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  • Is the system $y[n] = x[n] - x[n-1]$ stable or unstable ?
  • If unstable, can you please give some examples of input signals for which the system becomes unstable?
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    $\begingroup$ Homework? Please show what you have tried so far. $\endgroup$
    – Atul Ingle
    Feb 17 '17 at 3:53
  • $\begingroup$ Differentiation and Integration in continuous time are unstable. Summation in discrete time also unstable so I am guessing the difference equation also an unstable one. $\endgroup$
    – METALHEAD
    Feb 17 '17 at 3:56
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    $\begingroup$ metal, your first comment is factually incorrect. (or half incorrect.) oh, i guess a continuous-time differentiator is BIBO stable only if the bounded input is also continuous. but does that continuity mean anything for discrete-time bounded signals? $\endgroup$
    – robert bristow-johnson
    Feb 17 '17 at 4:41
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    $\begingroup$ well, the point was that the discontinuity of the input is solely to blame for the unbounded output of a continuous-time differentiator. so it is not BIBO stable. but discontinuity doesn't have meaning in discrete-time signals. so the discrete-time differentiator cannot be BIBO unstable for the same reason as the continuous-time differentiator is. and, as it turns out, the discrete time differentiator is trivially shown to be BIBO because the difference of two finite numbers is also finite. $\endgroup$
    – robert bristow-johnson
    Feb 17 '17 at 6:52
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    $\begingroup$ Just for the record, any FIR filter must be stable. $\endgroup$
    – Matt L.
    Feb 17 '17 at 9:18
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For BIBO stability in the case of discrete time, there is a necessary and sufficient condition given by $\sum |h[n]| < \infty$ that is if the impulse response is absolute summable then the system is BIBO stable. Clearly $$h[n] = \delta[n] - \delta[n-1]$$ and it has a finite support, the impulse response is absolutely summable and therefore BIBO stable.

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    $\begingroup$ I don't think "support" is the correct term to use in this argument. the impulse response for a simple first-order LPF: $$ h[n] = a^n \ u[n] $$ where $|a|<1$ and $u[n]$ is the discrete unit step function, that $h[n]$ does not have finite support yet it is BIBO stable. $\endgroup$
    – robert bristow-johnson
    Feb 17 '17 at 7:15
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    $\begingroup$ oh, i see, Arka, you're not saying that "finite support" is necessary, just that it is sufficient. okay, i agree. $\endgroup$
    – robert bristow-johnson
    Feb 17 '17 at 7:17
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Stability has different meanings depending on the system you are observing. Here, you have a discrete system, which is linear, and time-invariant. The most common stability criterion in this case is the so-called bounded-input, bounded-output (BIBO) stability.

The question of stability is thus: if an input is bounded in amplitude, is the output always bounded too?

In other words, if $|x[n]| \le B$, do we always have some (other) bound $B'$ such that $|y[n]| \le B'$?

Here, basic inequalities give the response: $$|y[n]| = |x[n]-x[n-1]| \le |x[n]| +|x[n-1]| \le 2B$$

As said by @robertbristow-johnson, this can be generalized to FIR (finite impulse response) filters, since you will have:

$$|y[n]| \le B\sum|h[n]| $$ with a finite sum of terms, always bounded, for instance by the number $L$ of non-zero terms (always finite for a FIR system), times the maximum amplitude coefficient $h_{M} = \max |h[n]|$:

$$|y[n]| \le BL h_{M} $$

But remember that there exist infinite-support systems which are BIBO too.

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