1
$\begingroup$
  • 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?
Improve this question
$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Just for the record, any FIR filter must be stable. $\endgroup$ – Matt L. Feb 17 '17 at 9:18
4
$\begingroup$

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.

Improve this answer
$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.