0
$\begingroup$

I'm currently reading "Reflections: Transmission Lines and Antennas" by M. Walter Maxwell, W2DU. This book explains well why the energy is not lost because of high SWR during transmitting, assuming the feedline and ATU are lossless. It also provides plots that allow determining the actual loss when the feedline has losses.

Related links:

What I don't quite understand though is what happens to the received signal level in the high SWR case.

Let's say the antenna impedance on a given frequency is 100 Ohm, the feedline is lossless 50 Ohm, the transceiver input impedance is 50 Ohm. Between the antenna feed point and the feedline SWR = 2, 11% reflected power. Case a) there is no ATU. Case b) there is a lossless LC-match between the transceiver and the feedline that matches seen impedance to 50 Ohm. How much of the received power will be delivered to the transciever in cases a) and b)?

Improve this question
$\endgroup$
0

3 Answers 3

Reset to default
4
$\begingroup$

Let's say the antenna impedance on a given frequency is 100 Ohm, the feedline is lossless 50 Ohm, the transceiver input impedance is 50 Ohm. Between the antenna feed point and the feedline SWR = 2, 11% reflected power.

This is a bit of dangerous thinking, because the distance between the antenna feedpoint and the feedline is zero. As such there can be no standing waves, and no ratio between the maximum and minimum RMS voltage.

A 50 ohm feedline terminated in a 50 ohm load (the receiver) is indistinguishable from a 50 ohm load without a feedline. So really, what you are asking is what happens when a 100 ohm source is connected to a 50 ohm load. The feedline is irrelevant.

schematic

simulate this circuit – Schematic created using CircuitLab

Is this necessarily bad? True, if you can't change R1, then R2 should also be 100 ohms if you want to maximize the power in R2. But if R2 is zero ohms, current is maximized. And if R2 is open, voltage is maximized. What do you want to maximize, power, current, or voltage?

For a receiver, the answer is probably none of these. Instead, you want to maximize the signal to noise ratio. Extracting the maximum power from the antenna might be important, especially as noise power from sources internal to the receiver, such as thermal noise, become significant relative to what's available from the feedpoint. But this is only one of many considerations. For example, the receiver likely has some passive filter which was designed with a particular source impedance assumed. When the source impedance deviates from the design value, the filter performance may degrade.

So, your question is actually quite difficult to answer. Receivers aren't simply resistive loads: they also have filters and active, nonlinear electronics, and the power extracted from the antenna is only indirectly relevant to receiver performance. Ultimately the answer depends on the receiver details.

Improve this answer
$\endgroup$
1
  • $\begingroup$ You are completely right that the receiver is designed for a specific source impedance and that changing it will affect the performance of the filters and other parts of the receiver. I didn't consider that! $\endgroup$
    – Aleksander Alekseev - R2AUK
    Apr 13, 2021 at 17:24
2
$\begingroup$

I think you've answered your own question.

In case a, with an antenna with 2:1 SWR and an otherwise lossless system, 11% of the power will be reflected and re-radiated by the antenna, so 89% makes it to the receiver.

In case b, all of the power will be delivered to the receiver.

Antennas are reciprocal, so all losses in transmit are the same as losses in receive. We usually calculate for the transmit case, the sequence of mismatches and losses is different in the receive case but the answer will be the same.
The matching circuit that makes the 100 ohm antenna look like 50 ohms, will also make the 50 ohm coax look like 100 ohms, for the antenna.

Some interesting caveats:

  • The receiver input impedance may not be 50 ohms. Professional equipment is usually well matched so that its behaviour is predictable, but amateur transceivers might not be, and at least at HF, there is no strong reason to be at 50 ohms on receive.
  • antennas, even a dipole well matched to the transmission line, still scatter a lot of the power. They have a significant radar cross section; the currents generated by the incident field will radiate as well as feeding the coax.
  • so, saying "All the power" isn't all that clear; all of what power? It's really just the best case against which other scenarios can be measured.
Improve this answer
$\endgroup$
1
  • $\begingroup$ I wish I could mark two answers as "accepted". Your answer is correct as well, but as Phil, W8II pointed out my original model was inaccurate. I choose his answer since I believe this was a very important observation. $\endgroup$
    – Aleksander Alekseev - R2AUK
    Apr 13, 2021 at 20:06
1
$\begingroup$

If the antenna isn't lossless (e.g. not superconducting), then some of the received RF energy reflected back into the antenna due a feedpoint mismatch will eventually be dissipated as heat in the antenna, or re-radiated as RF. But for received signals this is possibly only a loss of nanoWatts or picoWatts.

However, not only is reflected signal energy dissipated or lost, but so is reflected noise (within the bandwidth of interest). Thus, the signal-to-noise ratio at the receiver will stay roughly the same. As long as they are above the receiver's input noise floor, the receiver AGC or RF gain can compensate for this slight signal level difference, and the lost energy due to the mismatch won't be noticed.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .