The actual cutoff frequency of an LC low pass filter

Discussion in 'General Electronics Chat' started by omar-rodriguez, May 16, 2018.

  1. omar-rodriguez

    Thread Starter Member

    Jun 24, 2015
    66
    2
    In this thread I'm going to show you that the cutoff frequency of an LC low pass filter is not [​IMG] as always people and even teachers says, I want to see your comments.

    I'm going to start by finding the transfer function of the filter with a resistive load
    [​IMG]
    [​IMG]

    From the previous equation you can see that if R is too big, then the equation simplifies to the transfer function of an LC filter without load

    [​IMG]

    Therefore the resonance frequency of the filter happens when the amplitude is maximum, and denominator becomes zero and R is too big.

    One thing is the resonance frequency and another is the cutoff frequency, so in order to find the cutoff frequency of the filter, the first thing to do is find the magnitude of the transfer function:

    [​IMG]
    Then by definition the cutoff frequency is reached when there is a drop of -3dB from the maximum gain, in this case 0dB.

    [​IMG]

    So the angular cutoff frequency w is given by the next equation

    [​IMG]

    This is too hard to solve by hand so I use Matlab to find the angular cuttoff frequency w, and this is what I found:

    [​IMG]

    [​IMG]

    I did a couple of simulations using matlab to see if it is true, so L=100mH, C=10nF and R=1kΩ replacing these values in the equation I get a cuttoff frequency of f=1.7659kHz, and the same in the simulation look:
    [​IMG]
    But when R is too big there is resonance and the formula doesn't work anymore, look at the simulation for L=100mH, C=10nF and R=10MΩ , fres=5kHz
    [​IMG]
    As I have shown, the cutoff frequency of a LC low pass filter is not [​IMG] that expression is the resonance frequency. The cutoff frequency is much more complicated and is given by [​IMG] .
     
    Last edited: May 16, 2018
  2. MrChips

    Moderator

    Oct 2, 2009
    16,887
    5,200
    If R is relatively large compared to the reactance of L and C, then you can eliminate R from the circuit.
    You are left with an L-C circuit which will have a resonance frequency (not a cut-off frequency).

    \omega^2 = \frac{1}{LC}
     
  3. omar-rodriguez

    Thread Starter Member

    Jun 24, 2015
    66
    2
    "Therefore the resonance frequency of the filter happens when the amplitude is maximum, and denominator becomes zero and R is too big."

    that's what I explained before
     
  4. BR-549

    AAC Fanatic!

    Sep 22, 2013
    4,011
    1,032
    The first formula you post...in post #1....was never taught to me as a cutoff frequency formula.

    That is for resonant frequency.

    They have roll-off formulas for RC filter low roll-off. And for high roll-off.
     
  5. omar-rodriguez

    Thread Starter Member

    Jun 24, 2015
    66
    2
    Even in a texas instruments application note appears that the cutoff frequency of the LC filter is[​IMG] , but why?

    [​IMG]
     
  6. omar-rodriguez

    Thread Starter Member

    Jun 24, 2015
    66
    2
    Finally after a while, I realized that there is only one particular case for which the resonance frequency and the cutoff frequency are equal. It is when the pass band is maximally flat, and the frequency respond of the filter is a Butterworth response, in other words when the quality factor is:

    [​IMG]

    From the magnitude of the transfer function, I will arrive to the result replacing R^2=L/(2*C)

    [​IMG]
    [​IMG]

    Finding the angular cuttoff frequency at -3dB

    [​IMG]

    [​IMG]

    [​IMG]

    [​IMG]

    Bingo!!!

    I think there is no explanation like this in any book
     
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