# How does one describe the “frequency response” of a mixer?

Suppose one uses a vector network analyzer to measure the $$S_{21}$$ parameter of an RF mixer over a frequency range. Would it be appropriate to call this the “frequency response” of the mixer? I find this confusing because mixers are not LTI systems, and it is therefore meaningless to speak of the frequency response of a mixer.

But, from what I understand, experimentally measuring the $$S_{21}$$ parameter through a LTI system such as a lowpass filter, amplifier, or cable precisely yields the frequency response of the system. In that case, what does the $$S_{21}$$ parameter through a mixer (not LTI) correspond to?

• – MBaz Mar 26 '19 at 14:24

A mixer is time-varying, but an ideal mixer is linear. In radio usage, a mixer is periodically varying, which actually does give you an avenue to evaluate it in terms of a frequency response.

In the time domain, in a superheterodyne or direct conversion receiver, an ideal mixer takes a received signal $$r_{rf}(t)$$ and multiplies it by a sine wave to produce an intermediate-frequency signal $$r_{if}(t) = A \cos \left (\omega_{lo} t \right)r_{rf}(t)$$. Note that the fact that this is a simple multiplication of the input by some independent quantity makes the system linear; the fact that the independent quantity isn't constant makes it time-varying.

You can take the Fourier transform of the above equation:$$R_{if}(\omega) = \mathcal{F}\left \lbrace r_{if}(t) \right \rbrace = \mathcal{F}\left \lbrace A \cos \left (\omega_{lo} t \right)r_{rf}(t) \right \rbrace$$

The Fourier transform of a multiplication is the convolution of the Fourier transforms of the multiplicands, so $$R_{if}(\omega) = \mathcal{F}\left \lbrace A \cos \left (\omega_{lo} t \right) \right \rbrace * R_{rf}(\omega)$$

The Fourier transform of a cosine wave at $$\omega_{lo}$$ is a pair of Dirac impulses at $$-\omega_{lo}$$ and $$\omega_{lo}$$, so the output of the mixer is, in turn, $$R_{if}(\omega) = \frac{A}{2} R_{rf}(\omega + \omega_{lo}) + \frac{A}{2} R_{rf}(\omega - \omega_{lo})$$

You can extend this for a mixer that multiplies by something other than a sine wave, such as a chirp or (commonly for actual RF circuits) a square wave.

So all of this was to get you ready for -- tada! -- the frequency response of a mixer. The ideal mixer responds as above. A real mixer has a frequency response that depends on what it does to the RF signal before it hits the mixing circuit, and what it does to the LO signal after it hits the mixing circuit. So -- at this level of analysis -- you can expect the output of a more real mixer to actually look something like $$R_{if}(\omega) = H_h(\omega) R_{rf}(\omega + \omega_{lo}) + H_l(\omega) R_{rf}(\omega - \omega_{lo})$$

In the nomenclature I'm using within this answer, the $$H_h(\omega)$$ or $$H_l(\omega)$$ would be your "frequency response of the mixer", depending on whether you were interested in the high-side or low-side mixing product.

The one thing that you can't do on a simple spectrum analyzer is just get the $$S_{21}$$ figure, at least assuming that the analyzer is looking at the inputs and outputs around the same frequency. You need to input one frequency, and then look for the output at the correct shifted frequency.