From: ming@enel.ucalgary.ca (Yang Ming)
Subject: Re: SWR Tuning and PLL
Date: Wed, 06 Jan 93 17:43:05 GMT
Organization: ECE Department, U. of Calgary, Calgary, Alberta, Canada

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.

> [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'.

>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.

>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.

>>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.

> [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. 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.

Ming.