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Finding half-power frequency.

  1. May 31, 2012 #1


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    Hi all. I'm trying to do a circuit problem in my book. The circuit consists of a resistor, capacitor, inductor and a voltage source all in series. The first part of the question says: Find the resonance frequency and half-power frequencies. My book does go on and solve this problem with two different methods using given formulas for series RLC circuits, however, I would like to solve this circuit without formulas (if possible). My problem is calculating the half-power frequency not the resonance frequency.

    R = 2 Ohms
    L = 1 mF
    C = .4 microH

    My approach:
    Transfer function: H(s) = R + sL + 1/sC => H(ω) = j(ω*R*C) + (1 - [ω^2 * L * C])

    Magnitude of Transfer function: |H(ω)| = √( (ω*R*C)^2 + (1 - [ω^2 * L * C])^2 )
    Set Magnitude of Transfer function equal to 1/√2 or set Magnitude of Transfer function squared equal to 1/2:
    |H(ω)| = 1/√2 or |H(ω)|^2 = 1/2
    Solving for this I get ω1 = -27k or ω2 = -65k

    Obviously this is wrong due to the negative ω's. Not only is the sign wrong but also the magnitude. The book achieved the answers: ω1 = 49k or ω2 = 51k

    Can anyone please tell me what I'm doing wrong (don't tell me to use formulas please)?
    Thank you for your time.

    PS: This is NOT a homework problem.
    Last edited: May 31, 2012
  2. jcsd
  3. May 31, 2012 #2


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    Resonance happens when the reactances of the capacitor and inductor are equal.

    The Q (or Q factor) of a circuit is equal to the reactance of the inductor at resonance divided by the resistance of the resistor.

    So, you may like to look up Q factor on Wikipedia or a text book to see how to derive the bandwidth.
  4. May 31, 2012 #3


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    Can you get the halfpower frequency from the transfer function?
  5. Jun 1, 2012 #4
    At DC your power dissipation is 0 (cap is blocking current). At very high freq, power dissipation is 0 (inductor is blocking current). At resonance your power dissipation is
    (V^2)/R since the LC is series resonant and behaves like a short, all of the input voltage is across the resistor.

    There will be two half power frequencies, one above and one below resonance, where the circuit power dissipation is (V^2)/2R, or the voltage across the resistor is 0.707Vin.

    Your transfer function needs to be in the form of output/input.
    In this case, since we are finding power consumption, which only occurs in the resistor, our output is the voltage across the resistor. Vr/Vin = R/(R + sL + 1/sc).
    Next, set Vr/Vin = 0.707 and solve for freq.
  6. Jun 1, 2012 #5


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    => H(ω) = 1/(jωC) ( j(ω*R*C) + (1 - [ω^2 * L * C]))
    Obviously can't be right.
  7. Jun 27, 2012 #6


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    Thank you. Now I know that in order to solve for the cutoff frequencies my transfer function has to be in the form of Vo/Vi and not in the form of impedence as I had tried. Using Vo/Vi works. Thanks again and sorry for the super late reply.
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