I have a system of a series RC circuit. The input is DC source power, The output signal is the voltage on the capacitor. My question Is this system is a closed loop or open loop? What is the control block diagram that represent this system?enter image description here


1 Answer 1


Well thinking of a circuit as feedback ruled or not is just a matter of "How would you like to see it".

For sure most of us will note use feedback theory to analize that circuit but none the less it can be done and I believe it is also quite educative.

First start from equations governing the circuit.


simulate this circuit – Schematic created using CircuitLab

First you can write capacitor CV law $$ v_\text{out}=\frac{1}{C}\int i\,\text{d}t $$

and then one simple Ohm's law

$$ i=\frac{1}{R}\left(v_\text{in}-v_\text{out}\right) $$

which combined give us

$$ v_\text{out}=\frac{1}{RC}\int \left(v_\text{in}-v_\text{out}\right)\,\text{d}t $$

This is plain sailing and nothing new nor exciting but this relation can be read to introduce feed back in the analisys.

$$ v_\text{out}=\underbrace{\frac{1}{RC}\int }_\text{forward gain} \; \underbrace{\left(v_\text{in}-\underbrace{v_\text{out}}_\text{unit feedback}\right)}_\text{summing node}\,\text{d}t $$

So we have:

  • \$ \left(v_\text{in}-v_\text{out}\right) \$ Vin voltage is compared against feedback in summing node, i.e we have a voltage feedback series compared.
  • We also note that Vout has unity coeffcient, i.e. \$\beta=1\$ unit feedback
  • Finally we have type 1 forward chain \$v_\text{out}=\frac{1}{RC}\int [\cdot]\text{d}t \$ or in Lapalce's domain \$A=1/sRC\$

in a simple diagram


simulate this circuit

In Laplace's domain you can easily calculate closed loop gain

$$ \frac{V_\text{out}}{V_\text{in}}=\frac{A}{1+\beta A}=\frac{1/sRC}{1+1/sRC}= \frac{1}{1+sRC} $$

as expected from "classic" circuit theory


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