Circuit Theory/2Source Excitement/Example46

Find the time domain expression for io given that Is = 1&mu;(t) amp.

Steady State Particular Solution
After a long time, the cap opens and inductor shorts putting the two resistors in parallel and splitting the current making io 1/2 amp.

Transient Particular Solution
Start writing a node equation:
 * $$ I_s = i_c + i_r + i_o$$

Substitute voltage terminal relationships:
 * $$ I_s = C {d V \over dt} + \frac{V}{R} + i_o$$

Find V in terms of io through the L R branch:
 * $$ V = L {d i_o \over dt} + R i_o$$

Substitute to get Is in terms of io:
 * $$ I_s = C {d (L {d i_o \over dt} + R i_o) \over dt} + \frac{L {d i_o \over dt} + R i_o}{R} + i_o$$
 * $$ I_s = CL {d^2 i_o \over dt^2} + CR {d i_o \over dt} + \frac{L}{R} {d i_o \over dt} + i_o + i_o$$

Substituting numbers from the problem:
 * $$ I_s = {d^2 i_o \over dt^2} + 2 {d i_o \over dt} + 2 i_o $$

Time constant
Guess: $$i_o(t) = Ae^{st}$$ Substituting:
 * $$ I_s = {d^2 Ae^{-st} \over dt^2} + 2 {d Ae^{-st} \over dt} + 2 Ae^{-st} $$
 * $$ I_s = s^2Ae^{-st} + 2sAe^{-st} + 2Ae^{-st}$$

Does this equal zero?
 * $$ s^2Ae^{st} - 2sAe^{st} + 2Ae^{st} = 0$$
 * $$ s^2 + 2s + 2 = 0$$

No. Rats. Need to evaluate the above quadratic in order to guess another solution.
 * $$ s_{1,2} = -1-j, -1+j = \sigma \pm j\omega$$
 * $$\sigma = -1$$
 * $$\omega = 1$$

So the next guess is:
 * $$ i_o = e^{-t}(A_1\cos t + A_2\sin t) + C_1$$

Finding the Constants
After a very long time, the capacitor is going to open and the inductor is going to short. This leaves two equal resistors in parallel that are going to split the current in half.
 * $$ i_o(\infty) = \frac{1}{2} = C_1$$

So now the expression for io is:
 * $$ i_o = e^{-t}(A_1\cos t + A_2\sin t) + \frac{1}{2}$$

Initially the current through the conductor is 0, so io(0+) = 0:
 * $$ i_o(o_+) = \frac{1}{2}+ A_1 = 0$$

Which means that:
 * $$A_1 = -\frac{1}{2}$$

The other initial condition affecting io is the voltage across the inductor .. which is zero. We can find an expression for VL:
 * $$V_L = L {d i_0(t) \over dt} = L (- e^{-t}(A_1\cos t + A_2\sin t) + e^{-t}(- A_1\sin t + A_2\cos t))$$

Setting all this equal to 0 at t=0 yields:
 * $$0 = -A_1 + A_2 $$

So:
 * $$A_2 = A_1 = -\frac{1}{2}$$

Thus io is:
 * $$ i_o = \frac{1}{2} + e^{-t}(-\frac{1}{2}\cos t - \frac{1}{2}\sin t) = \frac{1}{2}(1 - e^{-t}(\cos t + \sin t))$$

This solution is used to find io for a complicated source using the convolution integral.