I am trying to calculate the transfer function of this circuit, with the small-signal equivalent circuit below:
It's a frequency doubler with both the gate and the source of the HEMTs driven 180° out of phase, and it says the resulting transfer function of this circuit is:$$F(\omega)=\frac{V_{gs}}{V_s}=-\frac{2-L_gC_{gd}\omega^2}{1-L_g(C_{gs}+C_{gd})\omega^2}$$Also this function is given as a middle step:$$V_g=-V_s\frac{1+L_gC_{gs}\omega^2}{1-L_g(C_{gs}+C_{gd})\omega^2}$$But I don't understand how the author calculates that from the equivalent circuit. I do however understand that you can leave out L_s and the current source of the HEMTs at nodes S1 and S2, since they are directly connected to them.
How I tried to recalculate this equation is that I tried to find the "symmetry line" of the circuit, like you do to find the transfer function in a differential circuit, and then replace the cut connections with grounds.So cutting from the lower left to the upper right, I ended up with a circuit like this:
simulate this circuit– Schematic created using CircuitLab
However, this is a voltage divider, and therefore the transfer function is this (the minus is because the V_s and V_g are driven 180° out of phase):$$\frac{V_g}{V_s}=\frac{\frac{1}{j \omega C_{gd}+\frac{1}{j \omega L_g}}}{-(\frac{1}{j \omega C_{gd}+\frac{1}{j \omega L_g}}+\frac{1}{j \omega C_{gs}})}=-\frac{L_gC_{gs}\omega^2}{L_g(C_{gd}+C_{gs})\omega^2-1}$$And this is not equal to the "middle step" function mentioned above.So how is the middle step and the transfer function calculated?