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Semi Fourier series question (small)by toneboy1
Tags: fourier series 
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#1
Nov2512, 07:26 AM

P: 174

1. The problem statement, all variables and given/known data
Hi, I have x(t) = 1/2 + cos(t) + cos(2t) so I can see that a0 = 1/2 and that it is an even function so there is no bn Also that T = 2pi so an = 2/2pi ∫02pi x(t).cos(nω0t) dt but when I integrate this I get an = 0 yet I've been told that the answer is x(t) = 1/2 + Ʃn = 12 cos(nω0t) which would mean that an = 1, but I'm not sure how. 


#2
Nov2512, 08:33 AM

Mentor
P: 11,928

Your notation looks strange.
You should not get 0 in the integral for n=1, n=2. x(t) = 1/2 + 1*cos(t) + 1*cos(2t) + 0*cos(3t) +... is expressed as fourier series already, you can simply read off the parameters. 


#3
Nov2512, 12:06 PM

P: 296

But as mentioned, x(t) is already expressed as a fourier series  a sum of harmonically related sinusoids. If you proceeded with your integrals anyway, you would find that the integral was zero for all terms except the one cosine term that matched the frequency of cos(nw0t) and therefore only those an would be nonzero (and would be equal to the coefficients you see in x(t) already). This property is what the fourier series depends on: the area under the product of two sinusoids of different frequency over one period is zero. 


#4
Nov2512, 08:30 PM

P: 174

Semi Fourier series question (small)
Oh, ok, so what is the tell tale sign it is already a fourier series, if it a signal was just x(t) = cos(t) how would I know, without other terms to sum? P.S could you please have a look here: http://www.physicsforums.com/showthread.php?t=654650 Thanks!! 


#5
Nov2512, 08:54 PM

P: 174

If you could, could you please take a look at this: http://www.physicsforums.com/showthread.php?t=654650 Thanks very much! 


#6
Nov2512, 09:52 PM

P: 174

Ok that got me nowhere fast. I still couldn't get it,
how would one integrate: a_{n} = 2/2pi ∫_{0}^{2pi} (1/2 + cos(t) + cos(2t)).cos(nω0t) dt ? Many Thanks again 


#7
Nov2612, 12:09 AM

P: 296

x(t) = 1/2 + cos(t) + cos(2t) cos(t) has period 2∏ cos(2t) has period ∏ The period of x(t) will be least common multiple of ∏ and 2∏, so 2∏. So the fundamental frequency of x(t) will be w_{0} = 2∏f_{0} = 2∏/T = 2∏/2∏ = 1 This means your fourier coefficients will be found from: [itex]a_{n} = \frac{1}{\pi}\int^{2\pi}_{0}(\frac{1}{2} + cos t + cos 2t) cos (nt) dt[/itex] The integral of the product of two sinusoids over one period is zero unless their frequencies are the same. So n=1 will only produce one nonzero term of x(t) [itex]a_{1} = \frac{1}{\pi}\int^{2\pi}_{0}(\frac{1}{2} + cos t + cos 2t) cos (t) dt[/itex] [itex]a_{1} = \frac{1}{\pi}\int^{2\pi}_{0}cos (t) cos (t) dt[/itex] [itex]a_{1} = \frac{1}{\pi}\int^{2\pi}_{0} cos^{2}t dt[/itex] [itex]a_{1} = \frac{1}{\pi}\int^{2\pi}_{0} \frac{1+cos(2t)}{2} dt[/itex] which is nonzero and equal to the coefficient of cos(t) in the x(t) equation. Likewise there will be a nonzero a_{2} term but all other terms n > 2 will be zero. 


#8
Nov2612, 12:23 AM

P: 174

So to varify I understand what you mean, the integral of (cos(x))/2 and (cos(x).cos(2x)) is zero over one period but because (cos(x))^2 is the same frequency of 1/2pi. Likewise integrating: sin(t).(sin(2t))^2 would be zero? THANKS P.S I can understand if I'm being impetuous by bring this up again but did you get a chance to look at: http://www.physicsforums.com/showthread.php?t=654650 thanks anyway for your time. 


#9
Nov2612, 12:55 AM

P: 296

You may have some doubt in considering whether sin(x)cos(x) will be zero if integrated over a period. Both sinusoids have the same frequency but you can see the integral will be zero by considering that sin(2x)=2sin(x)cos(x). Cross products of sin and cos will be zero if integrated over a period. This property of sinusoids is called orthogonality and that term is no accident. There is an analogy between orthogonal vectors and orthogonal functions where products of orthogonal functions integrated over a period equalling zero is analagous to taking the dot product of two vectors and coming up with zero. Just like you can expand vectors as the sum of orthogonal basis vectors (i, j, k eg) you can expand functions as the sum of orthogonal functions. In fourier series those functions are sums of harmnocially related sinusoids. That is not the only way to expand functions. Another way is via power series where a Taylor series represents a function as a sum of powers in x. There are infinitely many possible basis functions  sinusoids in fourier series and powers in x in taylor series are not the only ones. 


#10
Nov2612, 01:07 AM

P: 174

Ok, I'll need to do some more digging into orthogonality, is that like with the wronskian? I'm a bit confused by that first paragraph you wrote, so you're saying because sin.cos is = sin(2x)/2, when integrated over a period ∏(?) it's zero? Ok so sinusoids^{2} don't really come up in Fourier, so if I get one I'm probably doing something wrong. Out of curiosity, you said sinusoids squared are tricky (always positive) so integrating like cos^{3}(t) over a period will be positive or negative, but not zero? Thanks! 


#11
Nov2612, 01:24 AM

P: 296

I suspect there is a general result here if these things are expressed as exponentials but I have never investigated this. 


#12
Nov2612, 01:40 AM

P: 174

I see. Could I just look at (sin(x))^3 and say 'well that's an odd function, so over a period it's zero'?
I think I now understand what and how to take into concideration, thanks. (I hope I explained what I did in evaluating that transform in the other thread adoquately). Thanks heaps. 


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