Finding the Fourier Series of a periodic rectangular wave

[Adir_Sh]

Homework Statement

The problem/question is attached in the file called "homework". In the third signal (the peridic rectangular wave), I am requested (sub-question b) to find the Fourier series of the wave.

Homework Equations

The file called "solution" presents a detailed solution to the problem. What I don't understand is why the phase (page 2 in that "solution" document) $\phi_{n}$ equals $-\pi$ for even values of $n$. What I know is that when the argument of the arctan function is undefined (since it's divided by 0), it means that the angle is $\pi/2$, because only then the tan would be undefind! So why $\phi_{n} = -\pi$ then? does someone have an idea?
I do understand the $\pm\pi$ addition for the odd values of $n$ (i.e $n=3,7,11,...$), but that $\phi_{n} = -\pi$ is not completely clear to me (and quite annoying to be honest...).

The Attempt at a Solution

Couple of hours of thinking, going back to the basics of polar representation of numbers in the x-y plane (in fact, it's the $a_{n}-b_{n}$ plane here) but nothing practical really... any suggestions?
Thanks!

 

[rude man]

Something wrong here somewhere.

If tan^-1(b_n/a_n) = -π for all even n then b_n = 0 for all even n. But b₂ = 2V_m/πn sin^2(nπ/4) = V_m/π ≠ 0.

So there is something wrong with the submitted solution.

 

[Adir_Sh]

Something wrong here somewhere.

If tan^-1(b_n/a_n) = -π for all even n then b_n = 0 for all even n. But b₂ = 2V_m/πn sin^2(nπ/4) = V_m/π ≠ 0.

So there is something wrong with the submitted solution.

Thanks for responding rude man.
So would you then consider $n=\pi/2$ as a solution for $\phi_{n}$ for all even n? or maybe you suggest a different solution?
The way I see it, for all even n you have $a_{n}=0$ and $b_{n}$ has a value which gets lower as n grows, but also $b_{n}=0$ for $n=0,4,8,...$.

How would you fix the submitted solution?

 

[rude man]

Thanks for responding rude man.
So would you then consider $n=\pi/2$ as a solution for $\phi_{n}$ for all even n? or maybe you suggest a different solution?
The way I see it, for all even n you have $a_{n}=0$ and $b_{n}$ has a value which gets lower as n grows, but also $b_{n}=0$ for $n=0,4,8,...$.

How would you fix the submitted solution?

I guess I would just start over:

V(t) = a₀/2 + Ʃ{a_ncos(nωt) + b_nsin(nωt)} from n = 1 to ∞.

Your a_n coefficients are

a_n = (V_m/π)∫cos(ωt)d(ωt) from 0 to π/2
b_n = (V_m /π)∫sin(ωt)d(ωt) from 0 to π/2

Notice that d(ωt) = ωdt. It's easier to integrate w/r/t ωt rather than t. So the limits of integration change from 0 → T/4 to 0 → π/2.

To express this series in the form V(t) = a₀/2 + Ʃc_ncos(nωt - ψ) from n = 1 to ∞,

c_n = √(a_n² + b_n²)
and ψ_n = tan^-1(b_n/a_n).

This is all standard textbook stuff.

In computing ψ be sure to retain the signs of a and b separately. For example, if a = 1 and b = -2, tan^-1(1/-2) ~ π - 0.46 whereas if a = -1 and b = 2, tan^-1(-1/2) ~ -0.46. These angles are different and are separated by π.

Happy computing!

 

[rezeew]

The phase in a Fourier series represents the shift or delay of the sinusoidal component in the signal. In a periodic rectangular wave, the signal is symmetric about the y-axis, meaning that it has no shift or delay. This results in a phase of 0 for all even values of n.

When solving for the Fourier coefficients, the arctan function is used to determine the phase. For even values of n, the argument of the arctan function is undefined, as you mentioned. This means that the phase could technically be any value, including -π. However, since the signal is symmetric, it makes sense to choose a phase of 0 (or -π, which is equivalent in this case) for simplicity.

In summary, the phase being equal to -π for even values of n in the Fourier series of a periodic rectangular wave is a result of the signal's symmetry and the choice of a reference phase. It is not a fundamental property of the signal itself.