# Finding original signal $x(t)$

For given 4 conditions, I have to find out what is $$x(t)$$ with period of 3, and I don't know if $$x(t)$$ is real or not.
For fourier coefficients $$x_k$$, $$1.\ x_k=x_{k+2}$$ $$2.\ x_k=x_{-k}$$ $$3.\ \int_{-0.5}^{0.5}x(t)\, \mathrm dx\ = 1$$ $$4.\ \int_{1}^{2}x(t)\, \mathrm dx\ = 2$$

from condition 1 and 2, I've got this equations
fourier series expansion of $$x(t)$$: $$x(t)=\sum_{k=-\infty}^{\infty}\ x_k \exp(j{2\pi \over 3}kt)$$ by condition 2, $$x(t)=x_0+\sum_{k=1}^{\infty}x_k(\exp(-j{2\pi \over 3}kt)+\exp(j{2\pi \over 3}kt))$$ and as $$\exp(-j2\pi ft)+\exp(j2\pi ft)=2\cos(2\pi ft)$$, $$x(t)=x_0+2\sum_{k=1}^{\infty}x_k\cos({2\pi \over 3}kt)$$ by condition 1, $$x(t)=x_0+2\sum_{m=1}^{\infty}\ (x_1\cos(2\pi{(2m-1) \over 3}t)+x_0\cos(2\pi{2m \over 3}t))$$

For condition 3,4 when I integrate $$x(t)$$, it gives me too complex sums.

$$\int_{-1/2}^{1/2}\cos(2\pi{2m-1 \over 3}t)\, \mathrm dt\ = {3 \over 2\pi (2m-1)}2\sin(\pi{2m-1 \over 3})$$ $$2\sin(\pi{2m-1 \over 3})={\sqrt3, 0, -\sqrt3, \sqrt3, 0, -\sqrt3, ...}$$ so, $$\int_{-1/2}^{1/2}\sum_{m=1}^{\infty}\cos(2\pi{2m-1 \over 3}t) \mathrm dt\ ={3 \over 2\pi}\sum_{n=1}^{\infty}({\sqrt3 \over6n-5}+{0 \over6n-3}+{-\sqrt3 \over6n-1})=0.75$$

This kind of summation comes out over and over. Of course, calculator can solve it because it converges. But this was problem from my mid-term exam and calculator was not allowed. This thought process(?) complicates the problem too much, so I wondered and asked here if there could be other way to solve this problem much simpler than this.

• (1) How is $a_k$ related to $x(t)$?, and (2) what is the signal processing context? This sounds like a math problem, which should be asked at one of the math stackexchange sites.
– MBaz
Commented Apr 13, 2021 at 13:10
• @MBaz $a_k$ is fourier coefficient of $x(t)$, so I asked it here because it is fourier series problem. Commented Apr 13, 2021 at 13:52
• OK. What about $b_k$? Are they zero or are you free to define them? In any case: the sum of an infinite number of cosines is an impulse. I fail to see how to arrange things to get the two integrals, especially since the integral away from the origin is larger.
– MBaz
Commented Apr 13, 2021 at 14:31
• The OP has chosen to withhold far too much information about the problem asked, not just the values of the $b_k$'s (coefficients of the sine terms of the Fourier series) as complained about by @MBaz but also the source of the $\frac 13$ terms in the arguments of the cosines. Is there a hidden meaning to that $\frac 13$ or are we supposed to deduce that it was given in the problem statement that the fundamental frequency of $x(t)$ is $f_0 = \frac 13$ Hz but the OP felt that was a trivial detail not worth mentioning? -1 pending emendations to the question asked. Commented Apr 13, 2021 at 15:35
• @DilipSarwate, Sorry for confusion, I omitted too much information, I changed $a_k$ to $x_k$ to show that I didn't mean that $a_k$ is coefficient of cosine terms. I also added period of $x(t)$ and more explanations. I'm so sorry for confusion Commented Apr 14, 2021 at 0:19