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

series LRC circuit ... find voltage across the resistor

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

This is the VL solution.

Outline:

## Contents

### Transfer Function

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

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

### 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:

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

### 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.

$V_{L_p} = 0$

This also means that C1 has to be zero.

### Initial Conditions

So far the full equation is:

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

Initial voltage is all across the inductor.

$V_L(0) = 1 = A$
$A = 1$

At this point will have to do integral .. to get to the current. There is no other way to use the known initial conditions: current (initially zero), and VC (initially zero). Will have to introduce integration constant and then evaluate that. More chance of mistakes, more complex, so start over with something else.

$i(t) = \frac{1}{L}\int_0^t V_L(t) dt = \int_0^t (e^{-2x} - B(x)e^{-2x}) dx$
f := (exp(-2*x) - B*x*exp(-2*x));
S :=int(f,x=0..t)

$i(t) = B*(\frac{e^{-2t}(2t+1)}{4}-\frac{1}{4}) - \frac{e^{-2t}}{2} + \frac{1}{2} + C_1$
$i(0) = 0 = C_1$

Ok so C1 is zero. Now need to find B. Find B by doing another integral to get VC:

$V_C(t) = \frac{1}{C}\int_0^t i(t) dt = 4*\int_0^t (B*(\frac{e^{-2t}(2t+1)}{4}-\frac{1}{4}) - \frac{e^{-2t}}{2} + \frac{1}{2}) dx$
f := (4*(B*(exp(-2*x)*(2*x+1)/4 -1/4) - exp(-2*x)/2 + 1/2));
S :=int(f,x=0..t)

$V_C(t) = B + 2t +e^{-2t} - Bt - Be^{-2t} - Bte^{-2t} - 1 + C_1$
$V_C(0) = 0 = B + 1 - B - 1 + C_1$
$C_1 = 0$

Still doesn't help us find B. Guess B = 2 since (2t-Bt) has to equal zero if VC is going to converge on 1. Then see if VC(∞) = 1:

$V_C(t) = 2 +e^{-2t} - 2e^{-2t} - 2te^{-2t} - 1$

At t = ∞ what is the 2te-2t term's value?

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


$V_C(\infty) = 2 - 1 = 1$

Yes! B = 2 works ... looks like the only thing that works .... So:

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


This means that VR is:

$V_R(t) = 4 * i(t) = 4te^{-2t}$

### 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.