On Jun 17, 4:45*pm, Roy Lewallen wrote:
Recalling my earlier example of a 100 watt transmitter connected to a
half wavelength 200 ohm transmission line and then to a 50 ohm resistive
load, the "forward power" on the line was 156.25 watts and the "reverse
power" or "reflected power" 56.25 watts. The subject of this topic asks
where the "reflected energy" goes. Does it go back into the transmitter?

------Z01=50------+------Z02=200------load

Pfor1 = 100w, Pref1 = 0w
Pfor2 = 156.25w, Pref2 = 56.25w
rho^2 = 0.36, rho = 0.6

That's obviously a Z0-match to 50 ohms. There are, as usual, four
wavefront components:

Pfor1(rho^2) = 36w, the amount of forward power on the 50 ohm line
that is reflected back toward the source.

Pfor1(1-rho^2) = 64w, the amount of forward power on the 50 ohm line
that is transmitted through the impedance discontinuity toward the
load.

Pref2(rho^2) = 20.25w, the amount of reflected power on the 200 ohm
line that is re-reflected back toward the load by the impedance
discontinuity.

Pref2(1-rho^2) = 36w, the amount of reflected power on the 200 ohm
line that is transmitted through the impedance discontinuity toward
the source.

Pref1 = Pfor1(rho^2) + Pref2(1-rho^2) - total destructive interference

Pref1 = 36w + 36w - 2*SQRT(36w*36w) = 0

Pfor2 = Pfor1(1-rho^2) + Pref2(rho^2) + constructive interference

Pfor2 = 64w + 20.25w + 2*SQRT(64w*20.25w)

Pfor2 = 64w + 20.25w + 72w = 156.25w

Note that the 72 watts of total destructive interference energy toward
the source has changed directions and joined the forward wave toward
the load as constructive interference.

No need to haul out a wattmeter or SWR meter. A pencil, paper, and
calculator is all one needs PLUS an understanding of the rules of
physics governing energy flow at a Z0-match.

Actually, you didn't even need to tell me what the Pfor2 and Pref2
values are. As you can see from the calculations, they are easy to
calculate even when they are unknown.
--
73, Cecil, w5dxp.com