Steps in applying the Laplace Transform:
- Transform the circuit from the time domain to s-domain.
- Solve the circuit using any circuit analysis techniques applied for DC Circuits.
- Take the inverse Laplace transform of the solution and thus obtain the solution in the time domain.
The elegance of using the Laplace transform in circuit analysis lies in the automatic inclusion of the initial conditions in the transformation process, thus providing a complete solution.
With the s-domain equivalents, the Laplace transform can be used to solve first-order and second-order circuits.
For a resistor
\(\displaystyle v(t) = R\,i(t)\)
Taking the Laplace transform,
\(\boxed{V(s) = R\,I(s)}\)
The impedance \(Z(s)\) of the resistor is
\(\boxed{Z(s) = \frac{V(s)}{I(s)} = R}\)
For an inductor
\(\displaystyle v(t) = L\frac{di(t)}{dt}\)
Taking the Laplace transform,
\(\displaystyle V(s) = L\left[s\,I(s) - i(0^-)\right] = sL\,I(s) - L\,i(0^-)\)
Since the inductor current cannot change instantaneously, \(i(0^-) = i(0^+) = i(0)\)
\(\boxed{V(s) = sL\,I(s) - L\,i(0)}\)
\(\boxed{I(s) = \frac{V(s)}{sL} + \frac{i(0)}{s}}\)
The impedance \(Z(s)\) of the inductor at zero initial condition \(i(0) = 0\) is
\(\boxed{Z(s) = \frac{V(s)}{I(s)} = sL}\)
For a capacitor
\(\displaystyle i(t) = C\frac{dv(t)}{dt}\)
Taking the Laplace transform,
\(\displaystyle I(s) = C\left[s\,V(s) - v(0^-)\right] = sC\,V(s) - C\,v(0^-)\)
Since the capacitor voltage cannot change instantaneously, \(v(0^-) = v(0^+) = v(0)\)
\(\boxed{I(s) = sC\,V(s) - C\,v(0)}\)
\(\boxed{V(s) = \frac{I(s)}{sC} + \frac{v(0)}{s}}\)
The impedance \(Z(s)\) of the capacitor at zero initial condition \(i(0) = 0\) is
\(\boxed{Z(s) = \frac{V(s)}{I(s)} = \frac{1}{sC}}\)
