# KL7AJ quick quiz 21/02/2016

Eric posed a quick quiz for the masses to test their knowledge under his heading “Do you really understand impedance matching?”

For your convenience, I will quote his challenge here. All connections are made with low-loss coaxial cable. The antenna tuner is high quality with negligible losses.
According to standard conversion charts, we find that 4:1 SWR will give us 36% reflected power. Keep that number handy.
Now, we set up the experiment. First, set the slugs on BOTH wattmeters to read REFLECTED power.
Turn on the transmitter, and adjust the antenna tuner for zero reflected power on Bird #1. Switch to forward power, and set transmitter output to exactly 100 watts. Readjust antenna tuner if necessary to achieve zero reflected power, while maintaining 100 watts forward.
Go to Bird #2 and confirm that reflected power is 36 watts.
Question: What is the FORWARD power on Bird Wattmeter #2? How you answer this question determines if you understand the conjugate match theorem or not.

Let us assume that the transmission lines are 50Ω, and that the Bird wattmeters are calibrated for 50Ω.

So, to extract the key information, we have a lossless system (KL7AJ is a lossless kind of guy) and the load is stated to be VSWR=4.

That means that Pf/Pr at Bird #2 MUST be 1/0.36=2.778, ie Pf=2.778Pr.

Also stated is that Bird#2 reads Pr=36W.

That means that Pf at Bird #2 MUST be 36*2.778=100W. Further, we can calculate that the power delivered to the load is simply Pf-Pr=64W (Duffy 2010).

Also stated is that the transmitter delivers 100W to this lossless system.

It is inconsistent that a system is lossless when 100W input results in just 64W output.

KL7AJ might not understand this stuff fully.

Where did he go wrong?

Well, starting at the top, it was ok until he said:

Go to Bird #2 and confirm that reflected power is 36 watts.
Question: What is the FORWARD power on Bird Wattmeter #2? How you answer this question determines if you understand the conjugate match theorem or not.

No, if the power into the lossless system is 100W, the power into the load is also 100W, and given that Pf-Pr=100 and Pr=0.36Pf (as stated), then 100=Pf-0.36Pf, so Pf=100/(1-0.36)=156.3W.

As I said, his statement of the problem indicates he might not understand this stuff fully.

Had he said Go to Bird #2 and confirm that reflected power is 36% of forward power, or if we simply dismiss as inconsistent that 36W figure (it is not needed to solve the problem), the problem is solveable and the answer is 156.3W.