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Allpass filter phase

Discussion in 'Electronic Design' started by garyr, Oct 24, 2012.

  1. garyr

    garyr Guest

    One version of a second-order allpass filter has a recursion equation:

    y(n) = a*x(n) + x(n-2) - a*y(n-2)

    and transfer function

    H(z) = (a + z**-2) / (1 + a*z**-2)

    The phase of the filter can be obtained by substituting z = exp(jw) in the
    above and computing the arctan of the ratio of the imaginary and real parts
    of H(w).

    The phase is also given by the equation

    1) theta(w) = -2*arctan{ (1 - a) * tan(w) / (1 + a) }

    and by the more cumbersome equation

    2) theta(w) = arctan[(a**2-1)*sin(2*w) / { (a**2+1)*cos(2*w)+2*a}]


    I derived the second equation by substituting exp(jw) for z in H(z) and
    multiplying the numerator and denominator by the conjugate of the
    denominator; the phase angle is then the arctan of the ratio of the
    imaginary and real parts of the numerator. I thought perhaps eq. 2 could
    be transformed into eq. 1 by substituting the trig identities for
    sin(2w) and cos(2w) but that led nowhere. There evidently is a way of
    arriving at eq. 1 but I haven't found it. Any suggestions? I'm not looking
    for the complete derivation, just the general approach.

    Thanks in advance.
    Gary Richardson
     
  2. I've long forgotten all my trig tricks. But have you plugged a few
    numbers into each equation.. just to check that you get the same
    answer? (A reality check.)

    George H.
     
  3. garyr

    garyr Guest

    I've long forgotten all my trig tricks. But have you plugged a few
    numbers into each equation.. just to check that you get the same
    answer? (A reality check.)

    George H.

    Yes, all three return idential values for 0 <= w <= pi/2.
     
  4. miso

    miso Guest

    It has been some time since I worked in Z domain, but wouldn't an all
    pass have poles and zeros on the unit circle, forming a square.

    Maybe the OP is over thinking this.
     
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