Circuit Theory/Phasors/Examples/example11/laplace solution

Laplace Solution


Can not get Mupad to evaluate the laplace transform of both functions, but can get the laplace transform to evaluate within MatLab:
 * $${V_s}(t) \rightarrow \mathcal{L}(V_s)$$ = s/(s^2 + 1)
 * $${I_s}(t) \rightarrow \mathcal{L}(I_s)$$ = -(2*((3^(1/2)*s)/2 - 1/2))/(s^2 + 1)

The equations can be solved in the Laplace domain:
 * $$ \mathcal{L}(V_1) = R_1 * \mathcal{L}(I_1)$$
 * $$ \mathcal{L}(V_2) = R_2 * \mathcal{L}(I_2)$$
 * $$\mathcal{L}(I_C) = C * s \mathcal{L}(V_C)$$
 * $$\mathcal{L}(V_L) = L * s \mathcal{L}(I_L)$$
 * $$ \mathcal{L}(I_1) + \mathcal{L}(I_s) - \mathcal{L}(I_2) - \mathcal{L}(I_c) = 0$$
 * $$ \mathcal{L}(I_c) - \mathcal{L}(I_s) - \mathcal{L}(I_L) = 0$$
 * $$ \mathcal{L}(V_1) + \mathcal{L}(V_2) - \mathcal{L}(V_s) = 0$$
 * $$ \mathcal{L}(V_c) + \mathcal{L}(V_L) - \mathcal{L}(V_2) = 0$$

The Laplace numerical results look as complicated as the symbolic phasor results. The Laplace solutions are too complicated for matLab or MuPad to get back into the time domain ... even when solved numerically. This is why people use phasors.