# Circuit Theory/Convolution Integral/Examples/example49/Vc

Given that the source voltage is (2t-3t^{2}), find voltage across the resistor.

This is the V_{c} solution.

Outline:

## Contents

### Transfer Function[edit]

simplify((1/(s*0.25))/(4 + s + 1/(0.25*s)))

### Homogeneous Solution[edit]

solve(s^2 + 4.0*s + 4.0,s)

There are two equal roots at s = -2, so the solution has the form:

### Particular Solution[edit]

After a long time attached to a unit step function source, the inductor has shorted and the capacitor has opened. All the drop is across the capacitor. The current is zero.

This also means that C_{1} is still unknown.

### Initial Conditions[edit]

So far the full equation is:

At t = ∞ what is the B term's value?

limit(B*t*exp(-t),t = infinity)

Mupad says 0. So This means that at t = ∞:

Initial voltage across the capacitor is 0 so:

Initial current through the series leg is zero because of the assumed initial conditions of the inductor. This means i(0) = 0, so:

This means that i(t) =:

### Impulse Solution[edit]

Taking the derivative of the above get:

### Convolution Integral[edit]

f := (4*exp(-2*(t-x)) - 8*(t-x)exp(-2*(t-x)))*(2*x-3*x^2); S :=int(f,x=0..t)

There will not be any constant since again, V_R(t) = 0 after a long time ... and the capacitor opens.