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

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

Here focused on finding current first:

## Contents

### Transfer Function[edit]

simplify(1/(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.

### Initial Conditions[edit]

So far the full equation is:

Initial current through the series leg is zero because of the assumed initial conditions of the inductor. This means:

Assuming the initial voltage across the capacitor is zero, then initial voltage drop has to be across the inductor.

After a long period of time, the current still has to be zero so:

This means that:

The 4 is lost in the numerator of the transfer function if a transfer function is written for V_{r} initially. The 4 does not make it into the homogeneous solution. In second order analysis, never write a transfer function for a resistor.

### 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.