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I'm struggling to understand the "meaning" of VSWR in the context of power delivered to a load.

Say I have a transmitter operating at 10MHz, a short (~1m) piece of assumed-lossless transmission line, and a load whose impedance is not well-matched to the line.

If I measure forward and reflected power at the transmitter, I'll see some reflected power due to the impedance mismatch. But I understand this gets reflected "again" at the transmitter back to the load and is eventually transferred to the load (it does not get dissipated in the transmitter). So for a lossless transmission line, what measurement is the power dissipated by the load? It is simply the forward power? It would seem like it's not the (forward minus reflected) power, so does that quantity (forward minus reflected) have any physical meaning?

Ultimately, I'm trying to measure the power delivered to this load at 10MHz, and don't quite understand if I can just measure voltage and current and multiply them (accounting for any phase shift), of it I need to account for SWR and/or reflected power in some way...

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  • \$\begingroup\$ At 10 MHz, the wavelength is 30 metres and so a 1 metre length of cable is not of any significance. \$\endgroup\$
    – Andy aka
    Feb 2 at 16:09
  • \$\begingroup\$ In other words, VSWR is not that meaningful on cables that have a fractional wavelength. \$\endgroup\$
    – Andy aka
    Feb 2 at 16:45
  • \$\begingroup\$ If the wire is lossless, then it's obvious that all the energy that comes out of the transmitter either goes back into the transmitter, goes into the antenna, or builds up in the wire until it bursts like a balloon \$\endgroup\$
    – user253751
    Feb 2 at 16:47
  • \$\begingroup\$ i.e. the wave is identical on a short wire, so the max/min ratio called VSWR=1 so this doesn’t tell you about mismatch attenuation. \$\endgroup\$ Feb 2 at 17:21
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Power transmission in a cable:

You have a 2 parallel wire cable between the signal source and the load. You can have power meter inserted at some point of the cable (practically it's at the source or the load end of the cable). It gives to you 2 numbers: Power towards the load and power towards the source. The first one can have text label "transmitted" and the second one can have text label "reflected". But those labels are only text, the measured things are net power flow towards the load and net power flow towards the source.

The momentary energy flow direction through the meter can change during the AC cycle. The direction is always the direction of the current in that wire which has the +polarity of the voltage. If voltage and current have phase difference there is a part of the time energy flow towards the load and a part of the time towards the source

Your bidirectional power meter calculates separately the average power when the flow is towards the load and when the flow is towards the source. There are numerous analog circuit constructions which can make it. At low frequencies, like at 10 MHz you can use transformers and rectifiers to make the separation. At microwaves you must use directional couplers to separate the waves which propagate to different directions.

If it happens that your load is mismatched, but the transmitter reflects back all of what returns from the load you have succesful or lucky matching circuit in the transmitter. Then your meter shows nothing coming back from the load because the net energy flow summed from the multiple reflections is only towards the load. Practical transmitters absorb a part of the returning signal. It's either dissipated in the transmitter circuit or directed back to the power supply through protection diodes.

VSWR is not practical thing to be measured directly as the voltage ratio Vmax/Vmin at 10 MHz because one needs a half wavelength piece of open cable where the voltage amplitude between the wires is possible to measure at every point. In shorter cables one cannot reliably detect a standing wave.

But it's common to calculate the theoretical VSWR from the reflection factor (=the VSWR which could be detected in a line which is at least half wavelength long and with the same mismatch as the actual line) because radio engineers and hobbyists are used to talk of VSWR as a measure for how well matching has succeeded. It's done even in cases when the cables are so short that no standing wave could be noticed. There's simple formula between that calculated VSWR and the ratio of powers to different directions.

The powers themselves cannot be decided from VSWR - only their ratio, the power reflection factor. It has formula Preflected/Pforward/=((VSWR-1)/(VSWR+1))^2 That formula isn't valid if there's nonlinearity which generates harmonic components that contain tens of percents of the power. That's because the reflection happens differently in different frequencies and the standing waves of the harmonic components are generally placed differently.

ADD: Practical transmitters have often non-linearly working output amplifier. The amp can be linear in cases it must be able to handle AM or SSB signal. Even linear amb starts to behave non-linearly with mismatched reactive-looking load because the voltage or current limits of linear operation range are exceeded or some protection circuits can start rectify excessive voltage and direct it back to power supply. The exact behaviour can be analyzed by making simulations. Ideal AC voltage source is poor model for real transmitter which has reactive load because it totally skips nonlinear effects.

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  • \$\begingroup\$ Interesting, thank you! I guess I'm not clear how it is that a transmitter absorbs a part of the returning signal. Say the load is open circuit, then we have infinite SWR, but clearly the transmitter does not dissipate infinite (or even significant) power? \$\endgroup\$
    – Dan
    Feb 3 at 0:11
  • \$\begingroup\$ Infinite SWR means 100% reflection. The open wire is reactive load. It surely charges some amount of energy to the electric and magnetic fields of the cable (both exist) All that energy returns within the same AC cycle back to the transmitter. Is it dissipated in the output stage transistors or returned to the power supply capacitors or is it directed to special dummy load like in radars or does it get reflected back to the line depends on the construction of the transmitter. Many transmitters cannot stand high reflection because it either causes overcurrent or overvoltage => auto-shutdown. \$\endgroup\$
    – user287001
    Feb 3 at 0:27
  • \$\begingroup\$ (continued) the power reflection factor is ((VSWR-1)/(VSWR+1))^2 The actual power inputted to the reactive load and reflected back depens on the load reactance and the inputted voltage. It's related with the reactive power. See this: electronics.stackexchange.com/questions/488482/… \$\endgroup\$
    – user287001
    Feb 3 at 0:35

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