Does someone know how to do the Fourier Transform of the signal

$$x(t) = t \cdot \frac{\sin^2(t)}{(\pi t)^2}$$

My first thought was: $$x(t)= \frac{t}{\pi^2} \cdot \frac{\sin^2(t)}{t^2} = \frac{t}{\pi^2} \cdot \operatorname{sinc}^2(t)$$

and try it with the convolution:

$$X(j \omega) = \frac{1}{2 \pi} \cdot F\left(\frac{t}{\pi^2}\right) * F(\operatorname{sinc}^2(t))$$

But the Fourier Transform of $t$ doesn't exist I think. How can I go from here?


The solution says

$$X(j \omega) = \frac{j}{2 \pi}\quad\text{for}\quad-2 <\omega<0$$


$$X(j \omega) = \frac{-j}{2 \pi}\quad\text{for}\quad0 <\omega<2$$

and $0$ everywhere else. But I have no idea how to get there.

  • $\begingroup$ Do you know anything about the "Dirac delta"? $\endgroup$ Jan 3, 2016 at 19:14
  • $\begingroup$ @Omnomnomnom yes, but I'm not sure where it would come into play here. $\endgroup$ Jan 3, 2016 at 19:15
  • $\begingroup$ Noting that $t=\int_0^t 1 dt$, perhaps you can find $F(t)$ using $F(1)$. $\endgroup$ Jan 3, 2016 at 19:21
  • $\begingroup$ Or, perhaps it is to say $$x(t)= C\sin(t) \operatorname{sinc}(t)$$ (for the right constant $C$). $\endgroup$ Jan 3, 2016 at 19:23
  • $\begingroup$ I'm so lost right now. $\endgroup$ Jan 3, 2016 at 19:27

2 Answers 2



From the Convolution Theorem, we have

$$\int_{-\infty}^\infty f(t)g(t)\,e^{i\omega t}\,dt=\frac{1}{2\pi}\int_{-\infty}^\infty F(\omega-\omega')G(\omega')\,d\omega'$$

Setting $f(t)=g(t)=\frac{\sin(t)}{\pi t}$ reveals

$$\int_{-\infty}^{\infty}\left(\frac{\sin(t)}{\pi t}\right)^2e^{i\omega t}\,dt=\frac1{2\pi}\int_{-\infty}^{\infty}F(\omega-\omega')F(\omega')\,d\omega'$$

where $ F(\omega)=\int_{-\infty}^{\infty}\frac{\sin(t)}{\pi t}e^{-i\omega t}\,dt=\text{rect}(\omega/2)$, where $\text{rect}(x)$ is the Rectangle Function.

Finally, note that if $F(\omega)=\int_{-\infty}^\infty f(t)e^{i\omega t}\,dt$, then $F'(\omega)=i\int_{-\infty}^\infty tf(t)\,e^{i \omega t}\,dt$.

SPOLIER ALERT Scroll over the highlighted area to reveal the solution

We have $F(\omega)=\text{rect}(\omega/2)$. Then, the convolution $\frac1{2\pi}\{F*F\}(\omega)$ becomes $$\frac1{2\pi}\{F*F\}(\omega)=\frac1{2\pi}\int_{-\infty}^\infty \text{rect}((\omega-\omega')/2)\,\text{rect}(\omega'/2)\,d\omega'=\begin{cases}\frac1{2\pi}(\omega +2)&,-2\le \omega <0\\\\\frac1{2\pi}(2-\omega)&,0<\omega\le 2\\\\0&,\text{elsewhere}\end{cases}$$The derivative of the convolution is given by $$\frac{d}{d\omega}\left(\frac1{2\pi}\{F*F\}(\omega)\right)=\begin{cases}\frac{1}{2\pi}&-2<\omega<0\\\\-\frac{1}{2\pi}&,0<\omega<2\\\\0&,\text{elsewhere}\end{cases}$$Finally, multiplying by $-i$ yields the Fourier Transform on interest $$\int_{-\infty}^\infty t\,\left(\frac{\sin(t)}{\pi t}\right)^2\,e^{i\omega t}\,dt=\begin{cases}\frac{-i}{2\pi}&,-2<\omega<0\\\\\frac{i}{2\pi}&,0<\omega<2\\\\0&,\text{elsewhere}\end{cases}$$

  • $\begingroup$ Thanks for the help, even though I don't fully understand. The post of Omnomnomnom helped already. $\endgroup$ Jan 3, 2016 at 19:59
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    $\begingroup$ @canbus this approach is neat too! First, note that $rect(\omega)*rect(\omega)$ gives you a triangle. Then, he uses the differentiation property of the F-transform: $$ \mathcal F [t\cdot f(t)] = \frac{d}{d\omega} \mathcal F[f(t)] $$ $\endgroup$ Jan 3, 2016 at 20:06
  • $\begingroup$ That's clever, I will try this approach too. $\endgroup$ Jan 3, 2016 at 20:10

Hint: You split $x(t)$ in a non-helpful way. Instead, note that $$ x(t) = \frac{1}{\pi^2} \sin t \cdot \operatorname{sinc}(t) $$ Now, $x$ is the product of two functions whose Fourier transform you may compute.

  • $\begingroup$ Ah, now I get what you meant. I will try that. Thanks $\endgroup$ Jan 3, 2016 at 19:37
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    $\begingroup$ Yeah, I left out a word in the comment, and I figure the $C$ just confuses things anyway. Let me know if this doesn't work out. $\endgroup$ Jan 3, 2016 at 19:39
  • $\begingroup$ I get $ \frac{j}{2 \pi} ( rect(\frac{\omega + \omega _{0}}{2 \pi}) - rect(\frac{\omega - \omega _{0}}{2 \pi})$. Sounds kind of right, if I look at the solution. Thanks for the quick help. $\endgroup$ Jan 3, 2016 at 19:57
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    $\begingroup$ I think in this context we should have $\omega_0 = 1$, no? $\endgroup$ Jan 3, 2016 at 19:59
  • $\begingroup$ You're right, but if we don't shift the rect pulse, they both stay at $\omega = 0$, no? The solution I was given says that there's a constant factor of $\frac{j}{2 \pi}$ for $-2 < \omega < 0$ and $\frac{-j}{2 \pi}$ for $0< \omega < 2$ $\endgroup$ Jan 3, 2016 at 20:04

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