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Finding the Fourier Series of a periodic rectangular wave

  • Thread starter Adir_Sh
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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) [itex]\phi_{n}[/itex] equals [itex]-\pi[/itex] for even values of [itex]n[/itex]. 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 [itex]\pi/2[/itex], because only then the tan would be undefind! So why [itex]\phi_{n} = -\pi[/itex] then? does someone have an idea?
I do understand the [itex]\pm\pi[/itex] addition for the odd values of [itex]n[/itex] (i.e [itex]n=3,7,11,...[/itex]), but that [itex]\phi_{n} = -\pi[/itex] is not completely clear to me (and quite annoying to be honest...:uhh:).


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 [itex] a_{n}-b_{n}[/itex] plane here) but nothing practical really... any suggestions?
Thanks! :tongue:
 

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Answers and Replies

  • #2
rude man
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Something wrong here somewhere.

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

So there is something wrong with the submitted solution.
 
  • #3
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Something wrong here somewhere.

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

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

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

How would you fix the submitted solution?
I guess I would just start over:

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

Your an coefficients are

an = (Vm/π)∫cos(ωt)d(ωt) from 0 to π/2
bn = (Vm /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 + Ʃcncos(nωt - ψ) from n = 1 to ∞,

cn = √(an2 + bn2)
and ψn = tan-1(bn/an).

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!
 

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