Circuit Theory/Convolution Integral/Examples/example49/Vc

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series LRC circuit ... find voltage across the resistor

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

This is the Vc solution.

Outline:

Transfer Function[edit]

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[edit]

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) = Ae^{-2t} + Bte^{-2t} + C_1

Particular Solution[edit]

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[edit]

So far the full equation is:

V_C(t) = 1 + Ae^{-2t} + Bte^{-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* {d V_C(t) \over dt} = \frac{1}{4}((-2A+B)e^{-2t} -2Bte^{-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)te^{-2t})=te^{-2t}
V_R(t) = 4*i(t) = 4te^{-2t}

Impulse Solution[edit]

Taking the derivative of the above get:

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

Convolution Integral[edit]

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.