pole-zero plot with zeros |a|

> So we have (z^8-a^8)/z^7/(z-a) = 1/z^7*(z^8-a^8)/(z-a).  So the
> first term gives the 7 poles at 0.
>
>

Ah, its ok - I got it now!

I think you've accidentally pressed / instead of *

(z^8-a^8)/[(z^7)*(z-a)]

I get the rest of it - thanks a lot!

Phil

>>>>> "Philip" == Philip Newman <nojunkmail@ntlworld.com> writes:

    >> So we have (z^8-a^8)/z^7/(z-a) = 1/z^7*(z^8-a^8)/(z-a).  So the
    >> first term gives the 7 poles at 0.
    >> 
    >> 

    Philip> Ah, its ok - I got it now!

    Philip> I think you've accidentally pressed / instead of *

    Philip> (z^8-a^8)/[(z^7)*(z-a)]

Isn't a/b/c the same as a/(b*c) because a/b/c means (a/b)/c?

At least that's how I learned the precedence rules way back in grade
school. 

But anyway, I'm glad everything is cleared up now. :-)

Ray