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I am using a double stacked Yagi-Uda antenna with about 57 Ω impedance. The 50 Ω feed line (coaxial) cables are combined using a tee connector. I observed, at the output of the tee connector, an impedance of about 28 Ω while transmission and hence I used a single short-end stub matching technique to transform this impedance into 50 Ω at the transceiver end.

Neglecting the practical terms such as attenuation, will this technique still cause reflections at the tee connector output? The wave shall see 28 Ω again from the stub to the tee connector output, so wouldn't there be a reflection again?

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Yes, you have correctly analyzed the situation, and there will be reflections. But you ask if there will be any loss.

Reflections don't cause loss directly; they just traverse the same piece of transmission line many times and the loss in the line adds up. (Or to take a sum-of-all-the-reflections perspective: They cause increased current in the transmission line (relative to what's going in and out the ends), and therefore additional loss due to resistance.)

Another effect of reflections is to add a time delay to part of the signal, like multipath, but this rarely matters, and it is not loss per se — all the energy in still goes out.

So, “neglecting the practical terms such as attenuation”, there is no loss, but in practice there is. You can reduce the coax loss by minimizing the length between the stub and tee, of course.

What would theoretically make sense here is a tee with two 50 Ω ports and one 25 Ω port, but those don't exist since they'd need specialized connectors. Instead, you would build a custom matching circuit which just keeps everything inside as short as possible or, for higher frequencies, has impedance-matched PCB traces.


Note that what you're effectively building here is a power divider (a.k.a. splitter in cable-TV land). Therefore, you could purchase it as a standard part instead of building your own, and it would be already designed to minimize reflections and loss. Then the worst problem you'd have is the 57 Ω - 50 Ω mismatch.

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Wherever there is an impedance discontinuity, there are reflections. So indeed, even with your stub there are still reflections.

However, the goal of a matching network, be it stubs, discrete components, or anything else, is to confine these reflections and the energy they contain to the smallest area possible. By reducing the amount of transmission line with reflections (and thus, a VSWR greater than 1:1), we minimize the loss.

Stubs do this (when they are the right length) by creating another reflection that cancels the initial reflection. So the stub operates with an increase VSWR (and thus, greater current, and greater loss), but since the stub is in practice less than a quarter wavelength long, the loss isn't much.

Compare this to the case with no stubs: the entire transmission line from the transmitter to the antenna operates at a high VSWR. Since this feedline is usually pretty long, this can add up to a lot of loss.

With discrete components the problem is the similar: that reflected energy is cancelled by reactive energy alternately stored and released in either the magnetic field of a coil, or the electric field of a capacitor. This flow of energy into and out of these reactive components is subject to loss when we use real components, but this loss is ostensibly less than the loss that would occur were that energy to travel the whole length of the feedline multiple times.

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There is ALWAYS SOME loss when adding coax ANYWHERE! Inthis case the solution is to use 1/4 wave (or odd multiples of 1/4 wave) lengths of 75 Ohm coax from the antennas to the T to make a "coax transformer", thus reducing the loss and removing the need for the stub. This the "standard" way to connect 2 50 ohm antennas to a 50 ohm radio.

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  • $\begingroup$ Why would the loss from a 75 ohm section of cable + connections be less than the loss of a stub? $\endgroup$ – Phil Frost - W8II Sep 21 '15 at 17:10

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