Circuit Theory/Convolution Integral/Examples/example49

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

series LRC circuit ... find voltage across the resistor

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

Outline:

Transfer Function

edit
 
simplify(4/(4 + s + 1/(0.25*s)))
 

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:

 

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.

 

This also means that C1 has to be zero.

Initial Conditions

edit

So far the full equation is:

 

Initial current through the series leg is zero because of the assumed initial conditions of the inductor. This means VR = 0:

 
 

Assuming the initial voltage across the capacitor is zero, no current is flowing so the drop across the resistor is zero.

 
 
 
 
 

Impulse Solution

edit

Taking the derivative of the above get:

 

Convolution Integral

edit
 
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.