I need to use a L-R-C tank in my design and I'm having trouble with the calculations. And I think it is because of the phase shift I didn't take into account.
Here you can see there is a differential equation introduced to extract the current-wave-charge equation. To my calculations the frequency is correct but I also need to know the phase shift \$ \theta \$ as well.
This is supposed to be the voltage wave of my LRC circuit:
ww = 660.764; vc = 30; slip = 0.99; result =
Cos[ww*t]*
vc*(Exp[-t*((0.461 + (ww*slip)*
ww*1715.38/(66564 +
6648.72*(ww*
slip)^2))/(2*(134*10^-5 + (5420 + (ww*
slip)^2*5.385)/(66564 +
6648.72*(ww*slip)^2))))]); Plot[result, {t, 0, Pi/ww}]
However, as you can see, the cosine wave is slightly shifted and it messes with my further calculations. So, what is the equation for the phase shift \$ \theta \$ ?
simulate this circuit – Schematic created using CircuitLab
I added the example schematic. In the initial condition, VC is 30V. Then I showed the graph of VC. VC is charged up to 30V in initial state and then slowly dies.
This above graph is taken from circuit lab simulation of the circuit above. This is the correct waveform with no phase shift because the simulator included the phase shift into equation. How can I so?
Best Answer
The takeaways you should get from this answer for an underdamped sinusoidal 2nd order response are;
You should expect capacitor current phase to always lead voltage phase by 90 deg. This is due to the current being the rate of change of voltage property. Your IL(t) reference direction was reversed such that discharging capacitor current was shown positive but should have been negative.
$$I_C=C\cdot dV/dt$$ By convention a negative slope for capacitor voltage decay creates a negative current. So your trace pairs looked like they were (mis) leading phase instead of lagging.
The frequency of the underdamped decay is; $$f=\dfrac{1}{2\pi\sqrt{LC}}$$
However in your formula, since the initial condition for Ic current was 0, thus θ = 0.
Sorry for the pun.
Falstad is consistent with your inverted plot above.