Circuit Theory/Convolution Integral/Examples/example49/VL
Given that the source voltage is (2t-3t2), find voltage across the resistor.
This is the VL solution.
simplify(s/(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.
This also means that C1 has to be zero.
So far the full equation is:
Initial voltage is all across the inductor.
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.
f := (exp(-2*x) - B*x*exp(-2*x)); S :=int(f,x=0..t)
Ok so C1 is zero. Now need to find B. Find B by doing another integral to get VC:
f := (4*(B*(exp(-2*x)*(2*x+1)/4 -1/4) - exp(-2*x)/2 + 1/2)); S :=int(f,x=0..t)
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:
At t = ∞ what is the 2te-2t term's value?
limit(B*t*exp(-t),t = infinity)
Mupad says 0.
Yes! B = 2 works ... looks like the only thing that works .... So:
simplify(2*(exp(-2*t)(2*t+1)/4-1/4) - exp(-2*t)/2 + 1/2)
This means that VR is:
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.