There’s an interesting feature of trigonometric functions that I came up with while dealing with the WFM of resolvers: When a trigonometric function $\cos (Ax + \varphi)$ is sampled $B$ times within $x \in [0, 2\pi)$ with equal time interval, the necessary and sufficient condition for the sum of $N$ consecutive samples to be zero is that $NA \mid B$ and $A \nmid B$ ($A,B,N \in \mathbb{Z}$).

It is somehow intrinsic once mentioned, but it really took me a while to prove. I guess that’s a hint for me that I should never get involved in math.

Prove:

For any $\varphi \in [0, 2\pi)$:

If $NA \mid B$ and $A \nmid B$, then $1- e^{2\pi j \frac{NA}{B}} = 0$ and$1- e^{2\pi j \frac{A}{B}} \neq 0$, thus ${\mathop{\rm Re}\nolimits} \left( {{e^{j\phi }}\left( {\frac{{1 - {e^{2\pi j\frac{{NA}}{B}}}}}{{1 - {e^{2\pi j\frac{A}{B}}}}}} \right)} \right) \equiv 0$, and the sum of $N$ consecutive samples is always zero.

Otherwise, if $NA \nmid B$ as well as $A \nmid B$, for any $\phi$ that satisfies:

there is always a $\Delta \varphi \neq n\pi (n \in \mathbb{Z})$ that makes

which means the sum of $N$ consecutive samples is determined by $\phi$.

If $A \mid B$, assume that $A = nB (n \in \mathbb{Z})$,

which means the sum of samples equals zero only with $\varphi = \frac{\pi}{2}$ or $\varphi = \frac{3\pi}{2}$.

Q.E.D.

Let’s say such conclusion is somehow useful. E.g., if the induced voltage of a resolver has a constant-amplitude component of:

where $n_{si} = N_s \cos (\omega_s \theta_i)$, $n_{ei} = N_e \cos (\omega_e \theta_i)$, and $\theta_i = \frac{2\pi i}{\omega_t}$.

What are the requirements to have $v_s \equiv 0$? In this example, $A = \omega_e \pm \omega_s$, $B = \omega_t$ and $N = \omega_t$. So the necessary and sufficient condition will be $\omega_e \pm \omega_s \nmid \omega_t$.