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It really boils down to aliasing. In continuous-time, if you have any two signals $x_1(t) = \sin(2 \pi F_1 t)$ and $x_2(t) = \sin(2 \pi F_2 t)$, then as long as $F_1$ and $F_2$ are distinct, the signals are, too.
But consider sampling at some time interval $T_s$, so that the sampled signals are $x_1(k) = \sin(2 \pi F_1 T_s k)$ and $x_2(k) = \sin(2 \pi F_2 ...
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Is the rate of 2B exclusive?
Yes. The sampling theorem states that the signal must be band limited to half the sample rate. That implies that the energy at the Nyquist frequency must be zero. In practice you need a healthy margin between the highest usable frequency and the Nyquist frequency. There is always some "transition band" that you need to get the ...
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The SNR is the ratio of $S$, the power in the signal $s(t)$, and $N$, the power in the noise $n(t)$. So, $$\text{SNR}_\text{dB} = 10\log_{10}\frac{S}{N}.$$
When using RMS values, we have $S = s_\text{RMS}^2/R$ and $N = n_\text{RMS}^2/R$. The resistances cancel out and then $$\text{SNR}_\text{dB} = 20\log_{10}\frac{s_\text{RMS}}{n_\text{RMS}}.$$
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Without more details, when a signal is properly recorded at $F_n$ Hz, one is entitled to expect that all information below the half of it, namely $F_n/2$ Hz, can be recovered, somehow. This is an application of the classical Shannon Sampling Theorem. Twice the maximum frequency is sufficient to sample a signal... in theory. The question says: max frequency ...
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First of all, as you said the sampling rate is probably $12$ Hz, rather than $12$ kHz, and perhaps they want to demonstrate an aliasing example.
Given a bandlimited continuous-time periodic signal
$$x(t) = \cos(16\pi t + \phi)$$
the samples taken at the rate $F_s = 12$ Hz will be denoted as $x[n]$ and will be obtained by via $x[n] = x(t_n)$ with $t_n = n ...
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Look at your signal and figure out what sampling rate you need to avoid aliasing. In this case, you are sampling at twice the maximum frequency component so you should not expect aliasing, but you work leads to aliasing components showing up at plus/minus $\frac{3\pi}{5}$.
Remember that the DTFT is periodic with period $2\pi$, and try re-working the ...
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Approaching The Sampling Theorem as Inner Product Space
Preface
There are many ways to derive the Nyquist Shannon Sampling Theorem with the constraint on the sampling frequency being 2 times the Nyquist Frequency.
The classic derivation uses the summation of sampled series with Poisson SummationFormula.
Let's introduce different approach which is more ...
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