Hello again,
Here's the thing, when you have an RLC circuit not excited by any forcing function you get a certain characteristic equation. When you apply a forcing function that is a step you get the same characteristic equation. Same characteristic equation, same roots.
With no input, we get (s*R)/L+1/(C*L)+s^2 as the denominator, and with a step input we get the same denominator but the numerator changes.
Same denominator, same roots.
Here's a full time domain solution for both cases (exponential solution cases):
Code:
No step:
-(C*R*e^((t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L)))/(2*L*sqrt(C^2*R^2-4*C*L))+e^((t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L))/(2*L)+(C*R*e^(-(t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L)))/(2*L*sqrt(C^2*R^2-4*C*L))+e^(-(t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L))/(2*L)
With a step:
(C*E*e^((t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L)))/sqrt(C^2*R^2-4*C*L)-(C*E*e^(-(t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L)))/sqrt(C^2*R^2-4*C*L)
Notice we get the same value for the exponents in both cases (positive or negative).
t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L)
t*sqrt(C^2*R^2-4*C*L))/(2*C*L)-(t*R)/(2*L)
One more easier to read example:
y''+A*y'+B*y=0
we get:
y=e^(-(x*A)/2)*(K1*sin((x*sqrt(4*B-A^2))/2)+K2*cos((x*sqrt(4*B-A^2))/2))
and for:
y''+A*y'+B*y=1
we get:
y=e^(-(x*A)/2)*(K1*sin((x*sqrt(4*B-A^2))/2)+K2*cos((x*sqrt(4*B-A^2))/2))+1/B
What's the difference? The second one has an added 1/B. That 1/B doesnt change the frequency in any way.
Unless i dont understand your question.