GREPhysics.NET: GR9277 #39

GR9277 #39

Problem

GREPhysics.NET Official Solution

\prob{39}

If n is an integer ranging from 1 to infinity, $\omega$ is an angular frequency, and t is time, then the Fourier series for a square wave, as shown above, is given by which of the following?

$V(t)=\frac{4}{\pi}\sum_1^\infty \frac{1}{n}sin(n\omega t)$
$V(t)=\frac{4}{\pi}\sum_0^\infty \frac{1}{2n+1}sin((2n+1)\omega t)$
$V(t)=\frac{4}{\pi}\sum_1^\infty \frac{1}{n}cos(n\omega t)$
$V(t)=\frac{4}{\pi}\sum_0^\infty \frac{1}{2n+1}cos((2n+1)\omega t)$
$V(t)=-\frac{4}{\pi}+\frac{4}{\pi}\sum_1^\infty \frac{1}{n^2}cos(n\omega t)$

Advanced Topics $\Rightarrow$ }Fourier Series

There's no need to go through the formalism of integrating out the coefficients.

One can tell by inspection that the function is odd. Thus, one would use the Fourier sine series. This leaves choices (B) and (A).

Choice (A) is trivially zero since for all integer n, $\sin(n\omega t)=0$ . Choice (B) remains.

Alternate Solutions

h.fei10 2012-11-04 04:20:23	The problem is how to find a correct one from A and B. Try this, move the origin to the right or left a little bit to make it symmetric about the V(t) axis, for example, let the new origin be $t=\pi/2\omega$ . The potenial then should be an even function. Replace $t$ with $t-\pi/2\omega$ , you can turn all the sine functions into cosine functions in B, but you can't do this in A. So B is the only correct answer. Reply to this comment
rbot 2012-07-30 20:29:17	My solution is simple, but lengthy when put into words. I hope it helps! We know that both A and B satisfy the condition that $V(\pi/\omega)$ and $V(2 \pi/\omega)$ are zero. Another condition is that for all the points between $t=\pi/\omega$ and $t=2\pi/\omega$ , V(t) must be -1. For $t=3\pi/2\omega$ (the middle point between $\pi/\omega$ and $2\pi/\omega$ ), $sin( n \omega 3\pi/2\omega ) = -1$ and $sin( (2n+1) \omega 3\pi/2\omega ) = -1$ . This would make it seem like both (A) and (B) are still in BUT also know that at points between $\pi/\omega$ and $3\pi/2\omega$ the signal is still -1. If we try $5\pi/2\omega$ for all possible n, $sin( n \omega 5\pi/2\omega ) = +1$ . Clearly that is not -1. However $sin( (2(1)+1) * \omega 5\pi/2\omega ) = -1$ , which satisfies the signal depicted, and so on... Reply to this comment
packinski 2011-08-01 16:29:30	Here is my solution: For $t=\pi/\omega$ we have that $V(t)=0$ . This, necessarily, has to be satisfied by the Fourier coefficients of as well. Answer (B) is the only one that satisfies this. Reply to this comment
diane 2010-01-05 21:39:32	First change the variable from x to $\omega t$ with dx = $\omega \;$ dt. Then the $\omega$ 's cancel when you do the integral. $f(\omega t)={\Sigma}_1^{\infty} A_n sin \; n \omega t$ Use orthonormality and multiply both sides by sine and integrate: $A_n {\int}_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}}{sin}^2 n \omega t \;dt = {\int}_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}}f(x)sin \; n \omega t dt$ $A_n (\pi) = \frac{1}{n } [(cos n \omega t)_{-\frac{\pi}{\omega}}^0 + (-)(cos n \omega t)_0^{-\frac{\pi}{\omega}}$ $A_n = \frac{1}{\pi} [(cos n \omega t)_{-\frac{\pi}{\omega}}^0 + (-)(cos n \omega t)_0^{-\frac{\pi}{\omega}}$ For even n, 1-1 =0. For odd n, replace n with 2n-1 or change the sum to start at zero instead of one, and replace n with 2n+1. $A_n= \frac{4}{\pi(2n+1)}$ Reply to this comment
Simplicio 2009-03-27 09:46:19	After establishing that the function is odd and left with (A) and (B), the best way to proceed is to ask: "How do I build a square wave using sine waves?" Consider sin(wt), it should be obvious that the period is t = $\frac{2\pi}{w}$ which is what we want. Next consider sin(2wt), which has a period of t= $\frac{\pi}{w}$ . If you add this to the first sin(wt), it adds in phase for the first half of the square (0 < t < $\frac{\pi}{2w}$ ) and anti-phase for the second half ( $\frac{\pi}{2w}$ ) < t < $\frac{\pi}{w}$ ) ), so it cannot produce a square-shape. If you repeat this, you find that only the odd sin(wt), sin(3wt), sin(5wt) etc ... would work hence (B) is the answer! Reply to this comment
Maxwells_Demon 2008-09-21 05:29:19	The expressions in (A) and (B) are the same (confirmed by my math, my textbooks and online) http://mathworld.wolfram.com/FourierSeriesSquareWave.html" target="_blank">Link #1 http://www.sfu.ca/sonic-studio/handbook/Fourier_Theorem.html" target="_blank">Link #2 (A) and (B) are ultimately the same. The real answer lies in the region you are integrating. If you are integrating from -L to L, you will get the expression in (B) and if you integrate from 0 to 2L, you will get the expression in (A) (where L = pi/w in this case). Do the integration for each to see :). Since the graph crosses the V(t)=0 axis you should integrate from -L to L and arrive at the answer (B) because that is what the graph tells you. If there was no square wave on the negative t-axis then the answer would be (A). After exhausting this problem I'm not sure how else to solve it to make it easier in a test-setting situation - but the integration does give you the correct solution integrating it from -L to L as the V(t) graph implies. Reply to this comment

Comments

ahmedali
2017-08-30 13:35:15

here is a valid solution but kinda lengthy. \r\nas all us thought the solution should be A, or B because they bare both odd functions.\r\nnow to choose between them, both A,and b give zero if we put t=(Pi/Omega), but here is the thing, we can take the limit as n goes to zero for Sin(2n+1)Pi/(2n+1)pi, that would give non zero, and that is the only term can satisfy this relation n=0, however we cannot do that in A because the series starts from n=1 and we do not have the n=0 term. so the answer should be B

enterprise
2018-03-31 20:18:51

I can\\\\\\\'t understand what you are trying to say here. My impression is that this is complete nonsense.

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