# Fourier transform of triangular function

Determine $X(\omega)$.

1. $g(t)$: I understand how to create a box from [-1,1] of amplitude 1/2.
2. $x(t) = g(t) * g(t)$
3. $X(\omega) = G(\omega)G(\omega)$

the solution I am seeing says that $G(\omega) = \frac{2\sin(\omega)} {2\omega}$

I don't understand where $\sin$ came from and that the values of the 2s correlate to. I've seen proofs, but can someone provide a simple explanation of what the variables are. Thanks

## 3 Answers

A triangular function can be generated by convolving two box functions as shown below.

This is where your Step 2 comes from.

The fourier transform of a convolution $g(t) \ast g(t)$ can be calculated by multiplying the fourier transform of $g(t)$ with itself, i.e. $G(\omega)G(\omega)$.

Recall that the Fourier transform of a box function is a Sinc function ($\textrm{sinc}(x) = \frac{\textrm{sin}(x)}{x}$).

Hence, $G(w)$ is some scaled version of a sinc function, and the Fourier transform of the triangular function is $G(w)^2$.

OK, so you understand that the signal $x(t)$ is given by the convolution of two rectangular functions extending from $-1$ to $1$ with a height of $1/2$. The only thing that is left to do is determine the Fourier transform of this rectangular function. You can do this very easily by applying the definition of the Fourier transform:

$$G(\omega)=\int_{-\infty}^{\infty}g(t)e^{-j\omega t}dt=\frac12\int_{-1}^{1}e^{-j\omega t}dt$$

I'm sure you can solve this integral yourself. The sine function comes into play because

$$\sin\omega = \frac{e^{j\omega}-e^{-j\omega}}{2j}$$

Finally, the Fourier transform of $x(t)$ is given by

$$X(\omega)=G^2(\omega)$$

The basis functions in Fourier Transform are Sine and Cosine. You shouldn't really be surprised that Sin function appeared in your analysis of a complex signal.