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I worked out this output of a sequence:

$$Y(z) = \frac{(z-1)(z+\frac{3}{2})}{z^2 (z-\frac{j}{2})(z+\frac{j}{2})(z-\frac{1}{2})}$$

My problem is the $z^2$. $$z^2 = z * z = (z - 0)(z-0)$$

So I think that are 2 poles in the center.
But does that have a special meaning or so ?

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This is only one pole at $z=0$, though it has an order of $2$. –  Ilya Jun 26 '11 at 12:25
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up vote 2 down vote accepted

It's just a double pole at the origin (you can think of it as two poles that coincide). Regarding the meaning: in general, it's nothing remarkable; but to get a definite answer you should specify what you are speaking of. Is this the Z transform of a discrete signal? If so, the normal convention is to write it in terms of $z^{-1}$, instead of $z$ - and the mere formula (or pole-zero spec, if it's a rational function) is not enough, you need to specify the region of convergence.

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Yes, it is a Z transform. I just have to plot the pole-zero diagram, so I think it's enough to draw the pole at the origin. –  madmax Jun 27 '11 at 10:43
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