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

series LRC circuit ... find voltage across the resistor

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

This is the Vc solution.

Outline:

## Contents

### Transfer Function

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

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

### 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:

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

### 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.

$V_{C_p} = 1$

This also means that C1 is still unknown.

### Initial Conditions

So far the full equation is:

$V_C(t) = 1 + Ae^{-2t} + Bte^{-2t} + C_1$

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 = ∞:

$1 = 1 + C_1$
$C_1 = 0$

Initial voltage across the capacitor is 0 so:

$0= 1 + A + C_1$
$A = -1$

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

$i(t) = C* {d V_C(t) \over dt} = \frac{1}{4}((-2A+B)e^{-2t} -2Bte^{-2t})$
$0 = \frac{1}{4}(-2A+B)$
$B = 2A$
$B = -2$

This means that i(t) =:

$i(t)= \frac{1}{4}((-2(-1)+(-2))e^{-2t} -2(-2)te^{-2t})=te^{-2t}$
$V_R(t) = 4*i(t) = 4te^{-2t}$

### 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.