4
\$\begingroup\$

I am building several types of radio antennas, and I would like to test which is the most effective. I am wondering: What is the best way to measure the strength of the received signal?

It is at ~151 MHz and 2 watt transmission power, and I will be testing over a distance of at least a couple of hundred feet. The results don't have to be calibrated, only correct relative to one another. Any ideas?

Edit as per comment request:

I am transmitting from a handheld unit which is capable of 2, 4, and 8 watt transmission power, but I am using the 2 watt setting to comply with MURS regulations.

The ideal measurement device would be low cost or home made, such as the one described in the answer below.

Thanks in advance.

\$\endgroup\$
5
  • \$\begingroup\$ You could run the receiver antenna output into a 50-ohm 2W resistor to ground and measure the RMS power dissipated in the resistor. I could be way wrong, which is why I am commenting rather than answering. \$\endgroup\$
    – vicatcu
    Commented Feb 24, 2019 at 23:54
  • \$\begingroup\$ Tell us more - what transmitting and receiving equipment do you have on hand? And how would you define "best?" Are the transmitter and receiver fixed in place, or handheld? \$\endgroup\$
    – tomnexus
    Commented Feb 25, 2019 at 18:23
  • \$\begingroup\$ @tomnexus : Thanks for the advice. I have added the information that you suggested. \$\endgroup\$
    – Danny
    Commented Feb 25, 2019 at 19:50
  • \$\begingroup\$ and what receiver do you use, or is that not under your control? I ask because without access to real test gear, the best thing is probably to use the receiver you already have, to compare your different antennas. Making something accurate for 150 MHz will not be simple. \$\endgroup\$
    – tomnexus
    Commented Feb 25, 2019 at 19:55
  • \$\begingroup\$ @tomnexus : I have a second handheld unit, but that is it. Both are Baofeng GT-3TP Mark III. I am researching the circuit that is outlined in the answer below, what are your thoughts on that? \$\endgroup\$
    – Danny
    Commented Feb 25, 2019 at 20:04

1 Answer 1

1
\$\begingroup\$

All you need is a tuned whip or dipole, a matched tuned amplifier with "hot-carrier" or Schottky diode, 1k ~10k load and DMM to measure square law power. Then use Friis Loss calculator to compare results with distance.

Two rooftops worked for me at work in '76 when I measured my antenna polar pattern with a turntable to plot polar gain and thus beamwidth. I had enough Tx power so that an Rx amp was not necessary.

schematic

simulate this circuit – Schematic created using CircuitLab

A half-wave dipole should be fed using a balanced transmission line matching its typical 65–70 Ω input impedance.

You do not have to tune the Rx antenna since you are only interested in relative gain.

But if desired to measure matched frequency and impedance tuning, I used a directional coupler or RF splitter to measure the reflected power using the same diode detector for null voltage. Then nearby motion reflections from people and static reflective walls will affect the result within 10 wavelengths or so declining with distance. I used a sweep generator to find resonant point then trimmed antenna at desired f. A ground plane and vertical or horizontal polarization are considerations depending on what you want to test.

In the field, for calibrated measurements, I have used omni-frequency calibrated biconical dipole antenna in an open field site with quasi-peak measuring spectrum analyzers.

If your background RF levels are greater than your received signal you can also use a radio with AGC to record levels or an SDR or use an LC tank at 131 151 MHz with Zc= 10% of R load e.g. near C~11pF 100nH 1k.

\$\endgroup\$
8
  • \$\begingroup\$ Thanks so much for the quick response! This is super helpful, I will definitely do some research on that type of circuit. \$\endgroup\$
    – Danny
    Commented Feb 25, 2019 at 3:26
  • \$\begingroup\$ I wonder who the "driveby" troll -1 was? \$\endgroup\$
    – D.A.S.
    Commented Feb 26, 2019 at 19:09
  • \$\begingroup\$ So is the power just V^2? \$\endgroup\$ Commented Oct 2, 2020 at 17:17
  • \$\begingroup\$ @LewisKelsey P= V^2/R \$\endgroup\$
    – D.A.S.
    Commented Oct 4, 2020 at 0:53
  • \$\begingroup\$ Well yes but there's no R in this case; it measures a voltage, but what R? If the DMM is after the antenna then the voltage drop over the antenna is already factored in \$\endgroup\$ Commented Oct 4, 2020 at 16:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.