# Tag Info

## Hot answers tagged sampling

4

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 ...

3

Is this an entirely stupid idea? No, but you've just came to the conclusion that instead of sampling complex, with Nyquist rate being the bandwidth, you should do twice as many samples. That simply means you're not doing IQ sampling, but low-IF or direct-RF sampling. Mix the signal with a higher intermediate frequency and filter it in such a way that ...

3

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 ...

3

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 ...

2

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 ...

2

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 ...

1

As Marcus commented, it certainly does follow the Nyquist criterion. However you just need to keep in mind that I/q sampling is complex so each complex sample can be considered 2 real independent samples and Nyquist will still apply. Also take a look at this answer "Complex sampling" can break Nyquist?

1

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 ...

1

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 ...

1

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 ...

Only top voted, non community-wiki answers of a minimum length are eligible