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 V_c solution.

Outline:

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:

$V_{C_h}(t)=A{e}^{-2t}+Bt{e}^{-2t}+C_1$

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:

$V_C(t)=1+A{e}^{-2t}+Bt{e}^{-2t}+C_1$

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:

$i(t)=C*\frac{d{V}_C(t)}{dt}=\frac{1}{4}((-2A+B)e^{-2t}-2Bt{e}^{-2t})$

$0=\frac{1}{4}(-2A+B)$

$B=2A$

$B=-2$

This means that i(t) =:

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

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

Impulse Solution

Taking the derivative of the above get:

$V_R\delta (t)=4e^{-2t}-8t{e}^{-2t}$

Convolution Integral

$V_R(t)=\underset{0}{\overset{t}{\int }}(4e^{-2(t-x)}-8(t-x)e^{-2(t-x)})(2x-3x^2)dx$

f := (4*exp(-2*(t-x)) - 8*(t-x)exp(-2*(t-x)))*(2*x-3*x^2);
S :=int(f,x=0..t)

$V_R(t)=8-8e^{-2t}-10t{e}^{-2t}-6t$

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