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For example, let's say a causal LTI System is described by the following equation:

$$y[n] - ay[n-1] = x[n] - bx[n-1],\quad n \in Z$$

Is there a way to determine (in this case) the stability of the system without using the $\mathcal Z$-transform?

(Solving the problem using the $\mathcal Z$-transform would be a classical easy problem - verify that ROC of $H(z)$ contains the unit circle, I know how to do that)

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Yes you can. Just find the roots of the charactheristic polynomial associated with the LCCDE. Then for the causal system to be stable, all of those roots should be inside the unit circle; i.e., $|r_k| < 1$ where $r_k$ is the k-th root.

The characteristic polynomial for the above causal LCCDE is: $$1 - az^{-1}$$ and its single root is $r = a$, hence for this LCCDE to be stable you need $|a|<1$.

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  • $\begingroup$ So... LCCDE means using Laplace to my equation and getting H? But laplace for discrete systems isn't the same with Z-transform? $\endgroup$
    – S'en dou
    Commented Feb 2, 2018 at 11:11
  • $\begingroup$ In this context LCCDE means Linear Constant Coefficient Difference Equation and Z-transform applies only. Laplace transform is for continuous time systems where LCDDE means linear constant coefficient differential equations. $\endgroup$
    – Fat32
    Commented Feb 2, 2018 at 11:12
  • $\begingroup$ So how did you determine the characteristic polynomial? $\endgroup$
    – S'en dou
    Commented Feb 2, 2018 at 11:17
  • $\begingroup$ @S'endou characteristic equation associated with $y[n]+a y[n-1] +b y[n-2] + c y[n-3]$ is $1 + a z^{-1} + b z^{-2} + c z^{-3}$... $\endgroup$
    – Fat32
    Commented Feb 2, 2018 at 12:23
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    $\begingroup$ I get it now. So you care only about the left hand, you ignore $x[n] - bx[n-1]$. Thanks $\endgroup$
    – S'en dou
    Commented Feb 2, 2018 at 19:43
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It boils down to the same thing, with or without $\mathcal{Z}$-transform. You determine the roots of the characteristic equation, which in your case is

$$\lambda=a\tag{1}$$

For stability, the magnitude of the roots of the characteristic equation must be smaller than $1$. So for your example you get the condition $|a|<1$ for stability.

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  • $\begingroup$ happened again... :-) $\endgroup$
    – Fat32
    Commented Feb 2, 2018 at 11:10
  • $\begingroup$ I am not sure I understand. What is the characteristic equation in this case and why? $\endgroup$
    – S'en dou
    Commented Feb 2, 2018 at 11:16
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    $\begingroup$ @S'endou: Did you check the link in my answer? That should make it clear. $\endgroup$
    – Matt L.
    Commented Feb 2, 2018 at 11:42

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