# Z-transform of difference equations and stability of a process

According to this paper:

$y(t)$ is stationary if all of the roots (of characteristic equation) lie outside the unit circle

Here, $y(t)$ is causal.

To me it seems the case is exactly the opposite, the roots should be inside the unit circle for causal process to be stable.

Example, consider the difference equation:

$$y_t-5y_{t-1}+6y_{t-2}=0$$

The characteristic equation is then: $$r^2-5r+6=0$$

Solution for characteristic roots: $$r=2, r=3$$

Solution for the difference equation (assuming 1 as initial conditions): $$y(t)=2^t+3^t$$

Clearly the roots are outside the unit circle (as 1<2,3), and the process unstationary (as it grows with t). To me it seems that the 3rd equation in the paper should have negative exponent for zs (in which case the solution would be the same as mine, even if arrived through the z transform).

Yet, the claim appears in many places so it can't be wrong (I doubt the paper is wrong either). However, for LTI system to be stable the poles must be inside the circle, a other claim that is repeated everywhere. Is the statement just plain wrong or are these things somehow not equivalent?

## 1 Answer

The paper deals in the lag operator $L$, rather than $z^{-1}$ as DSP people tend to use.

As a result, all the DSP results you know are inverted. So "inside the unit circle" becomes "outside the unit circle", black is white, and ... avoid zebra crossings.

• K, is this the same then, as stating that the system is anti-causal? – Dole Apr 20 '16 at 15:23
• A causal system expressed in terms of $L$ will have its impulse response with only non-negative exponents (i.e. zero or positive). $L^n$ terms in the system impulse response with $n<0$ will be anti-causal. – Peter K. Apr 20 '16 at 15:34
• K, Makes sense, but I still don't quite see why the characteristic roots are "upside down", due to use of lag operator. I mean, I could just take the normal z-transform of the equation without the use of L operator and then say the roots should be inside the circle... If this is indeed just a matter of expression, isn't it a bit too limiting to say they should be outside the circle? – Dole Apr 20 '16 at 16:18
• How is it more limiting than saying they should be inside the unit circle? The substitution is $L = z^{-1}$. The two ways of looking at it are precisely equivalent. There is nothing more complex than that. Statisticians / econometrists sometimes use the lag operator instead of what DSP people consider the standard $z^{-1}$ operator. In some ways, they are right: $z$ is really an "advance" operator, which is somewhat nonsensical given the way equations are usually written in terms of past values of the input and output (for causal systems). – Peter K. Apr 20 '16 at 16:22
• K I suppose limiting is not the correct term. It's a bit confusing given the same profs of statistics/econometrics only talk about the before mentioned solution, not bringing up that there is lag operator transformation involved when bringing up the quote above... Thank you for the explanation! – Dole Apr 20 '16 at 16:53