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Relation between RBW, Noise Floor and Sweep Time

Question asked by fieldfox_user on Aug 10, 2010
Latest reply on Aug 11, 2010 by tabbott
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!