# Circuit Theory/Convolution Integral/Examples/example49

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:

### 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.

$i(t)={\frac {V_{R}}{4}}$ $V_{L}(t)=L{di(t) \over dt}={\frac {1}{4}}((-2A+B)e^{-2t}-2Bte^{-2t})$ $V_{L}(0)=1={\frac {B}{4}}$ $B=4$ $V_{R}(t)=4te^{-2t}$ ### 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.