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Bandwidth at resoance and peak dB level

  1. Feb 24, 2013 #1
    Hi,

    I'm investigating what happens to bandwidth at resonance of a series RLC circuit when each component is doubled.

    i have simulated the circuits and created a bode plot. You can see from the photos that at resonance the gain peaks at -20.82 dB. I would like to know how to calculate this value.

    i have an input voltage of 10Vp and the voltage at resonance across the resistor is 6.428V. i understand that the voltage across both resistors is 7.07V because the multimeter is measuring RMS but how do i calculate the db level.
     

    Attached Files:

  2. jcsd
  3. Feb 24, 2013 #2
  4. Feb 25, 2013 #3
    Thanks for the reply but that still doesn't tell me how to calculate the -22 dB.
     
  5. Feb 25, 2013 #4
    You have a voltage divider with impedances:

    Z1 = R1 + j*omega*L + 1/(j*omega*C) = R1 + j*(omega*L - 1/(omega*C) )
    Z2 = R2

    Your measured voltage, call it Vout, is then given by:

    Vout = Z2/(Z1 + Z2) = R2/(R1 + R2 + j*(omega*L - 1/(omega*C) ) )

    with magnitude:

    |Vout| = R2/sqrt( (R1 + R2)^2 + (omega*L - 1/(omega*C) )^2 )

    Find peak value by finding roots of the derivative of |Vout| with respect to omega and substitute into Vout, you probably know the drill.

    My solution gives:

    Peak value = R2/sqrt( (R1+R2)^2 + (L/sqrt(L*C) - sqrt(L*C)/C)^2 ) = 0.0909 V = -20.8279 dBV

    for R1 = 10 ohm, R2 = 1 ohm, L = 70 mH, C = 110 uF.
     
  6. Feb 25, 2013 #5
    Thanks,

    Once you're broke down the equations its made it easier to see what is going on.

    Thanks for the help :)
     
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