Laplace transform of capacitor functions and initial conditions

Question

I may be asking something trivial, but unfortunately, I could not find an answer so far.

Suppose an AC circuit. The voltage of the capacitor is given by

$$v_C(t)=Q(t)/C = 1/C [ \int_0^t i_C(τ)dτ + v_C(0)]$$

and the current is

$$i_C(t) = C \frac{dV_C(t)}{dt}$$

Their Laplace transforms are: $$V_C(s) = \frac{I_C(s)}{Cs}+\frac{V_C(0)}{s}$$ and $$I_C(s)=sCV_C(s)-CV_C(0)$$ According to Ohm's Law: $$Z_C(s) = V_C(s)/I_C(s)$$ By substituting, $$Z_C(s)=\frac{1}{Cs-\frac{V_C(0)}{V_C(s)}}$$

It is known that $$Z_C(s)=\frac{1}{Cs}$$ which obviously occurs from the above equation if \$\small V_C(0)=0\$, but that is not always the case. Could someone explain this issue? What happens if my capacitor's initial condition is not zero?

Answer 1

View 1

When solving circuits using Laplace transform, one method commonly taught is to replace a capacitor with an initial voltage with a capacitor with zero initial voltage and a special voltage in series with it. Slide 25 and 26 give an example. The equivalent circuit in \$s\$ domain has a capacitor \$C\$ with impedance \$1/(sC)\$ and a voltage source \$v(0)/s\$ in series. This equivalent circuit has the same characteristics as the original capacitor.

(Image taken from above link.)

View 2

For AC circuits, the AC impedance is given as \$dv/di\$ (ratio of change in voltage to change in current). A similar argument is used to arrive at the formula \$1/sC\$ in the Wikipedia article for impedance

As such, I do not see any problem in expression you have derived; It's just that the initial condition is handled separately and impedance of capacitor is taken as \$1/sC\$. This turns out to be easier to work with in practice.