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I have a simple RLC circuit (100 ohm resistor, 10mH inductor and 1uf capacitor in series) and I need to measure the circuit's resonant/natural frequency and damping coefficient. The input to the circuit is a 2% duty cycle, 1khz 1V square wave (like the unit impulse).

Circuit Diagram

My understanding is that when the circuit resonates, the combined impedance of the inductor and capacitor is zero, so the circuit's impedance is at a minimum. If I measure the voltage across the resistor, this should be at a maximum (since the current is at a maximum).

I calculate the resonant frequency for this circuit to be ~1.6KHz however I don't measure any maximum or minimum voltage around this point. The voltage just keeps decreasing as the frequency increases.

Top graph is the output waveform

This seems to have something to do with using an impulse input because if I use a sine wave instead it is easy to see this effect. Why does it not resonate in this case?

I also calculated the damping coefficient to be 0.5. This should mean the system is underdamped and should oscillate. I haven't been able to work out how to measure this yet. Any suggestions?

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  • \$\begingroup\$ Did you look at the resulting waveform on a scope? \$\endgroup\$
    – jippie
    Aug 30, 2014 at 8:27
  • \$\begingroup\$ I don't see your problem, the results are what I would expect. All the resonant-gain-bodeplot thing works for pure tone inputs, aka sinusoids. If you sen in a SUM of sinusoids, e.g. a square wave, the output will be a sum of sinusoids, weighted by the RLC series. And that's not something naked eye can easily see. Try to make the FFT of the output, then sweep your square wave from 160Hz to 16kHz, looking at the first spike (fundamental harmonic). Do you see it now? \$\endgroup\$
    – Vladimir Cravero
    Aug 30, 2014 at 8:34
  • \$\begingroup\$ @jippie yes the signal on the scope is right (the impulse response of the system). \$\endgroup\$
    – geniass
    Aug 30, 2014 at 9:36
  • \$\begingroup\$ I think you improve your question with a circuit diagram, including AC source and your meter/scope. \$\endgroup\$
    – jippie
    Aug 30, 2014 at 9:51
  • \$\begingroup\$ @Vladimir can you explain why it doesn't oscillate /resonate with the square wave input? The fourier transform spike would make sense because the unit impulse includes all frequencies \$\endgroup\$
    – geniass
    Aug 30, 2014 at 10:49

2 Answers 2

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The problem is, I feel, in your thinking. The unit impulse does contain all frequencies and therefore, the RMS value of the particular frequency that corresponds with your filter's resonant frequency is infinitely small.

But, you might say that your filter's bandwidth is still quite wide so the energy of the spectrum around your resonance is not infinitely small. However, all the "in-band" frequencies that are "stimulating" your filter are incoherent and you can't expect to see those energies translated to one clear and obvious sinewave.

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  • \$\begingroup\$ I'm not expecting a sine wave output from an impulse input. I was expecting that same impulse response graph in the question but with a peak in the amplitude at a certain frequency. Also since the input isn't an ideal impulse, shouldn't it basically become a square wave at low frequency (duty cycle = 2% of period)? \$\endgroup\$
    – geniass
    Aug 31, 2014 at 7:09
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So you can see the resonance with a sine wave, that is good. When doing the short pulse response are you keeping the pulse width constant? Or do you keep the duty cycle constant? I was going to say if you keep the pulse width constant you should be able to see the resonance... but I'm not quite sure.

Re damping and oscillations: A damping factor of 0.5 is fairly well damped. (Isn't that a Q of 1 ?) You can see just a bit of ringing in your 'scope shot. That looks about right for a Q of 1.

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  • \$\begingroup\$ I'm keeping the duty cycle constant at 2% and varying the period. Also could measuring the damping coefficient have something to do with measuring the positive peak voltage and negative peak voltage? \$\endgroup\$
    – geniass
    Aug 31, 2014 at 7:12
  • \$\begingroup\$ Hi @geniass, OK is there anyway you can keep the width constant? Your impulse is not really a perfect impulse in that it was some width. If you keep changing the width that changes the "size" of the impulse. Re damping: If you write down all the equations and solve at resonance I think that you'll find that with a sine wave stimulus the ratio of output amplitude to input is equal to the Q. (Q = 1/(2*damping).. I think.) (I only remember the relation between Q and damping because for a Butterworth response both are equal to sqrt(2).) \$\endgroup\$
    – George Herold
    Aug 31, 2014 at 13:58

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