Regarding amplitude measurement from I and Q samples

[Maitry07]

Hello support team,

I have implemented direct conversion method to generate I and Q samples from the RF frequency. and then I am using FPGA IP core to convert I and Q samples to amplitude measurement. I am taking digital samples of measured amplitude.
My input RF-ADC has 14 bit resolution and having very high dynamic range.

When I am applying RF input power -15 dBm to -2 dBm, I am getting accurate amplitude measurement having accuracy of 0.11 dB within 1,00,000 samples.
As I am using function generator to generate 35 MHz RF input. In order to verify the lower amplitude , I have tried attenuator of 30 dB to generate -45 dBm RF input. But I have observed that with the lower power of -45 dBm, my amplitude measurement accuracy becomes poor near to 1.9 dB within 1,00,000 samples.

What should be the reason behind that? Can attenuator of 30 dB provide very high group delay which further reduce my amplitude accuracy?
because my RF-ADC has very high dynamic range. then why I am not getting good accuracy with lower amplitude?

Instead of using function generator+attenuator to generate lower RF amplitude, should I use stable signal generator like keysight that can directly provide stable lower amplitude RF signal?

Awaited your response.

 

[kaz1]

In order to simplify your debugging try avoid concept of dBm or dB. Just measure the mean of I^2+Q^2 and compare across RF levels. Then tell us what you are getting

 

[Maitry07]

I am getting SQRT(I^2+Q^2) which is Vpk ,

From which I am using formulla for getting it into dBm

PdBm = 10*log (10. (I^2+Q^2)

 

[kaz1]

I am getting SQRT(I^2+Q^2) which is Vpk ,

From which I am using formulla for getting it into dBm

PdBm = 10*log (10. (I^2+Q^2)

Avoid that calculation as well for Forum so we can help.
Just show us the mean value of I^2+Q^2

 

[albbg]

I am getting SQRT(I^2+Q^2) which is Vpk ,

From which I am using formulla for getting it into dBm

PdBm = 10*log (10. (I^2+Q^2)

You have to convert first the result from Vpk to Vrms (even if I expected I and Q to be already rms). In case of sinewave just divide by sqrt(2).
Then calculate the power in mW:

P(mW) = 1000*Vrms^2/R

then take the log to convert to dBm:

P(dBm) = 10*Log(1000*Vrms^2/R)

usually R=50 thus:

P(dBm) = 10*Log(20*Vrms^2) = 13 + 20*Log(Vrms)

 

[Maitry07]

Avoid that calculation as well for Forum so we can help.
Just show us the mean value of I^2+Q^2

For I^2+Q^2 , when I am applying -4.7 dBm RF input, below are the max and min value for the conseutive 1,31,072 samples.

Min value (I^2+Q^2) = 0.033015493
Max value (I^2+Q^2) = 0.033976138
Difference in measurement = 0.00096044

For I^2+Q^2 , when I am applying -40 dBm RF input using attenuator of 30 dB at the output of function generator, below are the max and min value for the conseutive 1,31,072 samples.

Min value (I^2+Q^2) = 7.8827*10^-6
Max value (I^2+Q^2) = 2.9508*10^-5
Difference in measurement = 2.1625 *10^-5

FYI: When I have verified function generator output with spectrum analyzer, I have observed that with lower amplitude from the function generator, I am getting unstable output at spectrum. so I am suspecting that function generator itself is providing unstable output at the lower amplitude.

 

[kaz1]

For I^2+Q^2 , when I am applying -4.7 dBm RF input, below are the max and min value for the conseutive 1,31,072 samples.

Min value (I^2+Q^2) = 0.033015493
Max value (I^2+Q^2) = 0.033976138
Difference in measurement = 0.00096044

For I^2+Q^2 , when I am applying -40 dBm RF input using attenuator of 30 dB at the output of function generator, below are the max and min value for the conseutive 1,31,072 samples.

Min value (I^2+Q^2) = 7.8827*10^-6
Max value (I^2+Q^2) = 2.9508*10^-5
Difference in measurement = 2.1625 *10^-5

FYI: When I have verified function generator output with spectrum analyzer, I have observed that with lower amplitude from the function generator, I am getting unstable output at spectrum. so I am suspecting that function generator itself is providing unstable output at the lower amplitude.

I will assume you have divided I/Q by 2^15 to get fractional values.
-4.7dBm => 10*log10(0.033496) = -14.75 dB (average)
-40dBm => 10log10(1.8695e-05) = -47.283 dB (average)
The change of dBm is 35.3 dB
The change of dB is 32.5 dB
not that bad and if I take minimum values only it gets better.

 

[albbg]

For I^2+Q^2 , when I am applying -4.7 dBm RF input, below are the max and min value for the conseutive 1,31,072 samples.

Min value (I^2+Q^2) = 0.033015493
Max value (I^2+Q^2) = 0.033976138
Difference in measurement = 0.00096044

For I^2+Q^2 , when I am applying -40 dBm RF input using attenuator of 30 dB at the output of function generator, below are the max and min value for the conseutive 1,31,072 samples.

Min value (I^2+Q^2) = 7.8827*10^-6
Max value (I^2+Q^2) = 2.9508*10^-5
Difference in measurement = 2.1625 *10^-5

Using the formula I gave, in the case Vrms = Vpk/sqrt(2):

P(dBm) = 10 + 20*Log(Vpk)

and since your values are already squared we can apply:

P(dBm) = 10 + 10*Log(I^2+Q^2)

With your figures:

P1 = 10 + 10*Log(0.033015493)=-4.8dBm
P2 = 10 + 10*Log(0.033976138)=-4.7dBm

P3 = 10 + 10*Log(7.8827e-6)=-41.0dBm
P4 = 10 + 10*Log(2.9508e-5)=-35.3dBm

At low amplitude there is a huge difference between max and min value. What about the waveform ?
In the first post you said the low amplitude was obtained by means of an attenuator, but in post #6 it seems that the low amplitude was obtained regualating the generator itself (and it seems it's no working properly). In this last case try to lock the output power of the generator and change the level by means of attenuators. Or, better, use another generator.

 

[FvM]

Possible reasons for measurement error with lower manitude are DC offset and noise. Analysing the output with zero input will clarify.

 

[kaz1]

Using the formula I gave, in the case Vrms = Vpk/sqrt(2):

P(dBm) = 10 + 20*Log(Vpk)

and since your values are already squared we can apply:

P(dBm) = 10 + 10*Log(I^2+Q^2)

With your figures:

P1 = 10 + 10*Log(0.033015493)=-4.8dBm
P2 = 10 + 10*Log(0.033976138)=-4.7dBm

P3 = 10 + 10*Log(7.8827e-6)=-41.0dBm
P4 = 10 + 10*Log(2.9508e-5)=-35.3dBm

At low amplitude there is a huge difference between max and min value. What about the waveform ?
In the first post you said the low amplitude was obtained by means of an attenuator, but in post #6 it seems that the low amplitude was obtained regualating the generator itself (and it seems it's no working properly). In this last case try to lock the output power of the generator and change the level by means of attenuators. Or, better, use another generator.

isn't your first formula saying:
P(dBm) = 13 + 20*Log(Vrms)
Why changed 13 to 10 ?

 

[albbg]

3dB are due to conversion from peak to rms

 

[psoc77]

It is paramount to remove any DC offset in the incoming signal. Usually, it has to be digitally filtered by a DC- blocker or some other algorithm. The simple one is to calculate a sum of all incoming data, and subtract +/-1 from the signal if the sum was positive/negative. It is very slow, but OK for continuous RF signal.
Eventually, I/Q imbalance is what determines sensitivity.