Electronic – phase shift in LRC circuit

analogcircuit analysisfilter

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.
RLC
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}]

Shift
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 \$ ?

schematic

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.

waveform

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;

  • there was no "error" in the question for the values or slopes or initial conditions or his calculations. It was his expectation that was perplexed.
  • The formula with some phase offset almost looked like the response but in fact, there is no initial phase shift.
    • It is not a cosine that starts at some lag angle for theta, so it= 0
    • It is a product of a decaying exponential \$e^{-t/\tau}\$ times a cosine wave so the slope just at t=0 for switch closure IMMEDIATELY starts to decay at some rate, \$V/τ=dV/dt=Ic/C\$
  • The 2nd plot shows I(L1) which was defined in the opposite direction of I(C1),
    • both dV/dt and Ic are negative. (-ve ) just after t=0
  • except for above, my simulation is exactly the same as given, only by plot scale & resolution

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.

enter image description here Sorry for the pun.

Falstad is consistent with your inverted plot above.

enter image description here

  • My Falstad Simulations show the peak levels on the left.
  • The oscillating cycle period is just < 10ms as expected.