# Why does an RLC filter circuit ring?

I have the circuit below. I understand the concept of time constant in both inductors and capacitors.

The literatures says ringing because the signal resign in the pass band which have slope very SHAPE SLOPE. Is thisbright or wrong? I don't understand. All I know is it causes only degradation of signal regarding to those freq response.

It says slow response of these circuits in the time domain. I'm not asking about explain in the frequency domain.

It does not happen with sinusoidal signals. It explains about the quality of time response of the filter.

I did not use math equation because it will blur understand of the visual event.

My literature says it has a lot of parameter include overshoot and settling time. • How does ringing in an RLC filter circuit happen? Is it related to the dlope of the pass band and stop band?

I have written the though from the literature: • If a hammer strikes a bell it rings. It rings at a certain frequency defined by the mass of the bell and the flexibility of the bell's material. Mass and flexibility are physical versions of capacitance and inductance and, the resonant frequency formula is the same format i.e. you'd recognize it. So, study why pitchforks/bells ring and apply that to RLC circuits. Note that bells don't ring when damping is too high just like RLC circuits with too much series resistance. yesterday

The series RLC circuit shown in the OP is described by a 2nd order differential equation. In this particular case, it is:$$\frac{d^{2}v_{C}}{dt^{2}}+\frac{R}{L}\frac{dv_{C}}{dt}+\frac{1}{LC}v_{C}=\frac{1}{LC}v_{s}$$

Using 2nd order differential equation avalysis reveals the characteristic equation:$$m^{2}+\frac{R}{L}m+\frac{1}{LC}=0$$

This equation can be solved to reveal three cases:

1. Over damped case: This case is sloe to respond to step changes.
2. Critically damped case: This case provides the fastest rise time with no over shoot of the final value.
3. Under damped case: This case will result in over shoot of the final value. If it is very underdamped, then the ringing that you see will occur

The cases may be separated by using the quadratic formula to obtain the zeros of the characteristic equation. There are two zeros: $$\z_1\$$ and $$\z_2\$$. So:$$z_{1},z_{2}=-\frac{R}{2L}\pm\frac{1}{2}\sqrt{\left(\frac{R}{L}\right)^{2}-4\frac{1}{LC}}$$

The cases may be determined by the sign of the expressionunder the square root as follows:

1. Over damped: $$\\left(\frac{R}{L}\right)^{2}-4\frac{1}{LC}>0\$$
2. Critically damped: $$\\left(\frac{R}{L}\right)^{2}-4\frac{1}{LC}=0\$$
3. Under damped: $$\\left(\frac{R}{L}\right)^{2}-4\frac{1}{LC}<0\$$

The oscilloscope display clearly shows the ringing of the underdamped case.

Increasing R will "damp" the ringing.

All of second order theory is too much to be contained within this answer. There are plenty of on-line and library resources available to study. It is very interesing.

• ooo ya youre right. i ve learn it in College it 25 years ago. Thanks ngab @RussellH 2 days ago
• @adhitronic It is fun to recall from long past. If this answers your question, it is customary to press the accept button. Cheers 2 days ago
• I forgot ngab. Accept button where ngab @ERussellH ? yesterday
• Click on the green checkmark below the vote buttons yesterday