Understanding $\mathcal Z$-transforms and pole locations I am trying to gain a better understanding of pole locations in the $z$-plane of a given discrete transfer function, $H(z)$.  I think I have a pretty good understanding of how to use the $\mathcal Z$-transforms for filter design, but there is something about the core concepts that still confuse me.

I understand that poles are locations in the $z$-plane (complex plane) where a specific value of $z$ results in $\left|H(z)\right|$ approaching infinity.  This happens when the denominator of a $\mathcal Z$-transform expression has a zero at $z=a$.  i.e. 

$$
H(z) =\frac{1}{z-0.5} 
$$

would have one pole at $z = 0.5$.  The inverse $\mathcal Z$-transform of this fraction would be

$$
h(n) = \mathcal{Z}^{-1}\left\{\frac{1}{z-0.5}\right\} = 2\cdot0.5^{n-1}
$$

Since $z$ is less than 1, this impulse response will decay as $n$ increases.  So the response is stable (which happens as long as the pole lies within the unit circle).

However, if we look at the definition of the $\mathcal Z$-transform, with $z$ written in complex exponential form, i.e. $z=re^{j\omega}$:

$$
H(z) = H\left(re^{j\omega}\right) = \sum_{n=-\infty}^{n=\infty} h(n)\left(re^{j\omega}\right)^{-n} = \sum_{n=-\infty}^{n=\infty} h(n)r^{-n}e^{-j\omega n},
$$

we can now express the $\mathcal Z$-transform for our pole directly using using $z=re^{j\omega}$ with $r=0.5$ and $\omega$=0:

$$
H(0.5) = \sum_{n=-\infty}^{n=\infty} h(n)0.5^{-n} = \infty
$$

We know this sum is infinity from our original transfer function.

Now here is where my question begins to form. When $h(n)$ is summed against the $0.5^{-n}$ function, we get an unbounded summation, or a pole.  Of course that can only happen if $h(n)$ is infinite (which it is), since the above summation for $H(0.5)$ could not be infinite otherwise.

But if $h(n)$ is infinite such that it sums to infinity multiplied by $0.5^{-n}$, it would clearly also sum to infinity against $0.58^{-n}$ or $0.25^{-n}$ for that matter.

So my question is, if there is a pole at 0.5, why aren't there poles all along the real axis (except for zero)?  Why is there only a pole at 0.5?

To make it explicit, we can use the specific $h(n)$ that was the inverse $\mathcal Z$-transform of $H(z)$, above:

$$
H\left(\frac 14\right) = \sum_{n=-\infty}^{n=\infty} \left[2\cdot\left(\frac{1}{2}\right)^{n-1}\right]\cdot \left(\frac{1}{4}\right)^{-n}
= \sum_{n=-\infty}^{n=\infty} 4\cdot 2^n = \infty
$$

So, this sum makes it look like there is a pole at $z=0.25$ for the $\mathcal Z$-transform of $h(n)$, but the transfer function says there is not!

In other words, the single pole makes sense to me looking at the $H(z)$ transfer function, but it doesn't make sense to me when I take it and use it directly in the $\mathcal Z$-transform expression.

What am I doing wrong?
The actual computation for the inverse Z transform is

$$\frac1{z-0.5}=\frac1z\cdot\frac1{1-\frac{0.5}{z}}=z^{-1}\sum_{k=0}^{\infty}(0.5)^k z^{-k}=\sum_{n(=k+1)=1}^\infty (0.5)^{n-1}z^{-n}$$

From that you see that the series is one-sided, $h(n)=0$ for $n&lt;1$. This should remove most of the infinities that you stumbled upon.



Note that this geometric series only converges for $\left|\frac{0.5}{z}\right|&lt;1$, that is $|z|&gt;0.5$. For applications one is only interested in the values of the Z transform on or close to the unit circle, $|z|\approx 1$, the eigenvalues and singular values of the sequence shift operator. The unit circle is clearly and comfortably contained in the region of convergence. 



So for instance, if the pole were at $z=2$, then the inverse Z transform would again use the Laurent series on and close to the unit circle, i.e., for $|z|&lt;2$, to give

$$\frac1{z-2}=-\frac12\,\frac1{1-\frac z2}=-\sum_{n=0}^{\infty}2^{(-n)-1} z^{-(-n)},$$

that is, $h(n)=0$ for $n&gt;0$ and $h(n)=-2^{n-1}$ for $n\le 0$