Circuit Theory/Convolution Integral/Examples/example49/VL

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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 VL solution.

Outline:

Transfer Function[edit]

H(s) = \frac{V_L}{V_S} = \frac{s}{4 + s + \frac{1}{0.25s}}
simplify(s/(4 + s + 1/(0.25*s)))
H(s) = \frac{s^2}{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_{L_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.

V_{L_p} = 0

This also means that C1 has to be zero.

Initial Conditions[edit]

So far the full equation is:

V_L(t) = Ae^{-2t} + Bte^{-2t}

Initial voltage is all across the inductor.

V_L(0) = 1 = A
A = 1

At this point will have to do integral .. to get to the current. There is no other way to use the known initial conditions: current (initially zero), and VC (initially zero). Will have to introduce integration constant and then evaluate that. More chance of mistakes, more complex, so start over with something else.

i(t) = \frac{1}{L}\int_0^t V_L(t) dt = \int_0^t (e^{-2x} - B(x)e^{-2x}) dx
f := (exp(-2*x) - B*x*exp(-2*x));
S :=int(f,x=0..t)
i(t) = B*(\frac{e^{-2t}(2t+1)}{4}-\frac{1}{4}) - \frac{e^{-2t}}{2} + \frac{1}{2} + C_1
i(0) = 0 = C_1

Ok so C1 is zero. Now need to find B. Find B by doing another integral to get VC:

V_C(t) = \frac{1}{C}\int_0^t i(t) dt = 4*\int_0^t (B*(\frac{e^{-2t}(2t+1)}{4}-\frac{1}{4}) - \frac{e^{-2t}}{2} + \frac{1}{2}) dx
f := (4*(B*(exp(-2*x)*(2*x+1)/4 -1/4) - exp(-2*x)/2 + 1/2));
S :=int(f,x=0..t)
V_C(t) = B + 2t +e^{-2t} - Bt - Be^{-2t} - Bte^{-2t} - 1 + C_1
V_C(0) = 0 = B + 1 - B  - 1 + C_1
C_1 = 0

Still doesn't help us find B. Guess B = 2 since (2t-Bt) has to equal zero if VC is going to converge on 1. Then see if VC(∞) = 1:

V_C(t) = 2  +e^{-2t}  - 2e^{-2t} - 2te^{-2t} - 1

At t = ∞ what is the 2te-2t term's value?

limit(B*t*exp(-t),t = infinity)

Mupad says 0.

V_C(\infty) = 2 - 1 = 1

Yes! B = 2 works ... looks like the only thing that works .... So:

i(t) = 2*(\frac{e^{-2t}(2t+1)}{4}-\frac{1}{4}) - \frac{e^{-2t}}{2} + \frac{1}{2} = te^{-2t}
simplify(2*(exp(-2*t)(2*t+1)/4-1/4) - exp(-2*t)/2 + 1/2)

This means that VR is:

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.