Forums

Sub-Nyquist Sampling

Started by Roush, Craig Ryan \UMKC-Student\ September 3, 2003
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



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 >
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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: 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
> >
> >
> >
> > ------------------------ Yahoo! Groups Sponsor
> > ---------------------~-->
> > Buy Ink Cartridges or Refill Kits for Your HP, Epson, Canon or Lexmark
> > Printer at Myinks.com. Free s/h on orders $50 or more to the
> > US & Canada. http://www.c1tracking.com/l.asp?cidU11
> > http://us.click.yahoo.com/l.m7sD/LIdGAA/qnsNAA/wHYolB/TM
> > --------------------------
> > -------~->
> >
> > _____________________________________
> > Note: If you do a simple "reply" with your email client, only
> > the author of this message will receive your answer. You
> > need to do a "reply all" if you want your answer to be
> > distributed to the entire group.
> >
> > _____________________________________
> > About this discussion group:
> >
> > To Join:
> >
> > To Post:
> >
> > To Leave:
> >
> > Archives: http://www.yahoogroups.com/group/matlab
> >
> > More DSP-Related Groups: http://www.dsprelated.com/groups.php3
> >
> > ">http://docs.yahoo.com/info/terms/
> >
> > _____________________________________
> Note: If you do a simple "reply" with your email client, only the author of
this message will receive your answer. You need to do a "reply all" if you want
your answer to be distributed to the entire group.
>
> _____________________________________
> About this discussion group:
>
> To Join:
>
> To Post:
>
> To Leave:
>
> Archives: http://www.yahoogroups.com/group/matlab
>
> More DSP-Related Groups: http://www.dsprelated.com/groups.php3
>
> ">http://docs.yahoo.com/info/terms/



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 can be
obtained because \tau is known. As long as the input frequency are sufficiently
separated, 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


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
> >