Second order filter - quadratic formula and standard normalized form

Discussion in 'Circuit Help' started by Jonathan56, Jan 4, 2017.

  1. Jonathan56

    Jonathan56

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    Hi,

    I have a question regarding corner frequency for second order filter.

    I don't understand how I can have two roots in the denominator which will indicate two corner frequencies each of them at different frequency and with 20db/decade and the same transfer function express in standard normalized form (express with the quality factor) indicate only one corner frequency (power of two <=> 40db/decade).
    2 corner frequency or one?

    Example with an LCR filter - two-pole low-pass filter
    s = jw and w = 2 x pi x F
    G(s) = 1 / (1 + sL/R + s²LC) = 1 / (1 + sA + s²B) with A = L/R and B = LC
    Quadratic formula
    G(s) = 1 / [ ( 1 - s/s1 ) (1 - s/s2) ]
    with s1/2 = -A/(2B) x [1 -/+ rsqt(1 - 4B/A)]

    ==> This indicates that there is two corner frequencies

    standard normalized form
    G(s) = 1 / [1 + s(Qw0) + (s/w0)²]

    ==> This indicates that there is only one corner frequency

    The example above is taken from the book:
    Fundamentals of Power Electronics SECOND EDITION
    Robert W. Erickson and Dragan Maksimovic
    page 282

    Thank you for your help.
    Jonathan
     
    Jonathan56, Jan 4, 2017
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  2. Jonathan56

    Harald Kapp Moderator Moderator

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    This is not really a contradiction.
    You can build a filter where the two corner frequencies are different, one for each order of the filter.
    You can also build a filter where the two corner frequencies are identical, looking like a single corner frequency.

    Think of a second order filter as a series connection of two first order filters where you can define the corner frequency of each filter individually.
     
    Harald Kapp, Jan 4, 2017
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  3. Jonathan56

    Jonathan56

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    Thanks, I am not sure I understand correctly your answer.

    I am talking about one same filter. LRC (L in serial with RC in parallel) Vout taken at RC

    according the standard normalized form
    Corner frequency = 1/ [2 x pi x sqrt (LC)] and Q = R x sqrt (C/L)
    according the factorized form with the two roots (using the quadratic formula)
    Corner frequency 1 = 1/(4 x pi x RLC) x [1 - sqrt(1 - 4R²C/L)]
    Corner frequency 2 = 1/(4 x pi x RLC) x [1 + sqrt(1 - 4R²C/L)]

    What is the corner frequency of the LRC filter?
     
    Jonathan56, Jan 4, 2017
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  4. Jonathan56

    LvW

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    Jonathan - why do you think that you would have two corner frequencies?
    Just because you have found two roots?
    That`s not logical.
    Corner frequencies are defined for the magnitude response of a second order filter.
    Now - you have to distinguish between two cases:

    (a) Two REAL roots (poles of the function): In this case, you can (can !!) realize the filter with two first-order sections having two different corner frequencies. However, also a realization in one 2nd-order filter block is possible.

    (b) When you decrease the degree of damping for the filter (smaller resistors) the two real poles approach each other until they form a double pole (Quality factor 0.5). If you further decrease damping the poles split up and form a conjugate-complex pole pair (equal real part and imag. part with different sign).
    This gives a second-order filter with rather good selectivity (Butterworth or Chebyshev response) and a pole quality factor Qp>0.5. Such a filter has only one single cut-off frequency (but a pair of conjugate-complex poles).
     
    LvW, Jan 4, 2017
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  5. Jonathan56

    Ratch

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    What makes you think that the "standard normalized form" has only one frequency?

    Jonathan56.JPG

    Ratch
     
    Ratch, Jan 4, 2017
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  6. Jonathan56

    LvW

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    Ratch - is this question directed to me?
    Are you saying that a second-order function (always) has TWO corner frequencies?
    It has (for Qp>0.5) two poles (conjugate-complex) and ONE SINGLE corner frequeny.
    Or did I misunderstood your reply?
     
    LvW, Jan 4, 2017
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  7. Jonathan56

    Ratch

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    Sorry, I meant for the reply to go to Jonathon56. I am saying that a quadratic equation has two solutions, not one as he stated.

    Ratch
     
    Ratch, Jan 4, 2017
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  8. Jonathan56

    Jonathan56

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    Hi LvW and Ratch.

    Thanks for your answers it makes me realise my mistake in my analyse.
    To answer to the question of Ratch, indeed you are right, I was just to focus on the assunptotic behaviour to sketch the magnitude bode plot (but without the magnitude of the transfert function...).
    I guess I was trying to take to much shortcut in my analyse. Lesson learned!

    Thanks.
     
    Jonathan56, Jan 5, 2017
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