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

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

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

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

### 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:

${\displaystyle i_{h}(t)=Ae^{-2t}+Bte^{-2t}+C_{1}}$

### 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.

${\displaystyle i_{p}=0}$

### Initial Conditions

So far the full equation is:

${\displaystyle i(t)=Ae^{-2t}+Bte^{-2t}+C_{1}}$

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

${\displaystyle i(0)=0=A+C_{1}}$

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

${\displaystyle V_{L}(t)=L{di(t) \over dt}=(-2A+B)e^{-2t}-2Bte^{-2t}}$
${\displaystyle V_{L}(0)=1=-2A+B}$

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

${\displaystyle C_{1}=0}$

This means that:

${\displaystyle A=0}$
${\displaystyle B=1}$
${\displaystyle i(t)=te^{-2t}}$
${\displaystyle V_{r}(t)=4te^{-2t}}$

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:

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

### Convolution Integral

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

${\displaystyle V_{R}(t)=8-8e^{-2t}-10te^{-2t}-6t}$

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