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Apart from practical errors related to the truncation of the infinitely-long energy signal when discretized to digital, can we find a UNIQUE mapping between the two signal forms? No. I think about this differently than Tim Wescott's in his excellent answer and I hope we can make a fun discussion out of this. Tim accurately describes that any bandlimited ...


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Apart from practical errors related to the truncation of the infinitely-long energy signal when discretized to digital, can we find a UNIQUE mapping between the two signal forms? Yes. Say you're starting with $x(t)$, which is perfectly bandlimited to less than $f_B = \frac{1}{2 T_s}$, and you take samples at every $t = T_s k$: $x_k = x(T_s k)$. All you need ...


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Given $ \left\{ x \left[ n \right] \right\}_{n \in M} $ where $ M $ is the set of indices given for the samples of $ x \left[ n \right] $. The trivial solution (Which it would be great to have a faster more efficient solution is what I'm looking for) would be: $$ \arg \min_{y} \frac{1}{2} \left\| \hat{F}^{T} y - x \right\|_{2}^{2} $$ Where $ \hat{F} $ is ...


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I've sampled it at a 1 bit resolution (as 8 bit oscilloscope readings & 0b1) No. Taking the LSB is NOT 1 bit quantization. A one bit quantizers are not trivial to design. For a zero mean signal that's reasonably symmetrically distributed over positive and negative values, you can just take the sign bit. For signals that are asymmetrically distributed ...


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The reason is very simple in the context of the DFT and Sampling Theorem. In that context the sampling duration is about the duration you have full knowledge of and able to reconstruct under the assumption of proper sampling. For discrete signals, in the context of the DFT, the model is about the signals being periodic. Hence teh last sample gives you the ...


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I feel that someone ought to advise the questioner against thinking that this process is automagically going to give an extra 4 bits of resolution. IF the noise on the DC signal is uncorrelated, gaussian-distributed noise, then yes it will. In the vast majority of real cases, the noise is not like that. In particular, for DC type signals with low-cost ADCs, ...


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I agree with your result for $Z(j\omega)$. Apart from scaling, what you have is the original spectrum with positive and negative frequencies swapped and multiplied with factors $1-j$ and $1+j$, respectively. In order to restore the original signal, we need to add right-shifted and left-shifted versions of the spectrum, while getting rid of the complex ...


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You seem to get hung up on that "distinction" whether or not there is overlap. This is actually irrelevant for the definition of aliasing, as correctly pointed out in myzz's answer. As explained in my answer to your previous question, aliasing means that frequencies are shifted to some place where they were not before the sampling process. This is ...


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There really is no practical difference between the two. The important thing to understand about aliasing is not the exact definition of the word, but rather the concept: when two frequency bands alias and there is a signal in one of these bands, after sampling you can't tell which band the signal came from. In a sense, by the time you're asking what is or ...


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I sadly don't have the reputation to comment on Fat32's answer, but let me try to answer directly instead: the frequency response of a digital filter is always relative to the processing sampling rate. So if you design a filter with cutoff at 100Hz and 1kHz sampling rate, then you are really designing for a normalized cutoff frequency $f/f_s$ of 1/10 and ...


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What you describe is indeed aliasing. If there are frequency components above half the sampling frequency then they will be folded back to the interval $[0,f_s/2]$. This folding back is called aliasing, no matter whether these aliased components overlap with other parts of the signal spectrum or not. Note that you can also use aliasing to your advantage if ...


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I'm not going to be formal about this, but the easy way to think about aliasing is in terms of Laplace vs. z-transform. The sampling process essentially substitutes $z\leftarrow e^{sT}$ (where $T$ is the sampling interval) while the reconstruction does the inverse $s\leftarrow \frac{\ln{z}}{T}$. The way to think about it is that the Fourier transform found ...


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My input signal is a dc signal (sensor output) and im getting kind of a headache to understand why oversampling can increase the resolution of dc signals. The ADC puts out integers. So, let $x$ be the integer that would come out of the ADC (100.3, say, or -333.3). Now let $y = \lfloor r \rfloor$ be the quantization operation, where you get straight ...


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