From: QRP RX:
On Sun, Oct 13, 2019 at 04:29 PM, W5DXP wrote: For maximum power transfer, the impedance looking back down the transmission line from the EDZ feedpoint needs to be 175+j1000 ohms, the conjugate of the feedpoint impedance.
That's not true.
I was just quoting the following from:
Maximum Power Transfer Theorem for AC Circuits: It can be stated as in an active network, the maximum power is transferred to the load when the load impedance is equal to the complex conjugate of an equivalent impedance of a given network as viewed from the load terminals. ("equivalent impedance" means the impedance transformed by the transmission line's SWR not being equal to 1:1.)
Here's a simple example based on TLDetails:
100w XMTR---50 ohm coax---1:1 balun==========59.5 deg 600 ohm feedline======== EDZ Zfp=189 - j1000 ohms
There is an SWR of ~12:1 on the 600 ohm feedline. The forward power is ~350w and the reflected power is ~250w. The SWR on the coax is 1:1, i.e. zero reflected energy is reaching the source. The power delivered to the antenna is 97.9 watts, i.e. 97.9% efficiency in power transfer. Where did the 250 watts of reflected power go? When it encountered a 50 ohm Z0-match at the balun, all the reflected power was redistributed back toward the antenna load through the processes of re-reflection and total destructive interference toward the XMTR. There's only 2.1 total watts lost in the 600 ohm feedline, i.e. 0.094 dB, mostly I^2*R losses.
If we replace the balun with a 50 ohm dummy load, disconnect the antenna, and measure the impedance looking back down the feedline, what do we measure? TLDetails says very close to 189+j1000 ohms, actually considering losses, 195+j994 ohms, i.e. a near-conjugate match.
The conservation of energy principle tells us that if zero reflected energy is reaching the source (XMTR) in a low loss system, that most of the reflected power is being redistributed back toward the load (antenna). That's what Z0-matches do.
More info at:
