"Cecil Moore" wrote in message
...
Reg Edwards wrote:
You will be pleased to hear I'm back on Sierra Valley, Californian Red
tonight.

Peter Vella Merlot for me. My doctor told me to drink two
glasses of red wine a day so I use ~400 ml iced-tea glasses.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil, if I drank that much wine every day I'd believe the SWR at the input of
the 72-ohm line is 1:1 too, but it ain't. I'm concerned with the guys who think
it is. So please let me be totally elementary in explaining why the SWR on the
72-ohm line is 1.44 everywhere on the line, including the input, yet still has a
50-ohm input impedance.

Step 1. We have a lossless 50-ohm line of random length feeding a 50-ohm source
to an antenna with terminal impedance of 50 + j0 ohms. The input voltage is 1 v
and the input current is 0.02 a. Z = E/R, and 1/0.02 = 50 verifying that the Zo
of the line is 50-ohms. The power absorbed by the antenna is IE = E*/R = 0.02
watts. The SWR is 1:1 everywhere.

Step 2. We now insert a lossless 72-ohm, 1/2 wl line between the antenna and the
output end of the 50-ohm line. Does the power absorbed in the antenna change?
NO. Do the forward voltage and current on the 50-ohm line change? NO. Are there
any reflections on the 50-ohm line? NO. Are the forward voltage and current in
the 75-ohm line 1 v and 0.02 a.? NO. Are there reflections on the 72-ohm line?
YES. Is the input impedance of the 72-ohm line 50 + j0 ohms? YES. Is the SWR at
the input of the 72-ohm line1:1? NO. Is the SWR at the input of the 72-ohm line
1.44:1? YES. Does the antenna still absorb 0.02 watts? YES.

Step 3. To understand why the input impedance of the 72-ohm line is 50 + j0 with
an SWR = 1.44 we need to understand the voltage and current reflection
coefficients, rho as well as the voltage and current transmission coefficients,
tau.

Step 4. In going to a higher line impedance the voltage reflection coefficient
rhoE is positive, and the current reflection coefficient rhoI is negative; The
reverse is true when going to a lower line impedance, as when going from the
72-ohm line into the 50-ohm load.

Step 5. In going to a higher line impedance the voltage transmission coefficient
tauE = (rhoE + 1) and the current transmission coefficient tauI = (rhoI - 1).
The reverse is true when going to a lower line impedance.

Step 6. For an SWR of 1.44, rho = 0.1803. Therefore, when going from a 50-ohm
line to a 72-ohm line, tauE = 1.1803 and tauI = 0.8197. Consequently, the
forward voltage in the 72-ohm line is 1.1803v and the forward current is 0.02 x
0.8197 = 0.01639 a. Note that 1.180/0.01639 = 71.996, which would be 72 if we
used more significant figures.

Step 7. When forward voltage 1.1803 v reaches the 50-ohm antenna load, the
reflection coefficient 0.1803 x 1.1803 = 0.2128 v, which subtracts from the
1.1803 v at the input of the 72-ohm line, making the total voltage at the input
0.9675 v.

Step 8. When forward current 0.01639 a reaches the 50-ohm antenna load, the
current reflection coefficient 0.1803 x 0.0.01639 a = 0.002956 a, which adds to
the 0.01639 a, making the total current at the input of the 72-ohm line 0.01935
a.

Step 9. Dividing the total voltage at the input of the 72-ohm line by the total
current, we get 0.9675/0.01935 = 50.01 ohms, the input resistance of the 72-ohm
line with a 1.44:1 SWR on the line.

Step 10. Consequently, we see that the addition of the reflected voltage and
current to the forward voltage and current yield an impedance at the input of
the 72-ohm line that is 50 ohms, not 72 ohms. It is the effect of the 1.44:1 SWR
on the line that has changed the line impedance (not the characteristic
impedance) to 50 ohms.

I hope this helps in understanding why the SWR at the input of the 72-ohm line
is 1.44:1, and NOT 1:1.

Walt, W2DU