Jeff wrote:

On 03/07/2015 08:29, rickman wrote:
On 7/3/2015 2:50 AM, Jeff wrote:

Are you suggesting that the conjugate match will reflect back to the
antenna 100% of the original reflected wave from the antenna?


Yes, it must.

For example with an external ATU that provides a conjugate match it is
clearly the case that if a 1:1 VSWR is achieved then no reflected power
reaches the TX. (as shown on an SWR meter between the Tx and ATU.)

I am very certain that this assumption is not correct. I wish I had the
math to back me up. The only total reflection I am aware of is an open
circuit which of course absorbs no power at all.

Here is a point. The VSWR only shows no power being sent back to the
txmt output. That does not mean no power is absorbed from the reflected
wave by the matching circuit. I believe the example you gave was Z of
1063 -j0. Isn't that a real impedance of 1063 ohms which is equivalent
to a resistor? Resistors dissipate power don't they?


The point can be easily proved with a lossy feeder, the lossier the
better. If your assumption is correct then the power delivered to the
antenna would be the Tx power less the cable loss less the reflected
power at the antenna mismatch, however it is the case that can be
measured and seen on computer simulation that all of the power is
delivered to the antenna except that which is dissipated in the cable
loss (for the multiple reflections). Also it can been seen that with
perfect components in the matching circuit no power is dissipated there.

The 1063 ohms that you refer to is not resistive so with perfect Ls & Cs
no power will be dissipated in it.

In the real world the Cs and Ls in the matching unit will have some loss
associated with them but that is a different story.

Jeff


While conjugate matching is the way to transfer the maximum power from a
voltage (or current) generator to a load, it is not the way power
amplifiers are set up. The transmitter normally does not present a
match to signals from the aerial, hence the re-reflection.


--
Roger Hayter