Understanding $\mathcal Z$-transforms and pole locations

Question

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?

Answer 1

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 $n<1$. This should remove most of the infinities that you stumbled upon.

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Note that this geometric series only converges for $\left|\frac{0.5}{z}\right|<1$, that is $|z|>0.5$. 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.

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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 $|z|<2$, 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 $n>0$ and $h(n)=-2^{n-1}$ for $n\le 0$

Answer 2

You appear to assume that

$\sum_{n=0}^{\infty}{2\cdot0.5^{n-1}\cdot a^{-n}}$

is infinite when $a\neq0.5$. This does not appear to be the case. When $a=0.5$, both exponents cancel, the result is always $2\cdot 0.5^{n-1-n}=1$. When $a\neq0.5$, the exponents don't cancel. Instead, you could write:

$\sum_{n=0}^{\infty}{2\cdot0.5^{n-1}\cdot a^{-n}}=\sum_{n=0}^{\infty}{2\cdot0.5^{n-1}\cdot 0.5^{-n}}\cdot 2^{-n}\cdot a^{-n}= \sum_{n=0}^{\infty}{2^{-n}\cdot a^{-n}} $

Which is finite. I just noticed I'm using the unilateral z-Transform which you don't. In this case I would justify that by the signal having to be causal. The bilateral version appears not to converge anywhere, at all.