Question: Expand the following periodic signal in Fourier series:

$$s(t)=2\sin (1000\pi t)+0.5\sin (500\pi t)+\cos (250\pi t), -\infty < t <+\infty.$$

Determine the basic period of that signal, mean power value by using Parseval's theorem and draw amplitude and phase spectra (draw a discrete signal, not continual).

Periodic signal can be expanded in trigonometric series:
$$s(t)=a\_0+\sum\_{n=1}^{\infty}a\_n\cos(n\omega\_0t)+\sum\_{n=1}^{\infty}b\_n\sin(n\omega\_0t),$$

where $a\_0,a\_n,b\_n\in\mathbb R$ are Fourier series coefficients. They are evaluated as follows:

$$a\_0=\frac{1}{T}\int\_{-T/2}^{T/2}s(t)dt, a\_n=\frac{2}{T}\int\_{-T/2}^{T/2}s(t)\cos(n\omega\_0t)dt, b\_n=\frac{2}{T}\int\_{-T/2}^{T/2}s(t)\sin(n\omega\_0t)dt.$$

1. How can we determine how many $a\_n$ and $b\_n$ terms are there (When they are becoming zero terms)?
2. How do we determine basic period of $s(t)$?
3. How do we determine mean power value by using Parseval's theorem?
4. How do we determine and draw an amplitude and a phase spectra (discrete signals)?

Answer: 
The signal is in the fourier series form by itself and there is no need to do any thing
1. Because your signal is already a sum of cosines and sines you can quickly determine how many $a\_n$ and $b\_n$ there are by counting directly. Here we have 3 terms, so we will have 3 nonzero $a\_n$ and $b\_n$ (combined, not individually). In general it is not so easy to determine how many you will have, some signals will have infinitely many nonzero and some will have all of them zero (eg no signal).
2. In this instance, to determine the smallest period of $s(t)$, we will look for the term which is oscillating the 'slowest'. This will be $\cos(250\pi t)$ because $250\pi < 500\pi < 1000\pi$. Since cosine ordinarily has a period of $2\pi$ we will divide this by the coefficient of $t$ in the argument, this gives a period of $\frac{2\pi}{250\pi}=\frac{1}{125}$. Imagine we didn't have the $\cos(250\pi t)$ term, then the slowest oscillating term would be $0.5\sin(500\pi t)$. Since the period of sine is also $2\pi$ we would do the same thing to get a period of $\frac{2\pi}{500\pi}= \frac{1}{250}$.
3. I am not sure what it meant here, parseval's theorem relates the coefficients $a\_n$ and $b\_n$ to the norm or energy or the signal $\|f\|$. Maybe if you are more specific about mean value I can help you.
4. This is also not so clear to me. The phases are the coefficients of the argument $t$ and the amplitudes are the coefficients like $a\_n$ and $b\_n$.