Understanding the mathematical proof for the alias frequencies in a sampled sine wave

Question

I'm struggling to get my head round the mathematical proof for the alias frequencies in a sampled sine wave.

I understand that sampling a sine wave of frequency $f_0$ every $t_s$ seconds gives you:

$$x[n]=\sin(2\pi f_0nt_s)$$

I also understand that, because the sine wave is periodic every $2\pi$, you can add any multiple of $2\pi$ to the angle and get the same values for the sine, i.e.,

$$\sin(2\pi f_0nt_s)=\sin(2\pi f_0nt_s+2\pi m) \quad\text{(where $m$ is any integer).}$$

The proof I'm looking at then factors out $2\pi$ and $nt_s$ to get:

$$\sin\left(2\pi(f_0+\frac{m}{nt_s})nt_s\right)$$

...but then it says to let $m$ be an integer multiple of $n$ so we can replace the $\frac{m}{n}$ ratio with an integer $k$.

I don't understand how $m$ can go from being "any integer" to "an integer multiple of $n$". If $m$ is any integer and $n$ is an integer then how can the ratio between them be an integer?

I know I'm missing something obvious here and I'm searching for that light-bulb moment but it's not happening. Because this is so fundamental to DSP I don't just want to accept the formula and move on without thoroughly understanding it.

Answer 1

The reason is that if it is true for any $m$, it is also true for $m=kn$.

I will sketch the proof in another way.

Call $f_s = 1/t_s$ sampling frequency where $t_s$ is sampling period, the two signal $x(t) = \sin(2\pi f_0 t)$ and $x_k(t) = \sin(2\pi (f_0 + k f_s) t)$ have the same values at sampling instants (aliasing), i.e. $x[n] = x_k[n]$.

Indeed,

\begin{align} x[n] &= \sin (2 \pi f_0 n t_s) \\ x_k[n] &= \sin \left(2 \pi (f_0 + k f_s) n t_s\right) \\ &= \sin (2 \pi f_0 n t_s + 2\pi k f_s n t_s) \\ &= \sin (2 \pi f_0 n t_s + 2\pi k n) = \sin (2 \pi f_0 n t_s)\\ &=x[n] \end{align}