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I'm having a pretty difficult time understanding how tuning works when using harmonics in spectrum analyzers. I'm reading about the topic from Keysight's AN 150.

Question 1

The following plot is showed in the application note: enter image description here

[...] we first plot the LO frequency against the signal frequency axis. Multiplying the LO frequency by two yields the upper dashed line of Figure 7-3. As we did for fundamental mixing, we simply subtract the IF (5.1 GHz) from and add it to the LO second-harmonic curve to produce the 2− and 2+ tuning ranges. Since neither of these overlap the desired 1− tuning range, we can again argue that they do not really complicate the measurement process. In other words, signals in the 1− tuning range produce unique, unambiguous responses on our analyzer display.

I don't understand why it states that they don't overlap. I clearly see that for a pretty big range of the vertical axis, there will be two image frequencies: one corresponding to 1− (the desired one) and another one corresponding to 2−. In other words, for any signal frequency between 1 GHz (approximately) and 3.6 GHz, there will be two possible possible responses. So why does the text state that they don't overlap?

Question 2

This is a much more basic doubt. I'm not sure I understand the basics of how internal harmonic mixing works. Is mixing with harmonics something that happens and is undesirable, or is it something deliberate that allows us to measure higher-frequency signals?

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  • \$\begingroup\$ Mixing is essential to have a constant BW resolution sweep at some IF. In every case the sum & diff. products are created by a mixer and by design they choose the upper or lower to match the desired IF filter. This applies to both up conversion and down conversion. Here they also select the LO 2nd harmonic to extend the range of input frequencies and limit the sweep range so that the purple and Red do not overlap.. \$\endgroup\$ Jul 23, 2018 at 20:27

1 Answer 1

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In this case, it's all about the filters. The mixer is followed by a relatively narrow bandpass filter. So long as all the images can be filtered out, the spectrum analyzer will be able to unambiguously tune across the range. If there was a point where two different input frequencies are mixed and end up at the same frequency as the first bandpass filter, then it's not possible to tune unabiguously. The 1- and 2- lines show what input frequencies will be tuned to align with the first bandpass filter. Since the lines never cross, then two different input frequencies will never be converted to the first IF at the same time.

To your point 2, harmonic mixing is absolutely used to extend the range of the analyzer to higher frequencies. However, it also has to be taken into consideration within each frequency band to ensure it won't cause any problems. So, the answer is both: it can be useful, and it can be undesirable. Just like many other physical phenonmena, including resistance, capacitance, and inductance. Sometimes it's what you need, sometimes you get it for free and you have to deal with it (usually the case with parasitics).

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  • \$\begingroup\$ I got the second point. However, I'm still not getting the first one. In the Figure 7-4 of the AN I mentioned in the OP, they trace a horizontal line and due to its intersection with both lines (1+ and 1-), they say that there is overlapping. The same happens in Figures 7-5 and 7-6. They solve this with a preselector, and that I understand. But in the case I mention in the OP, I see the same overlapping that is described as a problem in the other plots, but not in this one. This is what troubles me... \$\endgroup\$
    – Tendero
    Jul 23, 2018 at 22:42
  • \$\begingroup\$ For example, if the input signal is a sinusoid at 2.5 GHz, there would be a response when the second harmonic is at 3.75 GHz and when the fundamental is at 7.5 GHz (approx.). Why isn't this a problem? \$\endgroup\$
    – Tendero
    Jul 23, 2018 at 22:47
  • \$\begingroup\$ They won't sweep the oscillator below 5 GHz, because that corresponds to DC at the input. Hence if there is an overlap before 5 GHz, it isn't relevant. \$\endgroup\$ Jul 23, 2018 at 22:50
  • \$\begingroup\$ They probably should have marked that on the graph, though! A nice vertical line where the 1- line intersects the x-axis would make it much more obvious. \$\endgroup\$ Jul 23, 2018 at 23:03

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