Circuit Theory/TF Examples/Example33

Series RLC circuit with two initial conditions Example 33 for wikibook circuit theory

Find i_o(t) if V_s(t) = 1 + cos(3t).

Choose Starting Point

Because of the initial conditions, going to start with V_c(t) and then work our way through the initial conditions to i_o.

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

mupad screen shot leading up to computing constant B

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:

$V_c(\mathrm{\infty })=1=1+C\Rightarrow C=0$

Which implies that C is zero.

From the given initial conditions, know that V_c(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:

$i_T(t)=C\frac{d{V}_c}{dt}=\frac{11e^{-\frac{11t}{16}}}{16}(0.5\cos \frac{7t}{16}-B\sin \frac{7t}{16})+\frac{7e^{-\frac{11t}{16}}}{16}(0.5\sin \frac{7t}{16}+B\cos \frac{7t}{16})$

The loop equation can be solved for the voltage across the LR parallel combination:

$V_C+V_{LR}+R_{2}C\frac{d{V}_c}{dt}-V_s=0$

$V_{LR}=V_s-V_C-R_2{i}_t=1-V_c-1.5*i_t$

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 = ∞ 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 I_L equal to the initial condition of .2 amps and solve for B:

solve(IL=0.2, B)

And get that B is -0.2008928571 ...

mupad screen shot finding the desired output i_o

The desired answer is i_o which is just V_LR/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

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