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

${\displaystyle H(s)={\frac {V_{c}}{V_{s}}}={\frac {\frac {1}{sC}}{{\frac {1}{sC}}+{\frac {1}{{\frac {1}{R_{1}}}+{\frac {1}{sL}}}}+R_{2}}}}$

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:

${\displaystyle H(s)={\frac {8s+4}{8s^{2}+11s+4}}}$

Homogeneous Solution

Set the denominator of the transfer function to 0 and solve for s:

solve(8*s^2 + 11*s + 4)

Imaginary roots:

${\displaystyle s_{1,2}={\frac {-11\pm {\sqrt {7}}i}{16}}}$

So the solution has the form:

${\displaystyle V_{c_{h}}=e^{-{\frac {11t}{16}}}(A\cos {\frac {7t}{16}}+B\sin {\frac {7t}{16}})+C}$

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:

${\displaystyle V_{c_{p}}=1}$

Initial Conditions

mupad screen shot leading up to computing constant B

Adding the particular and homogenous solutions, get:

${\displaystyle V_{c}(t)=1+e^{-{\frac {11t}{16}}}(A\cos {\frac {7t}{16}}+B\sin {\frac {7t}{16}})+C}$

Doing the final condition again, get:

${\displaystyle V_{c}(\infty )=1=1+C\Rightarrow C=0}$

Which implies that C is zero.

From the given initial conditions, know that V_c(0₊) = 0.5 so can find A:

${\displaystyle V_{c}(0_{+})=0.5=1+A\Rightarrow A=-0.5}$

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:

${\displaystyle i_{T}(t)=C{dV_{c} \over dt}={\frac {11e^{-{\frac {11t}{16}}}}{16}}(0.5\cos {\frac {7t}{16}}-B\sin {\frac {7t}{16}})+{\frac {7e^{-{\frac {11t}{16}}}}{16}}(0.5\sin {\frac {7t}{16}}+B\cos {\frac {7t}{16}})}$

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

${\displaystyle V_{C}+V_{LR}+R_{2}C{dV_{c} \over dt}-V_{s}=0}$

${\displaystyle V_{LR}=V_{s}-V_{C}-R_{2}i_{t}=1-V_{c}-1.5*i_{t}}$

VLR := 1 - VC - 1.5*IT

We know from the inductor terminal relation that:

${\displaystyle i_{L}={\frac {1}{L}}\int V_{LR}dt+C_{1}}$

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

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