First of all, apologies if this question has been asked before.

I'm just wondering if there is a mathematical relationship between the resolution bandwidth (RBW) and the sweep time (ST); and between resolution bandwidth and noise floor (NF) considering all other parameters remain the same. I came across some Agilent documents on Spectrum Analyzer fundamentals which stated that traditional swept SAs, the relation is given by ST = K * (freq_span/RBW^2).

However, I did a number of experiments in my N9320B - the data does not fit in to this equation for any value of K. If I do a log-log plot of RBW vs sweep time, I see that in some parts of the plot, sweep time is inversely proportional to RBW, in some other parts it's inversely proportional to RBW^2.

Thanks a lot for any help in advance!

I'm just wondering if there is a mathematical relationship between the resolution bandwidth (RBW) and the sweep time (ST); and between resolution bandwidth and noise floor (NF) considering all other parameters remain the same. I came across some Agilent documents on Spectrum Analyzer fundamentals which stated that traditional swept SAs, the relation is given by ST = K * (freq_span/RBW^2).

However, I did a number of experiments in my N9320B - the data does not fit in to this equation for any value of K. If I do a log-log plot of RBW vs sweep time, I see that in some parts of the plot, sweep time is inversely proportional to RBW, in some other parts it's inversely proportional to RBW^2.

Thanks a lot for any help in advance!

It seems you are referring to application note AN-150 on spectrum analyzer basics.

The reason it doesn’t apply to your N9320B is because the N9320B has a digital IF with digital bandwidth filters that have a 5:1 shape factor while traditional analog filters are usually in the 12:1 to 14:1 shape factor.

Digital filters can be swept much faster than analog filters. The paragraph right after the one you are quoting mentions digital filters do not follow quite the same rules and most new spectrum analyzer have digital IF’s; i.e. the X-series. Sweep time is still inversely proportional to resolution bandwidth.

Regards -