How to do the Fourier Transform of bounded function?

Question

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.

Answer 1

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

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