Circuit Theory/Convolution Integral/Examples/example49

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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.

Can do focused on Vr or: current, Vc, or VL before converting to Vr .. Below is the VR solution.


Transfer Function[edit]

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[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_{R_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{R_p} = 0

This also means that C1 has to be zero.

Initial Conditions[edit]

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{d i(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[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.