- #1
bigguccisosa
- 24
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Homework Statement
Hi, I have to find the RMS value of the inifnite series in the image below.
Homework Equations
https://en.wikipedia.org/wiki/Cauchy_product
Allowed to assume that the time average of sin^2(wt) and cos^2(wt) = 1/2
The Attempt at a Solution
So to get the RMS value I think I have to square the series and then take the time average.
[tex] [\sum_j B_j\cos(\Omega_j t) + A_j\sin(\Omega_j t)]^2 [/tex]
[tex] = [\sum_i B_i\cos(\Omega_i t) + A_i\sin(\Omega_i t)][\sum_j B_j\cos(\Omega_j t) + A_j\sin(\Omega_j t)] [/tex]
Now using the formula from the link,
[tex] = \sum_k \sum_{l=0}^k [B_l\cos(\Omega_l t) + A_l\sin(\Omega_l t)] [B_{k-l}\cos(\Omega_{k-l} t) + A_{k-l}\sin(\Omega_{k-l} t)] [/tex]
So now since I want to square the series, I evaluate when l = k-l right? that means l = k/2. So FOILing the inside sum gives me (and the sin*cos terms cancel due to orthogonality). (Really iffy here about whether this is correct)
[tex] \sum_k \sum_{l=0}^k B_lB_{k-l}\sin(\Omega_lt)\sin(\Omega_{k-l}t) + C_lC_{k-l}\cos(\Omega_lt)\cos(\Omega_{k-l}t) [/tex]
[tex] \sum_k B_{\frac{k}{2}}^2\sin^2(\Omega_{\frac{k}{2}}t)+ C_{\frac{k}{2}}^2\cos^2(\Omega_{\frac{k}{2}}t)[/tex]
If I go ahead and take the time average now,
[tex] \sum_k B_{\frac{k}{2}}^2\frac{1}{2}+ C_{\frac{k}{2}}^2\frac{1}{2}[/tex]
[tex]\frac{1}{2} \sum_k B_{\frac{k}{2}}^2+ C_{\frac{k}{2}}^2[/tex]
Problem is I don't know if this is correct, I feel like the k/2 subscript should become k somehow as to matchup with the Fourier series shown here http://planetmath.org/rmsvalueofthefourierseries1
Any help is appreciated
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