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I am highly confused about the stability of $\cos(x[n])$.

If we provide a bounded input such as $x[n]=u[n]$, the output is bounded.

Now if we provide an unbounded signal $x[n]=\delta[n]$, the output oscillates between $-1$ and $1$. So I think it's marginally stable in this case.

Am I correct? I have this confusion because in my book it says it's unstable when $x[n]=\delta[n]$. So my questions are: Is the function BIBO stable and also is it stable for an unbounded input ?

Note: $u[n]$ and $\delta[n]$ are unit step and dirac delta functions respectively.


Correction: As mentioned in the comment, $x[n]=\delta[n]$ is bounded too.

But this points more strongly to $\cos(x[n])$ being stable, isn't it?

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closed as unclear what you're asking by Marcus Müller, lennon310, A_A, Matt L., MBaz Oct 10 '18 at 2:20

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ @MarcusMüller I am really sorry for being so a vague. I think your last sentence is my answer. As I want to know if it's BIBO stable, all I can observe is for bounded inputs like unit step or dirac delta, the output is bounded(which is enough for BIBO). I just wanted to know whether my thinking is correct or not, that's all. You may very well close the question as "unclear" if it's still not answerable or if it is answerable, I request you to simply verify my BIBO claim :) $\endgroup$ – paulplusx Oct 6 '18 at 12:22
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    $\begingroup$ Well, yes, but you really don't need me to confirm that, do you? $\endgroup$ – Marcus Müller Oct 6 '18 at 12:27
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    $\begingroup$ again, you know that, why are you asking me to confirm? You even have books where exactly that is written? $\endgroup$ – Marcus Müller Oct 6 '18 at 12:34
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    $\begingroup$ No need for apologies don't worry about it. The point here is to have a clear question that can (hopefully) be answered. Maybe posting the original passage from the book would help too (?) $\endgroup$ – A_A Oct 7 '18 at 7:24
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    $\begingroup$ I think we have an answer. The system just flagged this post as having too many comments.... $\endgroup$ – Peter K. Oct 7 '18 at 14:53
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The I/O relation $$y[n] = \cos(x[n])$$

represents a stable system.

Because by definition of $\cos(x)$, irrespective of the input $x[n]$, its output will always be bounded by $$-1 \leq \cos(x[n]) \leq 1 $$ for all $x[n]$ for all $n$. Even if $x[n]$ is undefined ( such as would be the case with $x[n] = 1/n$ at $n=0$ as Marcus pointed out) we for sure know that the system output is bounded (the value of input is undefined, but the output is known to be within $\pm1$.)

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