+ Post New Thread
Results 1 to 8 of 8

24th March 2020, 20:47 #1
 Join Date
 Dec 2019
 Posts
 11
 Helped
 0 / 0
 Points
 139
 Level
 1
How to estimate signal power from IQ samples?
Hi,
I am struggling with the problem on how to calculate power with given IQ samples. Everyone says different ways of doing it. I need just a value not, the PSD. My algorithm goes as follows:
1) Calculate the sum of absolute values of IQ samples: A = sum(abs(samples[i])) , where samples is the vector w N complex samples
2) Average and RMS calculation so: B = A/(N*sqrt(2)) , where N  amount of complex samples
3) B is Vrms so using P = (Vrms)^2/R formula, I am calculating the power: P = (B)^2/R
I am working as a programmer and during writing code I found different ways of calculating a power. Please help me to figure out proper way.
Thanks in advance
Marcin

Advertisement

25th March 2020, 00:22 #2
 Join Date
 Nov 2009
 Location
 Italy
 Posts
 1,196
 Helped
 406 / 406
 Points
 8,816
 Level
 22
Re: How to estimate signal power from IQ samples?
if samples(i) = I(i)+jQ(i) then you method is correct.
You calculate the mean rms voltage over a given number of samples and from it the power.
You could also calculate before the power "P(i)" of the ith sample then calculate the average of the vector P(i).
The first one should be less sensitive to isolated peaks over the I/Q samples.

25th March 2020, 02:26 #3
 Join Date
 Dec 2019
 Posts
 11
 Helped
 0 / 0
 Points
 139
 Level
 1
Re: How to estimate signal power from IQ samples?
Hi,
Thank you for your reply. But your second suggestion will give different result. I've tried this on 100 random samples in MATLAB. I believe that I understood you correctly, you suggested to calculate power for each sample and then sum all result and divide by N ? Kinda averaging but based on instantaneous sample power?
Regards

Advertisement

25th March 2020, 16:17 #4
 Join Date
 Jan 2008
 Location
 Bochum, Germany
 Posts
 46,597
 Helped
 14176 / 14176
 Points
 266,542
 Level
 100
Re: How to estimate signal power from IQ samples?
Averaging instantaneous signal magnitude as described under 1) would work if the magnitude is constant (CW or FM signal). It doesn't work for a modulated signal where you need to average instantaneous signal power.
1 members found this post helpful.

25th March 2020, 17:33 #5
 Join Date
 Dec 2019
 Posts
 11
 Helped
 0 / 0
 Points
 139
 Level
 1
Re: How to estimate signal power from IQ samples?
Thank you for reply,
Indeed I want to calculate the power of modulated signal. This is wideband OFDM modulation. So are you suggesting to calculate instantaneous power for each sample, and then sum
and average it?
(∑(I²+Q²)/100)/N, were N is the number of complex samples in one slot
I assumed that R = 50Ω
I am right now?

Advertisement

25th March 2020, 22:50 #6
 Join Date
 Nov 2009
 Location
 Italy
 Posts
 1,196
 Helped
 406 / 406
 Points
 8,816
 Level
 22
Re: How to estimate signal power from IQ samples?
Yes, FvM is right.
I wrongly presumed your signal was a constant envelope one. In that case the two methods lead to the same result, but using the first one the average over the voltages tends to filter out any spikes.
However it's not correct to use it with modulated signals.
1 members found this post helpful.

26th March 2020, 13:36 #7
 Join Date
 Dec 2019
 Posts
 11
 Helped
 0 / 0
 Points
 139
 Level
 1
Re: How to estimate signal power from IQ samples?
Indeed envelope is not constant so final equation for average power should be (R = 50Ω) :
Pavg = Σ (√I² + Q²)²/(2*R*N) = Σ(abs(z))² /(2*R*N) = Σ((I² + Q²)/100)
I am right?

21st May 2020, 16:31 #8
 Join Date
 Aug 2019
 Posts
 8
 Helped
 2 / 2
 Points
 186
 Level
 2
Re: How to estimate signal power from IQ samples?
For any digital signalL mean Amp^2 = mean(I^2+Q^2) i.e. sum of (I^2+Q^2)/number of samples.
This is passed through 1 Ohm becomes mean power since P = A^2 *R (Ohms law)
   Updated   
edit: For any digital signal mean Amp^2 = mean(I^2+Q^2) i.e. sum of (I^2+Q^2)/number of samples.
This if passed through 1 Ohm becomes mean power since P = Amp^2 *R (Ohms law)
+ Post New Thread
Please login