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I've found an interesting exercise which I have been trying to solve for a couple days, without success.

Let $x(t) \in \mathbb{R}$ be a periodic signal with fundamental period $T_0 = \tfrac{1}{4}$, mean value $\overline{x(t)} = 1$ and power $P_x = 10$. Knowing that $x(t)$ isn't affected by an ideal lowpass filter with band $B = 7 \ \mathrm{Hz}$, find the possible expressions of $x(t)$.

Since the signal is periodic and real, its mean value and power can be evaluated as: \begin{align*} \overline{x(t)} &= \frac{1}{T_0}\int_{-T_0/2}^{T_0/2}x(t)\,\mathrm{d}t = 4\int_{-1/8}^{1/8}x(t)\, \mathrm{d}t = 1 \\ P_x &= \frac{1}{T_0}\int_{-T_0/2}^{T_0/2}\left|x(t)\right|^2\,\mathrm{d}t = 4\int_{-1/8}^{1/8}x^2(t)\, \mathrm{d}t = 10 \end{align*} From the preceding equations, it is: \begin{align*} \int_{-1/8}^{1/8}x(t)\, \mathrm{d}t &= \frac{1}{4} \\ \int_{-1/8}^{1/8}x^2(t)\, \mathrm{d}t &= \frac{5}{2} \end{align*} In order to fullfill the conditions found, the simplest signal which comes to my mind is the rectangular pulse $\hat{x}(t) = A\mathrm{rect}_B(t)$, properly repeated: \begin{align*} \int_{-1/8}^{1/8}A\mathrm{rect}_B(t)\, \mathrm{d}t &= \frac{1}{4} \\ \int_{-1/8}^{1/8}A^2\mathrm{rect}_B(t)\, \mathrm{d}t &= \frac{5}{2} \end{align*} The coefficients $A$ and $B$ must be such as: \begin{align*} A B &= \frac{1}{4} \\ A^2 B &= \frac{5}{2} \end{align*} The coefficients are $A = 10$ and $B = \tfrac{1}{40}$, so the aperiodic signal is: $$ \hat{x}(t) = 10\mathrm{rect}_\tfrac{1}{40}(t) $$ and the periodic signal is: $$ x(t) = \sum_{k \in \mathbb{Z}}10\mathrm{rect}_\tfrac{1}{40}\left(t-\frac{k}{4}\right) $$ This signal, however, is affected by the LPF. Considering the Fourier's transform of the aperiodic signal, it is: $$ \hat{X}(f) = \mathscr{F}\{\hat{x}(t)\}(f) = \frac{1}{4}\mathrm{sinc}\left( \frac{f}{40} \right) $$ The Fourier's transform of the filter is: $$ H(f) = \mathrm{rect}_{14}(f) $$ The filtered signal is then given by: $$ \hat{X}(f) H(f) \neq \hat{X}(f) $$

I would like to understand how to tackle this kind of problems and how to find the correct signal, which could be a rect, a tri, a sinc or a composition of them, without "guessing", if possible.

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    $\begingroup$ A simple sine signal (plus DC offset 1) is out of question? If so, why? It would be an easy solution. rect and tri are impossible, because there spectrum is infinite, which violates the LP condition. A repetition of sincs could also be possible. $\endgroup$
    – Max
    Jul 7, 2022 at 12:38
  • $\begingroup$ @Max Thank you, using your hint I've managed to find the signal $x(t) = 1+3\sqrt{2}\cos(8\pi t)$, which should fullfill the constraints (the same goes for $x(t) = 1+3\sqrt{2}\sin(8\pi t)$, if I am not mistaken). I'd like to see how to get a suitable signal with a repetition of sincs. $\endgroup$
    – gwn
    Jul 7, 2022 at 19:23
  • $\begingroup$ @gwn: There's a simple way to parameterize all signals satisfying the given conditions; right now you just have two. $\endgroup$
    – Matt L.
    Jul 8, 2022 at 6:57
  • $\begingroup$ @MattL. Thank you, unfortunately I'm not being able to find other signals than those I've already found, I believe I'm following the wrong path if I want to generalize the result. Would you mind posting an answer explaining how to do that? $\endgroup$
    – gwn
    Jul 8, 2022 at 15:13
  • $\begingroup$ @gwn: I've added more information to my answer. $\endgroup$
    – Matt L.
    Jul 9, 2022 at 10:02

1 Answer 1

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HINT:

Think in terms of Fourier series. A periodic signal can be represented by its Fourier series. If it is not affected by a lowpass filter, then there are only finitely many non-zero Fourier coefficients. Given the period and the cut-off frequency, you can figure out how many harmonics there are. In your specific case the answer is pretty simple.

EDIT:

From the comments it appears that you've figured out that the solution must be a sinusoidal signal with a DC component, and you've also figured out the amplitude of the sinusoid to satisfy the power requirement. Consequently, the most general result is

$$x(t)=1+3\sqrt{2}\cos(8\pi t+\phi)\tag{1}$$

with an arbitrary phase value $\phi$. The two solutions you've already come up with are the ones with $\phi=0$ and $\phi=-\pi/2$.

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