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Consider a signal $$ x(t) = \cos(175\pi t) $$ which is sampled to produce discrete time signal $$ x[n] = x(nT_s) $$ The fundamental period of $x[n]$ is $$ N_0 = 7 $$

Given this, what is the smallest possible sample rate $T_s$? (Ans: 1.6327 ms).

I would assume that this is related to finding the Nyquist frequency. I was thinking:

Since, $$N_0 = 7 \implies f_0 = \frac17 \implies f_{\mathrm{Nyquist}} = 2 \frac17\implies T_s = \frac72 $$ However, this is obviously incorrect. I am not even using any information of the original signal. Any suggestions on what I could be doing wrong here?

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Remember that the cosine function repeats itself every $2\pi$. The discrete signal can be written as $$x(n)=\cos(175\pi n T_s)$$

So when the argument reaches $2\pi$, we'll have a period. As we know that the fundamental period is $7$, then

$$175\pi \cdot7T_s=2\pi \implies T_s =\frac{2}{7\cdot 175}=1.6327\cdot 10^{-3}$$

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For a pure sinusoid in zero noise, just 3 or 4 non-aliased samples (at almost any sample rate) will work. Here are derivations of two closed form solutions by C.Turner:

http://www.claysturner.com/dsp/3pointfrequency.pdf

http://www.claysturner.com/dsp/4pointfrequency.pdf

For finite sampling in noise, the required sample rate will depend on the duration of sampling, the S/N ratio, and the parameter (frequency, etc.) estimation accuracy required. It's not just 2X the frequency or slightly over (except for an infinite length of sampling).

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