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R (sampling rate) fs (signal being sampled fN (the Nyquist frequency) fa (aliased frequency)

I was hoping someone could explain to me why this is true with a more mathematical proof. it seems reasonable for the aliased frequency to be related to these parameters, but I would like a more solid rationale.

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Lets say you have a continuous sinusoidal signal $x(t) = \cos(2\pi f_o t)$, now we sample that signal by letting $t=nT=\frac{n}{f_s}$, where $f_s$ is the sampling frequency. This gives $x[n]= \cos(2\pi f_o nT) $. Because sinusoids are periodic every $2\pi$ we can write: $$\cos(2\pi f_o nT) = \cos(2\pi f_o nT + mn2\pi T ),$$ where we require $mnT$ to be an integer. $n$ is always an integer, and $T$ is determined by our sampling frequency. Therefore this is equivalent to requiring $m=\frac{p}{T}$ where $p$ is an integer - in other words $m$ must be an integer multiple of the sampling frequency. Given these conditions we can write: $$ \cos(2\pi f_o NT + mn2\pi T ) = \cos(2\pi (f_o +m)nT ) = \cos(2\pi (f_1)nT ),$$ where $f_1= f_o+m$, but remember that $m$ is a multiple of the sampling frequency.

Therefore any sampled sinusoid is equivalent to another sampled sinusoid with a frequency that is an integer multiple of sampling frequency away from the original frequency. Note the above formulation could be done equivalently with $\exp(j2\pi f_o t)$ instead of $\cos()$ to be a bit more general.

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This is incomplete but should point you in the right direction. Assume $k$ is an integer and $f_s=1/T_s$ is the sampling frequency. The signal being sampled is $$\cos(2\pi f_0t).$$ The sampled signal is $\cos(2\pi f_0kT_s)$. This signal has an alias at frequency $f_1=(f_s/2-f_0)+f_s/2=f_s-f_0$. To prove it: \begin{align*} \cos(2\pi(f_s-f_0)kT_s) &= \cos(2\pi f_skT_s-2\pi f_0kT_s) \\ &= \cos(2\pi k - 2\pi f_0kTs)\\ &= \cos(2\pi f_0kT_s). \end{align*} A similar argument can be made for all other aliases.

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