Re: Question on pole-zero cancellation


Oli Filth skrev:
Hi,

I have a conundrum which I'm tying myself in knots thinking about; if
anyone can shed any light it would be much appreciated!

Take as a simple example the moving-sum filter:

L-1
H(z) = Sum { z^-n }
n=0

= 1 - z^-L
--------
1 - z^-1

The numerator includes a zero at DC, the denominator gives a pole at DC,
and so these cancel, giving non-zero response at DC.

However, if we implement the filter recursively:
(I hope my ASCII art comes out ok...)

+------+
---+--->| z^-L |------+
| +------+ | -
| v
+----------------->0-----------------+----->
^ |
| +------+ |
+-----| z^-1 |<---+
+------+

This is equivalent to:

+------+
---+--->| z^-L |------+
| +------+ | -
| v
+----------------->0--->0-----------------+----->
^ |
| +------+ |
+-----| z^-1 |<---+
+------+

i.e. implementing the feedforward section before the feedback section,
and doing the two additions separately.

The feedforward section has already killed the DC component, so we can't
recover it.

So my question is, how does the recursive moving-sum filter manage to
work (and we know it does), if it appears to be exactly equivalent to
something that blocks DC?

Because the zero at DC is exactly cancelled by a pole. Note that the
second block scheme of yours is an expanded representation of the
total system, not the actual system. You can multiply any number by 1,
no matter how the factor 1 is represented, provided the representation
is exact.

Any transfer function H(z) can be expanded in a similar way:

H(z) = 1*H(z) = (1-z)/(1-z)*H(z)

(* means multiplication) and we have the same "paradox" you
found above.

Rune

.