Circuit Theory/TF Examples/Example33

Find io(t) if Vs(t) = 1 + cos(3t).

Choose Starting Point
Because of the initial conditions, going to start with Vc(t) and then work our way through the initial conditions to io.

Transfer Function

 * $$H(s) = \frac{V_c}{V_s} = \frac{\frac{1}{sC}}{\frac{1}{sC} + \frac{1}{\frac{1}{R_1} + \frac{1}{sL}} + R_2}$$

The MuPad commands are going to be: L :=1; R1:=.5; R2:=1.5; C:=.5; simplify((1/(s*C))/(1/(s*C) + 1/(1/R1 + 1/(s*L)) + R2))

Which results in:
 * $$ H(s) = \frac{8s + 4}{8s^2 + 11s + 4}$$

Homogeneous Solution
Set the denominator of the transfer function to 0 and solve for s: solve(8*s^2 + 11*s + 4) Imaginary roots:
 * $$s_{1,2} = \frac{-11 \pm \sqrt{7}i}{16}$$

So the solution has the form:
 * $$V_{c_h} = e^{-\frac{11t}{16}}(A\cos \frac{7t}{16} + B\sin \frac{7t}{16}) + C$$

Particular Solution
After a very long time the capacitor opens, no current flows, so all the source drop is across the capacitor. The source is a unit step function thus:
 * $$V_{c_p} = 1$$

Initial Conditions
Adding the particular and homogenous solutions, get:
 * $$V_c(t) = 1 + e^{-\frac{11t}{16}}(A\cos \frac{7t}{16} + B\sin \frac{7t}{16}) + C$$

Doing the final condition again, get:
 * $$V_c(\infty) = 1 = 1 + C \Rightarrow C = 0$$

Which implies that C is zero.

From the given initial conditions, know that Vc(0+) = 0.5 so can find A:
 * $$V_c(0_+) = 0.5 = 1 + A \Rightarrow A = -0.5$$

Finding B is more difficult. From capacitor terminal relation: VC := 1 + exp(-11*t/16)*(-.5*cos(7*t/16) + B*sin(7*t/16)) IT := diff(VC,t) The total current is:
 * $$i_T(t) = C{d V_c \over dt} = \frac{11e^{-\frac{11t}{16}}}{16}(0.5\cos \frac{7t}{16} - B \sin \frac{7t}{16}) + \frac{7e^{-\frac{11t}{16}}}{16}(0.5\sin \frac{7t}{16} + B\cos \frac{7t}{16})$$

The loop equation can be solved for the voltage across the LR parallel combination:
 * $$V_C + V_{LR} + R_2 C{d V_c \over dt} - V_s = 0$$
 * $$V_{LR} = V_s - V_C - R_2 i_t = 1 - V_c - 1.5*i_t$$

VLR := 1 - VC - 1.5*IT We know from the inductor terminal relation that:
 * $$i_L = \frac{1}{L} \int V_{LR} dt + C_1$$

IL := 1/.5 * int(VLR,t) At this point mupad gave up and went numeric. In any case, it is clear from t = &infin; where the inductor current has to be zero that the integration constant is zero. This enables us to compute B from the inductor initial condition. t :=0 Set the time to zero, set IL equal to the initial condition of .2 amps and solve for B: solve(IL=0.2, B) And get that B is -0.2008928571 ... The desired answer is io which is just VLR/R_1. To calculate need to start new mupad session because t is zero now. Start with: B := -0.2008928571; R1 :=0.5; Repeat the above commands up to VLR and then add: io = VLR/R1

$$i_o = 3e^{-\frac{11t}{16}}(0.0879\cos\frac{7t}{16} - 0.219\sin\frac{7t}{16}) - 0.0625e^{-\frac{11t}{16}}(0.5\cos\frac{7t}{16} + 0.201\sin\frac{7t}{16})$$