Randomly Adding Sinusoids

Let $f(x) = \sin(x) + \sin(x+\theta)$ . It follows that $f'(x) = \cos(x) + \cos(x+\theta)$ . To find the amplitude of the combined sine-function, one can use the fact that the amplitude is equal to $|f(x_0)|$ , where $x_0$ is chosen such that $f'(x_0) = 0$ . Therefore, we are interested in the following equation: $\cos(x) + \cos(x+\theta) = 0$ From this equation follows that $\cos(x) = -\cos(x+\theta)$ , which is equivalent to saying $\cos(x) = \cos(x+\theta+\pi)$ , from which we can see that $x = x + \theta + \pi + k\cdot 2\pi$ or $x = -x - \theta + \pi + l\cdot2\pi$ , where $k,l \in \mathbb{Z}$ . We are particularly interested in values for $x$ , not for $\theta$ , since $\theta$ is arbitrarily chosen, so we skip to the second equation. We see that $2x = l\cdot2\pi + \pi -\theta$ , which is equal to saying $x = l\cdot \pi + \frac{\pi-\theta}{2}$ . We choose $x_0 = \frac{\pi-\theta}{2}$ . This does not matter, because the function has a period of $2\pi$ and for odd $l$ equals $-f(x_0)$ and for the amplitude we are just interested in the absolute value of this. We find that $|f(x_0)| = |\sin(\frac{\pi-\theta}{2}) + \sin(\frac{\pi-\theta}{2}+\theta)| = |\sin(\frac{\pi-\theta}{2}) + \sin(\frac{\pi+\theta}{2})|$ . Furthermore, with the addition formula for sine, namely $\sin(u+v) = \sin(u)\cos(v) + \sin(v)\cos(u)$ , we can see that $\sin(\frac{\pi-\theta}{2}) = \sin(\frac{\pi+\theta}{2}) = \cos(\frac{\theta}{2})$ . So, the amplitude, $|f(x_0)|$ , is equal to $|2\cos(\frac{\theta}{2})|$ .

Now we use the fact that $\theta \sim U(0, 2\pi)$ . This means that the probability density function of $\theta$ , say $g(\theta)$ , can be written as follows: $g(\theta) = \begin{cases} \frac{1}{2\pi} \textrm{ if } 0 < \theta < 2\pi \\ 0 \textrm{ elsewhere.} \end{cases}$ With this, we can find the expected amplitude of the combined sine-function, as this is equal to $\int_{-\infty}^\infty \left|2\cos\left(\frac{\theta}{2}\right)\right|\cdot g(\theta) \textrm{ d}\theta,$ which is equivalent to saying the following $\int_{0}^{2\pi}\left|2\cos\left(\frac{\theta}{2}\right)\right|\cdot \frac{1}{2\pi} \textrm{ d}\theta \overset{\ast}= \frac{2}{2\pi}\int_{0}^{\pi}2\cos\left(\frac{\theta}{2}\right) \textrm{ d}\theta = \frac{2}{\pi}\int_{0}^{\pi}\cos\left(\frac{\theta}{2}\right) \textrm{ d}\theta = \left. \frac{2}{\pi}\left(2\sin\left(\frac{\theta}{2}\right)\right)\right|_{\textrm{ }0}^{\textrm{ }\pi} = \boxed{\frac{4}{\pi}}.$

$\ast$ ) This step can be derived from the fact that the function is even around $\theta = \pi$ .

Represent the sinusoids as two unit vectors separated by a random angle $\theta$ . The first can be taken as the angle reference. We want to find the expected magnitude of the following quantity.

$\large{v = 1 + e^{j \theta}}$

Real and imaginary parts:

$\large{v_R = 1 + cos \theta \\ v_I = sin \theta}$

Magnitude:

$\large{M = \sqrt{v_R^2 + v_I^2} = \sqrt{2 + 2 cos \theta} = \sqrt{2} \sqrt{1 + cos \theta}}$

Take an angle-weighted average:

$\large{M_{av} = \frac{\sqrt{2}}{2 \pi} \int_0^{2 \pi} \sqrt{1 + cos \theta} \, d \theta = \frac{4}{\pi} \approx 1.2732}$