Here's Craig's reply, I'm assuming he meant to post it to the
group... > -----Original Message----- > From: Jeff Brower [mailto:] > Sent: Thu, September 04, 2003 11:41 AM > To: Egler, Mark > Cc: Craig Roush, Craig Ryan (UMKC-Student); > Subject: Re: [matlab] Sub-Nyquist Sampling > Mark- > > If you get a private reply from Craig about the reference > URL, can you post? It's an > interesting topic. > > Jeff Brower > system engineer > Signalogic > > "Egler, Mark" wrote: > > > > Craig, > > > > Could you elaborate on this a little? It sounds too good to > be true. I would > > have said it's impossible to distinguish the 8 aliased > frequencies in your > > example, before I read the paragraph about phase > differences, etc. What > > "delayed and undelayed signals" are you referring to? Can > you provide a > > reference? > > > > I thought I understood sub-Nyquist sampling to mean: > sampling a band-pass > > signal with part or all of its passband above the Nyquist > frequency (Fs/2), > > with the requirement that each frequency in the analog > pass-band maps > > uniquely to a frequency in the sampled base-band. If any > two frequencies in > > the analog pass-band both alias to the same base-band > frequency, you've lost > > the information to discriminate between them unambiguously. > Right? And, of > > course, the total bandwidth has to be less Fs/2. > > > > I guess you could have multiple analog pass-bands that all map to > > mutually-exclusive regions of the base-band, but your > example does not fit > > that model. > > > > Mark > > > -----Original Message----- > From: Roush, Craig Ryan (UMKC-Student) [mailto:] > Sent: Thu, September 04, 2003 9:41 AM > To: Egler, Mark > Subject: RE: [matlab] Sub-Nyquist Sampling > Mark, > This analysis is coming from "Digital Techniques for Wideband > Receivers" by James Tsui 1995 > ISBN 0-89006-808-9 > > To give you a bit more background: > > Consider a instantaneous frequency measurement receiver > (IFM). The main portion of this receiver is a correlator, > the input to the correlator is a delayed and undelayed copy > of the input signal. Thus you end up w/ > E = \sin(2 \pi f \tau ) and F = \cos(2 \pi f \tau) > \tau is the delay time and f is frequency. > > the freq can be obtained by \theta = \tan^-1 ( \frac{E}{F} ) > = 2 \pi f \tau, > however there is the following restriction \theta < 2 \pi, if > \theta > 2\pi, there becomes this "ambiguity" problem. The > maximum unambiguous bandwidth obtained is \Delta B = \frac{2 > \pi }{2 \pi \tau} = \frac{1}{\tau}. > > This only relation only limits the bandwidth , but not the > center frequency, thus, the center freq can be any value, ie > ( if \tau = .5 ns, the unambiguous bandwidth is 2 GHz, the > freq range can be either 0 to 2 GHz or 2 to 4 GHz, or any > other values as long as the bandwidth is 2 GHz. > > Now for the sub-Nyquist approach the input signal is divided > into 2 paths, and undelayed and one and a delayed one. > > After digitization, the digitized outputs can be processed > through an FFT operation. Let X_{ru}(k) and X_{iu}(k) > represent the real and imaginary components of the undelayed > case and let X_{rd}(k) and X_{id}(k) represent the delayed case. > The amplitude is obvious found from > > X_u (k) = [X_{ru}(k)^2 and X_{iu}(k)^2 ]^0.5. > > The delayed path has the same amplitude components, X_u(k_m) > represents the maximum amplitude of the frequency component > from the undelayed path, thus, X_u(k_m) can represent the > input frequency. The phase difference between the delay and > undelayed path can be written as : > \theta = \theta_d - \theta_u = 2\pi f \tau > where > \theta_d = \tan^-1(\frac{X_{id}(k_m)}{X_{rd}(k_m)} ) > \theta_u = \tan^-1(\frac{X_{iu}(k_m)}{X_{ru}(k_m)} ) > > From the phase difference \theta, the frequency of the input > signal canb be obtained because \tau is known. As long as > the input frequency are sufficeiently sperated, the input > freqs can be identified by observing those frequency bins > whose magnitude exceed a threshold. > > The Tsui leads into the aforementioned example > > hope that helps you understand, I pretty much copied straight > from chapter 5 of the book for you, the notation is in LaTeX > in case you have trouble reading it. > > -Craig > > > ________________________________ > > From: Egler, Mark [mailto:] > Sent: Thu 9/4/2003 8:38 AM > To: Roush, Craig Ryan (UMKC-Student); > Subject: RE: [matlab] Sub-Nyquist Sampling > > Craig, > > Could you elaborate on this a little? It sounds too good to > be true. I would > have said it's impossible to distinguish the 8 aliased > frequencies in your > example, before I read the paragraph about phase differences, > etc. What > "delayed and undelayed signals" are you referring to? Can you > provide a > reference? > > I thought I understood sub-Nyquist sampling to mean: sampling > a band-pass > signal with part or all of its passband above the Nyquist > frequency (Fs/2), > with the requirement that each frequency in the analog pass-band maps > uniquely to a frequency in the sampled base-band. If any two > frequencies in > the analog pass-band both alias to the same base-band > frequency, you've lost > the information to discriminate between them unambiguously. > Right? And, of > course, the total bandwidth has to be less Fs/2. > > I guess you could have multiple analog pass-bands that all map to > mutually-exclusive regions of the base-band, but your example > does not fit > that model. > > Mark > > > -----Original Message----- > > From: Roush, Craig Ryan (UMKC-Student) [mailto:] > > Sent: Wed, September 03, 2003 10:30 AM > > To: > > Subject: [matlab] Sub-Nyquist Sampling > > > > > > I am having some difficulty understanding the sub-nyquist sampling > > theory. > > > > For example: > > if you have an ADC that can only operate at 250MHz, then the max > > bandwidth is 125MHz. However, if I wanted 1000MHz and I want to use > > the same ADC, the bandwidth would "fold" into 125MHz output > band. So > > I could potentially have 8 solutions if I injected in a > 40MHz signal. > > 40, 210, 290, 460, 540, 710, 790, and 960MHz. > > > > What I do not follow is how we can determine which zone the > signal is > > in. From my reading it appears that you take the phase difference > > between the delayed and undelayed signal input, and since > you know the > > what the delay is, you can extract the true frequency. The delayed > > and undelayed phase can be determined straight from the > arctan of the > > ratio of I and Q data. And you can determine the max unambiguous > > bandwidth by 1/tau (where tau is the delay). > > > > I am just not clear why this works, and how to prove that > this works, > > I realize there are some fine points about this, but I want to > > understand the basic concept then move forward with the remaining > > concepts. > > > > Thanks, > > -Craig > > |