# What is modulator?

Discussion in 'Electronic Basics' started by boki, Sep 15, 2003.

1. ### Kevin AylwardGuest

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

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.
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.

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?

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

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

2. ### John FieldsGuest

John Fields wrote:

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

tHEN, On Sun, 21 Sep 2003 09:23:43 +0100, "Kevin Aylward"

---
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.

3. ### Kevin AylwardGuest

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.
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.
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

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.