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*Subject*: [amsat-bb] Coax Attenuation*From*: "william washburn" <billwashburn@xxxxxxxxxxxx>*Date*: Sun, 8 Jul 2001 13:33:28 -0700

Folks, Thanks to Howard Long and Bob Nielsen for helping me understand how to calculate power loss in coax and the implied power/dB relationship that goes along with it. My mistake was reasoning that if 100 feet of 3dB cable lost half of the input power, that 50 feet would lose only half as much power as the full 100 foot length. I routinely use sma fixed attenuators for testing and never thought twice about putting two 3dB attenuators together to get 6dB of attenuation. By looking at the power loss with the two attenuators connected together and with an input of one watt, I finally saw what was happening. The first 3dB passes 0.5012W, a power loss of 1.0W - 0.5012W or 0.4988W. The next 3dB stage attenuates the 0.5012W from the first stage by a factor of 0.5012 passing on 0.2512W to the output. This is what one would expect from a 6dB attenuator. The loss in the second stage is 0.5012W - 0.2512W or 0.2500W. The sum of the two losses is 0.4988W + 0.2500W or 0.7488W total. Using my original thinking, the power lost in the first stage would equal half of the full power loss or 0.7488/2 which is 0.3744W. >From the first calculation, however, the true power loss in the first stage is 0.4988W. Obviously, my half power idea was wrong. I now agree that the way to find the loss of a given length of coax is to simply scale the loss in dB. After all, disconnecting my two 3dB attenuators is the equivalent of cutting a 6dB length of cable in half. Thanks, guys Bill, WA6QGR ---- Via the amsat-bb mailing list at AMSAT.ORG courtesy of AMSAT-NA. To unsubscribe, send "unsubscribe amsat-bb" to Majordomo@amsat.org

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