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21. Re: [Amps] Measuring RF Power (score: 1): Author: Gary Schafer <garyschafer@comcast.net>; Date: Tue, 22 Mar 2005 16:30:31 -0500; No fault found with why average is not zero. This is the part I was responding to: I guess you meant instantaneous VOLTAGE rather than power? Otherwise you would be finding the RMS value of power. 73; /archives//html/Amps/2005-03/msg00503.html (9,438 bytes)
22. Re: [Amps] Measuring RF Power (score: 1): Author: Bill Fuqua <wlfuqu00@uky.edu>; Date: Tue, 22 Mar 2005 16:22:38 -0500; Now, how about the next example. RMS of a sine wave using only 4 samples but this time use 170 volts peak value and 144 ohms (100 watt light bulb). I have rounded the following to the nearest tenth t; /archives//html/Amps/2005-03/msg00504.html (9,373 bytes)
23. Re: [Amps] Measuring RF Power (score: 1): Author: Dennis12Amplify@aol.com; Date: Tue, 22 Mar 2005 16:51:30 EST; is No fault found with why average is not zero. This is the part I was responding to: I guess you meant instantaneous VOLTAGE rather than power? Otherwise you would be finding the RMS value of power.; /archives//html/Amps/2005-03/msg00505.html (9,060 bytes)
24. Re: [Amps] Measuring RF Power (score: 1): Author: TexasRF@aol.com; Date: Tue, 22 Mar 2005 17:16:05 EST; I power is No fault found with why average is not zero. This is the part I was responding to: I guess you meant instantaneous VOLTAGE rather than power? Otherwise you would be finding the RMS value o; /archives//html/Amps/2005-03/msg00506.html (9,976 bytes)
25. Re: [Amps] Measuring RF Power (score: 1): Author: Gary Schafer <garyschafer@comcast.net>; Date: Tue, 22 Mar 2005 17:37:36 -0500; As an additional note it is interesting that 120 volts DC on that same light bulb gives the same 100 watt power as the average AC power. The statement that you often see, "RMS has the same heating ef; /archives//html/Amps/2005-03/msg00507.html (11,287 bytes)
26. Re: [Amps] Measuring RF Power (score: 1): Author: Gary Schafer <garyschafer@comcast.net>; Date: Tue, 22 Mar 2005 17:53:21 -0500; When adding PEP to the mix it is vital that average power be understood properly. Best to put RMS power out of mind as it is good for nothing but confusion. Reading the FCC definition of PEP it goes; /archives//html/Amps/2005-03/msg00508.html (8,668 bytes)
27. Re: [Amps] Measuring RF Power (score: 1): Author: gdaught6@stanford.edu; Date: Tue, 22 Mar 2005 16:01:51 -0800; No, not all. What about P=V*I*cos(alpha) where alpha is the power factor? < snip, etc.> Yes! 73, George T. Daughters, K6GT _______________________________________________ Amps mailing list Amps@conte; /archives//html/Amps/2005-03/msg00509.html (9,215 bytes)
28. Re: [Amps] Measuring RF Power (score: 1): Author: "G0RUZ Conrad" <conrad@g0ruz.net>; Date: Wed, 23 Mar 2005 00:25:15 -0000; All this makes a good argument for measuring the change in temperature in a (known) load over time, it's far easier to understand :-) Conrad G0RUZ Next question what is a known load etc, etc! You jus; /archives//html/Amps/2005-03/msg00510.html (8,894 bytes)
29. Re: [Amps] Measuring RF Power (score: 1): Author: G3rzp@aol.com; Date: Wed, 23 Mar 2005 02:25:12 EST; Reading the FCC definition of PEP it goes something like: Peak envelope power is the AVERAGE power in one RF cycle at the crest of the modulation envelope. It is actually the ITU definition - if I re; /archives//html/Amps/2005-03/msg00514.html (8,609 bytes)
30. Re: [Amps] Measuring RF Power (score: 1): Author: R.Measures <r@somis.org>; Date: Wed, 23 Mar 2005 03:58:26 -0800; Agreed The question is indeterminate unless the waveform is stated. Not unless the waveform is stated. For example, the RMS potential of a square-wave is not .707 of E-pk. PEP measurement as used for; /archives//html/Amps/2005-03/msg00518.html (10,430 bytes)
31. Re: [Amps] Measuring RF Power (score: 1): Author: R.Measures <r@somis.org>; Date: Wed, 23 Mar 2005 04:04:48 -0800; My sis-in-law's electric water heater uses 240V-rms at c. 30A-rms. If there is no RMS power, why does the water get hot? Richard L. Measures, AG6K, 805.386.3734. www.somis.org _______________________; /archives//html/Amps/2005-03/msg00519.html (10,922 bytes)
32. [Amps] Measuring RF Power (score: 1): Author: R.Measures <r@somis.org>; Date: Wed, 23 Mar 2005 04:12:11 -0800; AsI recall, for a sinewave, the average potential is 0.636 of E-pk and the root-mean-square value is 0.707 of e-pk. AsI recall, for a sinewave, the average current is 0.636 of I-pk and the root-mean-; /archives//html/Amps/2005-03/msg00520.html (10,858 bytes)
33. Re: [Amps] Measuring RF Power (score: 1): Author: R.Measures <r@somis.org>; Date: Wed, 23 Mar 2005 04:15:59 -0800; How many watts of heat would a key-down 1500w RMS CW signal produce in a R? Richard L. Measures, AG6K, 805.386.3734. www.somis.org _______________________________________________ Amps mailing list Am; /archives//html/Amps/2005-03/msg00521.html (9,611 bytes)
34. Re: [Amps] Measuring RF Power (score: 1): Author: Gary Schafer <garyschafer@comcast.net>; Date: Wed, 23 Mar 2005 12:48:45 -0500; The water heater gets hot with 7200 watts average power. 73 Gary K4FMX _______________________________________________ Amps mailing list Amps@contesting.com http://lists.contesting.com/mailman/listin; /archives//html/Amps/2005-03/msg00527.html (10,162 bytes)
35. Re: [Amps] Measuring RF Power (score: 1): Author: Gary Schafer <garyschafer@comcast.net>; Date: Wed, 23 Mar 2005 12:52:29 -0500; If it were 1500 watts average power cw signal it would produce 1500 watts heat in the R. I don't know what you would do with 1500 watts RMS. 73 Gary K4FMX ____________________________________________; /archives//html/Amps/2005-03/msg00528.html (9,305 bytes)
36. Re: [Amps] Measuring RF Power (score: 1): Author: R.Measures <r@somis.org>; Date: Wed, 23 Mar 2005 10:11:51 -0800; Calculate P by measuring the peak-V with a NBS traceable o'scope, square E-pk, divide that by 2 x R and use it to calibrate a wattmeter. Richard L. Measures, AG6K, 805.386.3734. www.somis.org _______; /archives//html/Amps/2005-03/msg00529.html (9,969 bytes)
37. Re: [Amps] Measuring RF Power (score: 1): Author: Gary Schafer <garyschafer@comcast.net>; Date: Wed, 23 Mar 2005 13:31:06 -0500; A sine wave is considered in all of the above calculations. A sine wave is considered in all of the above calculations. Yes it is AVERAGE power. Look at the FCC's definition of peak envelope power. I; /archives//html/Amps/2005-03/msg00531.html (11,644 bytes)
38. Re: [Amps] Measuring RF Power (score: 1): Author: Gary Schafer <garyschafer@comcast.net>; Date: Wed, 23 Mar 2005 13:35:12 -0500; And that would give you average power to calibrate your watt meter at. 73 Gary K4FMX _______________________________________________ Amps mailing list Amps@contesting.com http://lists.contesting.com/; /archives//html/Amps/2005-03/msg00532.html (10,621 bytes)
39. Re: [Amps] Measuring RF Power (score: 1): Author: David Kirkby <david.kirkby@onetel.net>; Date: Fri, 25 Mar 2005 22:00:37 +0000; Rich, I think you are wrong. Here's my reasoning - I may well be wrong, as maths never was a good point of mine. You need to understand calculus to properly work out the RMS of an arbitrary waveform,; /archives//html/Amps/2005-03/msg00557.html (12,125 bytes)
40. Re: [Amps] Measuring RF Power (score: 1): Author: R.Measures <r@somis.org>; Date: Fri, 25 Mar 2005 14:26:07 -0800; The right answer is that RMS current x RMS potential = AC P Richard L. Measures, AG6K, 805.386.3734. www.somis.org _______________________________________________ Amps mailing list Amps@contesting.co; /archives//html/Amps/2005-03/msg00558.html (13,170 bytes)

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