I have a triangular pulse given by $$x\left(\frac{t}T\right) = \begin{cases} 1-\frac {|t|}T, & \text{if $T\ge t$} \\ 0, & \text{otherwise} \end{cases}$$

Given that $F\left(\operatorname{rect}\left(\frac{t}T\right)\right)=T\operatorname{sinc}(fT)$

where $F$ denotes a Fourier transformation

$$\operatorname{rect}\left(\frac{t}T\right)=\begin{cases} 1, & \frac{T}{2} \ge t \ge \frac{-T}{2} \\ 0, & \text{otherwise} \end{cases}$$

Prove that $F\left(x\left(\frac{t}T\right)\right)=T\operatorname{sinc}^2(fT)$.

As my knowledge, I prove above equation is $T^2\operatorname{sinc}^2(fT)$, instead of $T\operatorname{sinc}^2(fT)$.

We have $x\left(\frac{t}T\right)=\operatorname{rect}\left(\frac{t}T\right)*\operatorname{rect}\left(\frac{t}T\right)$ Then \begin{align} F\left(x\left(\frac{t}T\right)\right) & =F\left(\operatorname{rect}\left(\frac{t}T\right)\right)\cdot F\left(\operatorname{rect}\left(\frac{t}T\right)\right)\\ & =T \operatorname{sinc}(fT) \cdot T \operatorname{sinc}(fT) =T^2\operatorname{sinc}^2(fT). \end{align}

What is wrong in my solution? How to correct it?

  • $\begingroup$ Do you also have $\operatorname{sinc}^2(fT)$ instead of $\operatorname{sinc}^2(f)$? Also, what is $f$? $\endgroup$ – Silvia Ghinassi Oct 12 '15 at 3:04
  • $\begingroup$ Sorry. It must be $fT$. I corrected it $\endgroup$ – Jame Oct 12 '15 at 3:08
  • 1
    $\begingroup$ $f$ is frequency in Fourier domain $\endgroup$ – Jame Oct 12 '15 at 3:12

For $t=0$, we have $x(t/T)=1$ but $\operatorname{rect}(t/T)*\operatorname{rect}(t/T)=T$. This, and the difference between your answer and the correct one, should give you a hint of what is wrong.

  • $\begingroup$ Could you based on my solution and correct it. I have no idea to prove my problem $\endgroup$ – Jame Oct 12 '15 at 3:14

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