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  1. #1
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    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

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  2. #2
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    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 i-th 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.



  3. #3
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    Re: How to estimate signal power from IQ samples?

    Quote Originally Posted by albbg View Post
    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 i-th 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.
    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



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    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.


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    Re: How to estimate signal power from IQ samples?

    Quote Originally Posted by FvM View Post
    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.
    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?
    (∑(I+Q)/100)/N, were N is the number of complex samples in one slot
    I assumed that R = 50Ω
    I am right now?



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    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.


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    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 (√I + Q)/(2*R*N) = Σ(abs(z)) /(2*R*N) = Σ((I + Q)/100)
    I am right?



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    Re: How to estimate signal power from IQ samples?

    Quote Originally Posted by Marcinstein View Post
    Indeed envelope is not constant so final equation for average power should be (R = 50Ω) :
    Pavg = Σ (√I + Q)/(2*R*N) = Σ(abs(z)) /(2*R*N) = Σ((I + Q)/100)
    I am right?

    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)



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