Regarding power estimation from I and Q samples

[Maitry07]

Hello support team,

I am using high speed data converter , in which there is an inbuilt RF-ADC along with DDC. which can directly digitize RF input and generate I and Q samples accordingly at DC.
Below is the example.
RF frequency=35 MHz, NCO frequency =34 MHz, DDC output = I and Q samples at 1 MHz.
Now I am capturing these I and Q samples in signed decimal format. and stored it in the excel.

My RF input power: -25 dBm
DDC output I and Q sample example when my RF input is -25 dBm
I_sample (signed decimal value) = 1785
Q_sample (Signed decimal value) = 687
Pdbm = 10log(I^2+Q^2) = 65.6 dBm

So, is there any calculation I am missing? PdBm should come -25 dBm. My DDC output is in the FIX16_0 format, from which I have taken the signed decimal values. could you please guide on this?

Is there any requirement to scale down the I and Q samples by fixed scaling factor to get the correct pdBm result?

 

[KlausST]

Hi,

Makes sense now.

Btw: don't use 7+ decimal digits on dB values.
Even the 16 bit result gives just 5 decimal digits of resolution. This also leads to 5 decimals in db value resolution.
Best is to adjust the number of digits to the expected precision, not the resolution.

What precision do you expect from your measurement equippment?

Klaus

 

[kaz1]

Hello support team,

I am using high speed data converter , in which there is an inbuilt RF-ADC along with DDC. which can directly digitize RF input and generate I and Q samples accordingly at DC.
Below is the example.
RF frequency=35 MHz, NCO frequency =34 MHz, DDC output = I and Q samples at 1 MHz.
Now I am capturing these I and Q samples in signed decimal format. and stored it in the excel.

My RF input power: -25 dBm
DDC output I and Q sample example when my RF input is -25 dBm
I_sample (signed decimal value) = 1785
Q_sample (Signed decimal value) = 687
Pdbm = 10log(I^2+Q^2) = 65.6 dBm

So, is there any calculation I am missing? PdBm should come -25 dBm. My DDC output is in the FIX16_0 format, from which I have taken the signed decimal values. could you please guide on this?

Is there any requirement to scale down the I and Q samples by fixed scaling factor to get the correct pdBm result?

dBm is not same as dB. If your RF is -25dBm then expect digital power to be different,
your equation 10log(I^2+Q^2) = 65.6 dBm should be dB not dBm.

If you are after digital power mapping to dBm then you need to average it then compare provided you ADC is swinging correctly.

 

[Maitry07]

Hi,

Makes sense now.

Btw: don't use 7+ decimal digits on dB values.
Even the 16 bit result gives just 5 decimal digits of resolution. This also leads to 5 decimals in db value resolution.
Best is to adjust the number of digits to the expected precision, not the resolution.

What precision do you expect from your measurement equipment?

Klaus

Hi,

Actually, my digital output is 16 bits ( out of which 1 sign bit, 1 integer bit, 14 fraction bits). Based on which I was calculating.

I am planning to provide these digital information to be interface with at least 14 bit DAC.
So, for the 14 bit DAC , so if my DAC is having 0 to 10 V range , then 10/2^14 = 0.61 mV

Again , as now my above understanding for calculating dBm power from measured amplitude is correct, Then I need to use digital gain compensation, right?

 

[Maitry07]

Is there any feedback on the above?

 

[Maitry07]

Your full scale amplitude is 32767 = -2.2 dBm.
I = 1785 and Q = 687 is amplitude 1913.
dB from fullscale is 20*log(1913/32767) = -24.7 dB
If full scale is -2.2 dBm, your signal is -2.2 - 24.7 dBm = -26.9 dBm

@std_match , I have found one observation by using https://www.tek.com/en/blog/calculating-rf-power-iq-samples, that formula for the conversion of Vpk (amplitude) to PdBm is
PdBm = 10* log( 10 * (I^2+Q^2) ) , multiplication of 10 is inside the log .

PdBm = 10* log ( 10* 0.003407162213568) = -14.6 dBm . So this does not make any sense. why I am getting this kind of result?

0.0034071 is the square of 1913/32767 (SQRT(I^2+Q^2)). Could you please review, where is the mistake?