Can someone help me on the LC circuit?

Discussion in 'General Electronics Chat' started by lxt971, Mar 7, 2013.

  1. lxt971

    Thread Starter New Member

    Mar 7, 2013
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    Hi,
    I am studying LC circuit now. I have a question, hope someone can help.

    1. A series LC circuit with a signal source. Measure the voltage on the capacitor.

    [​IMG]

    2. if the input is a step function, the output is a sinusoid whose frequency is LC's resonance frequency.

    http://bbs.elecfans.com/data/attachment/forum/201303/07/081922t3aazv5ez9eg98a9.png

    3. if change the the input to a sinusoid whose frequency is LC's resonance frequency. the output goes to infinate. I understand in this case the LC's impedance is 0 at this frequency.
    http://bbs.elecfans.com/data/attachment/forum/201303/07/082136xd4dp48zbue3zp3p.png

    4. But i checked the fourier transform of the the step function. It's spectrum covers the whole frequency (from 0- infinate). To my understanding it contains many many sinusiods , including the resonance frequency. And step 3(sinusoid input)'s response should be a part of the the step response.
    how come the step response is amplitude limited, but sinusoid input response is infinate.

    Where is my mistake, could someone tell me? Thanks a lot.


    http://bbs.elecfans.com/data/attachment/forum/201303/07/082329wzdyl2lda22zdfq3.png
     
  2. Brownout

    Well-Known Member

    Jan 10, 2012
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    The circuit is a filter, rejecting all frequencies except for it's own fundamental frequency. However, the step function does show up in the system's transient response.
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Firstly this isn't a filter which rejects all but the resonant frequency.
    If you do a steady state analysis at a non-resonant frequency sinusoidal excitation there will still be a voltage across both L&C.

    I think the confusion partly arises due to a lack of appreciation of the transient nature of the case of circuit excitation with the sinusoidal source.

    Both the step response & the sinusoidal responses are going to be transient in nature. A confounding factor is the lack of an energy dissipative component in the circuit which is a nonsense in a practical situation. In that context the term transient takes on a certain ambiguity. It becomes difficult to differentiate between transient & steady state.

    The interpretation of the step input as an infinite frequency band source takes no account of the specific energy at each "quantum" of frequency. It rather begs the question - If the step is an infinite continuous summation of frequency terms what is the energy content of an infinitesimal quantum of frequency band?

    This is a thought provoking & intelligent question by the OP. It is unusual to to see such significant potential for excellence in forum newcomers.
     
    Last edited: Mar 8, 2013
  4. Brownout

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    Jan 10, 2012
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    Which is true of any filter. Having voltage at a non-resonant doesn't disqualify it as a filter. In fact, if you did the analysis, you would observe an infinite response at the resonant frequency, and infinitely smaller response at all non-resonant frequencies.
     
  5. t_n_k

    AAC Fanatic!

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    Yes you are correct. The ideal circuit shown has infinite "steady-state" response at resonance and less than infinite @ any other.
    However does the circuit qualify as a filter with respect to our usual understanding of that term?
    Not sure if that explains the OP's dilemma.
     
    Last edited: Mar 8, 2013
  6. Brownout

    Well-Known Member

    Jan 10, 2012
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    Ok, my terminology needs work. It's been 20 years since I had to think about such networks. The passband for this 'filter' is the frequency range below it's resonate frequency, and the reject back is the frequencies above it. It's not practical as a filter to be used in an actual circuit. However, networks that have responses at resonance many times higher than at other frequencies are useful in certain types of applications.

    That said, the infinite response at the circuit's resonate frequency is the reason for the observed waveform past the transient interval. The circuit's attenuation of frequencies higher than resonance is responsible for the transient response to those frequencies of the reject band.

    Sorry if my earlier reponse, or this one for that matter, caused any confusion.
     
  7. t_n_k

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    On the contrary, your comments are not at all confusing. They are quite pertinent and welcome. It would be good if the OP re-engages but that may not happen. Thanks.
     
  8. nigelwright7557

    Senior Member

    May 10, 2008
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    A series LC circuit is an accepter circuit and so shorts out the resonant frequency.
    1/ (2 pi SQUARERROOT(LC))

    A parallel LC circuit is a rejector circuit which allows the resonant frequency across it but virtually shorts out any other frequency.
     
  9. t_n_k

    AAC Fanatic!

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    That's true but in this case the output [according to the OP's original post] is viewed as the capacitor voltage.
     
  10. BillB3857

    Senior Member

    Feb 28, 2009
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    For cap voltage to be greater than inductor voltage, the input frequency would need to be lower than resonance. Above resonance, the inductor would develop the higher voltage. At resonance, Xl=Xc. Voltages would read equal but phases would be different. At lease that's the way I remember it from over 50 years ago.
     
  11. lxt971

    Thread Starter New Member

    Mar 7, 2013
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    Thanks a lot for your help. But i am still confused. :)
    I simplified it by changing the excitation to an impulse.

    1. in time domain, the output is still a sinusoid. The energy is 1/2*CV^2 (V is max voltage on Cap). Cause it's a lossless circuit, so I know the impulse's energy is also 1/2*CV^2 . it's a energy limited input. I think it consists with t_n_k's explanation.

    2. in frequency domain. impulse's fourier transform is 1. According to Parseval's Theorem, the energy will be infinate.(integrate 1 from -OO to +OO)
    [​IMG]
    I checked wiki, it says Parseval's Theorem's precondition is f(x) is two square integrable. my math is not very good, i don't know if impulse's square is integrable or not. when i start to think about it, i can see stream is coming out of my head.
    So i'd like to ask, is Parseval's Theorem not applicable to impulse or step? if it isn't, what's the physical meaning of the amplitude in their fourier transform.
    Thanks again.
     
  12. t_n_k

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    Irrespective of the forcing function chosen, I don't know if a rigorous proof exists that allows a verification of the readily derived time domain solution with sole reference to a Fourier decomposition of the forcing function. Keep in mind that (ignoring the more useful Laplace transforms) the frequency domain approach requires one to determine the response not only at the known resonant frequency (magnitude & phase) but at all component frequency values which we believe to be an infinite & continuous distribution.

    How would one do that? Perhaps the more skilled mathematicians in the Forum can assist one way or the other by indicating if a method exists or not.
     
  13. Brownout

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    Jan 10, 2012
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    No. The sinusiod only captures energy at it's specific frequency from the impulse.
     
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