Did I multiply this infinite series correctly?

1. Jul 4, 2016

bigguccisosa

1. The problem statement, all variables and given/known data
Hi, I have to find the RMS value of the inifnite series in the image below.

2. Relevant 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

3. 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.
$$[\sum_j B_j\cos(\Omega_j t) + A_j\sin(\Omega_j t)]^2$$
$$= [\sum_i B_i\cos(\Omega_i t) + A_i\sin(\Omega_i t)][\sum_j B_j\cos(\Omega_j t) + A_j\sin(\Omega_j t)]$$
Now using the formula from the link,
$$= \sum_k \sum_{l=0}^k [B_l\cos(\Omega_l t) + A_l\sin(\Omega_l t)] [B_{k-l}\cos(\Omega_{k-l} t) + A_{k-l}\sin(\Omega_{k-l} t)]$$
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)
$$\sum_k \sum_{l=0}^k B_lB_{k-l}\sin(\Omega_lt)\sin(\Omega_{k-l}t) + C_lC_{k-l}\cos(\Omega_lt)\cos(\Omega_{k-l}t)$$
$$\sum_k B_{\frac{k}{2}}^2\sin^2(\Omega_{\frac{k}{2}}t)+ C_{\frac{k}{2}}^2\cos^2(\Omega_{\frac{k}{2}}t)$$
If I go ahead and take the time average now,
$$\sum_k B_{\frac{k}{2}}^2\frac{1}{2}+ C_{\frac{k}{2}}^2\frac{1}{2}$$
$$\frac{1}{2} \sum_k B_{\frac{k}{2}}^2+ C_{\frac{k}{2}}^2$$
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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Last edited: Jul 4, 2016
2. Jul 4, 2016

On this one I think you can use $A \cos(\omega t) +B \sin(\omega t) =\sqrt{A^2+B^2} cos(\omega t-\phi)$ where $\phi=arctan(B/A)$. I think you will find the mean square of a given frequency which is $(A^2+B^2)/2$ is not affected by that from another frequency. Essentially this says that the total energy is the sum of the energies of the different frequency components. To get the r.ms. you simply take the square root of the total mean square=sum of the mean squares. I'd enjoy seeing if you and/or others concur, but I think this is correct. I think they might even be looking for you to compute/prove what I just stated, but it can help to know what you need to compute.

Last edited: Jul 4, 2016
3. Jul 4, 2016

bigguccisosa

I haven't seen that identity before actually, but it does help to simplify things a bit. But how do you know what the mean square will be? With this formula I still see myself having to do a Cauchy product, which I think is where I am messing up with the indices.

4. Jul 5, 2016

That identity is quite useful for writing the form $y(t)=Acos(\omega t) +B sin(\omega t)$ as a single cosine term. The easiest way to derive it is to simply factor out $\sqrt{A^2+B^2}$ to get $y(t)=\sqrt{A^2+B^2} ((A/\sqrt{A^2+B^2})cos(\omega t)+(B/\sqrt{A^2+B^2})sin(\omega t))$. Then let $cos(\phi)=A/\sqrt{A^2+B^2}$ and $sin(\phi)=B/\sqrt{A^2+B^2}$. You should recognize the terms as $cos(\omega t-\phi)$ using the simple $cos(\theta-\phi)$ identity. With a single $\cos^2(\omega t-\phi)$ term, the mean is of course equal to 1/2. (This is because $cos^2(\omega t)=(cos(2 \omega t)+1)/2$. The $cos(2 \omega t)$ averages to zero. )When you have the sum of a series of squares of different frequencies along with the cross terms , I don't know how thoroughly you are expected to prove that the cross terms don't generate any non-zero interference terms, but when you have $cos(\omega_1 t)cos(\omega_2 t)$, it will give sum and difference of frequency terms each of which averages to zero for $\omega_1$ not equal to $\omega_2$.