# Circuit Theory/TF Examples/Example34/io

## Contents

## Starting Point[edit]

Starting point of i_{o} looks good. Take integral to visit V_{c} initial condition. Then get expression for V_{total}, then take integral of that to visit I_{L} initial condition. Lots of integrals.

## Transfer Function[edit]

L :=1; R1 := 1/2; C := 1/2; R2 := 1.5; simplify((1/(R2 + 1/(s*C))/(1/(s*L) + 1/R1 + 1/(R2 + 1/(s*C))))

## Homogeneous Solution[edit]

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

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

So the solution is going to have the form:

## Particular Solution[edit]

After a very long time the inductor shorts, all the current flows through it so:

## Initial Conditions[edit]

Adding the particular and homogeneous solutions, get:

Doing the final condition again, get:

Let's try for V_{c} first. From the terminal relation for a capacitor:

io := exp(-11*t/16)*(A*cos(7*t/16) + B*sin(7*t/16))

VC := 2* int(io,t)

We know that initially V_{c} = 1.5 so at t=0 can find equation for A and B:

t :=0; solve(1.5 = VC)

At this point mupad goes numeric and get this equation:

Need another equation. Can find V_{t} by adding V_{r} and V_{c}. Then from V_{t} can find expression for the current through the inductor and visit it's initial condition. Need to start over in MuPad because t=0 has ruined the current session. So repeating the setup of V_{C}:

io := exp(-11*t/16)*(A*cos(7*t/16) + B*sin(7*t/16))

VC := 2*int(io,t)

The integration constant is going to be zero because after a long time V_{C} is zero (the inductor shorts).

VT := VC + io*1.5

From the terminal equation for an inductor:

IL := int(VT,t)

Mupad goes numeric.

At this point have to figure out the integration constant. After a long time, the inductor's current is going to be 1 because it shorts the current source. Looking at IL in the mupad window can see that every term is multiplied by e^{-0.6875t} which is going to zero as t goes to ∞. This means the integration constant is 1.

So add 1 to IL, then set t=0 and I_{L} = 0.5 and again solve for A and B:

t :=0; solve (IL + 1 = 0.5)

Get this equation:

Now need to solve the two equations and two unknowns:

solve([A = - 0.6363636364*B - 0.7244318182,A = 6.273453094*B + 1.802644711],[A,B])

Get:

So now have time domain expression for i_{o} step response:

## Impulse Response[edit]

The impulse response is the derivative of the step response:

i_u := exp(-11*t/16)*(-0.4916992187 * cos(7*t/16) - 0.3657226563*sin(7*t/16))

i_s := diff(i_u,t)

## Convolution Integral[edit]

The first step is to substitute into i_s for t:

i_sub := subs(i_s, t = y-x):

Now form the convolution integral:

f := i_sub*(1 + 3*cos(2*x)): io := int(f,x = 0..y)

Replacing y with t:

i_o :=subs(io, y=t)

There is going to be an integration constant. This value can not be determined because the driving function oscillates. The initial conditions of the inductor and capacitor have already been visited. More information (like a specific value at a future time) is needed in order to compute an integration constant.

Thus the final answer is:

The answer indicates that i_{o} = 0 when t=0. The exponential terms die after 5/0.6875 = 7.27 seconds leaving a sinusoidal function at more than a 90° phase shift from the current source with a DC level of about 0.5 amps.