I noticed that if I set IF bandwidth low, the measures done with a vector analyzer are better (with higher SNR) but it is slower.

Precisely, what does it happen when I change it? Can you explain me simply what it represents?

  • 2
    \$\begingroup\$ Set IF bandwidth high and the SNR goes up and sweep time goes down? Are you sure it's not the opposite (smaller IF bandwidth increases sweep time and SNR)? \$\endgroup\$
    – Shamtam
    Nov 17, 2019 at 1:58
  • 2
    \$\begingroup\$ SNR is usually 1/sqrt(BANDWIDTH). \$\endgroup\$ Nov 17, 2019 at 2:02
  • \$\begingroup\$ Sorry, I wrote it in the opposite way \$\endgroup\$
    – Kinka-Byo
    Nov 17, 2019 at 3:18

2 Answers 2


Many receivers, such as your vector analyzer, use a superheterodyne receiver (https://en.wikipedia.org/wiki/Superheterodyne_receiver) architecture. These are distinct from other receivers in that they use an Intermediate Frequency (IF), or multiple IFs, as opposed to either just directly using the RF coming in, or down converting the signal to baseband (i.e. from DC, or near DC, and up rather than bandpass around some carrier frequency).

Some of the key advantages of doing this are: filters can be made more selective at lower frequencies, it's easier to make tuneable oscillators than it is amplifiers and filters, and lower frequency components tend to be cheaper and perform better. These advantages also apply to direct conversion down to baseband, but there are other challenges there.

So in your VNA:

What it does is measure the response of your network over a bit of the spectrum at a time.

It sends out RF signals at some frequency and analyzes the reflections. A tuneable oscillator is fed into a mixer with the incoming reflections. This down-converts the reflections from their RF frequency to a fixed IF.

This allows filters and amplifiers designed to operate at a fixed IF to be used across the whole range. Making the bandwidth of this IF section smaller lets you look at a smaller piece of the spectrum at a time. This has two impacts:

  1. Generally less spectrum will have less noise in it
  2. To look through the whole range you want to it, it will take more time to do the sweep

Edit: I suppose also the rise time of the IF signal's envelope would be slower with a narrower bandwidth.

  • \$\begingroup\$ Thank for you answer. Can you give me the precise definition of sweep time? \$\endgroup\$
    – Kinka-Byo
    Nov 18, 2019 at 10:38
  • 1
    \$\begingroup\$ Don’t guess, rise time will be (10~90%) = 0.35 x 2 / IF BW \$\endgroup\$ Nov 18, 2019 at 14:11
  • \$\begingroup\$ "It sends out RF signals at some frequency and analyzes the reflections". Not always. What you're describing is an S11 measurement. VNAs are also used to analyze the through path characteristic, or S21. \$\endgroup\$
    – SteveSh
    Sep 12, 2022 at 23:08

SNR is a power ratio of Ps/Pn, but as the voltage ratio it would be Vs²/Vn².

If we look a carrier sweeping thru a BPF, we take one half of -3dB BW to get the single sided Low Pass equivalent time response. enter image description here

This may be simplified and another way of looking at it is;

The Spectrum Analyzer Local Osc, LO Sweep rate = span/sweep time= RBW²/k

The IF is the final mixer Intermediate freq. selected.
RBW = Resolution BW
k = constant that affects accuracy.

further reading

  • \$\begingroup\$ Thank you very much, a last question. Which is the precise definition of sweep time? I'm not sure to know it correctly \$\endgroup\$
    – Kinka-Byo
    Nov 18, 2019 at 10:43
  • \$\begingroup\$ Time to sweep from start to end in ms \$\endgroup\$ Nov 18, 2019 at 14:07

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.