Fourier series analysis

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 1

The signal is in the fourier series form by itself and there is no need to do any thing

Answer 2

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 $b_n$.

Answer 3

Note that the function $s(t)$ is a sum of periodic functions.

If $\phi(t+A) = \phi(t)$ and $\psi(t+B) = \psi(t)$, then when $\sigma = \phi + \psi$ and $C =\text{lcm}(A,B)$, $\sigma(t+C) = \sigma(t)$. Further, if $A$ and $B$ are the base periods of $\phi$ and $\psi$ respectively, then $C$ is the base period of $\sigma$. If this isn't a known result, you should show it.

Using this and what we know about the base period of $\sin$ and $\cos$, we can get that the base period of $s$ is $\frac{1}{125}$, and so the angular frequency is $2\pi$ times the reciprocal of the base period, i.e. $\omega = 250\pi$.

We now have that the Fourier series is $s(t) = A_0 + \sum A_n \cos(250\pi nt) + \sum B_n \sin(250\pi nt)$, and by inspection we can see that $A_1 = 1$, $B_2 = 0.5$, and $B_4 = 2$, while the rest are $0$. To show this with the integral formulae, simply use orthogonality of $\{1,\cos(nx),\sin(mx)\}$.

Parseval gives you that the average power of the signal is equal to the sum of the squared absolute value Fourier coefficients $-$ from the exponential series. To get the same for the trigonometric series, take half of the square of the $\sin$ and $\cos$ coefficients (but not $A_0$ when it's nonzero), so $P_{avg} = [(1)^2+(\frac{1}{2})^2+(2)^2]/2 = \frac{21}{8}$.

For the last bit, $M_n = \sqrt{A_n^2+B_n^2}$ is the magnitude, and $\theta_n = \tan^{-1}(B_n/A_n)$ is the phase. Simply calculate these and plot them over the discrete $n-$axis.

Answer 4

Suppose that $t$ is the time in the seconds (s). Units of the amplitude can be different (meters, amperes, volts etc.) and presented in the linear scale (not watts, not decibels etc.), if we determine the energy parameters.

1. Real Fourier analysis have to determine three terms (harmonics),

cosine harmonic $\cos(250\pi\ t)$ with
the amplitude $b = 1,$ the period $T = \dfrac 1 {125}$s, and the frequency $f = \dfrac {1}{T} = 125$ Hz;

sine harmonic $2\sin(1000\pi t)$ with
the amplitude $a = 2,$ the period $T = \dfrac{1}{500}$ s, and the frequency $f = \dfrac{1}{T} = 500$ Hz;

sine harmonic $0.5\sin(500\pi t)$ with
the amplitude $a = 0.5,$ the period $T = \dfrac{1}{250}$ s, and the frequency $f = \dfrac{1}{T} = 250$ Hz.

2. The basic period of $s(t)$ is the period for the signal in general, and can be determined as the least common multiple (lcm) of the all harmonics, if it exists:

$T_B = \mathrm {lcm}(1/125, 1/500, 1/250) = 1/500\,\mathrm{lcm}(4, 1, 2) = 1/125.$

3. Mean power value given as the integral of the energy on the basic period:

$$P = \frac{2}{T_B}\int\limits_{-\frac{T_B}{2}}^{\frac{T_B}{2}} s^2(t) dt,$$ The square can be presented as the sum of the products of the orthogonal functions, and the integrals of the different functions products equals to zero. So we will have the sum of the harmonics squares: $$P = \frac2{T_B} \sum_i (a_i^2\dfrac12(1 + \cos\dots) + b_i^2(1 - \cos\dots)).$$ Integrals of the cosines on the period are zeros, so

$$P = \dfrac12\sum_i (a_i^2 + b_i^2) = \dfrac12(1^2 + 2^2 + 0.5^2) = 2.625.$$

Parsevals theorem proves this fact in the general case.

4. The terms of the signal amplitude and phase are based on the presentation of harmonics in the form of $$h(t) = A\sin(\omega t + \varphi) = A\sin\omega \cos\varphi + A\cos\omega \sin\varphi = a_n\sin\omega + b_n\cos\omega,$$ where $A$ is the signal amplitude and $\varphi$ is the signal phase.

Amplitude $$A = \sqrt{a_n^2 + b_n^2},$$ because $a_n^2 + b_n^2 = A^2.$

Easy to see that for sine harmonic $\varphi = 0$, and for cosine one $varphi = \frac\pi2.$

In the common case the function $\arctan\dfrac{b_n}{a_n}$ isn't correct when $a_n < 0$ and $b_n <0,$ so can not be recommended.

My recomendation is $$\varphi = 2 \arctan\dfrac{b_n}{1+a_n},$$ which is based on the identity $$\tan 2\varphi = \dfrac{\sin\varphi}{1 + \cos\varphi},$$ because it doesn't reqire any additional logics.

Amplitude and phase of $s(t)$ are discrete functions and looks up as bars.