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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 ...


2

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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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 ...


0

Lanczos2 has a frequency response peak of about -40 dB or 0.01 at about 0.9 times the sampling frequency, see this question. Your peak aliasing outside the main lobe is just left from where the aliasing goes to zero frequency so I think it is the same peak. The amplitude of the aliasing is also roughly as expected. So I don't see any error by you.


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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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Yes you can bandpass filter an adequate portion of a sampled (ideal impulse modulated) signal spectrum and still retain the same information of the lowpass filtered version. As you have stated, the sampled signal has a spectrum which includes shifted and weighted copies of the original (possibly baseband) signal. Assuming no aliasing occured during the ...


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Not sure what you mean by "wrong". The Nyquist criteria simply requires you to have "two samples per Hz of bandwidth". It doesn't have to be $[-f_{max},-f_{max}]$, it can be any frequency range that includes at least $2 \cdot f_{max}$ of bandwidth. However, for real signals you need to figure out what to do with the negative frequencies. For more info ...


3

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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This is just a matter of variable substitution in the integral. The inverse Fourier transform of $\delta(\Omega T)$ is $$\mathcal{F}^{-1}\big\{\delta(\Omega T)\big\}=\frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\Omega T)\,e^{j\Omega t}\,d\Omega\tag{1}$$ Substituting $\alpha=\Omega T$ gives $d\alpha=d\Omega\cdot T$, and $(1)$ becomes $$\mathcal{F}^{-1}\big\{...


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Beating just below the Nyquist frequency occurs when an attempt is made to reconstruct the time continuous signal without the use of sinc interpolation. This sinc reconstruction method and, in fact, requirement for the Nyquist–Shannon sampling theorem to hold true, is the Whittaker–Shannon interpolation formula.


3

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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