I believe I understand the role of the IF bandwidth (IFBW) in a VNA measurement:

The signal is mixed down to the intermediate frequency (IF) and the receiver detects the signal in a frequency band around that frequency whose width is the IFBW. Selecting a narrow IFBW increases the frequency selectivity of the measurement and thus reduces noise.

What I do not understand why this also decreases the measurement speed.

Does the receiver measure the time it takes for a certain amount of energy to be deposited? If that were so, then increasing the power of the probe signal should reduce the measurement time, which it does not.

What exactly is the reason that recording one data point takes the VNA longer when employing a smaller IFBW?

  • \$\begingroup\$ The tighter the filter's BW is the longer that filter takes to produce an output signal. \$\endgroup\$
    – Andy aka
    Sep 12, 2022 at 16:47

1 Answer 1


Filters take time to settle. This is a fundamental property of any frequency-selective device: the narrower the transition from low gain to high gain (and, by extension, the narrower a bandpass filter), the longer it takes for a filter to settle out after a change in input.

I'm not sure exactly what your measurement setup is, but it doesn't matter: a measurement is going to involve changing the signal that the IF "sees", waiting for the IF filters to settle, then measuring their output. For accuracy, that wait needs to be longer than the settling time of the filter -- and the settling time of the filter is going to be inversely proportional to the bandwidth.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.