Finding a periodic signal knowing its period, mean value and power

Question

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.

Answer 1

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 $\phi=-\pi/2$.