On page 367, while discussing about current-driven passive mixer, there is this saying:
the switches in Fig. 6.39(b) also mix the baseband waveforms with the LO, delivering the upconverted voltages to node A.
I am quite confused on where does this waveform come from. I seems like the output (i.e \(V_1-V_2\)) "reflects" back on input (i.e the source current).
I checked the reference [5], which doesn't explain the problem well neither.
According to reference [5],
at the RF side, the baseband voltage appears as \(+v_{BB}(t)/2\) or \(-v_{BB}(t)/2\), depending on which switch is ON.
In Fig.2.(a), \(v_{BB}(t)\) is the voltage between a differential pair. I agree on the fact that \(v_{BB}(t)\) oscillate between positive and negative values, due to commutating of switches. On the other hand, \(v_{RF}\) is a single end voltage, and it's continuous in time, because on-time of each switch are complementary and covers the entire time exclusively.
With this doubt, I turn to simulation for better understanding. This time I use Simulink.
In this simulation, \(\omega_{RF} = 550\text{MHz}, \omega_{LO} = 500\text{MHz}\). Therefore, we can find a down-converted shape at \(50\text{MHz}\) corresponds to the voltage of differential pair on the baseband side. To make the shape of \(Z_{BB}(f)\) more distinguishable, I model the load as RC combination in series. So the shape of the down-converted signal is defined by $$Z_{BB}(s) = R+\frac{1}{sC}$$
The input source current has a clean spectrum peaks at \(550\text{MHz}\).
There exists a up-converted \(Z_{BB}(f)\) shaped spectrum at \(\omega_{RF}\), along with a image around \(\omega_{LO}\) at \(450\text{MHz}\), which corresponds to the single-ended voltage at input node.
Now the doubt becomes clear. The explanation is quite straightforward. The voltage of the current source is the product of current it supplies and load impedance. The central frequency of the current is \(\omega_{RF}\), so the resultant voltage is actually the spectra of load impedance up-converted to \(\omega_{RF}\)