doesn't work because the original signal is simply not the integral of the signal you sketched below it. Just graphically integrate that signal to see this. Your second approach doesn't work for the very same reason. If you integrate the rectangular signal you don't get a signal that is zero for $t>1$. The value of the integral would remain at its final value and wouldn't change anymore because the original signal doesn't change for $t>1$.

That exercise can be solved in several ways, and you should try them all to get some practice. First, just compute the Fourier integral without trying any smart tricks. The signal is simple enough for this to be a good option, even if just as a reference.

Second, convince yourself that if $X(j\omega)$ is the Fourier transform of $x(t)$, then $jX'(j\omega)$ is the Fourier transform of $tx(t)$, where $X'(j\omega)$ is the derivative of $X(j\omega)$ with respect to $\omega$. Since the signal is defined as $t$ times some other signal (whose Fourier transform is very basic), using the above-mentioned property is quite obvious.

A third approach could be to define a signal $x_1(t)$ which satisfies

$$x(t)=\int_{-\infty}^tx_1(\tau)d\tau$$

Then compute the Fourier transform of $x_1(t)$ and use the integration property of the Fourier transform to derive $X(j\omega)$. This is what you actually tried to do, but your definition of $x_1(t)$ was wrong. As a hint, note that for the correct definition of $x_1(t)$ you need two Dirac delta impulses.

Obviously, if you do things right, all three results should agree.

$\endgroup$
1
  • $\begingroup$ I find this really insightful, thank you. $\endgroup$
    – Essam
    Apr 22 at 20:30

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.