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measuring cable loss
On Aug 9, 10:17 pm, "Jimmie D" wrote:
"Owen Duffy" wrote in message ... Jim Lux wrote in news:f9gg4i$7q9$1 @nntp1.jpl.nasa.gov: ... Jim good points and all noted. Jimmie hasn't give a lot of detail about the specification he is apparently trying to meet. Reading between the lines, it might be an EIRP, and assuming a given antenna gain, he is trying to calculate the permitted transmitter power output. Not only is the uncertainty of practical service equipment an issue in tenth dB accuracy, but no mention has been made of transmission line loss under mismatch conditions, and mismatch loss. Jimmie, if you have a plausible story to tell the regulator, then that might suffice. If you have assessed the Return Loss of a rho=1 termination, then you could use that and the measured Forward and Reverse power using say a Bird 43 at the transmitter end of that known line loss (being half the return loss) to calculate the power absorbed by the load. The calculator athttp://www.vk1od.net/tl/vswrc.phpdoes just that. The calculator at http://www.vk1od.net/tl/tllc.phpcould be used to calculate the expected RL of the o/c or s/c line section, just specify a load impedance of 1e6 or 1e-6 for each case. For example, at 1GHz, the RL of 200' LDF4-50A with a 1e-6 load is 8.9dB, and if you got much higher than that, you might suspect the cable to be faulty. Tenths of a dB, remember that most service type power meters are probably good for 6% to 10% of FSD, so I will go with Jim's 1dB accuracy. BTW, directional wattmeters for the ham market are often not capable of reasonable accuracy on loads other than the nominal 50 ohm load. There are a range of tests that such an instrument should satisfy, but for hams, it is usually considered sufficient if the "reflected" reading is approximately zero on a 50 ohm load. Owen I think I have given enough info. But I will try yo expess it in another way. Power delivered to the antenna but be maintained with in +- 1 db in this case that power is 100 watts. Power is normally checked at the TX and recorded after allowing for line loss as "power at the antenna". Power checks are done on a weekly basis. Once a year the line loss is measured and this value is used to subtract from the power at the transmitter for the rest of the year. With this in mind it would be most prudent to measure the cable loss accurately. to establish the annual benchmark. Considering the test equipment I have available to use in a temperature stablized building an Agilent network analyzer or use an old HP power meter at the top of the tower I am thinking that measuring rho of the cable while terminated in a short may be the more accurate way to go. Jimmie As I mentioned before, be sure the cable is really 50 ohm (assuming your instruments are calibrated to 50 ohms), or at least determine what it is. Make your rho measurement; at that length of line, you can adjust the frequency of measurement over a small range and get values for rho at angles of 0 degrees and at 180 degrees. I will assume that the cable is 50 ohms and the cable attenuation changes practically none between the two readings, so the readings will be the same. Now without changing anything, measure an attenuator with nearly the same attenuation your think the cable has, also open- circuited/shorted at the output. If the attenuator has the same attenuation as the line, you should get the same value. You can then have that attenuator calibrated at 1GHz to make sure it's correct. Because your measured attenuation is twice the line attenuation, you will get the line within 1dB if the measurement is within 2dB. It shouldn't be very expensive to get a couple attenuators that would bracket the line loss, and have them calibrated, and expect that they would hold the calibration for a relatively long time if they aren't mistreated. Seems like we never see much variation from one cal to the next of decent attenuators. As Jim noted, beware of environmental changes. I don't think that dimensional changes will much matter, but the copper resistance will, some. The effect, though, is not nearly as much as Jim suggested, because of skin effect: a 1 degree C change causes the DC resistance to change by 0.4%, but the AC resistance changes by only 0.2%. Since the dB attenuation due to copper resistance is linear with resistance, if the line attenuation is about 3.5dB, you'd need a 10% change in AC resistance to see an 0.35dB change in attenuation. That's a 50 degree C change, perhaps worth worrying about if you're in an extreme climate. Looking at it another way, it's about 0.007dB/degree C. It's probably worth making a point to measure the line loss at or near the temperature extremes it experiences, though that would mean climbing the tower at a couple times you might least like to. Be sure moisture doesn't get into the line! Cheers, Tom |
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