RobR wrote:> Well, this seems to be a hot topic. > I am not that much in signal theory. > But what I got so far is, that IQ sampling indeed holds twice the spectral > information that real sampling provides. > So one complex sample holds information of two real samples at twice the > complex sampling rate. > > One could also, maybe more theoreticaly, sample signal and first > derivative for the same effect. > > If I take signal, first and second derivative, then I would need at least > third of nyquist? This steps could be carried on to infinity derivatives, > where I would need zero samples (asymptotically), right?Right. I think R.B-J. wrote that in an earlier message in this thread. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Nyquist constrain and IQ represented signal
Started by ●November 14, 2007
Reply by ●November 14, 20072007-11-14
Reply by ●November 14, 20072007-11-14
Randy Yates wrote:> Jerry Avins <jya@ieee.org> writes:>>There's no free lunch here. A complex sample counts for two real >>samples.> I didn't intend to imply that there is a free lunch. When it comes > to a specific implementation, different tradeoffs must be performed > in order to compare real sampling and signal paths with their complex > counterparts. One of them is the number of words per sample interval, > as you've pointed out.(snip)> But, Jerry, I hope you get this one fact and get it good: In the set of > complex numbers, a single element counts as ONE sample. That is, even > though a complex number has two components (real and imaginary, or > magnitude and phase), the "thing" being sampled in a theoretical complex > sampler is the set of complex numbers, so one complex number counts as > one sample.One complex sample counts as one (complex) sample. In a large number of cases that is just as good as two real samples. There are popular FFT routines for transforming real data that accept 2N real points, convert to N complex points, do a complex FFT on that, convert to the appropriate complex result for the original data. N complex samples of a real signal are, of course, not worth 2N real samples. -- glen
Reply by ●November 14, 20072007-11-14
Jerry Avins <jya@ieee.org> writes:> [...] > BTW, how many bits are needed for a complex sample that stands in > place of two real 12-bit samples?What sample rate is required for a signal with a 100-Hz bandwidth? -- % Randy Yates % "The dreamer, the unwoken fool - %% Fuquay-Varina, NC % in dreams, no pain will kiss the brow..." %%% 919-577-9882 % %%%% <yates@ieee.org> % 'Eldorado Overture', *Eldorado*, ELO http://www.digitalsignallabs.com
Reply by ●November 14, 20072007-11-14
maybe i'll regret that i'm piping in on this... On Nov 14, 2:31 pm, "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote:> "RobR" <mas...@gmx.de> wrote in message > > news:-cydnXVvT4TpnqbanZ2dnUVZ_vOlnZ2d@giganews.com... > > > > there is a question that bugs me for quite a long time: > > > You can read about Nyquist constrain online, that to reconstruct all > > frequencies within a signal, it has to be sampled with at least twice > > the bandwidth _or_ maximum frequency. > > Maybe this _or_ is already the problem... > > > Let's see, > > I have an IQ branched digital signal. > > So my maximum positive signal is not equal to the bandwidth of the > > signal, > > since also the negative frequencies are present, so the bandwidth of the > > signal is twice the maximum positive frequency. > > > Now what is the Nyquist frequency for such a signal representation? > > Is it twice the max positive frequency or twice the bandwidth (4 times > > max pos. frequency)? > > > From what i have seen, it is twice the max positive frequency. > > But how can this be? Is it the IQ demodulation, that gives me twice the > > information in contrast to a signal in "as it is" representation (one > > real stream)? > > In fact, i need a clarification between nyquist adopted to a real signal > > and nyquist adopted to a complex IQ representation signal....> > First, try to keep it simple and also keep the terminology straight. > > I don't know what an IQ "branched" signal is.... anything would be > conjecture.i might conjecture that this is an IF signal, with finite bandwidth, that can go all the way down to virtually DC.> Start out with purely real signals. Let's call the bandwidth B - in the > generally accepted sense that it's generally nonzero from zero to B and > zero above B. Then you should know what the sampling criterion is for > that one - the sample rate needs to be *greater than* 2B.this we agree on. and remember that if the real signal has content from f=0 to f=B, it also has content from f=-B to f=0. that is why we need to sample at greater than 2*B.> When you need to deal with the negative frequencies, that's the time to > deal with negative frequencies.this is a tautology, like saying that the pope is Catholic. or that "W" stands for "Worthless". i might add that "W" is short for "Worthless piece of crap". these are all tautologies. :-)> Then, move to a complex signal. Let's assume the bandwidth of the real > part is B and let's assume that the bandwidth of the imaginary part is B. > The sample rate of each is as above. So, it's effectively doubled because > there are 2 sequences.this i agree with, too, because nothing else was said about the nature of the real and imaginary parts of the signal. but *if* an additional piece of information was there, specifically that the imaginary part was the Hilbert Transform of the real part (which is something i suspect might be the case for quadrature signal pairs), then sampling (with complex valued samples) at a rate just greater than B is sufficient. there would be no overlapping (and aliasing) in the frequency domain if you did that. but, still, if I is completely arbitrary, then Q is fully defined in terms of I, you still must sample Q along with I, and you still have 2 real sample values per unit of time 1/B (or just a millismidgen less than 1/B). no free lunch, essentially, in any case, you need just a teeny bit more than two real samples per 1/B amount of time. the amount of information is the same. of course, if I and Q are totally independent and assumedly unrelated, then you need four samples in an amount of time that is just under 1/ B.> Then move to sampling bandpass signals, etc.oh, wunnerful. :-) r b-j
Reply by ●November 14, 20072007-11-14
Randy Yates wrote:> Jerry Avins <jya@ieee.org> writes: >> [...] >> BTW, how many bits are needed for a complex sample that stands in >> place of two real 12-bit samples? > > What sample rate is required for a signal with a 100-Hz bandwidth?For the kind of signal that can be carried on a twisted pair, a rate in excess of 200 per second, whether the samples are real or complex if the signal is baseband. You asked to make a point; what point? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●November 14, 20072007-11-14
On Nov 14, 2:19 pm, Randy Yates <ya...@ieee.org> wrote:> Jerry Avins <j...@ieee.org> writes: > > [...] > > BTW, how many bits are needed for a complex sample that stands in > > place of two real 12-bit samples? > > What sample rate is required for a signal with a 100-Hz bandwidth? > -- > % Randy Yates % "The dreamer, the unwoken fool - > %% Fuquay-Varina, NC % in dreams, no pain will kiss the brow..." > %%% 919-577-9882 % > %%%% <ya...@ieee.org> % 'Eldorado Overture', *Eldorado*, ELOhttp://www.digitalsignallabs.comA signal with 100 Hz bandwidth requires greater than 200 samples per second. If the signal is real this might be done by collecting samples of the real channel at an interval of less than 5 milliseconds. If the signal is complex this might be done by collecting samples of the I channel at an interval of less than 10 milliseconds and samples of the Q channel at an interval of less than 10 milliseconds. The sampling times and intervals of the I and Q channels need not be the same. The sampling times can be alternated to share an ADC on the I and Q channels. It's usually easier to leave the rates the same. Dale B. Dalrymple
Reply by ●November 14, 20072007-11-14
dbd wrote:> On Nov 14, 2:19 pm, Randy Yates <ya...@ieee.org> wrote: >> Jerry Avins <j...@ieee.org> writes: >>> [...] >>> BTW, how many bits are needed for a complex sample that stands in >>> place of two real 12-bit samples? >> What sample rate is required for a signal with a 100-Hz bandwidth? >> -- >> % Randy Yates % "The dreamer, the unwoken fool - >> %% Fuquay-Varina, NC % in dreams, no pain will kiss the brow..." >> %%% 919-577-9882 % >> %%%% <ya...@ieee.org> % 'Eldorado Overture', *Eldorado*, ELOhttp://www.digitalsignallabs.com > > A signal with 100 Hz bandwidth requires greater than 200 samples per > second. If the signal is real this might be done by collecting samples > of the real channel at an interval of less than 5 milliseconds. If the > signal is complex this might be done by collecting samples of the I > channel at an interval of less than 10 milliseconds and samples of the > Q channel at an interval of less than 10 milliseconds. The sampling > times and intervals of the I and Q channels need not be the same. The > sampling times can be alternated to share an ADC on the I and Q > channels. It's usually easier to leave the rates the same.Dale, Here's a thought: since real numbers are merely a special case of complex numbers, we could do complex sampling of our 100 Hz baseband signal. For that case, we get by with something a little in excess of 100 complex samples per second. Moreover, we know /a priori/ that each sample will be of the form x[n] + j0, so we are free to merely imagine that we have taken the imaginary samples. It seems a great savings of sample count. Do you think the idea is patentable? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●November 14, 20072007-11-14
On Nov 14, 2:19 pm, Randy Yates <ya...@ieee.org> wrote:> What sample rate is required for a signal with a 100-Hz bandwidth?How long do you plan on sampling relative to the sampling error or noise level?
Reply by ●November 14, 20072007-11-14
Fred Marshall wrote:> "RobR" <masked@gmx.de> wrote in message > news:-cydnXVvT4TpnqbanZ2dnUVZ_vOlnZ2d@giganews.com... >> Hello, >> >> there is a question that bugs me for quite a long time: >> >> You can read about Nyquist constrain online, that to reconstruct all >> frequencies within a signal, it has to be sampled with at least twice the >> bandwidth _or_ maximum frequency. >> Maybe this _or_ is already the problem... >> >> Let's see, >> I have an IQ branched digital signal. >> So my maximum positive signal is not equal to the bandwidth of the >> signal, >> since also the negative frequencies are present, so the bandwidth of the >> signal is twice the maximum positive frequency. >> >> Now what is the Nyquist frequency for such a signal representation? >> Is it twice the max positive frequency or twice the bandwidth (4 times max >> pos. frequency)? >> >> From what i have seen, it is twice the max positive frequency. >> But how can this be? Is it the IQ demodulation, that gives me twice the >> information in contrast to a signal in "as it is" representation (one real >> stream)? >> In fact, i need a clarification between nyquist adopted to a real signal >> and nyquist adopted to a complex IQ representation signal. >> >> Best regards, >> Robert > > First, try to keep it simple and also keep the terminology straight. > > I don't know what an IQ "branched" signal is.... anything would be > conjecture. > > Start out with purely real signals. Let's call the bandwidth B - in the"Purely" real? Don't you mean degenerate complex signals, where the permissible set of values is highly constrained? :-) It puzzles me when people treat complex like its a special case. All numbers are complex. Reals are a subset of complex, where the j part is always zero. Integers are a further subset, where the fractional part is always zero. Cardinals are a further subset, which by act of faith believes negative values don't exist - what else would you expect from a Cardinal? :-). Steve
Reply by ●November 14, 20072007-11-14
On Nov 14, 3:25 pm, Jerry Avins <j...@ieee.org> wrote:> dbd wrote: > > On Nov 14, 2:19 pm, Randy Yates <ya...@ieee.org> wrote: > >> Jerry Avins <j...@ieee.org> writes: > >>> [...] > >>> BTW, how many bits are needed for a complex sample that stands in > >>> place of two real 12-bit samples? > >> What sample rate is required for a signal with a 100-Hz bandwidth? > >> -- > >> % Randy Yates % "The dreamer, the unwoken fool - > >> %% Fuquay-Varina, NC % in dreams, no pain will kiss the bro=w..."> >> %%% 919-577-9882 % > >> %%%% <ya...@ieee.org> % 'Eldorado Overture', *Eldorado*, ELOh=ttp://www.digitalsignallabs.com> > > A signal with 100 Hz bandwidth requires greater than 200 samples per > > second. If the signal is real this might be done by collecting samples > > of the real channel at an interval of less than 5 milliseconds. If the > > signal is complex this might be done by collecting samples of the I > > channel at an interval of less than 10 milliseconds and samples of the > > Q channel at an interval of less than 10 milliseconds. The sampling > > times and intervals of the I and Q channels need not be the same. The > > sampling times can be alternated to share an ADC on the I and Q > > channels. It's usually easier to leave the rates the same. > > Dale, > > Here's a thought: since real numbers are merely a special case of > complex numbers, we could do complex sampling of our 100 Hz baseband > signal. For that case, we get by with something a little in excess of > 100 complex samples per second. Moreover, we know /a priori/ that each > sample will be of the form x[n] + j0, so we are free to merely imagine > that we have taken the imaginary samples. It seems a great savings of > sample count. Do you think the idea is patentable? > > Jerry > -- > Engineering is the art of making what you want from things you can get. > =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF==AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF= =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF Jerry By your definition, x[n] + j0 is special case of the complex sample called a real sample, so it would take over 200 per second for a 100 Hz BW. Complex samples have the form I[n] + jQ[n] (if the I and Q channels are sampled simultaneously) where the I and Q are independent channels. If the I and Q are not independent it is just a sampled real channel, scaled by a complex gain. I wouldn't bet that any scheme couldn't be patented, but that says more about the patent system than the scheme. Dale B. Dalrymple
