LRC resonance question

Discussion in 'Homework Help' started by epsilonjon, Feb 10, 2012.

  1. epsilonjon

    Thread Starter Member

    Feb 15, 2011
    65
    1
    Question:

    Consider the circuit below in which R1=R2=2Ω. Select C and L to obtain a resonant frequency of 150rad/s.

    [​IMG]

    Attempt:

    First I try to find the equivalent impedance:

    Z = wLj + 1 + \frac{1}{wCj + 1}<br />
= wLj + 1 + \frac{1-wCj}{w^2C^2 + 1}<br />
= 1 + \frac{1}{w^2C^2+1} + wj(L - \frac{C}{w^2C^2+1})<br />
= \frac{2 + w^2C^2 + jw[(L-C)+w^2LC^2]}{w^2C^2+1}.

    This is where i'm not sure what to do next. In the book I am reading from it only covers resonance for parallel LRC loads connected to a current source and series LRC loads connected to a voltage source. In both of those the ω term only appears in the imaginary part so you just set that equal to zero to maximize the magnitude of the transfer function and find the resonant frequency.

    I read on Wikipedia that the resonant frequency is the one at which the reactance is zero, which was indeed true in the two simple cases in my book, but I don't understand why this is the condition for resonance. Surely the important thing is to maximize the magnitude of the transfer function?

    For example, for the circuit above, say the input was a current source and the output was the capacitor voltage. Then the transfer function is just Z, so my first thought would be that I need to find |Z|, then differentiate this wrt ω and set this to zero to find the maximum? But I don't need to do this? I can just find the ω that makes the reactance zero and i'm done? Why does this ω neccessarily maximize |Z|?

    Thanks for your help! :)
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    In your attempt you have the R values as 1Ω when you state that R1=R2=2Ω.

    In any event it is possible that for the "resonance" / zero reactance condition that the effective impedance in this situation is not a minimum. I say minimum impedance because the circuit is 'series' in nature rather than parallel. So the peak current draw from the input source wont necessarily occur at the zero series reactance condition.
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    With respect to finding the zero series reactance case I have the following relationship based on the reactances ...

    X^2_C-\frac{4}{X_L}X_C+4=0

    from this one can deduce

    X_C=\frac{2}{X_L}\pm 2\sqrt{\[ \frac{1}{X^2_L} \]-1}

    Setting XL=1 provides a simple solution.
     
    Last edited: Feb 11, 2012
  4. epsilonjon

    Thread Starter Member

    Feb 15, 2011
    65
    1
    Thanks for your reply :)

    Sorry yeah it should be 2Ω but I put 1Ω by accident.

    So just to clarify, the resonant frequency is the one at which the reactance is zero? That is the definition?

    I'm just wondering why this frequency is the special one, when you coud perhaps choose a different one for which the output is larger?

    Thanks again!
     
  5. crutschow

    Expert

    Mar 14, 2008
    12,981
    3,221
    The LC resonant point is always where the capacitance reactance equals the inductive reactance. At the point the series resonant current is maximum and the parallel resonant voltage is maximum.
     
  6. epsilonjon

    Thread Starter Member

    Feb 15, 2011
    65
    1
    But say we have C=1μF and L=0.1H in the equation for Z in my original post. Then the plot of |Z| against ω looks like

    [​IMG]

    So if I connect a voltage source and gradually increase ω, the total current drawn will keep increasing and looks like it will level off somewhere over ω=2*10^6 rad/s.

    But this if I calculate the ω that makes the reactance 0, this works out to be just under ω=10^6 rad/s. This ω will not give the maximum series current.

    Am I missing the point of what resonance is?

    Thanks.
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    With R1=R2=2Ω, L=6.666mH and C=3.333mF [milli-Farad] the zero series reactance [unity power factor load] frequency is 150 rad/sec.

    If one then simulates the frequency response with these values, the peak current occurs at 41.35 Hz or ~260 rad/sec.

    So clearly the peak current does not occur at the zero series reactance [unity power factor load] case. One might then be tempted to suggest the resonance condition is at 260 rad/sec given the peak current occurrence observed.

    However, if one measures (again by simulation) the total power absorption of the circuit at frequencies of interest there are some useful insights.

    At 150 rad/sec the power absorbed from a 1V rms source is 333.9mW

    At 260 rad/sec the power absorbed from a 1V rms source is 357.0 mW

    At 212.15 rad/sec the power absorbed from a 1V rms source is 363.5mW

    So there is a maximum power absorption at ω=212.15 rad/sec which also happens to be at ω=1/√(LC) where L=6.666mH and C=3.333mF as stated above.

    It would seem to me then, that a suitable criteria for resonance with lossy circuit elements (resistors R1 & R2) present would be the case where maximum power is drawn from the source driving the circuit. This is consistent with the assumption that resonance would occur at the typical condition of ω=1/√(LC).
     
    Last edited: Feb 11, 2012
  8. epsilonjon

    Thread Starter Member

    Feb 15, 2011
    65
    1
    I don't understand how you got that the peak current occurs at ~260 rad/s. The |Z| versus ω graph is the same shape as the one above, so does |I| not keep increasing as you increase ω (since |I|=|V|/|Z|)?

    Edit: I just realised why it wasn't making sense. I'd missed out an ω somewhere when I plotted the graph. Doh!! :D The graph should look like this:

    [​IMG]

    That looks better. Hopefully now I can understand what you said about the power being maximum, but I'm going for my dinner first :p
     
    Last edited: Feb 12, 2012
  9. mlog

    Member

    Feb 11, 2012
    276
    36
    Good answer. Resonant frequency, I'll call "w" is defined as:

    w^2 = 1 / (C * L)

    or C * L = 1 / (w^2)

    Given that w = 150 rad/sec, then substitute into the last equation.

    C * L = 1 / (150^2) = 1 / 22500 = 0.000044

    So if you know either the value of C or L, you can find the other. For example, if L = 1 Henry, then C = 44 microFarad.

    I suspect there are three points to get from this problem.

    1. There is a superfluous information provided. You really only need the frequency to find C and L.

    2. You need to know the resonance equation and how to use it.

    3. There is not a unique value of C or L based on the information given. The choice of one value depends on the other.

    The same resonance equation is used for parallel and series C-L circuits.
     
  10. epsilonjon

    Thread Starter Member

    Feb 15, 2011
    65
    1
    I can see why that would make sense, but i'm struggling to show it in general.

    Say you have some combination of resistors, inductors and capacitors connected across a voltage source, and let Z=R(ω)+jX(ω) be the equivalent impedance. Then the power delivered by the source is

    P=\frac{|V|^2R(w)}{R^2(w)+X^2(w)}

    so it's not obvious from this because you don't know how the ω value will affect R(ω)? I guess you need to show that however you arrange the circuit elements, Z(ω) will always take some special form which makes R(ω)/(R^2(ω)+X^2(ω)) maximum when X(ω)=0?

    I'm confused by this statement because I swear t_n_k just showed that this is not true for this circuit? Or am I misunderstanding what you mean by "series resonant current"?

    Thanks again.
     
    Last edited: Feb 12, 2012
  11. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I'm skeptical that one could find a general relationship for any arbitrary circuit topology where resonance is possible.

    As to the definition of the resonance conditions it's clear that the simple series LCR behavior can't be generalized for all topologies - it only takes one atypical case to prove the rule doesn't apply to all cases.

    This might lead to a more general discussion of what resonance constitutes.

    I would suggest a definition might have regard to the total stored energy per excitation cycle. One would assume resonance occurs when the excitation frequency is at a value which maximizes the stored energy per cycle for a given excitation magnitude. This might reinforce my observation that the maximum power draw from the source in your case corresponded to the resonant frequency of the LC pairing. Consider the circuit Q-factor. Q is a measure of the ratio of energy storage compared with energy loss per excitation cycle and is a constant for a particular circuit. So presumably as energy storage is maximized at resonance so energy loss is also maximized.
     
  12. mlog

    Member

    Feb 11, 2012
    276
    36
    If I haven't made any math errors, the resonant frequency in radians is:

    w = sqrt [ (R1 + R2) / (R2 * L * C) ]

    and the Q is:

    Q = {sqrt [(R2 * L * C) * (R1 + R2)]} / [(R1 * R2 * C) + L]

    I apologize for my previous post where I treated the resonant frequency as simply sqrt[1/(L*C)].
     
  13. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Hi mlog,

    I guess you set up the condition where the effective impedance of the circuit is purely resistive.

    Checking with a simulation set as follows

    R1=R2=2Ω , L=100mH & C=888.888uF

    giving ω=√(R1+R2)/(R2*L*C)=150 rad/sec per your equation

    A simulation with these values shows current and voltage at 150rad/sec input frequency aren't in phase - which may be of interest to you.

    In fact, I have the condition for purely resistive input as corresponding to a valid solution of

    X_C=\frac{R^2_2}{2X_L}\pm \frac{R_2}{2} \sqrt{\frac{R^2_2}{X^2_L}-4}

    which interestingly is independent of R1.
     
  14. mlog

    Member

    Feb 11, 2012
    276
    36
    I'm going to experiment and see how this turns out. I've never used LaTeX before.

     \large w= \sqrt{ \frac {R_1+R_2}{R_2LC}}

     \large Q= \frac {\sqrt {(R_2LC)(R_1+R_2)}} {(R_1R_2C)+L}
     
  15. mlog

    Member

    Feb 11, 2012
    276
    36
    I laboriously cranked out the transfer function by hand, and the value for frequency and Q were part of the numerator. That's why I said if I didn't make any math errors... :D
     
    Last edited: Feb 13, 2012
  16. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Hi mlog,

    I don't think there is any issue with calculating the condition at which the circuit impedance is purely resistive at a specified operating frequency [fr] - which one would normally assume is indicative of the resonance condition. The result is easily verified. But what is unusual in this case is that peak current draw from the source does not coincide with the aforementioned specific design frequency - rather it occurs at a different frequency which [by the way] is not the 'typical' ω =1/√LC value. It seems this peak occurs at √3 * 2*∏*fr rads/sec when R1=R2 - a point I've yet to confirm. If true the peak also then corresponds to √(3/2)*ω [where ω=1/√LC].

    What seems to be at issue in this thread is the defining circuit condition at 'resonance'. The problem with this particular circuit is that it doesn't follow the normal behaviour of [say] a simple series LCR circuit. The various discussion points in preceding posts address the problem and its possible resolution in some detail. The shunt resistor [R2] around the capacitor is the 'cause' of the discrepancy - if any discrepancy actually exits.

    My feeling is that resonance is indeed occurring at ω =1/√LC.
     
    Last edited: Feb 13, 2012
  17. mlog

    Member

    Feb 11, 2012
    276
    36
    It's getting late. Maybe I'll think about this more tomorrow. However, I have two quick comments before I hit the hay.

    1. I assume you're using a simulation program to plot the resonance. If so, could the generator source impedance be shitting the resonant frequency somewhat? It's possible that R1 and R2 will have some effect, so possibly the source impedance is causing a shift (effectively increasing the value of R1).

    2. I don't think I mentioned there is a pole in the transfer function caused by the parallel combination of R2 and C. Maybe it's having some effect. I would expect it to act like a 1st order low pass filter though.
     
    Last edited: Feb 14, 2012
  18. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Hi mlog,

    Re your comments ...


    1. I get the same results irrespective of whether I simulate with a Spice based application (transient or frequency response) or a Matlab clone application using Laplace forms. I'm confident there are no errors there.
    2. One gets a response [let's say it's the source current vs frequency for a constant source voltage] based on the values selected for the total circuit. Focusing on the behavior of specific parts of the circuit may give a false impression of the overall result.
    To be honest I'm not particularly interested in delving into this much further. The OP hasn't come back with any follow-up comments or questions. It's about then I lose interest.


    Thanks for your interest.
     
  19. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    As a final note it's worth checking Wikipedia

    http://en.wikipedia.org/wiki/RLC_circuit

    Interestingly, the notes suggest series resonance with a shunt resistance on the capacitor occurs at

    \omega_o=\sqrt{\( \frac{1}{LC}-\frac{1}{(RC)^2} \)}

    In this case the assumption is that the excitation current & voltage are in phase at resonance. This probably negates my assumption about peak loss always occurring at resonance when lossy elements are part of the total circuit.

    It's possibly correct in fact to consider the peak energy storage in the circuit as the determinant of resonance - rather than the peak power loss at resonant frequency.

    A study of this is informative for the case at hand.

    Applying a 200V p-p source to the simulated circuit [R1=R2=2Ω, L=6.666mH, C=3.333mF] I have the simulated peak energies as

    At ω=150 rad/sec [in-phase case] Peak Energy = 6.323 Joule
    At ω=1/√(LC)=212.13 rad/sec Peak Energy = 6.061 Joule
    At ω=259.81 rad/sec [peak current draw] Peak Energy=5.633 Joule

    So the in-phase case gives a 'convincing' result for the energy criterion. One would need to test with other values of L & C to confirm this is true in general or otherwise. For instance with L=4mH and C=1.1111mF say. The roots of the impedance Laplace form numerator function should probably be complex for a resonance condition - or alternatively the damping factor should be less than unity.

    Resonant Frequency Proof:

    Wikipedia does not show the proof, which can proceed as follows:-

    Z=jX_L-\frac{jRX_C}{R-jX_C}=jX_L-\frac{jRX_C \[ R+jX_C \]}{R^2+X^2_C}=jX_L-\frac{jR^2X_C}{R^2+X^2_C}+\frac{RX^2_C}{R^2+X^2_C}

    For zero phase shift the imaginary terms must vanish

    jX_L-\frac{jR^2X_C}{R^2+X^2_C}=0

    or

    X_L ( R^2+X^2_C)-R^2X_C=0

    substituting

    X_L=\omega_oL
    &
    X_C=\frac{1}{\omega_oC}

    leads to

    \omega^2_oR^2LC^2+L-R^2C=0

    or

    \omega^2_o=\frac{1}{LC}-\frac{1}{R^2C^2}

    and finally

    \omega_o=\sqrt{\frac{1}{LC}-\frac{1}{R^2C^2}}

    Note that the addition of a resistance in series with the inductor does not change the outcome as far as the resonant frequency ωo is concerned.
     
    Last edited: Feb 15, 2012
Loading...