Circuit Theory/Convolution Integral/Examples/example49/current



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

Here focused on finding current first:

Transfer Function

 * $$H(s) = \frac{i}{V_S} = \frac{1}{4 + s + \frac{1}{0.25s}}$$

simplify(1/(4 + s + 1/(0.25*s)))
 * $$H(s) = \frac{s}{s^2 + 4s + 4}$$

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:
 * $$i_h(t) = Ae^{-2t} + Bte^{-2t} + C_1$$

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.
 * $$i_p = 0$$

Initial Conditions
So far the full equation is:
 * $$i(t) = Ae^{-2t} + Bte^{-2t} + C_1$$

Initial current through the series leg is zero because of the assumed initial conditions of the inductor. This means:
 * $$i(0) = 0 = A + C_1$$

Assuming the initial voltage across the capacitor is zero, then initial voltage drop has to be across the inductor.
 * $$V_L(t) = L{d i(t) \over dt} = (-2A+B)e^{-2t} -2Bte^{-2t}$$
 * $$V_L(0) = 1 = -2A + B$$

After a long period of time, the current still has to be zero so:
 * $$C_1 = 0$$

This means that:
 * $$A = 0$$
 * $$B = 1$$
 * $$i(t) = te^{-2t}$$
 * $$V_r(t) = 4te^{-2t}$$

The 4 is lost in the numerator of the transfer function if a transfer function is written for Vr initially. The 4 does not make it into the homogeneous solution. In second order analysis, never write a transfer function for a resistor.

Impulse Solution
Taking the derivative of the above get:
 * $$V_R\delta (t) = 4e^{-2t} - 8te^{-2t} $$

Convolution Integral

 * $$V_{R}(t) = \int_0^t (4e^{-2(t-x)} - 8(t-x)e^{-2(t-x)})(2x-3x^2) dx $$

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.