Low pass filter

Question

A wave defined by $f(t)=a$ for $t\in (0,T)$ and $f(t)=-a$ for $t\in (-T,0)$ (the wave is $2T$ periodic) is input into a system that transmits angular frequencies $<\omega$ and absorbs those $>\omega$. How might I find the form of the output? Firstly, I am not quite sure what is going on here. Am I supposed to find the Fourier series for $f(t)$ then eliminate the terms whose arguments of the cosines or sines are $>\omega$? But how do I find/express the "form" of the output? Thanks in advance!

Answer 1

In order to get the Fourier series you'll need to impose the condition that the waveform is periodic, which means it's a square wave with period $2T$. The Fourier series of a square wave is well known:

http://en.wikipedia.org/wiki/Square_wave
http://mathworld.wolfram.com/FourierSeriesSquareWave.html

From that you just remove the terms of the series which have frequency components greater than the cutoff frequency $w$, which results in a truncated series. The resulting filtered waveform will not have the sharp transition at $t = 0$ that the original waveform had, but will instead have rounded edges and a slower transition from $-a$ to $a$.

Answer 2

Assuming the wave is $2T$-periodic, it is an odd function, so all of the cosine terms in the Fourier series are $0$. Thus $$ f(t)=\sum_{n=1}^\infty\;b_n \sin\left(\frac{n\pi t}{T}\right) $$ where $$ b_n=\frac{1}{T}\int_{-T}^T f(t)\sin\left(\frac{n\pi t}{T}\right)\;\mathrm{d}t $$ The frequency of each term is $\frac{n}{2T}$, so remove all terms of the sum for which $\frac{n}{2T}>\omega$. Your answer should be a finite sum of sine waves.

Spoiler: If $n$ is even, $b_n=0$. If $n$ is odd, then $b_n=\frac{4a}{n\pi}$. Therefore, $$ \tilde{f}(t)=\sum_{k=0}^{\lfloor T\omega-1/2\rfloor}\frac{4a}{(2k+1)\pi}\sin\left(\frac{(2k+1)\pi t}{T}\right) $$