How do you measure the two to one bandwidth of an antenna using a Bird Watt meter? Tom Planer KJ9P

  • $\begingroup$ Hello Tom, and welcome to ham.stackexchange.com! $\endgroup$
    – rclocher3
    Jun 22 at 15:03
  • $\begingroup$ Tom, a wattmeter is used with a transmitter, and you are asking about only using a receiver. I'm confused. Can you please clarify this? $\endgroup$
    – Mike Waters
    Jun 22 at 21:49

First, receivers aren't usually as picky about SWR as transmitters. An SWR mismatch to a receiver causes signal loss, but the signal and the noise are usually attenuated about the same, so all a receiver needs to do to make up the difference is apply more gain. Receivers have plenty of gain available, so there is usually no problem. On the other hand, an SWR mismatch causes more trouble for a transmitter, because it makes the final amplifier stage work harder for the same amount of output power, often at a higher temperature and voltage.

Next, there is no good way to measure the bandwidth of an antenna with just a receiver. On HF, one can often roughly tune a transmatch (antenna tuner) by adjusting the transmatch for the loudest signal in a receiver, but that procedure doesn't tell the operator anything about the bandwidth. To measure SWR, one must generally transmit.

Maybe you meant to say you wanted to check the SWR with a transceiver or a transmitter and a Bird wattmeter. I'll proceed assuming that's what you meant.

Many ham wattmeters read SWR directly, but Bird wattmeters don't. Not to worry, you can calculate the SWR from the forward and reflected power measured by your Bird.

SWR, more properly Voltage Standing Wave Ratio (VSWR), can be defined as follows:

$$ VSWR = \frac{1+\sqrt{P_r/P_f}}{1-\sqrt{P_r/P_f}} $$

You'd like to know what $P_r/P_f$ is for a VSWR of 2. Solving algebraically, when VSWR is 2, $P_r/P_f$ is 1/9. So when the forward power is nine times the reverse power, then your SWR is 2. So on the band of interest, transmit at various frequencies and measure the forward and reverse power. Find the frequency where the reverse power is lowest, which is where your SWR is lowest of course. Then tune away from that frequency in each direction until your reverse power has risen to one-ninth of your forward power, and record those two frequencies where your measured SWR is 2. The difference between those two frequencies is your 2:1 bandwidth.

Of course that assumes that your wattmeter is perfectly accurate, which it isn't. Bird wrote an article describing some of the issues involved. Their article doesn't seem to be available on their website any longer, but there's a copy of it here. One of the points that the article makes is that "downscale" measurements, that is measurements with the needle at the low end of the scale, are often not as accurate as "upscale" measurements, because typically the rated accuracy of a wattmeter that reads forward and reverse power directly is given as a percentage of the full scale, such as 5%. I'll add that the wattmeter accuracy is probably out of specification, worse than 5%, for a wattmeter and an insert that haven't been calibrated in many years, like a typical hamfest Bird model 43. So your forward power measurements may be acceptably accurate, but your reverse power measurements may be less accurate because they are usually "downscale" measurements made with the same insert used to measure the forward power.

For instance, say you're measuring 100 W of forward power and 11 W of reverse power, which should be an SWR of 2, with the same element that has a full scale of 100 W and a rated accuracy of 5% of full scale. (Let's assume that the calibration is good.) 5% of 100 W is 5 W, so your forward power could be anywhere from 95 W to 105 W and your reverse power could be anywhere from 6 W to 16 W. The actual SWR could be as good as 1.6 or as bad as 2.4. Or if the meter and the insert need calibrating, then the accuracy could be worse. So keep that in mind.

  • 1
    $\begingroup$ You could theoretically estimate the bandwidth in a method similar to tuning a loop. Assuming a uniform AWGN background (which is not a bad assumption on a narrow bandwidth), you take a bazillion measurements at various frequencies and average them out to get the smoothed rx strength based on the background noise. All theory though, not sure if it would actually work in practice. $\endgroup$ Jun 25 at 15:22

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