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 IngleFeb 12, 2017 at 15:40
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$\begingroup$ 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. $\endgroup$– Ankit SahayFeb 12, 2017 at 16:19
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1$\begingroup$ This is a memoryless system. It is static. In that sense, it is internally stable. $\endgroup$– Rodrigo de AzevedoFeb 13, 2017 at 1:02
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2 Answers
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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$
