# Allpass filter phase

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

1. ### garyrGuest

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.

Gary Richardson

2. ### George HeroldGuest

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

George H.

3. ### garyrGuest

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

George H.

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

4. ### misoGuest

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.  