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I was trying to solve a Fourier transform of a function using the properties of Fourier transforms. The function is given as:

$\frac{At}{2}$ for $-2<t<2$ and $0$ for all other $t$. Doing the integral from the definition of of a Fourier transform gives the result: -$Ai\left(\frac{\sin(2\omega)-2\omega\cos(2\omega)}{\omega^2}\right)$. While when I try to do it with derivative property of Fourier transforms I get a different answer as $\frac{2A\sin(2\omega)}{2i\omega^2}$.

The idea is that I express the above bounded function as the product of the function with a rectangular pulse that has a period of 4. Like this: $\frac{At}{2}p(t)$. Then I take the derivative of this as just $\frac{A}{2}p(t)$ and then take the Fourier Transform of $\frac{A}{2}p(t)$. For the Fourier transform of the main function I simply divide by $iw$.

The two results are completely different from each other. What's wrong with taking my approach?

P.S I am sorry for not being able to write the formulas in a very mathematical manner.

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    $\begingroup$ Why is the derivative of $At p(t)/2$ just $Ap(t)/2$? Shouldn't it be $Ap(t)/2 + At p'(t)/2$? $\endgroup$
    – GKH
    Feb 22, 2020 at 16:59
  • $\begingroup$ @GKH yes you might be correct. I thought the pulse wouldn't make any difference. $\endgroup$
    – David Prifti
    Feb 22, 2020 at 17:36
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    $\begingroup$ Swapping orders of derivation and integration requires some care, under some hypotheses. When the latter are not met, some integrals may be undefined, and one can get any (false) result. This is especially possible when it involves distributions or generalized functions. $\endgroup$
    – Laurent Duval
    Feb 22, 2020 at 18:32
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    $\begingroup$ @Laurent Duval right, but can you be more specific on this example? $\endgroup$
    – GKH
    Feb 22, 2020 at 19:52
  • $\begingroup$ @David Prifti, have you managed to get the correct answer? $\endgroup$
    – GKH
    Feb 23, 2020 at 8:21

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Well, imho, one good way to do it is to take the signal's derivative in time and make use of the property $$\frac{d}{dt}x(t) \longleftrightarrow j2\pi f X(f)$$

I'll try to show it analytically.

Since the signal is discontinuous at $t=\pm 2$ you should include Delta function in the derivative, that is, $$\frac{d}{dt}\Big[\frac{A}{2}t\:\mathrm{rect}\Big(\frac{t}{4}\Big)\Big]= \frac{A}{2}\mathrm{rect}\Big(\frac{t}{4}\Big) + \frac{A}{2}t\:\mathrm{rect}'\Big(\frac{t}{4}\Big)$$

Given that a unitary rectangular pulse can be written as a sum of two unit step functions $$\mathrm{rect}\Big(\frac{t}{T}\Big) = u\Big(t+\frac{T}{2}\Big) - u\Big(t-\frac{T}{2}\Big)$$ then the pulse's derivative is $$u'\Big(t+\frac{T}{2}\Big) - u'\Big(t-\frac{T}{2}\Big) = \delta\Big(t+\frac{T}{2}\Big) - \delta\Big(t-\frac{T}{2}\Big)$$

Going back, we get $$\frac{d}{dt}x(t)= \frac{A}{2}\mathrm{rect}\Big(\frac{t}{4}\Big) + \frac{A}{2}t\:\mathrm{rect}'\Big(\frac{t}{4}\Big) = \frac{A}{2}\mathrm{rect}\Big(\frac{t}{4}\Big) + \frac{A}{2}t\Big(\delta(t+2) - \delta(t-2)\Big)$$

The sampling property of a Delta function yields

$$t\delta(t \pm t_0) = t\Big|_{t=\pm t_0}\delta(t \pm t_0) = (\mp t_0)\delta(t\pm t_0)$$

and in your case

$$\frac{A}{2}t\Big(\delta(t+2) - \delta(t-2)\Big)=-A\delta(t + 2)-A\delta(t - 2) $$

and in total

$$\frac{d}{dt}x(t)= \frac{A}{2}\mathrm{rect}\Big(\frac{t}{4}\Big) + \frac{A}{2}t\:\mathrm{rect}'\Big(\frac{t}{4}\Big) = \frac{A}{2}\mathrm{rect}\Big(\frac{t}{4}\Big) -A\delta(t + 2)-A\delta(t - 2)$$

The Fourier Transform of the latter is

$$F\Big\{\frac{d}{dt}x(t)\Big\} = 2A\mathrm{sinc}(4f) - Ae^{j2\pi 2f} - Ae^{-j2\pi 2f} = 2A\mathrm{sinc}(4f) - 2A\cos(2\pi 2f)$$

and from the derivative property $$X(f) = \frac{1}{j2\pi f}\Big(2A\mathrm{sinc}(4f) - 2A\cos(2\pi 2f)\Big) = \frac{jA}{2\pi f}\Big(2\cos(2\pi 2f) - 2\mathrm{sinc}(4f) \Big)$$

which gives

$$X(f) = jA\Big(\frac{2\cos(2\pi 2f)}{2\pi f} - \frac{\sin(4\pi f)}{(2\pi f)^2} \Big)$$

Finally, with the $j\omega$ notation

$$X(j\omega) = jA\Big(\frac{2\omega\cos(2\omega)}{\omega^2} - \frac{\sin(2\omega)}{\omega^2} \Big) =jA\Big(\frac{2\omega\cos(2\omega) - \sin(2\omega)}{\omega^2} \Big)$$

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    $\begingroup$ I think it should be pointed out that computing $X(f)$ from the first equation in your answer only works if $X(0)=0$, which happens to be the case for the OP's example. In other words, dividing by $j2\pi f$ in the Fourier domain does not generally correspond to integration from $-\infty$ to $t$ in the time domain. $\endgroup$
    – Matt L.
    Feb 23, 2020 at 11:57
  • $\begingroup$ Indeed. The signal is odd, so its mean value is zero and that allows the property to work. $\endgroup$
    – GKH
    Feb 23, 2020 at 12:12

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