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Above is a circuit with L C R components to model ringing.

As you see a 1V 50Hz pulse is applied to the circuit. The ringing can be seen at the output pulse in blue color which has the resonance frequency of the parallel LC.

Even though the input signal is 50Hz, since it is a pulse(composed of harmonic sines) it generates ringing at the 5kHz component of the pulse(resonance freq. component).

My question is: Is there a remedy for this ringing at the output without affecting the pulse widths?


I added a series LC with the same resonance freq. in series with R as seen below:

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If I take the output from terminal of R1 the output (Out1 on circuit) is with a large spike(please left-click to enlarge): enter image description here

But if I take the output from the terminal of C2 the output (Out2 on circuit) is clean and with no ringing; with large rising and falling edges but this could be fixed with a comparator I guess. Here is Out2 on circuit:

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Here is the output from Andy aka's suggestion which completely removed ringing by keeping the pulses sharp:

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  • \$\begingroup\$ But this is only a simulation of ringing and consequently you know, a-priori, the L and C values and the configuration of these components that give rise to it. Therefore, of course you can cancel the ringing quite successfully. It's like saying I've got some random noise, but I know it; so I can invert it; and add it; and hey-presto, noise cancelled. However, in practice you don't know the noise a-priori. \$\endgroup\$ – Chu Jul 29 '16 at 0:11
  • \$\begingroup\$ But I can observe the ringing freq. on scope and set up an LC which gives the same resonant frequency. Do u agree? \$\endgroup\$ – user16307 Jul 29 '16 at 0:20
  • \$\begingroup\$ i don't get what you mean. as long as i observe the nature of ringing i can remove it by using any LC combination which has the same resonant freq. with ringing. that was in my mind with the current knowledge i have. what would be your remedy to ringing? \$\endgroup\$ – user16307 Jul 29 '16 at 1:17
  • \$\begingroup\$ If you choose different L and C values for the compensation, eg 10mH and 0.1uF, the result would not be the same \$\endgroup\$ – Chu Jul 29 '16 at 1:18
  • \$\begingroup\$ isn't resonant freq. 1/(2 * pi * sqrt(L * C)) ? why would it be different? \$\endgroup\$ – user16307 Jul 29 '16 at 1:19

The ringing is because the inductor and capacitor are acting as an underdamped resonant circuit. You can add more damping by placing a resistor in parallel with them both or a resistor in series with the inductor or capacitor. If you increase the value of series resistor (or decrease the parallel resistor) the ringing should start to subside but, you will never get a perfect rectangluar wave coming out when ringing is minimized.

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  • \$\begingroup\$ the thing is i'm afraid if the ringing oscillates large it might cause false triggering. the remedies you suggest sounds great but only if one knows the circuitry. sometimes the circuit is very complex and you have this type of ringing caused by lets say wrings stray capacitances ect. in that case one cannot place an extra resistor to add damping. i mean if you have an output with large ringing and you cannot interact with the circuit, is there a way to minimize this ringing as well? \$\endgroup\$ – user16307 Jul 28 '16 at 12:26
  • \$\begingroup\$ You could try slowing down the rise/fall time of the input waveform. You might even be able to add a series tuned circuit across the resistor that was resonant with the ringing frequency. \$\endgroup\$ – Andy aka Jul 28 '16 at 12:28
  • \$\begingroup\$ please see my edit. was that what you meant? \$\endgroup\$ – user16307 Jul 28 '16 at 13:00
  • \$\begingroup\$ No, I said a series tuned circuit across the resistor BUT it's always good to experiment with different configurations as a learning exercise. What you have done (OUT2) is used a damped low pass filter to remove the ringing (you can still see a little bit) but you might get steeper rise and fall by playing games with the values. Maybe lower C2 progressively or try what I originally suggested. \$\endgroup\$ – Andy aka Jul 28 '16 at 13:09
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    \$\begingroup\$ There's many ways of doing this and it looks like you found one way but, I have to be honest this was still not what I meant. Originally you had the resistor going to ground and my suggestion would have placed C2 and L2 (in series) across R1. I saw R1 as a load resistor and not to be moved from ground. You might find that if you added a load to your circuit it won't work as well. Keep messing with things dude. \$\endgroup\$ – Andy aka Jul 28 '16 at 14:17

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