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I have a basic question about sweep times and number of points.

  • My frequency span of interest is: $f_{span} = 30\ \rm MHz$
  • My resolution bandwidth is: $RBW = 10\ \rm kHz$
  • The minimum number of points is: $n = 2\frac{f_{span}}{RBW} = 6000\ \rm points$
  • My minimum dwell time per point is: $T_s = 15\ \rm msec$

My scan time for this measurement would be:

$$t_1 = n T_s = 6000 \cdot 15\ \rm msec = 90\ \rm sec$$

  1. If I exceed the number of points in the required span by 2, can I decrease my required dwell time by 2?

$$t_2 = 90\ \rm sec = 12,000 \cdot 7.5\ \rm msec$$

  1. Can these 2 measurements on a Spectrum Analyzer be considered "equivalent"?

  2. What are the trade offs?

To give some background, I am using a software tool that fixes the number of points in a span to default value.

Since I cannot set number of points to exactly what is required, I would like to set the sweep times to the smallest allowed time. Is my reasoning correct?

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  • $\begingroup$ Why choose 15 ms per point? It's not unreasonable, but with 10 kHz RBW it could be shorter and still work. If you have twice as many points with half the time on each, the trace will be noisier (sample or RMS) / higher (max) / lower (min) / broader (normal) depending on your detector. $\endgroup$
    – tomnexus
    Jan 12 at 18:48
  • $\begingroup$ @tomnexus Thank you for your input. I am not choosing the dwell time - it is a requirement. The detector is peak. $\endgroup$ Jan 12 at 18:54
  • $\begingroup$ What's the settling time of the synthesizer? It seems at least you'd lose that. And two 7.5 ms samples of noise will have quite different statistics to one 15 ms sample. But if you state what you're doing, as you snap from 6k to 12k points, the measurement is still correct, just a bit different. The user could always push it back to 6k later (and accept the change in sweep time). R&S spectrum analysers would allow you to set all four parameters, but mark the measurement if the sampling was "incomplete" $\endgroup$
    – tomnexus
    Jan 12 at 19:01
  • $\begingroup$ Hello and welcome to ham.stackexchange.com! $\endgroup$
    – rclocher3
    Jan 12 at 20:39
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    $\begingroup$ @rclocher3 Thank you for the welcome. $\endgroup$ Jan 13 at 0:05
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Old skool basics:

RBW sets detector time: 3 periods; there is a relation with the accuracy (exponential settling). RBW of 10 kHz requires about 300 uS for measurement time of level.

RBW sets als maximum frequency step size: half the RBW, otherwise you miss power in-between points. For 10 kHz RBW the maximum step is 5 kHz.

Number of steps for 30 MHz span is consequently N = 2 * span / RBW = 6000 points.

Settling of PLL is in a good design, and depending on the required frequency accuracy 30 to 40 periods of the reference frequency. For step of 5 kHz the reference frequency must be (still old skool!) maximum 5 kHz and frequency settling requires at least 30 periods or 6 mS.

Summed-up: 6000 points * (6 mS +0.3 mS) = 37.8 seconds


SDR type receivers with a microcomputer (sufficient calculation power) and FFT analysis can do this in a fraction of this time. Don't waste time with oldskool stuff.

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  • $\begingroup$ A proper old spectrum analyser has a ramp generator with no settling time :) and it would go 40 times faster than the PLL one. $\endgroup$
    – tomnexus
    Jan 12 at 23:26
  • $\begingroup$ @F. Sessnik Ok, you seem to be saying that the dwell time per point may be lower than 15 msec and still be accurate. My constraint is that I cannot set the number of points exactly - I am must exceed these requirements. Is this lower dwell time (6 mS + 0.3 mS) still accurate for a greater number of points because the RBW is the same & PLL settling time? $\endgroup$ Jan 13 at 0:01
  • $\begingroup$ @tomnexus and oogieoogieful : yes, ramp generator is real oldskool. So theoretical fastest solution (RBW=10 kHz and span = 30 MHz) without PLL is in the order of 3000 * 0.3 mS, (assumed Bessel filter 7th order including post-detection LPF) or one second. The noise averaging is not mentioned yet, but this is about the limit of oldskool spectrum analyzers. $\endgroup$
    – F. Sessink
    Jan 13 at 11:05

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