🚀 go-pugleaf

wilf@sce.carleton.ca (Wilf Leblanc) writes:

>jefft@phred.UUCP (Jeff Taylor) writes:

>>	[mock conversation about zeros of linear phase filter]

>>	"WHERE ARE THE ZEROS?

>>	"On the unit circle."

>"... but the zeros are on the unit circle" ??????

>	[counterexample deleted]

>The zeros are not on the unit circle.

To clarify: an FIR filter with coefficient symmetry (and therefore
linear phase) may only have zeros with the following forms, if the
coefficients are also assumed to be real:

1. zero at 1

2. zero at -1

3. zeros at e^{j\omega} and e^{-j\omega} (conjugate pair on the unit
	circle)

4. zeros at \alpha e^{j\omega}, \alpha e^{-j\omega}, 
	\alpha^{-1} e^{j\omega}, and \alpha^{-1} e^{-j\omega}
	(conjugate reciprocal quad)

5. zeros at \alpha and \alpha^{-1}

These forms may be derived from the following considerations:

a) To get real coefficients, any complex roots must occur in conjugate
pairs. 

b) The Z transform in this case is a symmetric polynomial in z^{-1}.
It can be shown that if z_0 is a root of a symmetric polynomial, then
so is 1/z_0.

So zeros of an FIR filter *can* appear on the unit circle, but not
always.
-- 
Steve Eddins	
eddins@brazil.eecs.uic.edu 	(312) 996-5771 		FAX: (312) 413-0024
University of Illinois at Chicago, EECS Dept., M/C 154, 1120 SEO Bldg,
Box 4348, Chicago, IL  60680