# A Summer 2015 Mathematics A To Z: z-transform

## z-transform.

The z-transform comes to us from signal processing. The signal we take to be a sequence of numbers, all representing something sampled at uniformly spaced times. The temperature at noon. The power being used, second-by-second. The number of customers in the store, once a month. Anything. The sequence of numbers we take to stretch back into the infinitely great past, and to stretch forward into the infinitely distant future. If it doesn’t, then we pad the sequence with zeroes, or some other safe number that we know means “nothing”. (That’s another classic mathematician’s trick.)

It’s convenient to have a name for this sequence. “a” is a good one. The different sampled values are denoted by an index. a0 represents whatever value we have at the “start” of the sample. That might represent the present. That might represent where sampling began. That might represent just some convenient reference point. It’s the equivalent of mileage maker zero; we have to have something be the start.

a1, a2, a3, and so on are the first, second, third, and so on samples after the reference start. a-1, a-2, a-3, and so on are the first, second, third, and so on samples from before the reference start. That might be the last couple of values before the present.

So for example, suppose the temperatures the last several days were 77, 81, 84, 82, 78. Then we would probably represent this as a-4 = 77, a-3 = 81, a-2 = 84, a-1 = 82, a0 = 78. We’ll hope this is Fahrenheit or that we are remotely sensing a temperature.

The z-transform of a sequence of numbers is something that looks a lot like a polynomial, based on these numbers. For this five-day temperature sequence the z-transform would be the polynomial $77 z^4 + 81 z^3 + 84 z^2 + 81 z^1 + 78 z^0$. (z1 is the same as z. z0 is the same as the number “1”. I wrote it this way to make the pattern more clear.)

I would not be surprised if you protested that this doesn’t merely look like a polynomial but actually is one. You’re right, of course, for this set, where all our samples are from negative (and zero) indices. If we had positive indices then we’d lose the right to call the transform a polynomial. Suppose we trust our weather forecaster completely, and add in a1 = 83 and a2 = 76. Then the z-transform for this set of data would be $77 z^4 + 81 z^3 + 84 z^2 + 81 z^1 + 78 z^0 + 83 \left(\frac{1}{z}\right)^1 + 76 \left(\frac{1}{z}\right)^2$. You’d probably agree that’s not a polynomial, although it looks a lot like one.

The use of z for these polynomials is basically arbitrary. The main reason to use z instead of x is that we can learn interesting things if we imagine letting z be a complex-valued number. And z carries connotations of “a possibly complex-valued number”, especially if it’s used in ways that suggest we aren’t looking at coordinates in space. It’s not that there’s anything in the symbol x that refuses the possibility of it being complex-valued. It’s just that z appears so often in the study of complex-valued numbers that it reminds a mathematician to think of them.

A sound question you might have is: why do this? And there’s not much advantage in going from a list of temperatures “77, 81, 84, 81, 78, 83, 76” over to a polynomial-like expression $77 z^4 + 81 z^3 + 84 z^2 + 81 z^1 + 78 z^0 + 83 \left(\frac{1}{z}\right)^1 + 76 \left(\frac{1}{z}\right)^2$.

Where this starts to get useful is when we have an infinitely long sequence of numbers to work with. Yes, it does too. It will often turn out that an interesting sequence transforms into a polynomial that itself is equivalent to some easy-to-work-with function. My little temperature example there won’t do it, no. But consider the sequence that’s zero for all negative indices, and 1 for the zero index and all positive indices. This gives us the polynomial-like structure $\cdots + 0z^2 + 0z^1 + 1 + 1\left(\frac{1}{z}\right)^1 + 1\left(\frac{1}{z}\right)^2 + 1\left(\frac{1}{z}\right)^3 + 1\left(\frac{1}{z}\right)^4 + \cdots$. And that turns out to be the same as $1 \div \left(1 - \left(\frac{1}{z}\right)\right)$. That’s much shorter to write down, at least.

Probably you’ll grant that, but still wonder what the point of doing that is. Remember that we started by thinking of signal processing. A processed signal is a matter of transforming your initial signal. By this we mean multiplying your original signal by something, or adding something to it. For example, suppose we want a five-day running average temperature. This we can find by taking one-fifth today’s temperature, a0, and adding to that one-fifth of yesterday’s temperature, a-1, and one-fifth of the day before’s temperature a-2, and one-fifth a-3, and one-fifth a-4.

The effect of processing a signal is equivalent to manipulating its z-transform. By studying properties of the z-transform, such as where its values are zero or where they are imaginary or where they are undefined, we learn things about what the processing is like. We can tell whether the processing is stable — does it keep a small error in the original signal small, or does it magnify it? Does it serve to amplify parts of the signal and not others? Does it dampen unwanted parts of the signal while keeping the main intact?

We can understand how data will be changed by understanding the z-transform of the way we manipulate it. That z-transform turns a signal-processing idea into a complex-valued function. And we have a lot of tools for studying complex-valued functions. So we become able to say a lot about the processing. And that is what the z-transform gets us.

## Author: Joseph Nebus

I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.

## 12 thoughts on “A Summer 2015 Mathematics A To Z: z-transform”

1. When I’m able to, yes! Fortunately work gives me occasional chances to revisit my ancestral homeland and from there it’s a quite reasonable drive to Asbury Park and the Silverball Museum. It’s a great spot and I recommend it highly.

There’s apparently also a retro arcade in Redbank, with a dozen or so pinball machines and a fair number of old video games. I’ve not been there yet, though.

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1. Here is a bit more.

z is used in dealing with recurrence relations and their active form, with input as well, in the form of “z transfer function:
a(n) is the input at time n, u(n) is the output at time n, these can be viewed as sequences
u(n+1) = u(n) + a(n+1) represents the integral/accumulation/sum of series for the input process
z is considered as an operator which moves the whole sequence back one step,
Applied to the sequence equation shown you get u(n+1) = zu(n),
and the equation becomes
zu(n) = u(n) + za(n)
Now since everything has (n) we don’t need it, and get
zu = u + za
Solving for u gives
u = z/(z-1)a
which describes the behaviour of the output for a given sequence of inputs
z/(z-1) is called the transfer function of the input/output system
and in this case of summation or integration the expression z/(z-1) represents the process of adding up the terms of the sequence.
One nice thing is that if you do all of this for the successive differences process
u(n+1) = a(n+1) – a(n)
you get the transfer function (z-1)/z, the discrete differentiation process.

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1. That’s a solid example of using these ideas. May I bump it up to a main post in the next couple days so that (hopefully) more people catch it?

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