FPGA Central

Hello,

there is a question that bugs me for quite a long time:

You can read about Nyquist constrain online, that to reconstruct al
frequencies within a signal, it has to be sampled with at least twice th
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 th
signal,
since also the negative frequencies are present, so the bandwidth of th
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 ma
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 th
information in contrast to a signal in "as it is" representation (one rea
stream)?
In fact, i need a clarification between nyquist adopted to a real signa
and nyquist adopted to a complex IQ representation signal.

Best regards,
Robert

"RobR" <[email protected]> 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.
--
% Randy Yates % "Ticket to the moon, flight leaves here today
%% Fuquay-Varina, NC % from Satellite 2"
%%% 919-577-9882 % 'Ticket To The Moon'
%%%% <[email protected]> % *Time*, Electric Light Orchestra
http://www.digitalsignallabs.com

Randy Yates wrote:
> "RobR" <[email protected]> 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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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,
> Robert

Perhaps this will clear up part of it:
http://www.wescottdesign.com/article.../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

Jerry Avins <[email protected]> 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..."
%%%% <[email protected]> % 'Waterfall', *Face The Music*, ELO
http://www.digitalsignallabs.com

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.

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

"RobR" <[email protected]> wrote in message
news:[email protected]..
> 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
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

dbd <[email protected]> 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."
%%%% <[email protected]> % 'Yours Truly, 2095', *Time*, ELO
http://www.digitalsignallabs.com

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 spectra
information that real sampling provides.
So one complex sample holds information of two real samples at twice th
complex sampling rate.

One could also, maybe more theoreticaly, sample signal and firs
derivative for the same effect.

If I take signal, first and second derivative, then I would need at leas
third of nyquist? This steps could be carried on to infinity derivatives
where I would need zero samples (asymptotically), right?



Best regards,
Robert

Randy Yates wrote:
> Jerry Avins <[email protected]> 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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Randy Yates wrote:

> Jerry Avins <[email protected]> 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

Jerry Avins <[email protected]> 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 %
%%%% <[email protected]> % 'Eldorado Overture', *Eldorado*, ELO
http://www.digitalsignallabs.com

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:[email protected]..
>
>
> > 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

Randy Yates wrote:
> Jerry Avins <[email protected]> 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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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 B. Dalrymple

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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?

Fred Marshall wrote:
> "RobR" <[email protected]> wrote in message
> news:[email protected]..
>> 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

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 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.
> ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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

dbd wrote:

...

> I wouldn't bet that any scheme couldn't be patented, but that says
> more about the patent system than the scheme.

Curses! Foiled again!

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

"Ron N." <[email protected]> writes:

> 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?

The question was intended to be within the context of an ideal
mathematical system using infinite-precision quantization, i.e.,
time-quantization only.
--
% Randy Yates % "Rollin' and riding and slippin' and
%% Fuquay-Varina, NC % sliding, it's magic."
%%% 919-577-9882 %
%%%% <[email protected]> % 'Living' Thing', *A New World Record*, ELO
http://www.digitalsignallabs.com

"Steve Underwood" <[email protected]> wrote in message
news:fhg74v$o3f$[email protected]..
> Fred Marshall wrote:
>> "RobR" <[email protected]> wrote in message
>> news:[email protected]..
>>> 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

Complex is a special case too, isn't it? i.e. what about i,j,k and other
higher order spaces? Take your pick. So, not so puzzling after all.

Fred

RobR wrote:
(snip)

> 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?

I would have said one, but zero is pretty close to one.

I once thought that the Weather Channel should, in addition to the
current temperature also show the derivative. Maybe also the
second derivative would be nice. And the third...

-- glen

Steve Underwood <[email protected]> wrote in
news:fhg74v$o3f$[email protected]:

> 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
>

There is a book that does a great job of explaining why imaginary numbers
are no more imaginary as real numbers. It's called "Asimov on Numbers".

Its a great book for people who have learned way to much math by rote, and
never really understood why. I gave my copy to a niece who was taking her
third high school math course.

Of course, what do we know; we all believe in imaginary numbers and
negative frequencies, etc.....

Al Clark
Danville Signal Processing, Inc.