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

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

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 $n\le 0$

• But it doesn't. Even using your 1-sided expression, take z=1/4, for example: $$\sum_{n=1}^{\infty}(1/2)^{n-1}(1/4)^{-n} = \sum_{n=1}^{\infty}2\cdot 2^n$$ That looks like an infinite sum to me. Jan 2, 2014 at 18:59
• You want to use the Z transform as a short cut to the Fourier series, most of the manipulations with z series are purely formal, algebraic. But in the end, one is only interested in values $z$ on the unit circle. So poles inside the unit circle lead to causal parts of the filter sequence, poles outside to anti-causal parts of the filter sequence. Jan 2, 2014 at 19:18
• True, but it doesn't answer the question. Why is there only one pole at z = 0.5 ? Jan 2, 2014 at 21:08
• z=0.25 is not in the region of convergence, since it is smaller than 0.5. The pole is a pole of the Z transform, a Laurent series that is defined on some small annulus containing the unit circle, in the extremal cases only on the unit circle. Jan 2, 2014 at 21:17
• I think this is the actual answer: The Z transform is not defined outside the region of convergence. It can only be calculated inside the region of convergence. Jan 2, 2014 at 22:07

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.

• But even using the one sided sum, is it finite for other values? For your expression, let a = 1/4: $$\sum_{n=0}^{\infty}2^{-n}(1/4)^{-n} = \sum_{n=0}^{\infty}2^n$$ That sum sure looks like it explodes to me. Jan 2, 2014 at 18:53
• No why? The limit of the sum of a geometric series is finite. en.wikipedia.org/wiki/Geometric_series Jan 2, 2014 at 19:22
• Nope: The sum is only finite, if the argument of the series has a magnitude of less than 1. That's not the case here Jan 2, 2014 at 21:07