Dear Philippe Sir, I have corrected the background subtraction of our data. I considered data for I_dc = 0.0 mA as the background data. Moreover we have not used any preamplifier except the internal-preamplifier of the spectrum analyzer. As far as I know the gain of internal-preamplifier need not be corrected as it is automatically corrected being internal component of the Spectrum analyzer.

Dear Dinesh.

From the conceptual point of view if you have certain waveform v(t) then the effective power is given by P= integral v(t)^2 dt /R divided by the integration interval T. If you analyse a sufficiently long section of the waveforrm with time T by frequency analyzer giving the frequency components V(f)  or the frequency spectrum of the waveform. Then the integration V(f)^2/2 R.df on the whole frequency range  is the power of the signal since.The two is added because the the waveforms out of the frequency analyzer will be sinusiods. 

v(t) = The summation of all frequency components.

Dear Zekry Sir, thanks for your precious time.

I assume symbol v(t) you have used is the time dependent voltage. However in our measurement technique using the spectrum analyzer the output is power in dBm not the voltage. What will be your formula in that case? Moreover no where you have mentioned role of Bandwidth of the measurement, which is I think is a very important parameter in the calculations.

Thanks.

Dear Dinesh,

Not at all. It is my pleasure to introduce help when it is possible.

In case of the spectrum analyzer you have V(f) since the spectrum analyzer gives it  in dbm. Yes, then you have to sum up all the powers from all frequency components displayed by the analyzer covering the whole frequency range. Please revert to the definition of power as stated above. Making the summation on the bandwidth means multiplying by it. 

Then you have only to make a summation of all power on the whole bandwidth you get the total power in dbm provided that the resistance is taken as you really have eg 50 ohm.

You can calculate the power in mw by Pmw= 10^X where x is the power in dbm, then you add all powers in mw and reconverting the total power to dbm by Pdbm= 10 log Pmw.

hope it helps

Dear Prof. Zekry,

Thanks for your help. Indeed thanks a lot.

Could you please provide me some reference (text book link or research paper) about power calculations so that I can validate everything before making the final call?

One thing to keep an eye on is the matter of temperature differences across instrument (or simulation) boundaries. For example, if a simulation of a low-noise amplifier driven from a presumed 50-ohm source is carried out, with the whole operating at some (say) elevated temperature, and then one attempts to calculate the noise figure (NF), it will be incorrect, because unless suitable arrangements are made to keep the source resistance fixed at 50 ohms as temperature is varied, the internal noise voltage of this source will also increase with root-T and thus invalidate the NF result.

Barrie

Dear DINESH,

It is my pleasure to introduce some help to you. I am sorry for the late return to the question. You can revert to this outstanding book: digital communications by Iva Glover and Peter Grant,. Especially chapter two will be useful to you.

wish you success