Circuit Theory/Convolution Integral/Examples/example49

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

series LRC circuit ... find voltage across the resistor

Can do focused on V_r or: current, V_c, or V_L before converting to V_r .. Below is the V_R solution.

Outline:

Transfer FunctionEdit

${\displaystyle H(s)={\frac {V_{R}}{V_{S}}}={\frac {4}{4+s+{\frac {1}{0.25s}}}}}$

simplify(4/(4 + s + 1/(0.25*s)))

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

Homogeneous SolutionEdit

solve(s^2 + 4.0*s + 4.0,s)

There are two equal roots at s = -2, so the solution has the form:

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

Particular SolutionEdit

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.

${\displaystyle V{R_{p}}=0}$

This also means that C₁ has to be zero.

Initial ConditionsEdit

So far the full equation is:

${\displaystyle V_{R}(t)=Ae^{-2t}+Bte^{-2t}}$

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

${\displaystyle V_{R}(0)=0=A}$

${\displaystyle A=0}$

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

${\displaystyle i(t)={\frac {V_{R}}{4}}}$

${\displaystyle V_{L}(t)=L{di(t) \over dt}={\frac {1}{4}}((-2A+B)e^{-2t}-2Bte^{-2t})}$

${\displaystyle V_{L}(0)=1={\frac {B}{4}}}$

${\displaystyle B=4}$

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

Impulse SolutionEdit

Taking the derivative of the above get:

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

Convolution IntegralEdit

${\displaystyle V_{R}(t)=\int _{0}^{t}(4e^{-2(t-x)}-8(t-x)e^{-2(t-x)})(2x-3x^{2})dx}$

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.

This article uses material from the Wikipedia article  Metasyntactic variable, which is released under the  Creative Commons Attribution-ShareAlike 3.0 Unported License.