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Ian White, G3SEK wrote:
V(v) and V(i) are *added* - vectorially, as two RF voltages connected in series - and then the sum voltage is detected. No logs are taken and they are not multiplied together in any other way. The detected voltage (now DC) is squared by the meter scale law to indicate power. Yes, when you add two values on a slide rule, you are accomplishing multiplication because the scales are logarithmic. When you add two values of voltage, you are multiplying them if the meter scale is properly calibrated. Log(V1^2) = Log(V1*V1) = Log(V1) + Log(V1) You were caught in a totally incorrect statement about the meter scale law, but you can't stop dodging. Just read again what you wrote: I still don't understand why log(V1)+log(V1) doesn't equal log(V1*V2) which doesn't equal log(V1^2). Here's an example: Let's say we have a slide rule where the distance between the '1' index and the '2' is one inch. We set the index on '2' and read the value of the product off the '2' on the other scale. All we have done is add one inch to one inch and everyone knows that is two inches. But because of the logarithmic scale we read '4' as the answer. Thus the answer to '2' times '2' is '4' even though all we did was physically add one inch to one inch. How can one inch plus one inch yield an answer of '4'? The same way one volt plus one volt can yield an answer of 200 watts. Let's say one volt inside the Bird equals 100 volts on the transmission line and one volt inside the Bird equals 2 amps on the transmission line. When the meter measures two volts, we calibrate the Bird at 200W. When the meter measures four volts, we calibrate the Bird at 800W. That's similar to the logarithmic calibration of the slide rule. We are adding one volt to one volt but we are reading the product off the wattmeter scale. are you saying that log(V^2) is not equal to 2*log(V)? I say log(V^2)=2*log(V), i.e. the square law power function is a linear function using logarithms. That's the basis of the Bird wattmeter design. I would certainly like to see a proof that log(V^2) is not equal to 2*log(V). (So would a lot of people wearing white coats. :-)) So "The square law power function is a linear function using logarithms". Of course. Log(V1^2)= Log(V1)+Log(V1) does it not? Is that not a linear function? That's how slide rules work. You add logarithms to accomplish multiplication. What am I missing? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 80,000 Newsgroups - 16 Different Servers! =----- |
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