# Consider an ideal low pass filter $H(\omega)$, and the input to this filter is the periodic square wave $x(t)$. Find the output $y(t)$ The solution to the problem is $$y(t) = 5 + \frac{20}{\pi} \sin(\pi t) + \frac{20}{3\pi} \sin(3 \pi t)$$ and to get that the solution says to find the Fourier series expansion of $x(t)$ and I am having trouble doing that. I find that $C_0 = 5$ and that $$C_k = \frac{5}{j k \pi} \left( 1 - e^{-j k \pi} \right)$$. I don't know how to compute $$C_0 + \sum_{k=1}^{\infty} C_k e^{j k w t}$$. Also since the cut off is $4 \pi$ so when the Fourier series expansion greater $4 \pi$ those values will not be included as seen in the solution...

So you actually solve(d) this problem by computing the CTFS of the input $x(t)$: $$C_k = 5 \frac{ 1 - e^{ - j k \pi } }{ j k \pi } = \begin{cases} 0 &, \text{ for k even , } k \neq 0 \\ 10/jk\pi &, \text{ for k odd} \end{cases}$$ which gives the coefficient associated with the complex exponential of frequency $\omega_k = k \omega_0$ for the fundamental frequency $\omega_0 = \frac{ 2 \pi }{T_0} = \frac{ 2 \pi }{2} = \pi$ , as the period of the input $x(t)$ is $2$ seconds.
Since the ideal lowpass filter passes those frequencies below its cutoff frequency $\omega_c = 4 \pi$, then we would only take those coefficiencs whose frequencies fall below $\omega_c$ yielding those values of $k = 0,\pm 1, \pm 3$ . Those indices higher than $\pm 5$ are filtered out by the ideal lowpass filter.
Therefore the output $y(t)$ is: $$y(t) = 5 + \frac{10}{j\pi} \left( e^{j\pi t} - e^{-j\pi t} \right) + \frac{10}{j3\pi} \left( e^{j3\pi t} - e^{-j3\pi t} \right)$$ which simplifies to: $$y(t) = 5 + \frac{20}{\pi} \sin(\pi t) + \frac{20}{3\pi} \sin(3\pi t)$$
• One can alternatively write down the FT of $x(t)$, i.e. for a periodic square wave, $$X(j\omega) = \sum_{k=-\infty}^\infty \frac{2 sin k \omega_0 T_1}{k}\delta(\omega - k\omega_0).$$ Similarly, when one computes $Y(j\omega) = X(j\omega) H(j\omega)$, only a couple of terms will be left. Inverting them would give the result. – odea Oct 29 '18 at 8:47