0
\$\begingroup\$

enter image description here

Consider the circuit in the picture, where we have a voltage source (input) \$V(t)\$ and a capacitor \$C\$. The goal is to derive a state-space model of this circuit. The circuit satisfies the following equations

$$ \dot{V}_C(t) = \frac{1}{C}i(t) ,\quad V_C(t) = V(t). (1) $$

My intuition is to define the voltage \$V_C\$ around the capacitor as the state variable, which means that I should have a state-space model in the following form:

$$ \dot{V}_C(t) = a_1 V_C(t) + a_2 V(t), (2) $$ i.e. writing \$\dot{V}_C\$ as a function of the state \$V_C\$ and the input \$V\$, for some parameters \$a_1\$ and \$a_2\$.

However, clearly it is not possible to reformulate (1) into the form of (2), because there is no way to eliminate the variable \$i\$.

My question is then how to formulate a state-space model in this example. Is the choice of the state variable wrong since \$V_C = V\$? Then does it mean that there is no state in this example?

\$\endgroup\$
1
  • 1
    \$\begingroup\$ An ideal voltage source has zero internal impedance therefore, the voltage across C will always be V, no matter what V is. The only thing that matters in this case is the current through C, and that only if V varies. \$\endgroup\$ Sep 5, 2022 at 16:30

1 Answer 1

1
\$\begingroup\$

This is a trivial system in that \$G(s) = 1\$. \$V_{C}\$ is not independent from \$V\$.

A better way is to start with a system with a pole, then take the limiting case.

For example:

schematic

simulate this circuit – Schematic created using CircuitLab

Then the state equation for \$V_{C}\$ as the state variable is,$$\dot{V_{C}}=-\frac{1}{RC}V_{C}+\frac{1}{RC}V$$ Taking the limit as \$R \rightarrow 0\$ cause the equation to become undefined.

To Answer your questions: The capacitor voltage can be considered a system state but it is completely dependent on the applied voltage.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.