On 24 Aug 2003 07:28:47 -0700, germain1_fr@yahoo.fr (seb) wrote:
(snipped)
>>
>> Hi Jer,
>>
>> Yep. To give some numbers for seb (original poster);
>> if the original analog bandpass signal had a bandwidth of
>> roughly 10 kHz, we know that normal real (bandpass)
>> sampling would require a sample rate just greater than
>> 20K samples/sec if you used a single A/D. With the
>> quad sampling scheme (also called "complex down-conversion"),
>> two A/D converters would each be running at a rate just
>> greater than 20K/2, or 10K samples/sec.
>>
>> In both case, 20K samples/sec would be generated.
>> In both cases we're satisfying the Nyquist Theorem
>> (named by the great Claude Shannon who died last
>> year with Alzheimer's disease.)
>>
>> [-Rick-]
>
>Thinks,
>
>It is well understood (excuse my poor english, technical english is
>easy to read but wrinting is more difficult).
>
>There is something else with this paper:
>There is an example of application. When plug the in-phase and
>quadrature-phase to the horizontaly and vertical input of a scope, we
>observe a circle. But the original input is not specify (the
>modulation is set to 1 Hz). I think that this input signal must be of
>a particular form. But which ?
>
>For example, if i take a single cosinus (with 0 phase : A*cos(2*pi*f*t
>+ 0) with the same frequency as the modulation frequence, the in-phase
>signal will be A and the quadrature-phase will be 0. So the scope
>displays just one motionless point (A, 0).
Hi,
That oscillosope example was my effort to give physical meaning
to the notion of a complex exponential signal, like e^j*2*pi*Fo*t,
and physical meaning to the 'j' operator.
Those two real analog signals applied to the oscilloscope were
merely a real cosine wave and a real sinewave of the same
frequency. That is, both real signals were sinusoids of the same
frequency, but they are exactly 90-degrees (pi/2 radians) out
of phase with each other.
Signal 1: A*cos(2*pi*f*t) = A*sin(2*pi*f*t + pi/2)
and Signal 2: A*sin(2*pi*f*t)
There were *NO* "modulation" issues associated with
that example.
Hope that makes some sense.
[-Rick-]
Reply by seb●August 24, 20032003-08-24
ricklyon@REMOVE.onemain.com (Rick Lyons) wrote in message news:<3f474377.69195531@news.west.earthlink.net>...
> On Fri, 22 Aug 2003 13:52:20 -0400, Jerry Avins <jya@ieee.org> wrote:
>
> >seb wrote:
> >>
> >> Hy,
> >>
> >> I recently read this good article of Richard Lyons but there is
> >> something i can not understand.
> >>
> >> Does someone could help me understand this difficulty ?
> >>
> >> It it says :
> >> "Each A/D converter operates at HALF the sampling rate of standard
> >> real-sampling" !!!
> >>
> >> How does it work ? It seems that using half the sampling rate of
> >> standard real-sampling, we obtain the same frequency definition as for
> >> a full sampling rate of a "normal" signal ?
> >
> >What matters is how many samples are collected in a given time. The
> >number of samples collected by a pair of A/Ds acting alternately in a
> >single analog signal (or simultaneously on a quadrature-shifted pair) is
> >the same as the number collected by a single A/D running twice as fast.
> >
> >Jerry
>
> Hi Jer,
>
> Yep. To give some numbers for seb (original poster);
> if the original analog bandpass signal had a bandwidth of
> roughly 10 kHz, we know that normal real (bandpass)
> sampling would require a sample rate just greater than
> 20K samples/sec if you used a single A/D. With the
> quad sampling scheme (also called "complex down-conversion"),
> two A/D converters would each be running at a rate just
> greater than 20K/2, or 10K samples/sec.
>
> In both case, 20K samples/sec would be generated.
> In both cases we're satisfying the Nyquist Theorem
> (named by the great Claude Shannon who died last
> year with Alzheimer's disease.)
>
> [-Rick-]
Thinks,
It is well understood (excuse my poor english, technical english is
easy to read but wrinting is more difficult).
There is something else with this paper:
There is an example of application. When plug the in-phase and
quadrature-phase to the horizontaly and vertical input of a scope, we
observe a circle. But the original input is not specify (the
modulation is set to 1 Hz). I think that this input signal must be of
a particular form. But which ?
For example, if i take a single cosinus (with 0 phase : A*cos(2*pi*f*t
+ 0) with the same frequency as the modulation frequence, the in-phase
signal will be A and the quadrature-phase will be 0. So the scope
displays just one motionless point (A, 0).
Reply by Rick Lyons●August 23, 20032003-08-23
On Fri, 22 Aug 2003 13:52:20 -0400, Jerry Avins <jya@ieee.org> wrote:
>seb wrote:
>>
>> Hy,
>>
>> I recently read this good article of Richard Lyons but there is
>> something i can not understand.
>>
>> Does someone could help me understand this difficulty ?
>>
>> It it says :
>> "Each A/D converter operates at HALF the sampling rate of standard
>> real-sampling" !!!
>>
>> How does it work ? It seems that using half the sampling rate of
>> standard real-sampling, we obtain the same frequency definition as for
>> a full sampling rate of a "normal" signal ?
>
>What matters is how many samples are collected in a given time. The
>number of samples collected by a pair of A/Ds acting alternately in a
>single analog signal (or simultaneously on a quadrature-shifted pair) is
>the same as the number collected by a single A/D running twice as fast.
>
>Jerry
Hi Jer,
Yep. To give some numbers for seb (original poster);
if the original analog bandpass signal had a bandwidth of
roughly 10 kHz, we know that normal real (bandpass)
sampling would require a sample rate just greater than
20K samples/sec if you used a single A/D. With the
quad sampling scheme (also called "complex down-conversion"),
two A/D converters would each be running at a rate just
greater than 20K/2, or 10K samples/sec.
In both case, 20K samples/sec would be generated.
In both cases we're satisfying the Nyquist Theorem
(named by the great Claude Shannon who died last
year with Alzheimer's disease.)
[-Rick-]
Reply by Jerry Avins●August 22, 20032003-08-22
seb wrote:
>
> Hy,
>
> I recently read this good article of Richard Lyons but there is
> something i can not understand.
>
> Does someone could help me understand this difficulty ?
>
> It it says :
> "Each A/D converter operates at HALF the sampling rate of standard
> real-sampling" !!!
>
> How does it work ? It seems that using half the sampling rate of
> standard real-sampling, we obtain the same frequency definition as for
> a full sampling rate of a "normal" signal ?
What matters is how many samples are collected in a given time. The
number of samples collected by a pair of A/Ds acting alternately in a
single analog signal (or simultaneously on a quadrature-shifted pair) is
the same as the number collected by a single A/D running twice as fast.
Jerry
--
Engineering is the art of making what you want from things you can get.
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Reply by One Usenet Poster●August 22, 20032003-08-22
seb wrote:
> Hy,
>
> I recently read this good article of Richard Lyons but there is
> something i can not understand.
>
>
> Does someone could help me understand this difficulty ?
>
> It it says :
> "Each A/D converter operates at HALF the sampling rate of standard
> real-sampling" !!!
>
> How does it work ? It seems that using half the sampling rate of
> standard real-sampling, we obtain the same frequency definition as for
> a full sampling rate of a "normal" signal ?
The Nyquist sample rate for real-valued signals is twice the highest
frequency. For complex signals, the Nyquist sample rate is *equivalent*
to the highest frequency.
OUP
Reply by seb●August 21, 20032003-08-21
Hy,
I recently read this good article of Richard Lyons but there is
something i can not understand.
Does someone could help me understand this difficulty ?
It it says :
"Each A/D converter operates at HALF the sampling rate of standard
real-sampling" !!!
How does it work ? It seems that using half the sampling rate of
standard real-sampling, we obtain the same frequency definition as for
a full sampling rate of a "normal" signal ?