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.

enter image description here

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?


2 Answers 2


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.

  • \$\begingroup\$ R-LC circuit I meant a band-pass filter. I didnt get what you mean at the first part of your explanation. But basically what Im asking was: 1-) Why when the input is a high freq. square-wave there is no ringing at the output(all square waves have steep edges why it causes ringing only at low square wave inputs)? 2-) If a square wave includes those extremely high freq. sinusoidal components why they exhibit themselves only at the edges even though they exist through the entire period of a pulse according to Fourier? \$\endgroup\$
    – user16307
    Commented Sep 23, 2016 at 23:57
  • 1
    \$\begingroup\$ If the square wave frequency is too high, it will mask the ringing effect but, it is still there but..... the next square wave shape comes along and resets things. Therefore, at low frequencies, this is easy to see. If the edge hasn't happened the LC circuit does not know that until it happens. Even if the edge has happened, the LC ringing may have died down before that next edge comes along. Damped LC circuits do not do perfect fourier! \$\endgroup\$
    – Andy aka
    Commented Sep 24, 2016 at 0:06
  • \$\begingroup\$ "at low frequencies, this is easy to see" - perhaps not that easy. 1V in, 45uV out! \$\endgroup\$ Commented Sep 24, 2016 at 1:44
  • \$\begingroup\$ The 45uV is because his circuit reacts to ~800-th harmonics of a 10 Hz square signal, which should be pretty small. A EE freshman should easily calculate its value. \$\endgroup\$ Commented Sep 24, 2016 at 1:55
  • \$\begingroup\$ @Andyaka I understand now thanks, the same oscillation occurs for different square wave freq. The only difference we cannot see the entire oscillation unless the square wave period is long enough. Here the equations also show the independence of the oscillation from the input wave's freq.: ocw.nctu.edu.tw/course/physics/physical_experiment/2-5.pdf \$\endgroup\$
    – user16307
    Commented Sep 24, 2016 at 15:22

This is your 389 question in 4+ years. It looks like this field is too challenging for you. But I will bite.

The circuit is a resonance tank with resonance frequency ~8 kHz, with a slight damping [sorry, not dumping nor demping] due to 33k resistor. The resonance function is a sharp decaying function on both sides of the 8kHz peak. Try to apply AC analysis function in the LTspice, (.ac dec 100 1 100000), to see its shape.

The output of first signal is NOT A SINUSOIDAL FUNCTION, it just looks pretty similar. The 10 kHz square wave is a bit above the resonance frequency, so the base harmonics is only slightly (I mean only 50X :-) attenuated, while all upper harmonics (third, fifth, etc.) are strongly attenuated by this sharp filter transfer function, with phase shifted, etc. That's why the output mostly contains the first harmonics, and looks like a sinusoid.

In second case the signal is way below the resonance. Since you are using an ideal pulse with no edge limit and therefore with infinite bandwidth, the ~801-th (or so) harmonics of the square wave gets into the circuit's resonance, and it "rings". If you would play with some deviation from 10 Hz, you will see different ringing amplitudes.

ADDITION: Yes, the square (10Hz) waveform can be expanded into Fourier series of sine functions, all with CONSTANT COEFFICIENTS (amplitudes). In this sense, this input signal does contain some small-amplitude continuous sine wave at about 8kHz. If one can build a perfect single-frequency filter, he/she would see a continuous sine wave. However, the simple LC is very far from the ideal single-frequency filter, and passes many other harmonics as well. The trick is that just as the input signal can be viewed as a SUM of all sine waves (with proper amplitudes), the output signal is also a SUM of all frequencies it passes. So the SUM of frequencies filtered by your simple LC filter gives you the waveform you see, and not a constant sine wave as you might expect.

  • \$\begingroup\$ "dumping"? Never heurd of sach a thing. \$\endgroup\$ Commented Sep 24, 2016 at 4:10
  • \$\begingroup\$ Ok, I had that coming. \$\endgroup\$ Commented Sep 24, 2016 at 4:20
  • \$\begingroup\$ sach sings thametimes okkurs \$\endgroup\$ Commented Sep 24, 2016 at 4:23
  • \$\begingroup\$ @AliChen are you saying that any other sinusoid inputs rather than resonance freq. will be an output with damped oscillation? does that mean even the square wave input is at resonance freq. the output will never be real sinusoid because it has other components besides the fundamental freq. ? \$\endgroup\$
    – user16307
    Commented Sep 24, 2016 at 9:23
  • \$\begingroup\$ Basically yes, because the simple LC circuit is not an ideal filter, and passes many other frequencies, sum of which forms the signal you see. \$\endgroup\$ Commented Sep 24, 2016 at 16:07

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