# Newbie ? about FFT Bins and Resolution BW

Started by September 17, 2008
```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.....

```
```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
```