"Quadrature signals : Complex, But not complicated" by Richard Lyons

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 ?

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

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.
�����������������������������������������������������������������������

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-]
 

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).

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-]