# Converting a triangle from the frequency domain to the time domain

I’ve been given a triangular signal that looks like this: $$X^{F}(\omega) = (2 -|\omega|) \cdot W_{[-2,2]}(\omega)$$ (this is just my interpretation of the signal from a picture I’ll add). I was asked to find it’s time domain representation. I tried expressing it differently, splitting the window into two heavisides, using the derivatives in the frequency domain to get $$t^2x(t)$$ in the time domain but i experienced difficulties and it didn’t quite work out. I recall that a triangle can be created from convolving two rectangles but am unsure how. If you could show me or hint at an easier method or set me on an elegant path to solving this i’d really appreciate it. thanks!

What the signal actually looks like:

• You’re correct, convolving two rectangles of the same length would give you a triangle. In the time domain that would correspond to multiplying two sinc functions together, resulting in a sinc squared (since both sincs would have the same argument). This is just intuition at this point. I’ll work it out for you later if someone hasn’t gotten to it yet ;), but that should get you started.
– Jdip
Jan 16 at 23:51
• I've a few hours where I can't attend to it, but I'll try to give it a shot myself if no one got to it first. - I see, thanks for the tip! Jan 17 at 6:16
• The Fourier Transform is typically a complex number. Your picture shows the magnitude but not the phase. Can you safely assume that the phase is zero ? Jan 17 at 9:29
• @Hilmar The OP has given an explicit formula for the Fourier transform which indicates that the transform is wholly real. Jan 17 at 15:20
• I think that had it not been real it would've been stated - and additiojal info given Jan 17 at 15:30

I recall that a triangle can be created from convolving two rectangles

That's correct. A rectangular function in the frequency domain inverse-transforms to a $$\texttt{normalized sinc}$$ in the time domain, and by the convolution theorem, convolution in the frequency domain translates to multiplication in the time-domain.

So, intuition tells us that convolving two rectangular functions in the frequency domain should give us the point-wise multiplication of two $$\texttt{normalized sinc}$$ functions in the time domain. Let's verify:

Let $$x(t) = \mathrm{sinc_{\pi}}(t) = \frac{\sin(\pi t)}{\pi t}$$ which, in the frequency domain, yields the well known $$x(t)\xrightarrow{\mathcal{F}}X(f) = \mathrm{rect}(f)$$ then gives: $$x^2(t) = \mathrm{sinc^2_{\pi}}(t) \,\,\xrightarrow{\mathcal{F}}\,\,\mathrm{rect}(f) * \mathrm{rect}(f) = \mathrm{triangle}(f)$$

• Hmmmm. I was always under the impression that the Fourier transform of $\operatorname{sinc}_{\pi}(t)$ was $\operatorname{rect}(f)$ which has value $1$ for $|f| = \left|\frac{\omega}{2\pi}\right| <\frac 12$, not $\operatorname{rect}(\omega)$ which has value $1$ for $|\omega| < \frac 12$. Could you please check? Jan 17 at 14:45
• You're absolutely correct, I'll edit.
– Jdip
Jan 17 at 16:26
• Nice. Now the OP can exercise his skills in applying your answer to his actual problem which involves rad/s instead of Hertzian frequency $f$. Jan 17 at 22:50
• I think a regular $\mathrm{sinc}(t) = \frac{\sin(t)}{t}$ might be more appropriate when dealing with angular frequencies... but since this looks like a homework question, I agree that the OP should try and figure this out by himself (which he pretty much has already)
– Jdip
Jan 18 at 1:30
• I'll do it - I just have a bunch of different stuff popping up - planning on doing it tommorow latest Jan 19 at 7:18

a proof that a convolution of 2 windows does indeed create $$X(\omega)$$: https://imgur.com/a/8qRjkJN (a server error occurred when I tried uploading the link) using this I now computed the triangular signal in time, where the fourier transform is defined as: $$\mathcal{F}\{x\}(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} \mathrm{d}t$$

I used the convolution - multiplication attribute and got the following: $$x(t) = \mathcal{F}^{-1} \{W_{[-1,1]} \ast W_{[-1,1]}\} = 2\pi \left[\frac{1}{\pi} \operatorname*{sinc}(t)\right]^2 = \frac{2}{\pi}\operatorname{sinc}^2(t)$$ I don’t really know if it’s equivalent to the expression posted in the answer above.