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

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series LRC circuit ... find voltage across the resistor

Given that the source voltage is (2t-3t2), find voltage across the resistor.

Here focused on finding current first:

## Contents

### Transfer Function

$H(s) = \frac{i}{V_S} = \frac{1}{4 + s + \frac{1}{0.25s}}$
simplify(1/(4 + s + 1/(0.25*s)))

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

### Homogeneous Solution

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


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

$i_h(t) = Ae^{-2t} + Bte^{-2t} + C_1$

### Particular Solution

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.

$i_p = 0$

### Initial Conditions

So far the full equation is:

$i(t) = Ae^{-2t} + Bte^{-2t} + C_1$

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

$i(0) = 0 = A + C_1$

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

$V_L(t) = L{d i(t) \over dt} = (-2A+B)e^{-2t} -2Bte^{-2t}$
$V_L(0) = 1 = -2A + B$

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

$C_1 = 0$

This means that:

$A = 0$
$B = 1$
$i(t) = te^{-2t}$
$V_r(t) = 4te^{-2t}$

The 4 is lost in the numerator of the transfer function if a transfer function is written for Vr 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

Taking the derivative of the above get:

$V_R\delta (t) = 4e^{-2t} - 8te^{-2t}$

### Convolution Integral

$V_{R}(t) = \int_0^t (4e^{-2(t-x)} - 8(t-x)e^{-2(t-x)})(2x-3x^2) dx$
f := (4*exp(-2*(t-x)) - 8*(t-x)exp(-2*(t-x)))*(2*x-3*x^2);
S :=int(f,x=0..t)

$V_R(t) = 8 - 8e^{-2t} - 10te^{-2t} -6t$

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