What is modulator?

Dbowey wrote:
jfields posted:
On 18 Sep 2003 16:15:44 GMT, dbowey@aol.com (Dbowey) wrote:

John Fields posted:
Never mind.... it didn't add anything to the discussion.

---
Well, you seemed to be laboring under the misapprehension that
amplitude modulation doesn't change the amplitude of the carrier, so
I thought I'd better set you straight.



I don't believe you set me straight....... Hmmmm....Nope. In my
humble opinion, you are not correct.
Click to expand...
And that opinion is demenstatable incorrect. I have given the
mathematical outline as to why. Indeed, I have explained in detail why
it is correct to state that the carrier amplitude varies an why it is
also correct to state its the sidebands that varies. It is trivially
obvious form the mathematical trigonometric identity relating products
of sines with sums of differences of sines. To repeat, if one writes

Vo = A(t).Sin(wt)

There is no real rational argument that can prevent Sin(wt) being
*defined* as the carrier, and A(t) *defined* as its amplitude. In
addition, I agree, that there is no real rational argument that can
prevent the alterative definition based on the "carrier" being constant.
What part of "mathematical identity" do you have trouble with?

Yes, sidebands can be generated by turning the carrier on and off or
changing its amplitude by changing the power supply voltage, but that
is not AM as it is generally used; it's closer kin is CW,
Click to expand...
This is a right daft arument.

as in
International Morse transmission.
Click to expand...
Unfortunately, this is a very constrained view of signal, or
mathematical analysis. What part of "There is nothing unique in the the
Time view or Fourier view" do you have problem with?

Indeed, as soon as it is brought to explicit attention that the Fourier
view is only one way of infinatly many ways to expand a function in an
orthogonal set, it should be immediately clear that "The universe, is
what we say it is" - James Burk - Connections" TV series.

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
 
Kevin, If you are satisfied with your math modeled view of AM and it serves
your needs then stick with it. As far as what is demonstrable, a math model
"demonstrates" nothing. On the other hand, it is simple, with a frequency
selective voltmeter or decent receiver, to demonstrate that the carrier
amplitude is constant during AM modulation.

Don
 
Dbowey wrote:
Kevin, If you are satisfied with your math modeled view of AM and it
serves your needs then stick with it.
Click to expand...
I am satisfied with *any* number of math models that can be used to
represent the real world. I am not stuck to just one model.

As far as what is
demonstrable, a math model "demonstrates" nothing.
Click to expand...
You obviously missed the point. I have been discussing other, equally
valid models such that their model parameters can be *physically*
measured in just as valid a manner as, for example, the non unique model
that uses sum and difference frequencies.

On the other
hand, it is simple, with a frequency selective voltmeter or decent
receiver, to demonstrate that the carrier amplitude is constant
Click to expand...
Have a little think about what actually happens in this measurement.

during AM modulation.
Click to expand...
You still fail to see the point. This is *deeper* then what you suppose
with such an elementary example. I don't know just what the problem is
as to why you don't understand the point being made. Are you familiar
with orthogonal expansions at all? You are making an assumption here
that, in this case, that this *particular* definition and measurement of
"carrier" amplitude is unique. It isn't.

In your example here, if one decides to use equipment specifically
designed to analyse a signal in the Fourier domain, that it will,
necessarily, pick out the Fourier co-efficients. However, this does not
mean that such measurement are any more real then, for example, using
equipment that picks out the Walsh function co-efficients. (Walsh
functions are an orthogonal set of square waves)

Read my posts again, and pay particular attention to the notions of "non
unique" and "equally valid physically"

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
 
"Dbowey" <dbowey@aol.com> wrote in message
news:20030919032412.13238.00001457@mb-m15.aol.com...
Kevin, If you are satisfied with your math modeled view of AM and it
serves
your needs then stick with it. As far as what is demonstrable, a math
model
"demonstrates" nothing. On the other hand, it is simple, with a frequency
selective voltmeter or decent receiver, to demonstrate that the carrier
amplitude is constant during AM modulation.
Click to expand...
More correctly, you should say that in standard AM
(without a supressed carrier) there is a constant-amplitude
component at the carrier frequency. That's not QUITE the
same thing.


Bob M.
 
On 19 Sep 2003 07:24:12 GMT, dbowey@aol.com (Dbowey) wrote:

Kevin, If you are satisfied with your math modeled view of AM and it serves
your needs then stick with it. As far as what is demonstrable, a math model
"demonstrates" nothing. On the other hand, it is simple, with a frequency
selective voltmeter or decent receiver, to demonstrate that the carrier
amplitude is constant during AM modulation.
Click to expand...
---
Let me dumb this down a little so it's easier to visualize, OK?

Using my previous example of a plate modulated linear final RF
amplifier, let's use 1Hz as the modulating frequency and connect the RF
amplifier's output to an oscilloscope (through an appropriate voltage
divider, of course). I believe that what you'll see when the carrier is
at 100% modulation is a 200VPP 1MHz signal when the amplitude of the
modulating signal is at its positive peak and 0V when the modulating
signal is at its lowest peak. Since the oscilloscope will be displaying
time in the X dimension and voltage in the Y dimension, it should be
readily apparent that the RF output signal's amplitude is varying in
time but, even more important, that it is varying in step with the
modulating signal. If you doubt that, then do this: Tune your
voltmeter to 1MHz, modulate the carrier with a 1Hz signal and then
report back as to how steady the voltmeter's needle (or display)
remained during the test.

As Kevin said, you need to think a little bit more about what's
happening when you make your measurements.

--
John Fields
 
Bob posted:
<< "Dbowey" <dbowey@aol.com> wrote in message
news:20030919032412.13238.00001457@mb-m15.aol.com...
Kevin, If you are satisfied with your math modeled view of AM and it
serves
your needs then stick with it. As far as what is demonstrable, a math
model
"demonstrates" nothing. On the other hand, it is simple, with a frequency
selective voltmeter or decent receiver, to demonstrate that the carrier
amplitude is constant during AM modulation.
Click to expand...
More correctly, you should say that in standard AM
(without a supressed carrier) there is a constant-amplitude
component at the carrier frequency. That's not QUITE the
same thing. >>

Yes, but that isn't the form of AM being discussed. However, just to cover the
bases........ And when suppressed Carrier is being utilized, even the low level
carrier leak will have constant amplitude.

Don
 
To dumb this down a bit, jfields posted:
<< On 19 Sep 2003 07:24:12 GMT, dbowey@aol.com (Dbowey) wrote:

Kevin, If you are satisfied with your math modeled view of AM and it serves
your needs then stick with it. As far as what is demonstrable, a math model
"demonstrates" nothing. On the other hand, it is simple, with a frequency
selective voltmeter or decent receiver, to demonstrate that the carrier
amplitude is constant during AM modulation.
Click to expand...
---
Let me dumb this down a little so it's easier to visualize, OK?

Using my previous example of a plate modulated linear final RF
amplifier, let's use 1Hz as the modulating frequency and connect the RF
amplifier's output to an oscilloscope (through an appropriate voltage
divider, of course). I believe that what you'll see when the carrier is
at 100% modulation is a 200VPP 1MHz signal when the amplitude of the
modulating signal is at its positive peak and 0V when the modulating
signal is at its lowest peak. Since the oscilloscope will be displaying
time in the X dimension and voltage in the Y dimension, it should be
readily apparent that the RF output signal's amplitude is varying in
time but, even more important, that it is varying in step with the
modulating signal. If you doubt that, then do this: Tune your
voltmeter to 1MHz, modulate the carrier with a 1Hz signal and then
report back as to how steady the voltmeter's needle (or display)
remained during the test.
---------

You have stated most of the obvious. Using a display in the time domain is an
ideal method for determining percent modulaton. It's just displaying the
algebraic sum of the voltages (carrier and sidebands). Good tool, but it
doesn't help decide if the carrier is constant or varying. You are making an
error in believing that it does.

However, when you observe the carrier and sidebands selectively, you WILL
observe that the carrier is indeed constant, and that is the point I am trying
to make.

When Kevin's math says one thing and observed facts show another, I believe one
should rethink their model.

Don
 
jfields also posted:

<< As Kevin said, you need to think a little bit more about what's
happening when you make your measurements. >>

I've thought about it quite a lot.. Turn the horizontal gain off and look at
that nice vertical line pumpng up and down. It's just a voltmeter.... you
think a bit more about that.

Don
 
John Fields wrote:
On 19 Sep 2003 07:24:12 GMT, dbowey@aol.com (Dbowey) wrote:

Kevin, If you are satisfied with your math modeled view of AM and it serves
your needs then stick with it. As far as what is demonstrable, a math model
"demonstrates" nothing. On the other hand, it is simple, with a frequency
selective voltmeter or decent receiver, to demonstrate that the carrier
amplitude is constant during AM modulation.

---
Let me dumb this down a little so it's easier to visualize, OK?

Using my previous example of a plate modulated linear final RF
amplifier, let's use 1Hz as the modulating frequency and connect the RF
amplifier's output to an oscilloscope (through an appropriate voltage
divider, of course). I believe that what you'll see when the carrier is
at 100% modulation is a 200VPP 1MHz signal when the amplitude of the
modulating signal is at its positive peak and 0V when the modulating
signal is at its lowest peak. Since the oscilloscope will be displaying
time in the X dimension and voltage in the Y dimension, it should be
readily apparent that the RF output signal's amplitude is varying in
time but, even more important, that it is varying in step with the
modulating signal. If you doubt that, then do this: Tune your
voltmeter to 1MHz, modulate the carrier with a 1Hz signal and then
report back as to how steady the voltmeter's needle (or display)
remained during the test.

As Kevin said, you need to think a little bit more about what's
happening when you make your measurements.

--
John Fields
Click to expand...
Take a look at the carrier of WWVB at 60 KHz. It is modulated by a
19 dB reduction in carrier level at a baud rate of 1. You can watch the
carrier level change on a slow sweep across the scope tube. I wound a
three foot square loop with 20 turns of 22 AWG wire, and used a couple
caps to tune it to 60 KHz. A couple op amps gave me around 9 volts P-P,
and you could watch the carrier level change.
--


Michael A. Terrell
Central Florida
 
On 19 Sep 2003 19:56:15 GMT, dbowey@aol.com (Dbowey) wrote:


However, when you observe the carrier and sidebands selectively, you WILL
observe that the carrier is indeed constant, and that is the point I am trying
to make.
Click to expand...
---
If you think the carrier amplitude remains constant then I suggest you
try the tuned voltmeter experiment and post what you find.

--
John Fields
 
On 19 Sep 2003 20:00:19 GMT, dbowey@aol.com (Dbowey) wrote:

jfields also posted:

As Kevin said, you need to think a little bit more about what's
happening when you make your measurements.

I've thought about it quite a lot.. Turn the horizontal gain off and look at
that nice vertical line pumpng up and down. It's just a voltmeter.... you
think a bit more about that.
Click to expand...
---
Try it with a 1Hz modulating signal and then think about why the line
_does_ pump up and down. Hint: At modulation frequencies above a
certain threshold, persistence of vision makes it impossible to follow
the bouncing ball. But it does bounce, as you'll see if you let the
beam play on a photodiode after your eyes fool you into thinking it's a
constant amplitude.

--
John Fields
 
Dbowey wrote:

Kevin, If you are satisfied with your math modeled view of
AM and it serves your needs then stick with it. As far as
what is demonstrable, a math model "demonstrates" nothing.
Click to expand...
Don,

As long as you think that mathematics is invalid as a means
to understand electronics, you're going to struggle to get
anywhere beyond the "Gee golly Mr. Wizard, that sure is
swell" level.

You're right about one thing, math doesn't "prove" stuff.
But PEOPLE prove stuff using mathematical arguments. Math
is a language, invented by people so that they can have
debates and discussions about how the universe works.

On the other hand, it is simple, with a frequency
selective voltmeter or decent receiver, to demonstrate
that the carrier amplitude is constant during AM
modulation.
Click to expand...
Talk about carrier waves and modulation and signal
processing on a mathematical level is interesting because it
leads to new discoveries, new techniques. I'd love to hear
about how your "frequency selective volt meter" works, and
how to interpret its output, but c'm on! I want to hear
about poles and zeroes or some equivalent explanation of how
it processes the signal. Otherwise, without the math, I
might just as well go downstairs and watch another episode
of Dexter's Lab.

-- Jim L.
 
On Fri, 19 Sep 2003 15:41:27 -0500, John Fields
<jfields@austininstruments.com> wrote:

On 19 Sep 2003 20:00:19 GMT, dbowey@aol.com (Dbowey) wrote:

jfields also posted:

As Kevin said, you need to think a little bit more about what's
happening when you make your measurements.

I've thought about it quite a lot.. Turn the horizontal gain off and look at
that nice vertical line pumpng up and down. It's just a voltmeter.... you
think a bit more about that.

---
Try it with a 1Hz modulating signal and then think about why the line
_does_ pump up and down. Hint: At modulation frequencies above a
certain threshold, persistence of vision makes it impossible to follow
the bouncing ball. But it does bounce, as you'll see if you let the
beam play on a photodiode after your eyes fool you into thinking it's a
constant amplitude.
Click to expand...
---
The bouncing ball analogy isn't a good one.

With the sweep off what will happen is that the carrier amplitude will
be represented by what looks like a vertical line (but what is, in
actuality, a spot on the screen oscillating vertically at the carrier
frequency) with a length proportional to the amplitude of the carrier.
The amplitude of the carrier will in turn be determined by the amplitude
of the modulating signal, which will either add to or subtract from the
CW length of the line. At high modulating frequencies the lengthening
and shortening of the line will not be discernible because of
persistence of vision, but at low modulating frequencies the length of
the line will be seen to change as the amplitude of the modulating
signal changes, and at a rate corresponding to the modulating frequency.

Since the length of the line represents the amplitude of the carrier and
it can be seen to vary as the amplitude of the modulating signal varies,
the carrier amplitude does _not_ remain constant as it is being
modulated.

--
John Fields
 
Jim Large posted, in part:
"I'd love to hear about how your "frequency selective volt meter" works...."

Google will offer up 2980 links for that. Have fun.

Don
 
jfields posted:
<< On 19 Sep 2003 19:56:15 GMT, dbowey@aol.com (Dbowey) wrote:


However, when you observe the carrier and sidebands selectively, you WILL
observe that the carrier is indeed constant, and that is the point I am trying
to make.
Click to expand...
---
If you think the carrier amplitude remains constant then I suggest you
try the tuned voltmeter experiment and post what you find.
What is there to experiment about? The measurement of the carrier separate
from the measurement of a sideband generated by a modulating tone (1000 or 1004
Hz), was a common set of tests on analog transmission systems.

Don
 
John Fields wrote:
On Fri, 19 Sep 2003 15:41:27 -0500, John Fields
jfields@austininstruments.com> wrote:

On 19 Sep 2003 20:00:19 GMT, dbowey@aol.com (Dbowey) wrote:

jfields also posted:

As Kevin said, you need to think a little bit more about what's
happening when you make your measurements.

I've thought about it quite a lot.. Turn the horizontal gain off
and look at that nice vertical line pumpng up and down. It's just
a voltmeter.... you think a bit more about that.

---
Try it with a 1Hz modulating signal and then think about why the line
_does_ pump up and down. Hint: At modulation frequencies above a
certain threshold, persistence of vision makes it impossible to
follow the bouncing ball. But it does bounce, as you'll see if you
let the beam play on a photodiode after your eyes fool you into
thinking it's a constant amplitude.

---
The bouncing ball analogy isn't a good one.

With the sweep off what will happen is that the carrier amplitude will
be represented by what looks like a vertical line (but what is, in
actuality, a spot on the screen oscillating vertically at the carrier
frequency) with a length proportional to the amplitude of the carrier.
The amplitude of the carrier will in turn be determined by the
amplitude of the modulating signal, which will either add to or
subtract from the CW length of the line. At high modulating
frequencies the lengthening and shortening of the line will not be
discernible because of persistence of vision, but at low modulating
frequencies the length of the line will be seen to change as the
amplitude of the modulating signal changes, and at a rate
corresponding to the modulating frequency.

Since the length of the line represents the amplitude of the carrier
and it can be seen to vary as the amplitude of the modulating signal
varies, the carrier amplitude does _not_ remain constant as it is
being modulated.
Click to expand...

I'll expand on this. In fact, it is not a simple matter to show that the
carrier
is constant or varies. Indeed, its not possible, in principle.

To determine that one is measuring the carrier frequency, rather than
its side
bands, one necessarily requires a small BW filter. However, a narrow
filter will
automatically filter out changes in signal level, even if they were
there, that
is the measurement itself, forces the carrier to be appear constant. If
the
carrier was changing at a 10Hz rate, it would not get through a 1Hz
filter.
However, if the BW is increased, then you don't know what frequency the
carrier
is, so you cant say much about it. It could be any value within that BW.
This
means that the so called sidebands could be measured as the carrier and
subsequently show variations in the carriers amplitude. This is
expressed in a
fundamental mathematical, provable, relation between F and T:

sigma_f.sigma_t >=1/2

That is, the product of the standard deviation of f(t) and the product
of the
standard deviation of its fourier transformis always greater then 1/2.
This
results shows that it is impossible, inprinciple, to know both the time
and
frequency distribution together to an
arbitrarily degree of accuracy.

Since the carrier amplitude has to vary at the modulating frequency, and
any
supposed sidebands are at this distance away from the carrier, its not
possible
to have a BW that allows possible carrier amplitudes to be measured, and
simultaneously determine that you are actually measuring the carrier
frequency.

To expand on this a bit more. One has to define what is referring to
when one
says things like "constant carrier amplitude". One needs to distinguish
between
averages, peaks, instantaneous etc. Just exactly what is being declared
constant? If one modulates a carrier with a fixed amplitude sine wave,
and does
a frequency sweep, with low BW, i.e. say, 1Hz resolution. Certain
amplitudes of
"sidebands" and "carrier" will be measured. If the modulation amplitude
is
increased, the "sidebands" will increase, and the "carrier" amplitude
will
remain invariant. So, in this average sense one might claims that the
carrier
amplitude remains constant. However, this is only achieved because the
BP filter
removes any potential "instantaneous" variations from the outset.

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
 
On 20 Sep 2003 03:14:14 GMT, dbowey@aol.com (Dbowey) wrote:

jfields posted:
On 19 Sep 2003 19:56:15 GMT, dbowey@aol.com (Dbowey) wrote:


However, when you observe the carrier and sidebands selectively, you WILL
observe that the carrier is indeed constant, and that is the point I am trying
to make.

---
If you think the carrier amplitude remains constant then I suggest you
try the tuned voltmeter experiment and post what you find.


What is there to experiment about? The measurement of the carrier separate
from the measurement of a sideband generated by a modulating tone (1000 or 1004
Hz), was a common set of tests on analog transmission systems.
Click to expand...
---
Yes, of course, but what was being measured was the average amplitude of
the carrier. By using a slowly varying modulating signal the amplitude
variations of the carrier caused by the amplitude variations of the
modulating signal can easily be seen.

As a matter of fact, changing the DC plate voltage of the final will
cause the RF output to change, proving that amplitude modulation _does_
cause the carrier amplitude to change.

--
John Fields
 
On Sat, 20 Sep 2003 09:15:42 +0100, "Kevin Aylward"
<kevin@anasoft.co.uk> wrote:

John Fields wrote:
On Fri, 19 Sep 2003 15:41:27 -0500, John Fields
jfields@austininstruments.com> wrote:

On 19 Sep 2003 20:00:19 GMT, dbowey@aol.com (Dbowey) wrote:

jfields also posted:

As Kevin said, you need to think a little bit more about what's
happening when you make your measurements.

I've thought about it quite a lot.. Turn the horizontal gain off
and look at that nice vertical line pumpng up and down. It's just
a voltmeter.... you think a bit more about that.

---
Try it with a 1Hz modulating signal and then think about why the line
_does_ pump up and down. Hint: At modulation frequencies above a
certain threshold, persistence of vision makes it impossible to
follow the bouncing ball. But it does bounce, as you'll see if you
let the beam play on a photodiode after your eyes fool you into
thinking it's a constant amplitude.

---
The bouncing ball analogy isn't a good one.

With the sweep off what will happen is that the carrier amplitude will
be represented by what looks like a vertical line (but what is, in
actuality, a spot on the screen oscillating vertically at the carrier
frequency) with a length proportional to the amplitude of the carrier.
The amplitude of the carrier will in turn be determined by the
amplitude of the modulating signal, which will either add to or
subtract from the CW length of the line. At high modulating
frequencies the lengthening and shortening of the line will not be
discernible because of persistence of vision, but at low modulating
frequencies the length of the line will be seen to change as the
amplitude of the modulating signal changes, and at a rate
corresponding to the modulating frequency.

Since the length of the line represents the amplitude of the carrier
and it can be seen to vary as the amplitude of the modulating signal
varies, the carrier amplitude does _not_ remain constant as it is
being modulated.


I'll expand on this. In fact, it is not a simple matter to show that the
carrier
is constant or varies. Indeed, its not possible, in principle.
Click to expand...
---
Beg to differ!^) If you use DC as the modulating signal it's not only
possible, it's extremely easy. For example:

ACIN---------------+
|
VMOD>-----+ |
| |
[LED]--> [LDR]
| |
| +----->ACOUT
| |
| [R]
| |
GND>------+--------+

This way it's only neccessary to change VMOD in order to cause the
amplitude of ACOUT to change, eliminating the problem of the pesky
sidebands interfering with the measurement. (As long as you wait long
enough after changing VMOD. ;^) For AC modulation:


ACIN---------------+
|
VMOD--[C]-+ |
| |
VBIAS-[R]-+ |
| |
[LED]--> [LDR]
| |
| +----->ACOUT
| |
| [R]
| |
GND>------+--------+

Certainly nothing elegant or linear about either circuit, just brute
force proof of principle.

--
John Fields
 
John Fields wrote:
On Sat, 20 Sep 2003 09:15:42 +0100, "Kevin Aylward"
kevin@anasoft.co.uk> wrote:


Since the length of the line represents the amplitude of the carrier
and it can be seen to vary as the amplitude of the modulating signal
varies, the carrier amplitude does _not_ remain constant as it is
being modulated.


I'll expand on this. In fact, it is not a simple matter to show that
the carrier
is constant or varies. Indeed, its not possible, in principle.

---
Beg to differ!^)
Click to expand...
Not wise John:)

If you use DC as the modulating signal
Click to expand...
You can't. If it is dc the carrier will *never* change. You can't just
ignore this inherent fact. This is the real world. It is ruled by the
laws of physics.

it's not only
possible, it's extremely easy.
Click to expand...

Nope:)

You obviously did not take in the sigma_f.sigma_t >=1/2. Its a
fundamental result. Not open to debate.

What you are now about to do, is use all those old arguments attempting
to show that one can determine position and momentum together. You will
fail. Trust me on this. Look up QM, e.g. wave-particle duality aspects.

For example:
ACIN---------------+
|
VMOD>-----+ |
| |
[LED]--> [LDR]
| |
| +----->ACOUT
| |
| [R]
| |
------+--------+

This way it's only neccessary to change VMOD in order to cause the
amplitude of ACOUT to change,
Click to expand...
Yep. Same old arguments.

Oh?. What happens to the carrier frequency while VMOD changes in time?

eliminating the problem of the pesky
sidebands interfering with the measurement.
Click to expand...
Nope. As soon as you start changing vmod, the frequency because
uncertain. You don't know that it is at the carrier frequency any more.
You have to measure it. To measure frequency necessarily takes a finite
tine, as noted in the Time-Frequency relation.

(As long as you wait long
enough after changing VMOD. ;^)
Click to expand...
You have to wait *for ever* before the carrier frequency can be shown to
the same as its original value. Therefore, you cant claim that you are
measuring the carrier, because you can't prove what its actual frequency
is with a finite wait time. The Time-frequency uncertainty relation will
bite you. There is no way around it.

For AC modulation:
ACIN---------------+
|
VMOD--[C]-+ |
| |
VBIAS-[R]-+ |
| |
[LED]--> [LDR]
| |
| +----->ACOUT
| |
| [R]
| |
------+--------+

Certainly nothing elegant or linear about either circuit, just brute
force proof of principle.
Click to expand...
Nope. *The* principle is that it is absolutely, theoretically and in
principle to simultaneously know a signals frequency spectrum and time
spectrum to an arbitrary degree of accuracy. Indeed, QM is just the
observation that position and momentum are related by Frequency-Time
fourier transform pairs as noted above.

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
 
On Sat, 20 Sep 2003 17:16:34 +0100, "Kevin Aylward"
<kevin@anasoft.co.uk> wrote:

John Fields wrote:
On Sat, 20 Sep 2003 09:15:42 +0100, "Kevin Aylward"
kevin@anasoft.co.uk> wrote:


Since the length of the line represents the amplitude of the carrier
and it can be seen to vary as the amplitude of the modulating signal
varies, the carrier amplitude does _not_ remain constant as it is
being modulated.


I'll expand on this. In fact, it is not a simple matter to show that
the carrier
is constant or varies. Indeed, its not possible, in principle.

---
Beg to differ!^)

Not wise John:)

If you use DC as the modulating signal

You can't. If it is dc the carrier will *never* change. You can't just
ignore this inherent fact. This is the real world. It is ruled by the
laws of physics.

it's not only
possible, it's extremely easy.


Nope:)

You obviously did not take in the sigma_f.sigma_t >=1/2. Its a
fundamental result. Not open to debate.

What you are now about to do, is use all those old arguments attempting
to show that one can determine position and momentum together. You will
fail. Trust me on this. Look up QM, e.g. wave-particle duality aspects.

For example:

ACIN---------------+
|
VMOD>-----+ |
| |
[LED]--> [LDR]
| |
| +----->ACOUT
| |
| [R]
| |
------+--------+

This way it's only neccessary to change VMOD in order to cause the
amplitude of ACOUT to change,

Yep. Same old arguments.

Oh?. What happens to the carrier frequency while VMOD changes in time?
Click to expand...
---
Well, it remains the same, of course, but the sidebands are created when
VMOD is changing.
---

eliminating the problem of the pesky
sidebands interfering with the measurement.

Nope. As soon as you start changing vmod, the frequency because
uncertain. You don't know that it is at the carrier frequency any more.
You have to measure it. To measure frequency necessarily takes a finite
tine, as noted in the Time-Frequency relation.
Click to expand...
---
Yes, of course, but that's not the point. The point is that the carrier
_amplitude_ changes during modulation. Using DC as the modulating
source allows carrier amplitude measurements to be made with different
_static_ values of modulating signal, eliminating the effects of the
sideband "interference".
---


(As long as you wait long
enough after changing VMOD. ;^)

You have to wait *for ever* before the carrier frequency can be shown to
the same as its original value. Therefore, you cant claim that you are
measuring the carrier, because you can't prove what its actual frequency
is with a finite wait time. The Time-frequency uncertainty relation will
bite you. There is no way around it.
Click to expand...
---
I see you saw the smiley but, again, I wasn't talking about frequency, I
was talking about amplitude. Check abse for some photos showing
ACOUT (f)VMODdc in about a half an hour.
---


For AC modulation:


ACIN---------------+
|
VMOD--[C]-+ |
| |
VBIAS-[R]-+ |
| |
[LED]--> [LDR]
| |
| +----->ACOUT
| |
| [R]
| |
------+--------+

Certainly nothing elegant or linear about either circuit, just brute
force proof of principle.

Nope. *The* principle is that it is absolutely, theoretically and in
principle to simultaneously know a signals frequency spectrum and time
spectrum to an arbitrary degree of accuracy. Indeed, QM is just the
observation that position and momentum are related by Frequency-Time
fourier transform pairs as noted above.
Click to expand...
---
Reminds me of that old Rogers and Hart song, "Where or When"...

Amplitude, Kevin, amplitude!-)


--
John Fields
 
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