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A system such that

$$y[n] = \log_{10}(|x[n]|) $$

was given as a stable one. I am not able to understand it though. When $x[n] = 0$ (which is bounded), the output $y[n]$ tends to get unbounded. So how is this system stable?

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  • $\begingroup$ Where "was it given"? I agree with your reasoning that it's not stable. Perhaps they are using a more restrictive definition of stability? $\endgroup$
    – Atul Ingle
    Feb 12, 2017 at 15:40
  • $\begingroup$ The question belongs from a book named "Signals and Systems" by Simon Haykin, although the solution does not come from a reliable source. Thank you for clearing the confusion. $\endgroup$ Feb 12, 2017 at 16:19
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    $\begingroup$ This is a memoryless system. It is static. In that sense, it is internally stable. $\endgroup$ Feb 13, 2017 at 1:02

2 Answers 2

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As correctly pointed out in a comment, this system is definitely not BIBO-stable. The output signal becomes arbitrarily large (in magnitude) when $x(t)$ approaches zero. So a bounded input signal can result in an unbounded output signal, and, consequently, the system is unstable.

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To check BIBO - stability , apply the definition.

Suppose there is an M$\in R$ such that: $|x(n)| <M \forall n \in N$ now you need to show that there is an $L \in R$ such that $|y[n]| < L \forall n \in N$

This is not a BIBO -stable system . Simple counterexample: $x(t) = |sint| \rightarrow |x(t) | <2 $ but $\lim_{t->0}|y(t)| = \lim_{t\rightarrow 0}|log(|sint|)|=\infty$
side note : $y(t)$ is well defined for $ : t \neq k\pi$

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