In the "ex_simrf_adc", there is a 50 Ohm connection with the voltage of 0.1 micro Volt input and it is converted to -142 dBm input and I am not able to get the exact power calculation. Could you please show me?

Hello,

In the "ex_simrf_adc", there is a 50 ohm connection with the voltage of 0.1 micro Volt input signnal and it is converted to -142 dBm power input signal. I am not able to get the exact power calculation. Could you please show me and what was the assumptions and how it was converted to -142 dBm input signal?

You can refer to the example simulink model ""ex_simrf_adc" provided by MATLAB.

Thank you,

Charles

Answer 1

[David Goodmanson]

Hi Charles,

I don't have simulink, but to convert from Vrms to dBm,

dBm = 10*log10(Vrms.^2/50) + 30;

Here Vrms.^2/50 is the power, and you use 10*log10 instead of 20*log10 because you are looking at power and not voltage which is the linear quantity. That's the answer in dB watts. To convert to milliwatts there is a factor of 10^3, which in terms of dB is extra contribution of 10*log10(10^3) = +30.

The expression gives the stardard result

1 Vrms = +13 dBm

In your case with .1 uV, things drop down by 10*log10( (10^-7)^2 ) = -140 for a final result of

.1 uV Vrms = -127 dBm

which is what the expression at the top produces.

That's 15 dB more than what ex_simrf_adc is saying, so how to explain the difference. This is a just guess, but the most likely explanation I can think of is that the voltage they are using is peak-to-peak (not rms voltage), and it's also open circuit voltage with a 50 ohm source, not voltage under the 50 ohm load. Going from open circuit voltage to loaded voltage with 50 ohms on each end drops the voltage by a factor of 1/2. Going from peak-to-peak to regular amplitude is another factor of (1/2), and from there to Vrms is a factor of 1/sqrt(2). All together, Vrms^2 is down by a factor of 32. And 10*log10(1/32) = -15 dB.