What effect does a tuner have at the antenna?

Discussion in 'General Technical Questions and Answers' started by W5DXP, Jun 4, 2008.

Not open for further replies.
1. W5DXPHam MemberQRZ Page

Let's make it a little more obvious. Zero reflected power makes it back to the antenna because both ends are Z0-matched to 50 ohms.

50 ohm RCVR--50 ohm coax--+--1/2WL Z0 feedline--+--50 ohm coax--50 ohm antenna

The SWR on all sections of coax is 1:1, i.e. there are no reflections on the coax. The forward power on the coax is 1uw from the antenna. The reflected power on the coax is zero. This is true for any value of Z0 for the 1/2WL of feedline. No reflected power ever makes it back to the antenna so zero reflected power is ever re-radiated. Your "subtraction of net power from the system" is impossible in a lossless conjugately matched system.

In all of the lossless examples, 1uw always makes it to the receiver and 1uw is always re-radiated in accordance with the rules for linear systems - no matter what the value of Z0 on the 1/2WL of feedline.

In a lossless conjugately matched system, all available power is delivered to the load. Hand waving will not change that fact of physics.

In fact, here is an example where the receiver load is replaced by a 50 ohm resistor and the antenna radiation resistance is replaced by a 50 ohm resistor. Which end is the source and which end is the load? Hint: No matter what the Z0 of the 1/2WL of lossless feedline, the system is perfectly symmetrical and conjugately matched for both transmit and receive.

50 ohm resistor--50 ohm coax--+--1/2WL Z0 feedline--+--50 ohm coax--50 ohm resistor

Last edited: Jul 2, 2008
2. W5DXPHam MemberQRZ Page

I recognize it as an irrelevant straw man diversion having nothing to do with the discussion topic. Feel free to discuss it without me. But, IMHO, it is foolish to discuss tuner components before you understand how a Z0-match at a transmission line discontinuity re-distributes 100% of the reflected energy back toward the load with no tuner and no lumped components in sight.

Because a 50 ohm Z0-match is established by reflections and wave cancellation. I have explained that multiple times before. You really need to take time to understand how and why that happens. Have you even looked at the graphic I posted? If not it is reproduced on my web page at: http://www.w5dxp.com/Z01Z02V.GIF

In the graphic, V3 and V4 are the two component reflected waves that cancel during destructive interference. Their component energies join the V1+V2 forward wave during constructive interference that sends all of the available energy toward the load.

The same graphic, except in s-parameter terms, is at: http://www.w5dxp.com/Z01Z02s.GIF

Lumped components not propagating waves would certainly be magic at work. Do you really want to present magic as your explanation for how things work? Do the lumped L and C models of transmission lines not allow waves to propagate? How do these latest examples cancel the reflections with no lumped components anywhere?

Last edited: Jul 8, 2008