# antenna gain, calculating voltage scaling

I am working on a software defined radio project, but I am kind of new to antennas. There's so much info out there on antennas that I'm having trouble finding the answer to my specific question. Hopefully asking here will speed up my understanding.

Suppose I have an Antenna connected to a proper termination load (50 ohms) with all the proper transmission lines with the proper characteristic impedance. We will assume just about everything is 100% efficient for simplicity. I have an ADC that is monitoring the voltage at that termination load. The ADC is much faster than the received carrier frequencies, so I can extract actual waveforms. I emit some RF energy into my antenna at a "good" azimuth angle, say 0 degrees, which should have a gain of 0 dB according to my antenna's gain plot. I measure the voltage from my ADC over time and it shows I have a 200mV peak sinusoid. I then rotate my antenna such that the emitter is at an angle, say 45 degrees, in which my antenna's gain plot says it should be -4dB. What is the amplitude of the waveform I will measure now?

Here is my guess: The power received from 0 degrees is : P0 = (0.2/sqrt(2))^2 / R. Note that I used the RMS voltage. My power gain in..non-log(?) will be : PG = (dB/10)^10 = (-4/10)^10 = 0.0001048576. That means the power at 45 degrees will be a scaled version of 0 degrees: P45 = PG*P0. Solve for voltage using power equation: Vans = sqrt(2*P45*R). The R's will cancel when you simplify. I got an answer of 2.048 mV peak sinusoid.

• Where did you get $(-4/10)^2=0.000105$? I get 0.16. Commented Mar 14, 2016 at 23:25
• $10^{-0.4}$ is even bigger than that. Commented Mar 14, 2016 at 23:26
• Sorry, I mistakenly wrote ^2 instead of ^10. I edited original post. Commented Mar 15, 2016 at 1:15
• You should be taking $10^{x/10}$ instead of $(x/10)^{10}$ to convert from dB to power ratios. Commented Mar 15, 2016 at 2:00

$10^{\frac{4}{20}}$ = 1.585