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
Nyquist constrain and IQ represented signal
Started by ●November 14, 2007
Reply by ●November 14, 20072007-11-14
"RobR" <masked@gmx.de> writes:> 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, > RobertHi Robert, The problem as I see it is the definition of bandwidth. Most folks would say that a signal that has a non-zero spectrum over -B to +B Hz has a bandwidth of B Hz. I challenge that definition and say it has a spectrum of 2B Hz. However, in making this new definition, you must realize that when a signal is real, half of this bandwidth is unusable. On the other hand, all the bandwidth is usable when the signal is complex. With that definition of bandwidth, I contend that sampling at Fs samples/second gives you Fs Hz of bandwidth, from -Fs/2 to +Fs/2. When the sampling is real, half that bandwidth is unusable. When it is complex, the entire bandwidth is usable. -- % Randy Yates % "Ticket to the moon, flight leaves here today %% Fuquay-Varina, NC % from Satellite 2" %%% 919-577-9882 % 'Ticket To The Moon' %%%% <yates@ieee.org> % *Time*, Electric Light Orchestra http://www.digitalsignallabs.com
Reply by ●November 14, 20072007-11-14
Randy Yates wrote:> "RobR" <masked@gmx.de> writes: > >> 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 > > Hi Robert, > > The problem as I see it is the definition of bandwidth. Most folks would > say that a signal that has a non-zero spectrum over -B to +B Hz has a > bandwidth of B Hz. I challenge that definition and say it has a spectrum > of 2B Hz. However, in making this new definition, you must realize that > when a signal is real, half of this bandwidth is unusable. On the other > hand, all the bandwidth is usable when the signal is complex. > > With that definition of bandwidth, I contend that sampling at Fs > samples/second gives you Fs Hz of bandwidth, from -Fs/2 to +Fs/2. When > the sampling is real, half that bandwidth is unusable. When it is > complex, the entire bandwidth is usable.There's no free lunch here. A complex sample counts for two real samples. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●November 14, 20072007-11-14
On Wed, 14 Nov 2007 08:17:24 -0600, RobR wrote:> 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, > RobertPerhaps this will clear up part of it: http://www.wescottdesign.com/articles/Sampling/sampling.html. I can't remember if it's in there, but the Nyquist-Shannon theorem says that the _overall_ sampling rate must be twice the maximum bandwidth -- it doesn't say that they all have to be taken the same way. You can use I and Q, or signal and 1st-derivative. In theory, you could sample a 1MHz signal by taking the signal value and its 2 million derivatives, once a second, but you may run into a few practical considerations if you try it. -- Tim Wescott Control systems and communications consulting http://www.wescottdesign.com Need to learn how to apply control theory in your embedded system? "Applied Control Theory for Embedded Systems" by Tim Wescott Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html
Reply by ●November 14, 20072007-11-14
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. My response to Robert was based the concept of the frequency domain that the Fourier transform provides us. In that domain, the relevent issue is the field over which the functional domain operates. And that is where the concept of bandwidth and its definition arises. In other words, I took Robert's question to be fundamentally a mathematical one, and I answered in the same spirit. Implementation issues had no part in it. 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. -- % Randy Yates % "So now it's getting late, %% Fuquay-Varina, NC % and those who hesitate %%% 919-577-9882 % got no one..." %%%% <yates@ieee.org> % 'Waterfall', *Face The Music*, ELO http://www.digitalsignallabs.com
Reply by ●November 14, 20072007-11-14
On Nov 14, 9:39 am, Randy Yates <ya...@ieee.org> wrote:> Jerry Avins <j...@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. > > My response to Robert was based the concept of the frequency domain that > the Fourier transform provides us. In that domain, the relevent issue is > the field over which the functional domain operates. And that is where > the concept of bandwidth and its definition arises. In other words, I > took Robert's question to be fundamentally a mathematical one, and I > answered in the same spirit. Implementation issues had no part in it. > > 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. > -- > % Randy Yates % "So now it's getting late, > %% Fuquay-Varina, NC % and those who hesitate > %%% 919-577-9882 % got no one..." > %%%% <ya...@ieee.org> % 'Waterfall', *Face The Music*, ELOhttp://www.digitalsignallabs.comBut, Randy, I hope you get this one fact and get it good: In the set of complex numbers, each single element counts as one sample, and a complex number is an ordered pair of elements. At least that's how complex_sample_Nyquist differs from complex_sample_Yates. The advantage of complex_sample_Nyquist is that it does not require the invention of the term "unusable bandwidth" to describe the missing information capacity of the real samples compared to complex_sample_Yates. Dale B. Dalrymple http://dbdimages.com http://stores.lulu.com/dbd
Reply by ●November 14, 20072007-11-14
"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, > RobertFirst, 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 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. When you need to deal with the negative frequencies, that's the time to deal with negative frequencies. 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. Then move to sampling bandpass signals, etc. Fred
Reply by ●November 14, 20072007-11-14
dbd <dbd@ieee.org> writes:> On Nov 14, 9:39 am, Randy Yates <ya...@ieee.org> wrote: >> Jerry Avins <j...@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. >> >> My response to Robert was based the concept of the frequency domain that >> the Fourier transform provides us. In that domain, the relevent issue is >> the field over which the functional domain operates. And that is where >> the concept of bandwidth and its definition arises. In other words, I >> took Robert's question to be fundamentally a mathematical one, and I >> answered in the same spirit. Implementation issues had no part in it. >> >> 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. >> -- >> % Randy Yates % "So now it's getting late, >> %% Fuquay-Varina, NC % and those who hesitate >> %%% 919-577-9882 % got no one..." >> %%%% <ya...@ieee.org> % 'Waterfall', *Face The Music*, ELOhttp://www.digitalsignallabs.com > > But, Randy, I hope you get this one fact and get it good: In the set > of complex numbers, each single element counts as one sample, and a > complex number is an ordered pair of elements.We must have studied an entirely different set of mathematics then. I come from texts such as: @book{herstein, title = "Topics in Algebra", author = "I.N. Herstein", publisher = "Wiley", edition = "second", year = "1975"} @book{durbin, title = "Modern Algebra: An Introduction", author = "John~R.~Durbin", edition = "3rd", year = "1992", publisher = "John Wiley \& Sons, Inc."} @book{artin, title = "Algebra", author = "Michael Artin", publisher = "Prentice Hall", year = "1991"} @book{hungerford title = "Algebra", author = "Thomas W. Hungerford", publisher = "Springer-Verlag", year = "1974"} @book{fraleigh, title = "A First Course in Abstract Algebra", author = "John B. Fraleigh", publisher = "Addison Wesley", edition = "second", year = "1976"} @book{bhattacharya, title = "Basic Abstract Algebra", author = "P.B.Bhattacharya and S.K.Jain and S.R.Nagpaul", publisher = "Cambridge", year = "1986"} In [herstein], for example, there is a single element in C called the multiplicative identity. In ordered-pair notation, that element is (1,0). The fact that it is an ordered pair of real values doesn't make it two elements of C - it is one "thing" - one element, of the set C of complex numbers. What texts have you studied? -- % Randy Yates % "Maybe one day I'll feel her cold embrace, %% Fuquay-Varina, NC % and kiss her interface, %%% 919-577-9882 % til then, I'll leave her alone." %%%% <yates@ieee.org> % 'Yours Truly, 2095', *Time*, ELO http://www.digitalsignallabs.com
Reply by ●November 14, 20072007-11-14
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? Best regards, Robert
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. > > My response to Robert was based the concept of the frequency domain that > the Fourier transform provides us. In that domain, the relevent issue is > the field over which the functional domain operates. And that is where > the concept of bandwidth and its definition arises. In other words, I > took Robert's question to be fundamentally a mathematical one, and I > answered in the same spirit. Implementation issues had no part in it. > > 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.I didn't think you might be confused, but I wanted to be careful that Robert didn't become so. As to whether (x, y) and/or a + iy represent one coordinate or two, that's a matter of semantics. BTW, how many bits are needed for a complex sample that stands in place of two real 12-bit samples? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
