Kevin Aylward wrote:

gwhite wrote:
Kevin Aylward wrote:

gwhite wrote:
Kevin Aylward wrote:


It can be done with class-A or class-B since the assumption is
that the amp will be driven (or nearly so) to the rails by the
carrier alone.

I agree, this is another method of generating x-product
multiplication terms. However, arguable, a class A amplifier is not
really a class A amplifier if it is driven to saturation. Its a
really a switching amp or, a pulse amplitude modulator if its rails
are varying.


Here's a "class-A" amplifier that can be amplitude modulated but yet
not saturated (assumes a constant load R):

Ho hummm...


V+
|
LC Tank
| AM RF_out
+--||--O
Carrier |
RF_in c
O--||---b Class A biased (no base bias details)
e
|
|
RF_Choke
Audio |
in c
O--||-- b Class A modulator (no base bias details)
e
|
GND


This is a single ended amplitude modulator. The top transistor could
be driven to the switch mode by the carrier, but this is not
necessary to produce AM. Practically, it will be driven to the
switch mode for efficiency reasons.

And you think that this is me to me? I suppose you aint read many of my
10,000+ posts.

I'm pretty sure I won't bother at this point.

Err..you've missed the cap from the top transistor emitter to ground. If
the bottom transistor circuit was a true current source at rf, the top
transistor could not effect the output current at all.

True. It should be there.

Oh, and its not very linear with required to audio input signal without
an emitter resister anyway. The distortion of an basic tranister amp for
2nd Harmonic = Vi(mv)%, that is 10 mv will get you 10% distortion.

The point wasn't to design a perfect amp. The point was to illustrate
you're wrong.

Amplitude modulation can be "made" via linear methods.

Nope. Not a chance.

Oh, but it can.

"Multipliers"
cannot be generally stated to be either linear or non-linear.

If one input of a multiplier is held constant, the other input has a
linear response. If the other input is a function of time, the response
to the first input is non-linear. That is, it dose *not* satisfy a(f(t))
= f(at).

No, false, or whatever negation pleases you best. You were already
given the answer. You confuse time-invariance with linearity. You need
not take my word for it. Consult any Signals and Sytems text, any
Linear Systems text, or any Communications text. IIRC, the following
was a homework problem in Stremler's text:

Determine linearity
Determine time-invariance

The System
+---------------+
| |
in | /¯¯¯\ | out
x(t) O--------( X )---------O y(t)
| \___/ |
| | |
| | |
| O |
| cos(w_c·t) |
+---------------+

http://www.amazon.com/exec/obidos/tg...glance&s=books

I suppose if you want to make up your own definition of linearity, you
can get whatever anwswer you wish.


A system which includes a multiplier must be put through the linearity
test to see if the configuration is linear or non-linear. IOW, it
can be either.

Ho humm again. You confuse where the term linearity is to be applied.

I defined the system quite clearly. I mean, I don't think it can be
made simpler.

You can stick with the formal definition given in pretty much every
Signals and Sytems text, any Linear Systems text, or any Communications
text, or you can make up your own and get the answer that pleases you.


+------+
x(t) O---| h(t) |---O y(t)
+------+


linearity:

a·x1(t) = a·y1(t)
b·x2(t) = b·y2(t)

if x(t) = a·x1(t) + b·x2(t)

then

y(t) = a·y1(t) + b·y2(t)

Linearity can more easily be expressed as:

a(f(t)) = f(at)

Except that isn't "the" definition (hey, but it is true if a = 1). It
doesn't even meet you own description:
"A linear system, cannot produce frequencies that are not in the input,
essentially, by definition." -- Kevin Aylward

I would like to see you apply this "definition."

If that is true, then the system is linear. This can be true for
systems with multipliers.

Nope. A signal being acted on by a multiplier is a non-linear system if
the second input is non constant with time. Your way of base on this
one.

Again, you confuse linearity and time-invariance.

This system is linear and has a multiplier (it is not time invariant):


Nope, its not.

Oh, but it is. Some rather trivial math can lead you to understand what
linear means.

The System
+---------------+
| |
in | /¯¯¯\ | out
x(t) O--------( X )---------O y(t)
| \___/ |
| | |
| | |
| O |
| cos(w_c·t) |
+---------------+


It produces a DSB signal (y(t)). w_c·t could be "added in later"
(linearly) to y(t) in the proper amplitude and phase and the resultant
signal would for all practical purposes be indistiguishable from
standard AM.

This is not a linear circuit. You need to understand what linear means.
A linear system, cannot produce frequencies that are not in the input,
essentially, by definition.

You don't know what linear means. There it is.

With all due respect, I would guess you
don't have an EE B.S. degree.

Cut the chest puffing.

This is all pretty basic stuff really.

On that much we certainly agree, Comm101 ought to do it.

No non-linear circuit was used but yet AM was produced.

Nonsense. Your pretty misguided on this.

Well I won't take you word for it, and you need not take mine. You
could make it easy just by applying the linearity test. That is,
linearity as it is defined in every Signals and Sytems text, any Linear
Systems text, or any Communications text. I'm not as "original" as you,
I simply trust the guys who wrote the books.

You can not achieve
multiplication without a non-linear circuit.

How about this one:

The System
+---------------+
| |
in | /¯¯¯\ | out
x(t) O--------( X )---------O y(t)
| \___/ |
| | |
| | |
| O |
| 2 |
+---------------+

It looks functionally to be an amp with a gain of two. Is a "gain of 2"
circuit non-linear? Isn't that a multiplier in there?

For example, Gilbert
multipliers use the fact that Id=Is.exp(vd/Vt).

I think that descibes pretty much every bipolar. So I guess transistor
amps cannot be made linear, or at least function sufficiently linear for
the purpose of electronic designers. That is an "interesting"
contention.

That is it logs, adds
and antilog. Balanced switching mixers use switches. Fet mixers use
their square law response.

Not convenient, but it does dispel the "non-linearity is required"
myth.

Its not a myth. I know of no way whatsoever to generate an analogue
multiplication x product terms without having a device satisfying the
property of a.f(t) != f(at), i.e. a non-linear device. Please feel free
to suggest one, but file your patent first.

Why not dispense with the snidery, and simply prove your contention by
applying the linearity test to this one:

The System
+---------------+
| |
in | /¯¯¯\ | out
x(t) O--------( X )---------O y(t)
| \___/ |
| | |
| | |
| O |
| cos(w_c·t) |
+---------------+

Hey, I'll even give you a hint.

Let x1(t) = a1·cos(w1·t)
x2(t) = a2·cos(w2·t)

Then y1(t) = a1·cos([wc + w1]·t)/2 + a1·cos([wc - w1]·t)/2
y2(t) = a2·cos([wc + w1]·t)/2 + a2·cos([wc - w1]·t)/2

?
If x(t) = x1(t) + x2(t) does y(t) = y1(t) + y2(t)
(This *is* the linearity test.)

?
[x1(t) + x2(t)]·cos(wc·t) = x1(t)·cos(wc·t) + x2(t)·cos(wc·t) = y(t)

The answer is yes: the system is linear, by the linearity test.

If x(t) = a·cos(w·t) is applied to the system,
and produces y(t), does x(t-to) applied to the
system result in y(t) except all the t's are
replaced by t-to? (May all the t's in
the y(t) case be replaced by t-to, and
have the system response reflect that the input
was x(t-to).) (This is the time-invariance test.)

y(t) = x(t)·cos(wc·t) = a·cos(w·t)·cos(wc·t)
= a·cos([wc + w]·t)/2 + a·cos([wc - w]·t)/2

Now put the "to" into the input and see what
comes out

What? = x(t-to)·cos(wc·t) = a·cos(w·[t-to])·cos(wc·t)
= a·cos([wc·t + w·[t-to])/2 + a·cos([wc·t - w·[t-to])/2
| |
+---------------------------+
not t-to,

So the system is *not* time-invariant.


This is Comm101 stuff, if you're ever interested enough to crack a book
open, and actually work the problems.


Oh, by the way...your trying to teach your granny to such eggs son.

You're wrong about that one too: we have no relationship. I hope that
comforts you, as I know it does me.