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What does Q have to do with frequencies?

  1. Sep 24, 2009 #1
    Could someone explain why Quality factor, which as I understand it, is 2pi *(maximum energy stored/energy dissipated per cycle) is related to the resonant frequency/bandwidth?

    I don't understand the connection between the two concepts.


  2. jcsd
  3. Sep 25, 2009 #2


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    Q is really related to the period T. It depends on the energy lost in one period oscillation.
    It then gets related to the frequency by f=1/T.
  4. Sep 25, 2009 #3
    I"m having trouble relating the concept of 2pi *(maximum energy stored/energy dissipated per cycle) to the concept of( Frequency of resonance)/(bandwidth of resonance).

    These two things seem very different yet they are related (in fact they're equal), is there an intutive explaination for why this is so?

  5. Sep 25, 2009 #4

    Andy Resnick

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    You are asking a good question, and I don't have a compelte answer, but the bandwidth of a resonance is related to the damping/dissipation/dephasing phenomena present. 1/Q, in optical systems, is equal to the (negative of the) absorptive susceptibility. That may be true for resonant systems in general, but I don't know.

    Seigman's book "Lasers" has some information, but the connection between the two is not spelled out in detail.
  6. Sep 25, 2009 #5
    I can understand how the damping effects the system at the limits. That is, if you have no resistance, i.e. no damping then you get an infinite spike at the resanant frequency. And, if you have no reactance just damping- resistance then you get a flat frequency response. I'm having trouble visualizing what happens as you go between these limits.
  7. Sep 25, 2009 #6

    Andy Resnick

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  8. Sep 25, 2009 #7
  9. Sep 25, 2009 #8
  10. Sep 25, 2009 #9
    Here is an alternate view. The Q of a resonance is actually just the ratio of the reactive impedance Z = sqrt(L/C) to the real (resistive) part = R.

    So Q = sqrt(L/C) / R

    The resonant frequency (radians per sec) = w0 = 1/sqrt(LC). Substituting in the above equation we get

    Q = w0 sqrt (LC) sqrt(L/C) / R = w0L/R

    [Edit] Note that w0L is just the reactive impedance of L at resonance. Note that w0L I2 is the maximum reactive power (stored), and R I2 is the maximum resistive power (lost), so Q is the ratio of reactive to real power.

    Bob S
    Last edited: Sep 25, 2009
  11. Sep 25, 2009 #10
    I think it's the next step where I'm missing something important. Now if you take that system and drive it at a frequency that is not w0 but somthing close how does the damping and Q effect the steady state solution? What I'm triyng to get at is, how does larger damping widen the resonance curve? of the forced response?

  12. Sep 25, 2009 #11
    Hi John-
    The attached thumbnail shows the forced resonance curve of [STRIKE]your[/STRIKE]a series LRC circiut vs. frequency using a constant voltage source (= 1 volt). The maximum resonant current is about 19 milliamps. The dotted line shows the phase angle. The phase angle is zero at about 55 Hz.

    [Edit] Here is another thumbnail showing the forced-resonance response of the same series LRC circuit using a 1-amp constant-current source as the driving signal. Note that at the minimum (the self-resonant frequency), the output voltage is 52.5 volts, exactly the circuit resistance times the driving current. At other frequencies the voltage is higher, due to the added reactive impedance of the L and the C. The L and the C reactances cancel each other out at resonance.
    Bob S

    Attached Files:

    Last edited: Sep 25, 2009
  13. Sep 26, 2009 #12
    The two definitions of resonance given by the OP do seem very different. One perspective that may help to unify them might be given by considering the impulse response.

    The decaying sine wave as defined by the impulse response goes down by a factor of 1/Q with each oscillation (give or take a factor of 2pi). What if we drive the system with wave that is slightly off resonance? We can calculate the sum by take the sum of an chain of impulses. Actually in the limit this is a convolution integral, but I'm not going to argue such a precise calculation. Let's keep it simple and drive the system with a train of impulses.

    As long as the impulses are at the exact resonance frequency, the system response will just build up. If the impulses are slightly off frequency, then eventually they will get 180 degrees out of phase and will interfere destructively with the initial response. But if the frequency discrepancy is very small, this doesn't matter. Why? because by the time the phase shift has accumulated to a sufficient degree, the original waves have already died down, and the new phase is the dominant one.

    The half-bandwidth point is significant because it means that by the time the phase has shifted 180 degrees, the original waves are still fairly strong. So the destructive interference starts to become significant. As the driving frequency gets more out of whack, you are basically generating multiple trains of sine waves that are randomly out of phase which each other to the point where they can't build up to any significant degree.
  14. Sep 27, 2009 #13
    I think I understand my question now. I'll try to explain.

    First, why is Q related to the resonant frequency/bandwidth?

    Well using a series LRC as the resonant application.

    Q is not directly related to the resonant frequency, the resonant frequency is set by LC (although somehow that doesn't seem exactly right because I somehow think that the damping factor should slightly lower f0)

    But Q has everything to do with the bandwidth and peak of the magnitude of the transfer function. The resonant frequency sets where the peak is and Q sets the height and width/shape of the curve.

    Getting to the second part of my question "which as I understand it, is 2pi *(maximum energy stored/energy dissipated per cycle)", well to understand how that is related to Q then you have to look into what Q is and then you find the terms for energy storage and energy dissipation- which is what Q is all about.

    Thanks for all the help folks!

    Last edited: Sep 27, 2009
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