Basic Signal Processing Question

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Mark wrote:

> The L and R are time division multiplexed by sampling each at 38 kHz.
> The L and R samples are interleaved and FM modulated onto the main RF
> carrier.  This is spectrally equivalent to what you described.  This is
> two different descriptions of the same thing, one description looking
> at the frequency domain, and one description looking at the time
> domain.   Actually there are two different design approaches to an FM
> MPX encoder, the TDM approach and the L-R matrix approach, both create
> the same output signal and are fully compatable.

I had thought that the subcarrier amplitude wasn't equal to the mono 
signal, but it seems that it is.  I believe, then, that it could be 
decoded in the way you say, but it would be difficult to encode that way.

After encoding it you would have to send it through a filter that
was fairly flat to 15kHz, dropped off pretty fast by 19kHz, back
up to the original level by 23kHz, flat from there to 53kHz, and
then fall off by 57kHz.  Maybe not so bad as a digital filter but
not at all easy to get right as an analog filter.  Then add the properly 
filtered and phased 19kHz pilot.

The decoder would at least need to filter anything over 57kHz
where SCA subcarriers could reside.  It would also need to filter
the pilot and generate the right phase for the sampling signal.

-- glen

Jerry Avins wrote:
> Mark wrote:
> > The L and R are time division multiplexed by sampling each at 38 kHz.
> > The L and R samples are interleaved and FM modulated onto the main RF
> > carrier.  This is spectrally equivalent to what you described.  This is
> > two different descriptions of the same thing, one description looking
> > at the frequency domain, and one description looking at the time
> > domain.   Actually there are two different design approaches to an FM
> > MPX encoder, the TDM approach and the L-R matrix approach, both create
> > the same output signal and are fully compatable.
>
> There is no sampling. 38 KHz is the frequency of the subcarrier that
> carries the difference signal. The sum signal is transmitted as in a
> mono signal, so that a mono signal is received as such and a mono
> receiver receives L+R. That's what's meant by compatibility. After the
> difference signal is recovered from by demodulating the subcarrier (with
> the aid of the 19 KHz pilot tone), the two signals are matrixed to
> create L and R. If the pilot tone is too weak, there is no matrixing.
>
> Which standard have you been reading?
>
> Jerry
> --


The 38 kHz sampling and the L-R matrix are different but EQUIVALENT
methods.

Read this then we'll talk again.

http://members.tripod.com/~transmitters/stereo.htm

Mark


> Engineering is the art of making what you want from things you can get.
> =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF

Mark wrote:
> 
> Jerry Avins wrote:
> 
>>Mark wrote:
>>
>>>The L and R are time division multiplexed by sampling each at 38 kHz.
>>>The L and R samples are interleaved and FM modulated onto the main RF
>>>carrier.  This is spectrally equivalent to what you described.  This is
>>>two different descriptions of the same thing, one description looking
>>>at the frequency domain, and one description looking at the time
>>>domain.   Actually there are two different design approaches to an FM
>>>MPX encoder, the TDM approach and the L-R matrix approach, both create
>>>the same output signal and are fully compatable.
>>
>>There is no sampling. 38 KHz is the frequency of the subcarrier that
>>carries the difference signal. The sum signal is transmitted as in a
>>mono signal, so that a mono signal is received as such and a mono
>>receiver receives L+R. That's what's meant by compatibility. After the
>>difference signal is recovered from by demodulating the subcarrier (with
>>the aid of the 19 KHz pilot tone), the two signals are matrixed to
>>create L and R. If the pilot tone is too weak, there is no matrixing.
>>
>>Which standard have you been reading?
>>
>>Jerry
>>--
> 
> 
> 
> The 38 kHz sampling and the L-R matrix are different but EQUIVALENT
> methods.
> 
> Read this then we'll talk again.
> 
> http://members.tripod.com/~transmitters/stereo.htm

That's an interesting page. I find a stated advantage of FM over AM, 
"More distance on the dial between FM stations", a bit nerdy, but a 
small point. What you loosely call "sampling" is called in the article 
"time-delay multiplexing" and "chopping", hardly the same thing. 
Moreover, the paper describes a stereo modulator, but I was describing 
the demodulation process. My mistake there; you had mentioned the 
encoder before I chimed in.

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

Jerry Avins wrote:
(snip regarding FM stereo multiplexing)

>> The 38 kHz sampling and the L-R matrix are different but 
 >> EQUIVALENT methods.

They are equivalent if you ignore the filtering problem.

It might be that today one could generate a nice digital filter
to do it, but that would not have been possible when the standard
was being written.

>> Read this then we'll talk again.

>> http://members.tripod.com/~transmitters/stereo.htm

> That's an interesting page. I find a stated advantage of FM over AM, 
> "More distance on the dial between FM stations", a bit nerdy, but a 
> small point. What you loosely call "sampling" is called in the article 
> "time-delay multiplexing" and "chopping", hardly the same thing. 
> Moreover, the paper describes a stereo modulator, but I was describing 
> the demodulation process. My mistake there; you had mentioned the 
> encoder before I chimed in.

I find it more believable as a decoder.  The filtering required for the
encoder would be very hard to do, except possibly with a digital filter 
at a much higher frequency.  Well, for a decoder the filtering isn't
easy, either.  You have to at least filter everything above 53kHz
sharp enough to keep the tails of the SCA subcarrier centered at 67kHz
away.  SCA is an FM subcarrier so it can have long tails.

The inputs to the encoder must be filtered to less than 15kHz, so
obviously sampling at 38kHz would be enough.  Balanced modulators were
pretty easy to make, even with vacuum tubes.  Sharp filters with a flat
passband were not.

For comparison purposes there are two descriptions of the NTSC color 
subcarrier, and the choice of subcarrier frequency.  In one there is a 
180 degree phase shift between successive lines and for a given line 
between successive frames.  This tends to reduce the visibility of the 
subcarrier on a monochrome receiver.  In the other description, assuming 
relatively low horizontal spatial frequencies in the video image, the 
luminance signal has peaks at multiples of the line rate and the 
chrominance signal has peaks in between those.

As with the FM stereo case, one is time domain and one frequency domain,
but the result is the same.

-- glen

Jerry Avins wrote:
> John Monro wrote:
> 
>   ...
> 
>> Venkat,
>>
>> As you have correctly identified, the DC component of the sampling 
>> waveform is responsible for the baseband component in the sampled 
>> signal.  However, you are incorrect in thinking that multiplying by 
>> the usual sampling waveform is in any way similar to multiplying by a 
>> square wave.
> 
> 
>   ...
> 
> John, unless by "DC component, you mean that portion of the signal 
> during which a sample is being taken, I don't know what you mean. If you 
> do mean that, I think you're wrong. Consider a successive-approximation 
> converter clocked by a square wave at 600 KHz. It takes 12 cycles to 
> convert a 10-bit output and two more cycles to sample and hold. It 
> provides a fresh sample every 20 microseconds.
> 
> The final value in the holding capacitor represents a window in time of 
> no more than a few nanoseconds. For high-speed use, it must be in the 
> picosecond range. The actual clock waveform has no influence on the 
> operation except that a clean waveform with sharp edges is needed to 
> minimize sampling jitter.
> 
> Maybe there's something I just don't get. If so, please enlighten me.
> 
> Jerry


Jerry,
I thought that remark might be controversial, but unfortunately I was 
not able to check for responses until just now, so apologies for the 
delay.

By 'DC component' I mean the DC level you would measure if you applied 
the sampling waveform to a LP filter having a corner frequency a long 
way below the sampling rate.

I don't know what point you are making when you mention the ADC, 
particularly as this process takes place after the sampling stage, and 
the discussion is about the spectrum of the signal going into this stage.

I think there may be some difference in our interpretation of what 
Vancat is using as his sampling waveform.  As far as his square-wave is 
concerned, I re-read his original postings and I am sure that his query 
relates to the use of a square-wave as the actual sampling waveform, and 
not merely as the clock driving a conventional, (nanosecond 
aperture-time) sampler circuit. Note that his specifically mentions 
multiplying his input sinusoidal signal by the square-wave, and 
expresses surprise (!!) when there is no baseband signal.

In his second post he said:
	
	"Let me clarify.

       	I am sampling a sine wave(Fin) and the clock used for 
sampling 			is a square wave(Fclk) (or pulse train with 50% duty 
cycle.Not 			an impulse train). The 50% duty cycle square wave clock 
do not 			have any DC components.

      	Now we multiply these two waveforms in time domain, we do not 		 
have the baseband signal!! The only components that I am getting 	are 
Fclk+/-Fin. The output spectrum does not have Fin at all. I 		know this 
is not true in the real world.

	What am I missing."

I appeared to me that Vencat was puzzled by the fact that when he used a 
square-wave with zero DC-component as his sampling signal he did not get 
any baseband component in the sampled signal. My answer was an attempt 
to explain that phenomenon.

I think that some of the others in this thread may have have focussed on 
the square-wave aspect of the question and assumed that Vencat was 
sampling using a very wide pulse of 50 per cent duty cycle, having 
normalised values 1.000, 0.000, 1.000, 0.000  etc.  That is, a 
square-wave of amplitude 0.500 peak, with a DC offset of +0.500, and for 
this situation   I would have no issue with their comments regarding the 
sinc() falloff etc.

That said, Vancat quite clearly emphasises that his sampling signal has 
no DC component at all, so there is no offset, and so the sampling 
waveform has values of: 1.000, -1.000, 1.000, -1.000  etc.  My previous 
comments and the following argument are based on this interpretation of 
the problem.

My argument is this:

1.  The sampling process can be regarded as multiplication in the 
time-domain of an input signal by the sampling waveform, frequency fs.

2,  A typical or 'usual' sampling waverform has an amplitude of 1.000 
for a few nanoseconds and 0.000 for the rest of the cycle.

3.   Vencat's sampling waveform is a square-wave that is 1.000 for half 
the cycle and -1.000 for rest of the cycle.

4.  The 'usual' sampling waveform has frequency components of DC, fs, 
2fs, 3fs, ...  (All the harmonics, including DC.)
5.  Vancat's square-wave sampling waveform has frequency components of 
fs, 3fs, 5fs ... (No DC component, and no even harmonics.)

6.  When the sampling process multiplies the input-signal by the 
sampling-waveform we can separately consider the effect of the input 
signal on each of the sampling waveform's frequency components.
We can say that each of these components becomes multiplied (or 
'modulated') by the input signal in the course of the sampling process, 
producing a range of products in the sampled signal.

7. For any sampling waveform: If the input signal happens to be a simple 
sine-wave of frequency fin, components are generated that are fin Hz 
above and below the sampling waveform's frequency components.  (In the 
more usual case where the input signal is a broad-band signal we get 
upper and lower 'sidebands' each side of the sampling waveform's 
frequency components.)

8.  For a 'usual' sampling waveform, it follows that the frequency 
components at the output of the sampling process are:
(0 + fin), (fs +/- fin), (2fs +/- fin) ...
(Each of the terms above also has its negative-frequency counterpart, 
but for brevity this is not shown.)
The (0 + fin)component is the baseband signal that we 			must preserve, 
and that  (fs - fin) is the cursed alias signal 			that ultimately gets 
filtered out after the A/D stage.

9.  For Vencat's square-wave sampling waveform, the frequency components 
at the output of the sampling process are:
(fs +/- fin), (3fs +/- fin) ...
Notice that there is no baseband signal, but there is plenty of alias 
signal :-)

So Jerry (and the group) I hope you will agree that the DC component of 
the sampling waveform is in fact responsible for the baseband component 
in the sampled signal.



Regards,
John

John Monro wrote:

(snip regarding DC component)

> I thought that remark might be controversial, but unfortunately I was 
> not able to check for responses until just now, so apologies for the delay.

> By 'DC component' I mean the DC level you would measure if you applied 
> the sampling waveform to a LP filter having a corner frequency a long 
> way below the sampling rate.

> I don't know what point you are making when you mention the ADC, 
> particularly as this process takes place after the sampling stage, and 
> the discussion is about the spectrum of the signal going into this stage.

> I think there may be some difference in our interpretation of what 
> Vancat is using as his sampling waveform.  As far as his square-wave is 
> concerned, I re-read his original postings and I am sure that his query 
> relates to the use of a square-wave as the actual sampling waveform, and 
> not merely as the clock driving a conventional, (nanosecond 
> aperture-time) sampler circuit. Note that his specifically mentions 
> multiplying his input sinusoidal signal by the square-wave, and 
> expresses surprise (!!) when there is no baseband signal.

> In his second post he said:

>     "Let me clarify.
> 
>   I am sampling a sine wave(Fin) and the clock used for 
> sampling is a square wave(Fclk) (or pulse train with 50% 
> duty cycle.Not an impulse train). The 50% duty cycle square 
> wave clock do not have any DC components.
> 
>  Now we multiply these two waveforms in time domain, we do 
> not have the baseband signal!! The only components that I am 
> getting are Fclk+/-Fin. The output spectrum does not have Fin at 
> all. I know this is not true in the real world.

I am not so sure that I understand all of the discussion, but multiply 
by a square wave and filter is a popular laboratory instrument called
a lock-in amplifier.   A waveform either generated externally or by the
lock-in amplifier is used to modulate some part of the signal, and the 
result is then detected with both an in-phase and quadrature component.

It can in some cases be used to find a small signal with a large 
background noise if the real signal can be appropriately modulated.

In one that I worked on once a sinusoidal voltage was supplied to
one terminal of a device, and the voltage across a small resistor 
measured at another terminal.  With the device considered as a capacitor 
in parallel with a resistor the quadrature signal measures the 
capacitance.  The capacitance may be much smaller than the capacitance 
to ground in the coax cable, yet it is very easy to measure.
There can also be a large amount of random noise that will tend
to average out if one waits long enough.

Part of the calibration should be removing the DC component from the
square wave.  If a DC voltage is applied to the input one would then
adjust until the output, suitably filtered, went to zero.

-- glen

glen herrmannsfeldt wrote:
> John Monro wrote:
> 
> (snip regarding DC component)
> 
>> I thought that remark might be controversial, but unfortunately I was 
>> not able to check for responses until just now, so apologies for the 
>> delay.
> 
> 
>> By 'DC component' I mean the DC level you would measure if you applied 
>> the sampling waveform to a LP filter having a corner frequency a long 
>> way below the sampling rate.
> 
> 
>> I don't know what point you are making when you mention the ADC, 
>> particularly as this process takes place after the sampling stage, and 
>> the discussion is about the spectrum of the signal going into this stage.
> 
> 
>> I think there may be some difference in our interpretation of what 
>> Vancat is using as his sampling waveform.  As far as his square-wave 
>> is concerned, I re-read his original postings and I am sure that his 
>> query relates to the use of a square-wave as the actual sampling 
>> waveform, and not merely as the clock driving a conventional, 
>> (nanosecond aperture-time) sampler circuit. Note that his specifically 
>> mentions multiplying his input sinusoidal signal by the square-wave, 
>> and expresses surprise (!!) when there is no baseband signal.
> 
> 
>> In his second post he said:
> 
> 
>>     "Let me clarify.
>>
>>   I am sampling a sine wave(Fin) and the clock used for sampling is a 
>> square wave(Fclk) (or pulse train with 50% duty cycle.Not an impulse 
>> train). The 50% duty cycle square wave clock do not have any DC 
>> components.
>>
>>  Now we multiply these two waveforms in time domain, we do not have 
>> the baseband signal!! The only components that I am getting are 
>> Fclk+/-Fin. The output spectrum does not have Fin at all. I know this 
>> is not true in the real world.
> 
> 
> I am not so sure that I understand all of the discussion, but multiply 
> by a square wave and filter is a popular laboratory instrument called
> a lock-in amplifier.   A waveform either generated externally or by the
> lock-in amplifier is used to modulate some part of the signal, and the 
> result is then detected with both an in-phase and quadrature component.
> 
> It can in some cases be used to find a small signal with a large 
> background noise if the real signal can be appropriately modulated.
> 
> In one that I worked on once a sinusoidal voltage was supplied to
> one terminal of a device, and the voltage across a small resistor 
> measured at another terminal.  With the device considered as a capacitor 
> in parallel with a resistor the quadrature signal measures the 
> capacitance.  The capacitance may be much smaller than the capacitance 
> to ground in the coax cable, yet it is very easy to measure.
> There can also be a large amount of random noise that will tend
> to average out if one waits long enough.
> 
> Part of the calibration should be removing the DC component from the
> square wave.  If a DC voltage is applied to the input one would then
> adjust until the output, suitably filtered, went to zero.
> 
> -- glen
> 


Glen,

It is interesting that with the lock-in amplifier the signal frequency 
(fin) is equal to the 'sampling frequency' (fs).  The signal component 
is aliased down to (fs - fin)= 0 Hz, and measured as a DC level .

If we compare this arrangement with Vencat's use of the square-wave as a 
sampling waveform, his 'baseband signal' would be those frequencies 
below fs/2.  These frequencies are very strongly rejected by the lock-in 
amplifier, verifying what V. observed, that there is no baseband 
component in the sampled signal when you sample with a square wave.

Thanks for the example.

Regards,
John

John,

Your thoughtful analysis inspired me to think a bit more, but we still 
diverge on many points. I'll address the differences as they arise.

Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
John Monro wrote:
> Jerry Avins wrote:
> 
>> John Monro wrote:
>>
>>   ...
>>
>>> Venkat,
>>>
>>> As you have correctly identified, the DC component of the sampling 
>>> waveform is responsible for the baseband component in the sampled 
>>> signal.  However, you are incorrect in thinking that multiplying by 
>>> the usual sampling waveform is in any way similar to multiplying by a 
>>> square wave.
>>
>>
>>
>>   ...
>>
>> John, unless by "DC component, you mean that portion of the signal 
>> during which a sample is being taken, I don't know what you mean. If 
>> you do mean that, I think you're wrong. Consider a 
>> successive-approximation converter clocked by a square wave at 600 
>> KHz. It takes 12 cycles to convert a 10-bit output and two more cycles 
>> to sample and hold. It provides a fresh sample every 20 microseconds.
>>
>> The final value in the holding capacitor represents a window in time 
>> of no more than a few nanoseconds. For high-speed use, it must be in 
>> the picosecond range. The actual clock waveform has no influence on 
>> the operation except that a clean waveform with sharp edges is needed 
>> to minimize sampling jitter.
>>
>> Maybe there's something I just don't get. If so, please enlighten me.
>>
>> Jerry

I agree that sampling is an on/off process and as such can be said to 
have a DC component. That assumes a mathematical model, but I see a 
physical one.

> Jerry,
> I thought that remark might be controversial, but unfortunately I was 
> not able to check for responses until just now, so apologies for the delay.
> 
> By 'DC component' I mean the DC level you would measure if you applied 
> the sampling waveform to a LP filter having a corner frequency a long 
> way below the sampling rate.

The sampling waveform I envision controls a switch. Depending on the 
switch design, it may go from 1 to 0, +1 to -1, +12 to +10, or any other 
two-level waveform.

> I don't know what point you are making when you mention the ADC, 
> particularly as this process takes place after the sampling stage, and 
> the discussion is about the spectrum of the signal going into this stage.

It seems to me that Vencat takes for granted that multiplication by the 
sampling waveform _is_ the sampling process,  and so necessarily part of 
the sampling stage. Agreed, quantization comes after.

> I think there may be some difference in our interpretation of what 
> Vancat is using as his sampling waveform.  As far as his square-wave is 
> concerned, I re-read his original postings and I am sure that his query 
> relates to the use of a square-wave as the actual sampling waveform, and 
> not merely as the clock driving a conventional, (nanosecond 
> aperture-time) sampler circuit. Note that his specifically mentions 
> multiplying his input sinusoidal signal by the square-wave, and 
> expresses surprise (!!) when there is no baseband signal.
> 
> In his second post he said:
>     
>"Let me clarify.
> 
>I am sampling a sine wave(Fin) and the clock used for sampling             
> is a square wave(Fclk) (or pulse train with 50% duty cycle.
> Not an impulse train). The 50% duty cycle square 
> wave clock do not have any DC components.
> 
> Now we multiply these two waveforms in time domain, we do not          
> have the baseband signal!! The only components that I am getting     
> are Fclk+/-Fin. The output spectrum does not have Fin at  all. I        
> know this is not true in the real world.
> 
>     What am I missing."
> 
> I appeared to me that Vencat was puzzled by the fact that when he used a 
> square-wave with zero DC-component as his sampling signal he did not get 
> any baseband component in the sampled signal. My answer was an attempt 
> to explain that phenomenon.

What Vencat described is not a sampler, but a balanced modulator. The 
result of the operation is exactly what one would expect from a balanced 
modulator: sidebands without carrier, at every harmonic of the carrier.

> I think that some of the others in this thread may have have focussed on 
> the square-wave aspect of the question and assumed that Vencat was 
> sampling using a very wide pulse of 50 per cent duty cycle, having 
> normalised values 1.000, 0.000, 1.000, 0.000  etc.  That is, a 
> square-wave of amplitude 0.500 peak, with a DC offset of +0.500, and for 
> this situation   I would have no issue with their comments regarding the 
> sinc() falloff etc.

He isn't sampling at all. As I wrote above, he's modulating. Elsewhere 
in this thread someone equated chopping and time-division multiplexing 
to sampling. These processes create sidebands ad infinitum, but they are 
not the same process. (Balanced modulation can be seen as TDM of a 
signal and it's inversion.)

> That said, Vancat quite clearly emphasises that his sampling signal has 
> no DC component at all, so there is no offset, and so the sampling 
> waveform has values of: 1.000, -1.000, 1.000, -1.000  etc.  My previous 
> comments and the following argument are based on this interpretation of 
> the problem.
> 
> My argument is this:
> 
> 1.  The sampling process can be regarded as multiplication in the 
> time-domain of an input signal by the sampling waveform, frequency fs.

Not any waveform. Multiplying by a symmetric square wave yields AM 
suppressed carrier with harmonics that must be filtered in practice. 
Multiplying by a sine wave yields AM suppressed carrier with no 
harmonics. Multiplying by a train of impulses (all of the same sign) 
yields a train of samples. Confusing the physical implementation of 
sampling with that idealized mathematical representation started this 
thread.

> 2,  A typical or 'usual' sampling waverform has an amplitude of 1.000 
> for a few nanoseconds and 0.000 for the rest of the cycle.

Not in practice. Typically, one closes a switch between the signal and a 
holding capacitor for a substantial interval -- the acquisition time -- 
then opens it to lock in the sample. The opening is not instantaneous, 
so there is an aperture time. The few nanoseconds you refer to is the 
time for the switching element to disconnect.

Much of the discussion that follows is an accurate description based on 
the misconception I described above, so I snipped it. Sampling can be 
seen as a special case of amplitude modulation, but not every form of 
modulation (not even every form with a binary carrier) is sampling. I 
see the analysis on non-sampling modulation as irrelevant here.

   ...

> So Jerry (and the group) I hope you will agree that the DC component of 
> the sampling waveform is in fact responsible for the baseband component 
> in the sampled signal.

I do agree. I hadn't seen it in that light before.

Jerry,
You wrote:
> John,
> 
> Your thoughtful analysis inspired me to think a bit more, but we still 
> diverge on many points. I'll address the differences as they arise.
> 
I think we diverge mainly on matters of detail and definitions. In the 
following I have snipped out everything except your most recent comments.
> 
> I agree that sampling is an on/off process and as such can be said to 
> have a DC component. That assumes a mathematical model, but I see a 
> physical one.> 
(snip)
> The sampling waveform I envision controls a switch. Depending on the 
> switch design, it may go from 1 to 0, +1 to -1, +12 to +10, or any other 
> two-level waveform.
(snip)
But that is only the controlling waveform which, as you suggest, is 
device dependant.   I have no problem with the switch concept.  It can 
be seen as switching between (signal) and (no signal) as with the usual 
sampling arrangement, or between(signal) and (inverted signal) in the 
arrangement Vencat describes.

I recognise what you say about a physical model, but don't think it is 
unreasonable to look on the processes as multiplying the signal by 
values:(1.000 or 0.000) in the first case and (1.000 or -1.000) in the 
other.
  >
> He isn't sampling at all. As I wrote above, he's modulating. Elsewhere 
> in this thread someone equated chopping and time-division multiplexing 
> to sampling. These processes create sidebands ad infinitum, but they are 
> not the same process. (Balanced modulation can be seen as TDM of a 
> signal and it's inversion.)
> 
We agree that he is modulating, but where we seem to differ is that I 
claim that conventional sampling is also a modulating process, but using 
a very distinctive waveform.   I agree that Vencat's square-wave example 
can not be regarded as 'sampling' because his modulating signal does not 
meet the essential requirement of selecting the signal for a very brief 
time, and de-selecting it for the rest of the cycle.

(snip)

Jerry, you commented then on some of my numbered points:
>>
>> 1.  The sampling process can be regarded as multiplication in the 
>> time-domain of an input signal by the sampling waveform, frequency fs.
> 
> 
> Not any waveform. Multiplying by a symmetric square wave yields AM 
> suppressed carrier with harmonics that must be filtered in practice. 
> Multiplying by a sine wave yields AM suppressed carrier with no 
> harmonics. Multiplying by a train of impulses (all of the same sign) 
> yields a train of samples. Confusing the physical implementation of 
> sampling with that idealized mathematical representation started this 
> thread.
> 
I claim though that the representation that I used is a pretty accurate 
representation of the physical model you describe, and refers back, in a 
   non-mathematical way to the known characteristics of modulators.

There is no confusion!  The examples you give are exactly correct, 
including the generation of a train of samples.  If you look at those 
samples though, you see a train of pulses with successive pulses varying 
in amplitude in step with the input signal.  In other words, the input 
signal is modulating the sampling waveform.  This  produces a series of 
double-sideband suppressed carrier signals, each one centred on a 
multiple of fs, (including the fundamental, and DC).  I claim that this 
is a reasonable and accurate way of looking at the process.


>> 2,  A typical or 'usual' sampling waverform has an amplitude of 1.000 
>> for a few nanoseconds and 0.000 for the rest of the cycle.
> 
> 
> Not in practice. Typically, one closes a switch between the signal and a 
> holding capacitor for a substantial interval -- the acquisition time -- 
> then opens it to lock in the sample. The opening is not instantaneous, 
> so there is an aperture time. The few nanoseconds you refer to is the 
> time for the switching element to disconnect.
> 

Correct. For audio sampling the aperture time is microseconds, not 
nanoseconds. I have told myself a million times: "Don't exaggerate!"

> Much of the discussion that follows is an accurate description based on 
> the misconception I described above, so I snipped it. Sampling can be 
> seen as a special case of amplitude modulation, but not every form of 
> modulation (not even every form with a binary carrier) is sampling. I 
> see the analysis on non-sampling modulation as irrelevant here.
> 
Jerry, you don't seem to make allowance for the fact that I was 
responding to Vencat's query.  I was trying to give him some 
understanding of why his square-wave sampling waveform gave him the 
results he observed, and why the usual sampling waveform is the only one 
that is useful for sampling.  I tied that in to his observation that 
when there was no DC in a proposed 'sampling waveform' there was no 
baseband in the output.  I was not suggesting that people should use any 
sampling waveform other than the usual one.

>> So Jerry (and the group) I hope you will agree that the DC component 
>> of the sampling waveform is in fact responsible for the baseband 
>> component in the sampled signal.
> 
> 
> I do agree. I hadn't seen it in that light before.

It is something that does not seem to be mentioned in the text books. 
Since the sampling waveform is by its nature uni-polar, of course it 
must have a DC component, but it had not occurred to me until recently 
that the DC component alone determines the amplitude of the baseband 
signal in the sampled waveform.

Thanks for an interesting discussion,

Regards,
John