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I took a z transform and got a double pole at $z=1$, but I don't know if that's correct.

I'm lost because I don't know if $\cos(\theta)$ converges or diverges or what that means for $h[n]$ being absolutely summable.

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    $\begingroup$ assuming you mean "... $\cos\theta$ converges for $\theta \to \infty$: the cosine is an oscillation, right? What does "converge" mean to you? $\endgroup$ – Marcus Müller Mar 5 at 9:15
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    $\begingroup$ Please see this answer: dsp.stackexchange.com/questions/64349/… and try applying to your case. $\endgroup$ – Dan Boschen Mar 5 at 12:27
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    $\begingroup$ Are you saying that you have a system whose impulse response is $\cos\left (\pi \sqrt{n} \right ) $? And you want to know if that system is stable? $\endgroup$ – TimWescott Mar 7 at 6:17
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The system with impulse response given by $h[n] = \cos(\pi\sqrt{n})u[n]$ is BIBO-unstable because the sum $\sum_{n=-\infty}^\infty |h[n]]$ diverges instead of being convergent as is needed for BIBO-stability. Note that for all positive integers $k$, $h[k^2]=\cos(\pi k)$ has value $\pm 1$ and so $$\sum_{n=-\infty}^\infty |h[n]] = \sum_{n=0}^\infty |\cos((\pi\sqrt{n})|$$ is a sum that contains infinitely many $+1$ terms (and all the other terms are guaranteed to be positive too since $\cos(\pi r)=0$ if and only if $r = k+\frac 12$ where $k$ is an integer, and there is no integer $n$ whose square root is of the form $k+\frac 12$). So, $\sum_{n=-\infty}^\infty |h[n]]$ diverges, and the system is BIBO-unstable.

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  • $\begingroup$ +1 you could very simply prove that an infinite series does not converge by showing that the necessary condition $\lim_{n \to \infty} h[n] = 0$ does not hold, hence the series cannot be convergent... $\endgroup$ – Fat32 Mar 7 at 22:21
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The LTI system defined by the impulse response $$h[n] = \cos(\pi \sqrt{n} ) u[n] $$ is unstable, as the absolute sum of the impulse response does not converge and diverges to infinity instead, i.e.;

$$ \sum_{n=-\infty}^{\infty} |h[n]| = \sum_{n=0}^{\infty} |\cos(\pi \sqrt{n}) u[n]| \longrightarrow \infty $$

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  • $\begingroup$ +1 but you really should provide some areas as to why your assertion of the nonconvergence of $\sum_n |h[n]|$ is true. $\endgroup$ – Dilip Sarwate Mar 7 at 21:23
  • $\begingroup$ @DilipSarwate in absolute value I've trusted ;-) $\endgroup$ – Fat32 Mar 7 at 22:18
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The summation of $\cos(\pi (z^{n^(0.5)})) z^{-n}$ should converge for the $z$ transform to converge and it is stable if the ROC includes the unit circle. Since you got double poles at $|z|=1$, the causal ROC will lie outside the largest pole and hence it won't include $|z|=1$ and hence it cannot be stable.

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  • $\begingroup$ Hi Anushka Shanbang! Great to have you here. Sadly, I don't really understand your answer. It seems to be one runaway sentence. Can you explain what "the summation of $\cos$ with $z^{-n}$" means, mathematically? Because literally, that would be $\cos + z^{-n}$, and that makes no sense. $\endgroup$ – Marcus Müller Mar 5 at 12:38
  • $\begingroup$ the function should have a n^0.5 factor in the cosine argument, not z^0.5. And also, I'm not sure that the poles I got are correct?? Can someone confirm? Because, I know that if the pole is at z=1, it is marginally stable. $\endgroup$ – iplayball27 Mar 5 at 18:00
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    $\begingroup$ By absolute summability, this signal is not stable. Just look at the absolute value of each term and sum them. There is good chance that there are infinite number of 1's in it.Hence, if you do absolute sum of this signal, it is not guaranteed that the sum will be less than infinity. $\endgroup$ – jithin Mar 6 at 8:42
  • $\begingroup$ @jithinrj Not just a "good chance" that there are infinitely many 1's in the sum but a provable fact. See my answer to the question. $\endgroup$ – Dilip Sarwate Mar 7 at 21:48

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