What is modulator?

Kevin Aylward · Sep 21, 2003

John Fields wrote:

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?

---
Well, it remains the same, of course,

How do you know? This is an assumption. There is no "of course"
about it. As soon as vmod changes, the 0-xings of the carrier change,
so you don't know what its doing. The only way to determine what the
carrier is doing by a measurement, not by making assumptions. This is
what physics is all about.

but the sidebands are created
when VMOD is changing.

How do you know? You must measure it to know what its doing. This
measurement is inherently uncertain.

---

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.

---
Yes, of course, but that's not the point. The point is that the
carrier _amplitude_ changes during modulation.

But if the carrier frequency is not measured during the changes, we dont
know that the amplitude relates to the carrier.

As I noted, if we have a(t).sin(wt) we can define a(t) as the
"amplitude", or we can expand for a suitable input signal and instead
define the co-efficients of these sidebands as the amplitudes. However,
making measurements to pick out these factors is a bit more involved.

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".

Nope. Stop ignoring the fact that you cannot have a changing DC source.
A source either, never changes, or it changes. If it doesn't change, the
carrier cant change, period. If it does change, then the change has to
perturb the carrier frequency.

You still fail to understand the fundamental uncertainty relation
between frequency and time. Reread what I have wrote.

(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.

---
I see you saw the smiley but, again, I wasn't talking about
frequency, I was talking about amplitude.

But you didnt understand my point on this.

Frequency change is inherent when vmod changes, therefore you have no
way of knowing that the amplitude you are measuring is *at* the carrier
frequency. The only way to know that it is at the carrier frequency is
to use a small BW filter, but this will automatically filter out the
amplitude changes. The frequency uncertain ios directly related to rate
at which vmod changes.

What do I have to do, to get you to see this inherent conflict?

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.

---
Reminds me of that old Rogers and Hart song, "Where or When"...

Amplitude, Kevin, amplitude!-)

The measurement of amplitude is not guaranteed to be at the carrier
frequency. As I stated, once vmod changes, all bets are off.

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.

John Fields · Sep 21, 2003

The measurement of amplitude is not guaranteed to be at the carrier
frequency. As I stated, once vmod changes, all bets are off.

---
Come on, Kevin, now you're just being pig-headed.

Let's say that in the circuit above, VMOD is a stable DC voltage and
ACIN is a sinusoidal signal with a frequency of 1.000000MHz and an
amplitude of 10.000000VPP. Further, let's say that VMOD is steady and
that ACOUT is measured and found to be a sinusiodal signal with a
frequency of 1.000000MHz and an amplitude of 5.000000VPP.

Now let's say that it takes us a few seconds to adjust VMOD until ACOUT
has an amplitude of 2.500000V, and then a few seconds to measure the
frequency of ACOUT. Do you seriously doubt that the measurement will
yield a frequency of 1.000000MHz?

Of course, during the time that VMOD is being adjusted sidebands will be
generated, but once VMOD becomes stable the only thing remaining will be
the carrier, which will have been reduced in amplitude. It has to be,
since LDR and R are simply a resistive voltage divider and the point is,
as it has been from the beginning, that amplitude modulation changes the
amplitude of the modulated waveform.

--
John Fields

Kevin Aylward · Sep 22, 2003

John Fields wrote:

John Fields wrote:

The measurement of amplitude is not guaranteed to be at the carrier
frequency. As I stated, once vmod changes, all bets are off.

---
Come on, Kevin, now you're just being pig-headed.

Not really. I just don't think you see the, err... "deep" point being
made. I think maybe that it has still has not sunk in that in the real
world, there
are no sine waves. A sine wave exists for ever. You are assuming that
this is a mere detail, that can be ignored. I am pointing out that this
mere detail, has consequences sometimes missed.

Let's say that in the circuit above, VMOD is a stable DC voltage and
ACIN is a sinusoidal signal with a frequency of 1.000000MHz and an
amplitude of 10.000000VPP. Further, let's say that VMOD is steady and
that ACOUT is measured and found to be a sinusiodal signal with a
frequency of 1.000000MHz and an amplitude of 5.000000VPP.

Now let's say that it takes us a few seconds to adjust VMOD until
ACOUT has an amplitude of 2.500000V, and then a few seconds to
measure the frequency of ACOUT. Do you seriously doubt that the
measurement will yield a frequency of 1.000000MHz?

Not at all. A few second measurement time should result in better then 1
hz accuracy. Noting that from sigma_t.sigma_f >=1/2, the frequency
measurement accuracy of the carrier is a function of how long it is
measured for.

Of course, during the time that VMOD is being adjusted sidebands will
be generated, but once VMOD becomes stable the only thing remaining
will be the carrier, which will have been reduced in amplitude.

Once you wait long enough, in principle, sort of, and this is a
reasonable argument for saying that the carrier amplitude is varying.

Suppose one considers a pulsed sine wave on a scope. It would be hard to
deny that the "carrier" is changing its level. Yet, if feed trough a
spectrum analyser the "carrier" is constant. The thing to note is that
the Fourier transform is an integration of f(t)dt, so its forming a sort
of weighted average over time.

So, I agree, as I already pointed out, that in a manner similar to
measuring, defining and declaring that sidebands vary and the carrier
doesn't, one can alternatvly measure, define and declare that the
carrier varies. Both have their uses, under suitable conditions, but
strictly speaking, none are really correct. In a reality both models are
compromises. In a real dynamic situation, these simplified concepts
start to lose their individual meaning.

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.

What is modulator?

Kevin Aylward

Guest

John Fields

Guest

Kevin Aylward

Guest

Welcome to EDABoard.com

Sponsor

Online statistics

Forum statistics

What is modulator?

Kevin Aylward

Guest

John Fields

Guest

Kevin Aylward

Guest

Log in

Welcome to EDABoard.com

Sponsor