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}

Answer 2

I think the reason why the originally referenced proof feels unintuitive and unsatisfying is that upon first read it appears they are selecting any old integer that may satisfy the proof (the integer multiple of $n$ really bothered me when I first looked at). In actuality, however, they are selecting the exact and only integers that satisfy the proof for a given $k$ in $\sin(2 \pi (f_0 + f_s k)n / f_s)$. For each sample, the $f_0 + f_sk$ frequency goes through exactly $nk$ cycles more than $f_0$ does, and so reaches the same value at the sample.

In my opinion, graphing this really helps, so let's consider a situation where we are sampling $32$ times a second and start with a base frequency of $4$Hz. Taking $k=1$, we can graph frequency $f_0=4$ and $f_1=4 + 32 \cdot 1$, and mark the sample points:

Taking a look at the first two non-zero samples, you can see that $f_1 = 4 + 32\cdot 1$ goes though exactly $1$ and $2$ more cycles than $f_0$ respectively. This makes sense because $k=1$ and for the first sample $n=1$ and for the second $n=2$.

Next $f_0=4$ and $f_2=4 + 32 \cdot 2$:

Now $k=2$, and we see for $n=1$, $f_2= 4+32 \cdot 2$ goes through $2$ extra cycles for $n=1$ and $4$ extra cycles for $n=2$, and so on, all equal to $nk$. This is true for any integer $n$ and any integer $k$.

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