If i understand this correctly, the inverse of the sample time will give us the frequency resolution, which is the same as the resolution bandwidth on a spectrum analyzer, right? So if I'm going into an ADC, and i want the receive KTB to stay 14dB or so above the thermal and quantization noise of the ADC, then i can sample at 100MHz, and the band width can be at the Nyquist, or 50MHz. So then i will have a noise floor of "X" dBm/50MHz. But the problem is this "X" dBm/50MHz will be above the carrier tone of interest, of say "Y" dBm/Hz. Which is a very poor S/N ration. So can i just increase the sampled time period, so that the "X" dBm/50MHz noise floor will be reduced by 10*Log (50MHz- RBW)? So that if the sample time period is 1 second, then the noise floor will be at "X" dBm/Hz? (Normalized to 1 Hz BW) Or about a 10Log (50Mhz) = 77dB improvement? And again, this points to the FFT bin spacing as analogous to the RBW (resolution bandwidth) on a spectrum analyzer, right? Thanks much for any professional advice.....

# Newbie ? about FFT Bins and Resolution BW

Started by ●September 17, 2008

Reply by ●September 17, 20082008-09-17

On 18 Sep, 01:37, Paul <Quiller...@gmail.com> wrote:> If i understand this correctly, the inverse of the sample time > will give us the frequency resolution, which is the same as the > resolution bandwidth on a spectrum analyzer, right?Maybe. Since you are talking about an instrument, there might be differences between the theoretical foundation and the actual implementation. The term 'resolution' is also a bit ambiguous. With that in mind, I'd rephrase and say that the lobe width is given as 1/T where T is some observation duration.> So if I'm going into an ADC, and i want the receive KTB to stay > 14dB or so above the thermal and quantization noise of the ADC, > then i can sample at 100MHz, and the band width can be at > the Nyquist, or 50MHz. So then i will have a noise floor of > "X" dBm/50MHz. > > But the problem is this "X" dBm/50MHz will be above > the carrier tone of interest, of say "Y" dBm/Hz. Which is > a very poor S/N ration. > > So can i just increase the sampled time period, so that > the "X" dBm/50MHz noise floor will be reduced by 10*Log (50MHz- RBW)?In principle, yes. If the signal of interest is stationary and the noise is random, then you can increase the SNR of the spectrum by increasing the observation period. One explanation is that the energy of the signal of interest will be consentrated on a few spectrum bins, whereas the energy of the noise will be distributed over all the coefficients in the frequency band. Another explanation is that the signal energy integrates coherently over the duration of the obsrvation, while the noise integrates incoherently. The net effect is an increased spectral SNR.> So that if the sample time period is 1 second, then the > noise floor will be at "X" dBm/Hz? (Normalized to 1 Hz BW) > > Or about a 10Log (50Mhz) = 77dB improvement?Ouch! The numbers depend on what exactly goes on in the instrument you use and the settings etc.> And again, this points to the FFT bin spacing as analogous to > the RBW (resolution bandwidth) on a spectrum analyzer, right?Again, there are implementation issues with any particular technology and instrument, but the core principles are the same. But that can be said about a paper plane and a Boeing 747 too...> Thanks much for any professional advice.....Not making any claims about professionalism or usefulness... Rune