Understanding the behaviour of a spectrum analyzer

Question

I performed in the lab some experiments regarding a spectrum analyzer and I have a few questions about its behaviour. The analyzer used was this one

• At first I had to determine the spectrum of a 4kHz sinusoidal function. I obtained the expected spectrum (one stripe at 4k). I chose a sweep velocity of 0.5s/div and pass-band filter bandwidth of 100 Hz. Now I then varied the velocity, coming to the conclusion that a velocity that was less than 0.5s/div would not produce the correct spectrum, but any slower velocity would do it. At the time I didn't fully understand what the bandpass filter bandwidth meant. I just tried 100 Hz and it worked, while 300 Hz would produce a spectrum with a not so narrow stripe and 30 Hz wouldn't work. After the lab I did some research and understood that even if ideally the bandwidth of the bandpass filter is as narrow as possible, it is limited by the sweep time. Therefore, my question is: if I had increased the sweep time (less velocity), would I be able to produce correct results with a narrower bandwidth? Also, is there a mathematical expression that relates this two?

• Then I did the same work but with a rectangular pulse of 400 Hz, 20% duty-cycle. I had the same bandwidth but now with a sweep of 0.2s/div. This was the minimum sweep time I could define and again, 30 Hz bandwidth still not works. My question is the same as before but with a new factor: how does the frequency of a signal relates to the sweep time. Because now I could define a sweep time smaller than my previous case, so my guess is that obviously the frequency of the signal matters. A signal with a smaller (fundamental) frequency can be swept with a bigger velocity. Again, is there any expression that relates this two?

Thanks in advance!

Answer 1

This is actually said in a comment, but you have noticed in practice that feeding a sinusoidal signal through a bandpass filter generates an output sinewave which grows gradually. The frequency sweep mixing transfers a moving window of the analyzed frequency range to the passband of the I.F. filter. The faster is the sweep (Hz/second) the shorter burst of your signal goes through the I.F. filter and the lower is the built up max. oscillation amplitude at the output of the filter => the lower peak you see.

If you have some spare time you could easily find the phenomena with a circuit simulator (assuming you can construct there a bandpass filter). Feed a sinewave which starts at t=0 to a filter and see, how the output amplitude grows gradually to the final value. The growth rate depends in a complex way on the bandwidth and passband gain vs frequency curve of the filter, but the manual of your analyzer surely tells how many Hz/second the sweep rate can be if you have selected a certain resolution bandwidth and expect the measured amplitudes have a certain maximum error in decibels. Generally the allowed sweep rate decreases if you make the resolution bandwidth smaller or allow less amplitude error.

Answer 2

The envelope time response for a BPF is twice the -3dB BW of LPF .

So you would expect to use the same sweep rate for a LPF that is 1/2 of the IF BW.

There is a tradeoff between the fastest sweep time with an acceptable error and the IF + Resolution Filter BW. This ratio can be defined as a constant, k.

Below is an example of a low Q BPF filter
fo= 40 Hz, -3dB = 20 Hz.

The scope response was simulating a spectrum Analyzer with a 1 second sweep for a filter with a 20 Hz BW using a linear bidirectional sweep on a simulated scope.

The Spectrum Analyzer Local Osc settings

The IF is the final mixer Intermediate freq. selected. RBW = Resolution BW
k = constant that affects accuracy.
Bigger value = slower but more accurate.

IF BW = 20Hz (let's call this Resolution bandwidth or RBW)
Sweep range 20 to 80 Hz = 60 Hz Sweep time = 1 second
Sweep Rate = 60 cyc/sec²

Plugging in the numbers
LO Sweep rate = span/sweep time= RBW²/k

k = RBW²/ Sweep rate = 20²/60 = 0.33 [Hz²/cyc/sec²]= [cyc]

or k ~ 1/3 cycle or more for reasonable accuracy.

In your case , I have no sweep rate in Hz/s just 0.5s/division or 5 seconds per sweep time of 10 divisions. This implies you have excessive sweep range and that has a 2nd order effect on reducing sweep time.

RBW² = Sweep rate/k = Span/(10 * 0.5s/div) /k 100²Hz²= span /5 * 3

So my guess is your span was more than 10kHz or > 10000 * 5/3

If you have a separate RBW LPF on the result the formula looks like this.