Electronic – A question about ringing phenomena and resonance

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Below is an LC circuit fed by a 1V 10kHz square-wave. As you see the output is a sinusoid, which would be max. if it were produced by an input at resonance frequency.

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And here below the same circuit's output for a 10Hz square-wave input. As you see the output is a ringing signal which looks like a damped oscillation at resonance frequency.

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Why is ringing occurring here for square-wave inputs at only low frequencies? And is there a relation between Gibbs phenomena and ringing? And why ringing occurs near the rising and falling edges?

Best Answer

Below is an LC circuit fed by a 1V 10kHz square-wave. As you see the output is a sinusoid

A signal source that produces a 10 kHz square wave will continue to produce precisely that square wave when connected across a parallel tuned circuit. It does that because it is a signal source and has zero ohms output impedance and, if necessary, will supply infinite current to sustain itself.

It will not produce anything other than a 1V, 10 kHz square wave except when driving a short circuit and then the output is indeterminate.

If in fact the parallel tuned cicuit were pre-empted by a resistor of any value (33k in the question) then that makes absolutely no difference. It is a voltage source and will do what is has to do.

If, on the other hand, you are asking to consider what the LC voltage is after feeding with a voltage source in series with a 33k resistor then that is a different matter.

Why is ringing occurring here for square-wave inputs at only low frequencies?

Because it's not low frequency - the edge of the voltage (if infinitely steep) has contained inside it infinite harmonics and, one of those harmonics will be coincident with the LC resonant frequency and trigger a damped oscillation as seen in the 2nd waveform picture. Of course, the edge of the voltage only has to contain a harmonic coincident with the LC resonance for this to happen - it doesn't need to be infinitely fast.