I have to calculate the analytic expression of Fourier transform $$ x(t) = t{\rm rect} ( t- \frac{1}{2} ).$$ First I made the graph of these two signals and I obtained the graph I posted.

enter image description here

Now I wrote the analytic form of this graph that is $$ x(t) = t $$ for t from 0 to 1 and $$ x(t)=0 $$ for the rest. Now I should calculate the Fourier transform of $$ x(t) = t $$ but this should be $X(f)$ (?). My theory it’s completely wrong because my book obtained $$ X(f) = { \frac{1}{2} {\rm sinc} (f) + \frac{1}{i2\pi f}\left[({\rm sinc} (f )- \cos (\pi f ) \right]} e^{-i \pi f }. $$

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    $\begingroup$ This is probably the easiest if you do it the good old fashioned way and evaluate the actual Fourier integral. Simply adjust the integration interval to the range where x(t) is non-zero $\endgroup$ – Hilmar Apr 20 '20 at 14:30
  • $\begingroup$ Using your suggestion I understand that if $$ s(t)=\rect(t - \frac{1}{2} )$$ so $$ s’(t)= t \rect(t - \frac{1}{2} ) $$. So , since $$ x(t)=s’(t) $$ using derivative property of Fourier I obtain that $$ F [s’(t)] = i2 \pi f X(f) $$. Now I’m blocked :/ $\endgroup$ – Elena Martini Apr 20 '20 at 16:29
  • $\begingroup$ x(t) is a rect so I obtained $$ i2\pi f sinc(\pi f) e^{-i \pi f }$$ , I can write this as $$ i2\pi f sinc(\pi f) [ cos ( \pi f ) - i sen (\pi f )] $$ but this isn’t the result I should obtain $\endgroup$ – Elena Martini Apr 20 '20 at 16:36
  • $\begingroup$ I think you have your time-domain derivatives wrong. The derivative of rect() is NOT $ t \cdot rect$ and the derivative of $ t \cdot rect$ is NOT rect(). $\endgroup$ – Hilmar Apr 20 '20 at 16:55

If you take the derivative of the signal using the graph you should get $$x'(t) = \mathrm{rect}\Big(t-\frac{1}{2}\Big) - \delta(t-1)$$ because the derivative of the line $t$ in $(0,1)$ is its slope, $1$, but there is also a delta function "looking downwards" due to the discontinuity at $t=1$.

The Fourier Transform of the derivative is $$X_d(f) = e^{-j\pi f}\mathrm{sinc}(f) - e^{-j2\pi f}$$

The time differentiation/itegration property you get $$X(f) = \frac{X_d(f)}{j2\pi f} + \frac{1}{2}X(0)\delta(f) = e^{-j\pi f}\frac{1}{j2\pi f}(\mathrm{sinc}(f) - e^{-j\pi f}) + \frac{1}{4}\delta(f)$$

I don't see how your book's result is similar to this one. I've actually ran a simulation in Octave and this is what I get if I do an Inverse FT on the result. enter image description here

So maybe your book's result is incorrect (?)...

  • $\begingroup$ now I’ve better understand the process but I have a problem with the derivative. First of all I have $$ x(t) = t \rect ( t - \frac{1}{2} ) $$ , now $$ x’(t) = t’ \rect ( t - \frac{1}{2} ) + t \rect ‘( t - \frac{1}{2} ) $$ and this should be $$ x’(t) = \rect ( t - \frac{1}{2} ) + t \rect ‘( t - \frac{1}{2} ) $$ but $$ t \rect ‘( t - \frac{1}{2} ) = t [ \delta (t+0) -\delta (t-1) ] $$ so I obtained $$ x’(t) = \rect ( t - \frac{1}{2} ) + t \delta (t) - t\delta (t-1) $$ ( as always THANK YOU!) $\endgroup$ – Elena Martini Apr 21 '20 at 14:20
  • $\begingroup$ I write here the correct result because I wrote it wrong previously $$ (\frac{1}{2} sinc(f) + \frac{1}{i 2 \pi f}(sinc f - cos (\pi f ))) e^{- i \pi f} $$ $\endgroup$ – Elena Martini Apr 21 '20 at 14:32
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    $\begingroup$ Good! The sampling property of the delta function yields $$t\delta(t) = t\Big|_{t=0}\delta(t) = 0$$ and $$-t\delta(t-1) = -t\Big|_{t=1}\delta(t-1) = -1\delta(t-1) = -\delta(t-1)$$ Can you proceed from this now? $\endgroup$ – GKH Apr 21 '20 at 15:06
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    $\begingroup$ And yes, the new result is correct. ;) However, it seems to come from direct integration. Maybe that's what the example is looking for. My approach is a bit more elegant but maybe not what you're looking for. $\endgroup$ – GKH Apr 21 '20 at 15:12
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    $\begingroup$ I didn't get the first doubt. About the second doubt, I applied differentiation/integration property to find the answer in a more easy and convenient way. $X_d(f)$ is the Fourier Transform of the derivative of $x(t)$, that is, $x'(t)$. But that's not what you want. You want the FT of $x(t)$, which is $X(f)$. The aforementioned properties have $X(f)$ in their equations so you can solve for $X(f)$. It's like finding something else first that will help you find what you want in an easier way. $\endgroup$ – GKH Apr 21 '20 at 17:24

There is a Fourier transform property called as "differenciation in frequency domain" which is as follows:

If the Fourier transform of $x(t)$ is $X(\omega)$, then the Fourier transform of $tx(t)$ is as below: $$\mathcal{F}(tx(t)) = j\frac{d}{d\omega} (X(\omega))$$

Where $\omega = 2\pi f$,

I think you would be able to easily derive the answer you mentioned in your question now.

Hope that helps


The definition of the Fourier Transform is

$$F(\omega) = \int_{-\infty}^{+\infty} x(t) \cdot e^{- j \omega t} dt $$

Since you x(t) is only non-zero on $[0,1]$ you can simplify this to

$$F(\omega) = \int_{0}^{1} x(t) \cdot e^{- j \omega t} dt $$

Pop in the definition of x(t) between 0 and 1, solve the integral and you are done.

  • $\begingroup$ Using integration by parts and considering that the integral of a rect is 1 I have to solve $$ \int_{0}^{1} t e^{-i 2 \pi f t } $$ and I obtained $$ \frac{2(cos 2\pi f - i sen 2 \pi f ] -1}{-2 i \pi f } $$ $\endgroup$ – Elena Martini Apr 20 '20 at 17:51

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