# Transfer function of differential controller

I found an op amp circuit that is designed to be a derivative controller (D-controller)

simulate this circuit – Schematic created using CircuitLab

and the transfer function $$\frac{V_o}{V_i} = -s\cdot \frac{1}{4}RC$$ is given.

I tried to analyse the circuit with Kirchhoff's laws but the ground potential here is really confusing. How do I solve it?

These are the equations I have so far (assuming the reference potential of $$\V_{i}\$$ and $$\V_o\$$ is $$\\phi\$$ not ground):

$$I_1 = \frac{V_i}{R + \frac{4}{sC}}$$ $$\phi = I_1 \cdot \frac{1}{2}R + \frac{I_2}{sC}$$ $$\frac{I_2}{sC} = I_3 \cdot \frac{1}{2}R + V_o + \phi$$ $$I_1 = I_2 + I_3$$

The dc gain for this circuit is zero. So the transfer function is ac only. The reference voltage, $$\\phi\$$ is ac 0V. So proceed from there.
You can include $$\\phi\$$ in the analysis. It will cancel out eventually, but it is a distraction. Set it to zero then carry on.
Apply Thevenin to the feedback voltage divider from $$\V_o\$$ such that $$\Z_f=\frac{R}{2}+Z_{Th}\$$.
Your method will work with $$\\phi=0\$$. You want$$\Z_f\$$ and $$\Z_i\$$. Treat $$\I_2+I_3=I_f\$$ then $$\Z_f=\frac{V_o}{I_f}\$$.