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Well this seems like $1-|t|$ for $|t|<1$ and $0$ for $|t|>1$ . Taking the Fourier transform $$X(ω) = \int_{-\infty}^\infty(1-|t|)e^{-jωt}dt\\=\int_{-\infty}^\infty e^{-jωt}dt -\int_{-\infty}^\infty|t|e^{-jωt}dt\\=2\piδ(ω)-\int_{-1}^1|t|e^{-jωt}dt\\=2πδ(ω)-\int_{-1}^0-te^{-jωt}dt-\int_{0}^1te^{-jωt}dt$$

The infinity integral goes from -1 to 1 since the function is zero elsewhere. Now I'm having trouble continuing with this. It doesn't seem to give me the result of the problem's solution which is $sinc^2(\frac{ω}{2\pi})$ , the sampling function.

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  • $\begingroup$ The function is supported on $[-1,1]$, therefore the Fourier transform should be $X(\omega)=\int_{-1}^1 (1-|t|)e^{-j\omega t}dt$. $\endgroup$ – Frank Lu Jun 30 '17 at 17:01
  • $\begingroup$ $f(x) = 0$ outside of the interval $[-1,1]$ $\endgroup$ – Doug M Jun 30 '17 at 17:02
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The function is not $1-|t|$. That is only valid for $-1\le t\le 1$. Actually \begin{align} \int_{-\infty}^\infty f(t)e^{-i\omega t}\,dt &=\int_{-1}^1(1-|t|)e^{-i\omega t}\,dt =2\int_0^1(1-t)\cos\omega t\,dt\\ &=\frac{2}{\omega}\left[(1-t)\sin\omega t\right]_0^1 +\frac{2}{\omega}\int_0^1\sin\omega t\,dt\\ &=-\frac2{\omega^2}\left[\cos\omega t\right]_0^1\\ &=\frac{2(1-\cos\omega)}{\omega^2}=\frac{4\sin^2(\omega/2)}{\omega^2}. \end{align}

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We can directly use the fact that a Triangular function (same as one in your question) is a result when two rectangular pulses(with same width) gets convoluted... Now defining a rectangular function $$A.\Pi \left(\frac{t}{\tau}\right)=\left\{\begin{matrix}A~,-\tau/2<t<\tau/2 \\ 0~,~ ~~~~~~~\text{otherwise}\end{matrix}\right.$$ Like this one

Now it's FT is fiven by $$\mathcal{F}\left\{A.\Pi \left(\frac{t}{\tau}\right)\right\}=A\tau\frac{\sin(\omega\tau/2)}{\omega\tau/2}=\mathrm{sa}\left(\frac{\omega\tau}{2}\right)=A\tau. \mathrm{sinc}(f\tau)$$ And when two of these pulses are convoluted we can obtain a triangular pulse enter image description here

And it's Ft is given by $$\begin{align}\mathcal{F}\left\{A\mathrm{Trig}\left(\frac{t}{\tau}\right) \right\}&=A\tau.\frac{\sin^2(\omega \tau/2)}{\omega^2\tau^2/4}\\&=A\tau.\mathrm{Sa}^2 \left(\frac{\omega \tau }{2}\right)\\&=A\tau .\mathrm{Sinc}^2(f \tau)\end{align}$$ In this case, we've got $A=1$ and $\tau=1$ , so required FT is $$\begin{align}\mathcal{F}\left\{\mathrm{Trig}\left(t\right) \right\}&=\mathrm{Sa}^2 \left(\frac{\omega}{2}\right)\\&=\frac{\sin^2 (\omega/2)}{\omega^2/4}\\&=\mathrm{Sinc}^2(f)\\&=\mathrm{Sinc}^2 \left(\frac{\omega}{2\pi}\right)\end{align}$$ Hope this helps, please tell me if by any means i can improve my answer , thanks !

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