MIT 6.003 HW#8 Problem 4 - Fourier Coefficients of Triangle Wave

In the mentioned homework, part of the solution involves finding the Fourier coefficients of the triangle wave.

The solution mentions that we can express this function as follows:

What does that multiplication signal means?

I approached the problem from a completely different angle of viewing the triangle wave as the integral of a piecewise constant function defined as follows:

$$f(x)= \begin{cases} \frac{1}{2}&\text{if}\, -\frac{1}{4}\leq x\leq0\\ -\frac{1}{2}&\text{if}\, 0\leq x\leq\frac{1}{4} \end{cases}$$

Taking the integral similarly and then multiplying by $$\frac{1}{2\pi ik}$$ gives me a completely different answer of $$-\frac{1}{4\pi^2k^2}[\cos(\frac{\pi}{2}k)-1]$$.

What is wrong with my approach?

• Are you sure that's not the convolution operator? – Envidia Jan 19 at 0:34

The symbol $$\underset{T}{*}$$ in the solution refers to ($$T$$-) periodic convolution, which is defined as

$$y(t)=\int_{0}^Tx_1(\tau)x_2(t-\tau)d\tau\tag{1}$$

where $$x_1(t)$$ and $$x_2(t)$$ are $$T$$-periodic signals. If $$a_k$$ and $$b_k$$ are the Fourier coefficients of $$x_1(t)$$ and $$x_2(t)$$, respectively, then the Fourier coefficients $$c_k$$ of $$y(t)$$ are given by

$$c_k=Ta_kb_k\tag{2}$$

Your approach to the problem is valid for $$k\neq 0$$, and your result is actually pretty close because

$$\frac{1-\cos\big(\frac{\pi k}{2}\big)}{4\pi^2 k^2}=\frac{2\sin^2\big(\frac{\pi k}{4}\big)}{4\pi^2 k^2}=\frac{\sin^2\big(\frac{\pi k}{4}\big)}{2\pi^2 k^2}\tag{3}$$

so there's just some scaling problem that I'm sure you'll be able to fix (check your definition of $$f(x)$$!). Note that you need to compute the coefficient for $$k=0$$ separately.

• Thank you so much, this is extremely helpful. We have learned convolution but it was the first time they used that notation. I'm very happy that my solution was close, was starting to go a little crazy. – Abundance Jan 19 at 18:06