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Fourier coefficients

  • Thread starter magnifik
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  • #1
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I am trying to find the Fourier coefficients for the following signal:
wimwza.png


For some reason, I keep getting 0, which doesn't make sense to me.

I am even getting 0 for F0 even though there is clearly area under the curve. Here's my work for this part:
Period = T/2
Natural freq = 4pi/T
F0 = (2/T) int[sin(4pi/T)t dt] from 0 to T/2
= (1/2pi) [cos(4pi/T)t] from 0 to T/2
= 0

I am getting 0 for Fn as well. Any help would be appreciated!
 

Answers and Replies

  • #2
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The sinusoid finishes a half of a cycle every T/2, and a complete cycle is 2pi. Thus,

[tex]\frac{2 \pi}{2} = \frac{\omega T}{2}[/tex]
[tex] \omega =\frac{2 \pi}{T}\neq \frac{4 \pi}{T}[/tex]
 
  • #3
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The sinusoid finishes a half of a cycle every T/2, and a complete cycle is 2pi. Thus,

[tex]\frac{2 \pi}{2} = \frac{\omega T}{2}[/tex]
[tex] \omega =\frac{2 \pi}{T}\neq \frac{4 \pi}{T}[/tex]
I see. What I thought was that the natural frequency is equal to 2pi/T (general equation), but since the period is from 0 to T/2 then the period is T/2. So i did 2pi/(T/2) = 4pi/T
 
  • #4
vela
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Your function should be f(t)=sin 2πt/T. When t=T/2, the argument of the sine will be π, so it will have gone through a half cycle. Your function goes through a complete oscillation from 0 to T/2.
 
  • #5
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using f(t) = sin 2πt/T, i am getting infinity for F1

i got Fn = (-1/2pi(1-n))((-1)n - 1) - (1/pi(1+n))((-(-1)n - 1)

and plugging in 1, for the first part of the equation i am getting 0 in the denominator. i am also finding that for every odd n i am getting 0. this can't be correct, i assume?
 
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  • #6
vela
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Nope. You should find all the sine terms vanish since f(t) is an even function. If you can't figure it out, post more details of your calculation so we can see where the mistake is.
 
  • #7
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My calculations

27zgjsx.jpg
 
  • #8
vela
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You have to consider the cases for n=±1 separately. For example, when n=1, the integrand of the first integral is just 1, so the integral is equal to T/2.
 
  • #9
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You have to consider the cases for n=±1 separately. For example, when n=1, the integrand of the first integral is just 1, so the integral is equal to T/2.
is it possible to use the fact that Fn = 0 for odd n when f(t) = f(t+T/2)?? i'm not sure what you mean when you say you have to consider the cases for +1 separately since i thought the amplitude spectra is symmetric

when i do the coefficients for n = 3, 5, etc...i am getting 0
 
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  • #10
vela
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is it possible to use the fact that Fn = 0 for odd n when f(t) = f(t+T/2)??
Sure. Why not?
i'm not sure what you mean when you say you have to consider the cases for +1 separately since i thought the amplitude spectra is symmetric
Simply put, you can't divide by 0, so the way you integrated was incorrect. You have to evaluate the integral differently in those cases. It's just like you can't say

[tex]\int x^n\,dx = \frac{x^{n+1}}{n+1}+c[/tex]

when n=-1. You have to consider the n=-1 case separately. The way you integrated works when n doesn't equal 1 or -1. For those particular cases, you have to evaluate the integrals differently.
 

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