0
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ Where did you get \$(-4/10)^2=0.000105\$? I get 0.16. \$\endgroup\$ – The Photon Mar 14 '16 at 23:25
  • \$\begingroup\$ \$10^{-0.4}\$ is even bigger than that. \$\endgroup\$ – The Photon Mar 14 '16 at 23:26
  • \$\begingroup\$ Sorry, I mistakenly wrote ^2 instead of ^10. I edited original post. \$\endgroup\$ – user2913869 Mar 15 '16 at 1:15
  • \$\begingroup\$ You should be taking \$10^{x/10}\$ instead of \$(x/10)^{10}\$ to convert from dB to power ratios. \$\endgroup\$ – The Photon Mar 15 '16 at 2:00
1
\$\begingroup\$

-4 dB in power is also -4 dB in voltage (since you are comparing voltages across the same impedance, 50 ohms).

In converting a voltage ratio we use 20log. In converting a power ratio we use 10log. Thus the voltage will be reduced by a factor of 10^(-4/20) = 0.631. Thus the 200 mV peak sinewave, reduced by -4 dB, will become 126 mV peak.

\$\endgroup\$
  • \$\begingroup\$ +1. I hope you think my little tweak is an improvement. If not, feel free to revert. \$\endgroup\$ – davidcary Mar 15 '16 at 16:29
1
\$\begingroup\$

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?

If your unrotated-antenna signal is 200 mV then after rotation it is 4 dB lower. 4 dB is a numerical reduction of: -

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

So your 200 mV signal reduces to 126 mV

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.