# Circuit Theory/Convolution Integral/Examples/example49

series LRC circuit ... find voltage across the resistor

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

Can do focused on Vr or: current, Vc, or VL before converting to Vr .. Below is the VR solution.

Outline:

## Contents

### Transfer Function

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

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

### 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:

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

### 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.

$V{R_p} = 0$

This also means that C1 has to be zero.

### Initial Conditions

So far the full equation is:

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

Initial current through the series leg is zero because of the assumed initial conditions of the inductor. This means VR = 0:

$V_R(0) = 0 = A$
$A = 0$

Assuming the initial voltage across the capacitor is zero, no current is flowing so the drop across the resistor is zero.

$i(t) = \frac{V_R}{4}$
$V_L(t) = L{d i(t) \over dt} = \frac{1}{4}((-2A+B)e^{-2t} -2Bte^{-2t})$
$V_L(0) = 1 = \frac{B}{4}$
$B = 4$
$V_R(t) = 4te^{-2t}$

### 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.