Notes on power increase values
published on UK Free TV
Alan Tomlin asks: What is the relationship between increasing the power at the transmitter to the increased signal level you would expect at the receiving aerial in dB's. For example, the DTT power at switch over at the Winter Hill transmitter went up from 9166 watts to 100kw (an increase of 1000%). If for example the pre switch over received aerial signal at a particular site was 50dBuV, what signal level increase would you expect to see.
Mike Dimmick replies:
dB in power is 10 x log(10)(P1/P2). Because power is related to voltage with the equation P = VÃâÃÂ²/R, and squaring a value gives double the logarithm, dB in voltage is 20 x log(10)(V1/V2).
The Winter Hill increase was 10kW to 100kW, so the increase in dB is 10. (Brian's 'average' is an unweighted mean of the six muxes, and pretty meaningless - Mux 1 [BBC] was 10kW.)
Therefore you should expect 60 dBuV now. If you want actual values: it would be 50 dBuV into 75 ohms. The actual voltage is (10)^(50/20) = 10^2.5 = 316 uV. Power is 1.333 milliwatts.
Multiply by 10 for the increased power, multiply by 75 ohms and take the square root for the new voltage = 1mV. 1mV / 1uV = 1000, log(10)(1000) = 3, new dBuV = 20x3 = 60.
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