This is a follow-up to this question. It discusses the amplitude of a sum of $N$ functions of the form $a\sin(kx+c)$
$\sum_{i=1}^{N} a\sin(kx+c_i)$
with $a$ and $k$ constant and $c_i$ random numbers between $0$ and $2π$. Let's assume for the following that the $c_i$ are uniformly distributed within those limits. I am interested in the total result of that sum.
It is clear to me, according to for instance this link, that the sum of sine/cosine functions with the same frequency but different phases is again a sine/cosine function of the same frequency, but with different amplitude and phase. The expected value of the new amplitude amounts to $Na^2$. However, what happens to the phase of the sum? Intuitively, I would assume that for large $N$, the summation of all those sine/cosine functions with different phase tends to zero, as we will have all different shifts of them and there will always be pairs that completely cancel. This means, while we have a non-zero amplitude, the expected value of the phase should be $0$, $π$ or $2π$ when dealing with sine functions, and the total sum turns zero then. Is that assumption correct? How is it possible to calculate the expected value of the total phase, which, again from this link, can be calculated by
$\tan c=\frac{\sum_{i=1}^{N} \sin c_i}{\sum_{i=1}^{N} \cos c_i}$?
In addition, I am not sure if it would be the proper way to calculate the expected value of the sum of sines by considering resulting amplitude and phase separately. Wouldn't be the proper way to use something like the law of the unconscious statistician?
Any help and/or literature recommendations are greatly appreciated. I feel this is a rather common problem, but was not able to find useful references.
Best Answer
Lets start with an identity: $$a \sin(\theta)+b\cos(\theta)=r\sin(\theta + \arctan(b/a))$$ with $r=\sqrt{a^2+b^2}$
Now we look at $$\sum \sin(kx+c_i)=Im \sum e^{i(kx+c_i)}$$ $$=Im \left[ (\cos(kx)+i\sin(kx))\sum (\cos(c_i)+i\sin(c_i)) \right]$$ Collecting the imaginary part, we are left with $$\left[\sum \sin(c_i)\right] \cos(kx) + \left[\sum \cos(c_i)\right]\sin(kx)$$ using the first identity completes the job.