amplifiererror-amplifieroperational-amplifierpid controllertransfer function
Circuit below is an op-amp summer combined with PI controller.
simulate this circuit – Schematic created using CircuitLab
Notes:
V_ref = Reference voltage
V_act = Actual voltage
V_err = Amplified error voltage
Since the current through R2 is zero, Vo = Vx. You can calculate Vo (and Vx) just using the known expression of a voltage divider:
$$ V_o = V_x = \frac{\frac{1}{sC}}{\frac{1}{sC} + R_1} V_i= \frac{1}{R_1Cs + 1} V_i. $$
Let me label your circuit nodes for better understanding of your question
simulate this circuit – Schematic created using CircuitLab
To answer your question I must first clarify some definitions:
Now,
$$ y=k\cdot e $$ $$ e=r-y $$ $$ y=k\cdot (r-y) $$
So, the error gain function follows as
$$ k=\frac{y}{r-y} $$
So, you don't even need the transfer function to establish the value of k. You just need y and r.
Anyway the transfer function is
$$ TF=\frac{y}{r}=\frac{k}{k+1} $$
So, if you're given the transfer function (note that this is just a constant for this theoretical system) you can find k as
$$ k=\frac{TF}{1-TF} $$
Note also that, with negative feedback, in stable system conditions, and with finite k:
$$ 0\leq TF<1 $$
Best Answer
If you want a pure PI controller approaching a gain A=8 you have to set C1=0 and R4/R1=8. Then, the time constant of the integrating part is determined by the product Ti=R1C2. Now - if for a specific reason you want a gain below the value of "8" for very large frequencies (beyond the zero created by the product R4C2) you can add a small capacitor C1. Thus, the corresponding time constant R4C1 should be much smaller than R4C2. Hence, C1 < C2