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I have a pretty basic doubt about how tuning works in spectrum analyzers. I'm reading about the topic from Keysight's AN 150.

The problem arises in this paragraph in page 11:

We need to pick an LO frequency and an IF that will create an analyzer with the desired tuning range. Let’s assume that we want a tuning range from 0 to 3.6 GHz. We then need to choose the IF. Let’s try a 1-GHz IF. Because this frequency is within our desired tuning range, we could have an input signal at 1 GHz. The output of a mixer also includes the original input signals, so an input signal at 1 GHz would give us a constant output from the mixer at the IF. The 1-GHz signal would thus pass through the system and give us a constant amplitude response on the display regardless of the tuning of the LO.

The sentence in bold is what confuses me. As far as I know, the product of two sinusoids of different frequencies returns two sinusoids at the difference and the sum of the original frequencies. So how is it possible that the original input and LO frequencies are present after the mixing?

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  • \$\begingroup\$ Real world mixers aren't perfect. \$\endgroup\$
    – The Photon
    Jul 16 '18 at 23:09
  • \$\begingroup\$ @ThePhoton I know, I would expect them to present IMD: sinusoids at linear combinations of the harmonics of both inputs signals. But I don't understand why the exact original frequencies persist at the output. \$\endgroup\$
    – Tendero
    Jul 16 '18 at 23:11
  • \$\begingroup\$ Yes, but it also does contain the unmixed input. (by the way, that'd also be covered by linear combinations: 1a + 0b is still a linear combination of a and b. \$\endgroup\$ Jul 16 '18 at 23:19
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A RF mixer doesn't actually multiply two signals, at least not directly. Instead, the signals are first added and then passed through a nonlinear circuit, which ideally has a quadratic relationship between input and output.

We can therefore represent the mixer by a function:

f(x)=ax^2+bx+const

And putting the sum of two signals into that function gives us:

f(s_1+s_2) = a(s_1^2 + 2s_1s_2 + s_2^2) + b(s_1 + s_2) + const

This clearly contains the original input signals. Only the product of the two input signals is desired, the rest has to be filtered out.

Most mixers produce even more frequencies than those because they're not purely quadratic but also have summands with an even higher exponent. Third-order is especially nasty to filter out.

A typical mixer uses a diode as its nonlinear element, which has (roughly) the following function, ignoring some constant factors:

enter image description here

Computing its taylor series, we get:

Diode taylor series

This clearly has b != 0, so we get some of the original input signal at the mixer's output if we don't filter it out.

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  • \$\begingroup\$ Gilbert Multipliers have RF ports and LO/chopper ports. If the LO port has exactly the same frequency and phase as the RF port, the RF signal passes thru with full strength. \$\endgroup\$ Jul 23 '18 at 0:15
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That’s from LO leakage in the mixer, the IF is always above the input fmax of the analyzer . Also linearity of LO is improved if the deviation ratio is reduced, and also reduces LO 2nd harmonic leakage thru to IF.

I was sure they must have explained this, I checked and in their words

“Because there is finite isolation between the LO and IF ports of the mixer, the LO appears at the mixer output. When the LO equals the IF, the LO signal itself is processed by the system and appears as an ...output “

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