# Circuit Theory/TF Examples/Example33

Find i_{o}(t) if V_{s}(t) = 1 + cos(3t).

## Contents

## Choose Starting Point[edit]

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[edit]

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:

## Homogeneous Solution[edit]

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

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

Imaginary roots:

So the solution has the form:

## Particular Solution[edit]

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:

## Initial Conditions[edit]

Adding the particular and homogenous solutions, get:

Doing the final condition again, get:

Which implies that C is zero.

From the given initial conditions, know that V_{c}(0_{+}) = 0.5 so can find A:

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:

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

VLR := 1 - VC - 1.5*IT

We know from the inductor terminal relation that:

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 ...

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