Circuit Theory/Convolution Integral/Examples/example49/Vc
Given that the source voltage is (2t-3t2), find voltage across the resistor.
This is the Vc solution.
simplify((1/(s*0.25))/(4 + s + 1/(0.25*s)))
solve(s^2 + 4.0*s + 4.0,s)
There are two equal roots at s = -2, so the solution has the form:
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.
This also means that C1 is still unknown.
So far the full equation is:
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 = ∞:
Initial voltage across the capacitor is 0 so:
Initial current through the series leg is zero because of the assumed initial conditions of the inductor. This means i(0) = 0, so:
This means that i(t) =:
Taking the derivative of the above get:
f := (4*exp(-2*(t-x)) - 8*(t-x)exp(-2*(t-x)))*(2*x-3*x^2); S :=int(f,x=0..t)
There will not be any constant since again, V_R(t) = 0 after a long time ... and the capacitor opens.