I want to calculate the transfer function and the gain, zeros and poles of the transfer function, considering that: $$R_1=R_2=R_3$$ $$C_1=C$$ I want to know if this circuit is a P, or PI, or PD, or PID controller or a lag/lead compensator.

To get to the transfer function, I think AMPOP entries must be matched. I think that is a lead controller, but I'm not sure. How can I solve?

enter image description here

My work so far:

$$\cfrac{v_0-v_i}{R+\bigg(\cfrac{1}{R}+\cfrac{1}{Z_C}\bigg)^{-1}}=\cfrac{v_i}{R}\Leftrightarrow \cfrac{v_0-v_i}{R+\bigg(\cfrac{1}{R}+sC\bigg)^{-1}}=\cfrac{v_i}{R}\Leftrightarrow \cfrac{v_0-v_i}{R+\bigg(\cfrac{1+sCR}{R}\bigg)^{-1}}=\cfrac{v_i}{R}\Leftrightarrow \cfrac{v_0-v_i}{R+\cfrac{R}{1+sCR}}=\cfrac{v_i}{R}\Leftrightarrow \cfrac{v_0}{R+\cfrac{R}{1+sCR}}=\cfrac{v_i}{R}+\cfrac{v_i}{R+\cfrac{R}{1+sCR}}\Leftrightarrow v_0\Bigg(\cfrac{1}{R+\cfrac{R}{1+sCR}}\Bigg)=v_i\Bigg(\cfrac{1}{R}+\frac{1}{R+\cfrac{R}{1+sCR}}\Bigg)\Leftrightarrow \cfrac{v_0}{v_i}=\cfrac{\cfrac{1}{R}+\cfrac{1}{R+\cfrac{R}{1+sCR}}}{\cfrac{1}{R+\cfrac{R}{1+sCR}}}=\cfrac{\cfrac{1}{R}}{\cfrac{1}{R+\cfrac{R}{1+sCR}}}+1\Leftrightarrow \cfrac{v_0}{v_i}=\cfrac{1}{R}\times \big(R+\cfrac{R}{1+sCR}\big)+1=2+\cfrac{1}{1+sCR}=\cfrac{2+s2CR+1}{1+sCR}=\cfrac{3+s2CR}{1+sCR}$$


$$1+sCR=0\Leftrightarrow s=-\frac{1}{CR}$$


$$3+s2CR=0\Leftrightarrow s=-\frac{3}{2CR}$$


3 Answers 3


Well, that is more related to circuit theory than control. You probably know that:

  • the impedance of a capacitor can be written as \$ Z_C = \frac{1}{sC}\$,
  • the amp op will behave as if having a virtual ground between the two inputs,
  • that amp op configuration is an non-inverting amplifier.

So write down the KCL equations of the circuit, or just use the gain formula for non-inverting amplifiers (fixed in the current edit):

$$ v_i = \frac{R_3}{R_3 + R_1 + R_2||Z_C} v_o.$$

The transfer function will be

$$ \frac{v_i}{v_o} = H^{-1}(s) = \frac{R_3}{R_3 + R_1 + (1/R_2 +sC)^{-1}},$$

$$ H^{-1}(s) = \frac{R}{2R + (1/R +sC)^{-1}} = \frac{R(1/R +sC)}{2R(1/R +sC) + 1}, $$

$$ H^{-1}(s) = \frac{(1 +sRC)}{2(1 +sRC) + 1} = \frac{(1 +sRC)}{(3 +s2RC)},$$

$$ H(s) = \frac{(3 +s2RC)}{(1 +sRC)} = k_1 \frac{\frac{3}{2RC} +s}{\frac{1}{RC} +s} = k \frac{1 + \frac{2RC}{3}s}{1 + RCs},$$

Where \$k_1\$ is some gain, and \$ k = H(0)\$. This transfer function has a pole and a zero, and it is not of the form $$ PID(s) = k_p + k_i/s + k_d s,$$ since it has no purely derivative term (although in real systems they will never will due to sensor error amplification), nor it has an ideal integrator. In fact, it seems to be some compensator.

  • \$ \frac{1}{CR} > \frac{3}{2CR} \Leftrightarrow \$ lead compensator,
  • \$ \frac{1}{CR} < \frac{3}{2CR} \Leftrightarrow \$ lag compensator.

The gain will be \$k = H(0) = 3\$.

Obs: \$R_2||Z_C\$ is the equivalent of \$R_2\$ and \$Z_C\$ in parallel.

  • \$\begingroup\$ Thanks for the answer but my problem is that I don’t know the solution, so despite trying to do the exercise I’ll never know if I calculated it well. Could you tell me the final expression of the circuit, the gain, poles and zeros and the type of controller to try to reach the solution please and so I can confirm? \$\endgroup\$ Jan 25, 2020 at 0:43
  • \$\begingroup\$ Please, post the work you have done so far, so other people can also chime in. \$\endgroup\$
    – jDAQ
    Jan 25, 2020 at 0:44
  • \$\begingroup\$ I put in the post what I have done so far, but I don't know if I calculated it well and I don't know how to calculate the gain, nor know what type of controller it is. \$\endgroup\$ Jan 25, 2020 at 1:29

The transfer function of this circuit can be derived by inspection using the fast analytical circuits techniques or FACTs. Simply determine the time constant of the circuit in two conditions: when the stimulus is zeroed and when the output is nulled. Assemble the time constants and you have the transfer function in one shot. See the below sketch to follow the steps:

enter image description here

The dc gain \$H_0\$ is determined by opening the capacitor: you have a non-inverting op-amp whose gain is \$H_0=\frac{R_1+R_2}{R_3}+1\$. If all resistances are of equal value, \$H_0=3\$.

For the pole, reduce the input voltage to 0 V and replace the source by a short circuit. Now, "look" at the resistance offered by the capacitor's terminals when it is temporarily removed. This is the upper right sketch. The resistance is \$R_2\$ hence a time constant \$\tau=R_2C_1\$ and a pole \$\omega_p=\frac{1}{R_2C_1}\$. This pole becomes \$\omega_p=\frac{1}{RC_1}\$ when all resistances are \$R\$.

For the zero, either you calculate the gain \$H^1\$ of the circuit for \$s\$ approaches infinity (\$C_1\$ is replaced by a short circuit) or you apply a null-double injection or NDI (lower right sketch). Using either way, you find a zero located at \$\omega_z=\frac{1}{R_2||(R_1+R_3)C_1}\$. This zero becomes \$\omega_z=\frac{1}{\frac{2}{3}RC_1}\$ when all resistances are \$R\$.

The complete transfer function is thus: \$H(s)=H_0\frac{1+\frac{1}{\omega_z}}{1+\frac{1}{\omega_p}}\$.

Considering these expressions, the zero will always be higher than the pole. In my opinion, this circuit will always lag. The below Mathcad plot shows the derivation steps and plots the ac response for some given values of resistances:

enter image description here

It is possible to determine this transfer function without writing an equation, just by inspecting the schematic diagram. A quick SPICE simulation helps checking the values you have found are ok or not. It is then easy to fix the guilty part quickly.


Your transfer function, pole and zero = all OK.

I cannot imagine an application which needs a controller with DC gain this low (=3). It can well still exist.

Surely it can cause phase lag which can be applied in compensation. Maximum lag is about 12 degrees. It cannot make phase lead.

NOTE: This circuit isn't one which straight away brings up only a single meaningful application that must be in the field of control theory. I can as well be used for frequency response equalization.


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