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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...

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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: $$ y(t) = 5 + \frac{20}{\pi} \sin(\pi t) + \frac{20}{3\pi} \sin(3\pi t) $$

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  • $\begingroup$ Hi: is someone in the group, besides the person who asked, able to give a check to an answer ? It seems to me like, if the person who asked the question hasn't checked it by now, they're not going to !!!! and I bet the answer is correct :). $\endgroup$
    – mark leeds
    Aug 30, 2018 at 3:14
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    $\begingroup$ @markleeds Hi Mark! Unfortunately that's an unsolved problem as all stackexhange sites are full of questions being asked, then answered but never ever rated back. The habit is related with software which requires the very guy who asked the question to verify that the solution indeed worked. No one else could do it. But when it's a theoretical subject such as math.se or dsp.se, then someone else can also judge that an answer is correct, at least to indicate this to the community benefit. Yet it's not practiced. As a reaction, I delete my answers which get no votes or checks. $\endgroup$
    – Fat32
    Aug 30, 2018 at 22:01
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    $\begingroup$ thanks for info. I find it baffling that someone could ask a question, get an answer where time was put in to make it clear and then not show minimal thanks by just checking it. ??? $\endgroup$
    – mark leeds
    Aug 31, 2018 at 20:23
  • $\begingroup$ 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. $\endgroup$
    – odea
    Oct 29, 2018 at 8:47

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