From: gary@ke4zv.uucp (Gary Coffman)
Subject: Re: SWR Tuning and PLL
Reply-To: gary@ke4zv.UUCP (Gary Coffman)
Organization: Destructive Testing Systems
Date: Thu, 7 Jan 1993 23:13:23 GMT

In article <93Jan06.174305.37224@acs.ucalgary.ca> ming@enel.ucalgary.ca (Yang Ming) writes:
>In article <1993Jan5.181318.19138@ke4zv.uucp> gary@ke4zv.UUCP (Gary Coffman) writes:
>>>
>>>If you are talking about a linear amplifier, the question is how the
>>>vectorial combination of the two voltages can *change* the slope?
>>
>>The loadline slope is defined as E/I where E and I are the dynamic
>>voltages and currents at the output of the active device. Since
>>the reflected wave's voltage adds vectorially to the voltage
>>being generated by the active device, E changes. Therefore E/I
>>changes, and thus the output impedance changes.
>>
>
>OK. You ARE talking about linear circuits. You are right that E
>changes due to the reflected wave, but you forget that I changes
>as well, and E/I is constant is linear circuits.

No. If E/I is constant, you get an unvarying DC static line. That's
not true in a power amplifier, only in a small signal amplifier.
Also, the voltage and current in a reflected wave are sign reversed
from the forward wave. The return current is out of phase with the
forward current while the return voltage is in phase with the forward
current, plus line delay phase shifts of course. Maxwell covers this
in chapter four better than I can. 

>> [deleted]
>
>>The output impedance isn't imaginary in the sense of 0+jX, it's
>>a *virtual* value that's a combination of the dynamic voltages and
>>currents being generated by the active device with the voltages and
>>currents in the reflected wave. This is similar in concept to the
>>virtual ground at the input of an opamp. It's not a real ground,
>
>As far as I know, the vertual ground comes from the assumption
>that the opamp has infinite gain, which is not true in reality.
>And therefore this 'ground' is somewhat 'floating'.

Yes, but the difference is very small, in picoamps with most good
opamps. No operating circuit performs exactly like the simple idealized
models because practical parts have complex behavior, but the difference
is usually small enough to ignore for engineering analysis.

>>nor is it an imaginary ground in the sense of 0+jX, it's a *virtual*
>>ground caused by the summing of voltages in a node.
>>
>
>Seems to me that the vertual ground you talked about is at the
>node of the standing wave fromed by the forward and backward
>waves. This voltage node is 0v only if you have a total reflection.
>This way, these two waves are with the same amplitude.
>This is true only if the antenna does not absorb (radiate) any power,
>which is not true for *any* antennas. The antenna radiates some power
>and reflects some with lower amplitude.

The same analogy still holds, however, with reduced amplitude reflections.
In this case you just pair the reflected wave with a *part* of the forward
wave being generated by the source. If you didn't do that, there wouldn't
be any output from the amplifier, thus no reflected wave, etc. As long as
the reflected wave is smaller than the forward wave being generated by
the source, and it has to be, then the net power flow is forward in all
cases.

>>Let me give a crude analogy to illustrate what happens when a
>>reflected wave enters an operating amplifier's output port, think
>>of it as trying to piss into the nozzle of a operating firehose. 
>>It all gets reflected back to you. The conjugate mirror effect is 
>>what causes total re-reflection of the wave.
>>
>
>This is N/A. EM wave on the transmission line is not the same thing.

But EM waves trying to get into an operating *source* are the same
thing. You have to realize that the behavior at the output terminal
of an active device is not transmission line behavior. Let's try
another analogy to try to get the point across. Assume a 12 volt
battery feeding a 6 ohm load. There are two amps forward current
flowing. Now insert a 1.5 volt battery in series with the 12 volt
battery, but of reversed polarity. This is your reverse wave. No
power from the 1.5 volt battery is dissipated by the 12 volt battery.
The voltage can't overcome the forward voltage to get back in the
bigger battery. It is cancelled out by 1.5 volts of the forward
voltage leaving 10.5 volts for the load. The bigger battery now
sees a higher output impedance and supplies less current to the 
load. Like an amplifier, the current it *could* have supplied
before doesn't *back up* in the amplifer, it simply isn't drawn
from the power supply.

>>>b)	Linear circuts' operating points do NOT change with the
>>>	amplitude of the signal (even it is a combination of whatever
>>>	waves). This is the *definition* of linearity;
>>
>>No, you're confusing different things again. Linear circuit operating
>>points do not change with varying *input* signals. That's true. But
>>the reflected wave isn't an *input* signal going into the *input*
>>port of the device. Instead it's a rogue *output* signal that's trying
>>to squeeze back into the *output* port of the device. As I noted above,
>>the vector sum of the reflected wave and the forward wave from the
>>active device changes the loadline seen by the active device. An
>>amplifier is a 4 port, not a 2 port, network.
>>
>
>Doesn't matter even it is output, if it is linear. You know S-parameters
>that are used to describe a transistor? They are got from maesurement
>in linear region and used in amp design. They don't consider S22
>non-linear.

It does matter. Two port and four port networks have different behavior.

>> [deleted]
>
>>a physical resistor. The virtual resistance, the load line, the *output* 
>>impedance, all the same thing in a transmitter, changes dynamically under 
>>conditions of reflected waves and varying load. This makes the book equation 
>>too simple to describe the real behavior of transmitters. There are non-linear
>                                                                     ^^^^^^^^^
>>dependent variables in the equation describing the output behavior of a
>
>Ah. Finally I got the keyword that you ARE talking about nou-linearity.
>Fine, transistors are better described with nonlinear parameters,
>no doubt about it. But befor nonlinearity takes over, at small signals,
>they can be nicely described with linear terms. The point I have been
>trying to make is that reflected wave CAN be absorbed by the amplifier,
>partly or tatolly, depending on the whole network, while your point is
>that it is totally reflected. If you agree that there is a linear part
>in amplifier parameters, then you can't ignore that some reflected
>wave is absorbed by the amplifier.

Transmitter output stages are almost *by definition* large signal
devices. Small signal parameters are essentially worthless in
describing their behavior. Solving by parts is the only reasonable
way of dealing with the problem. As long as the reflected wave is
opposed by the forward wave *at the active device terminal* no
power can flow back into the active device.

>>transmitter. The only valid solutions require complex vector calculus.
>>I don't have the symbols for that on this terminal, look harder for
>>Maxwell's book, it's ISBN 0-87259-299-5.
>
>I certainly will keep an eye on this book.

Do that, it's very enlightning for power amplifier designers. Chris
Boswick also has a book out that's useful. Too often texts treat
parts of RF systems in isolation, sources, transmission lines, loads,
etc. You can't do that in the real world when any of the parameters
departs from ideal textbook properties. The behavior of transmitting
sources in the presence of reflected waves is one of those cases.

Remember that transmission line theory, that allows the forward
and reverse waves to pass each other unmodified, depends on the
fact that there is no dissipative resistance present at any point
along the line, or any generating source present at any point
along the line. Only at the terminals do real resistances and
sources exist. The behavior there is different than at other 
points out on the line.

Gary
-- 
Gary Coffman KE4ZV          |    You make it,     | gatech!wa4mei!ke4zv!gary
Destructive Testing Systems |    we break it.     | uunet!rsiatl!ke4zv!gary
534 Shannon Way             |    Guaranteed!      | emory!kd4nc!ke4zv!gary      
Lawrenceville, GA 30244     |                     | emory!ke4zv!gary@gatech.edu
Newsgroups: rec.radio.cb