I am trying to prove if $h_1[n] = 2^{-n} u[n]$ and $h_2[n] = (0.1)^n$ are BIBO stable or not, but I'm not sure if these analysis are correct. I'm having some trouble to understand this specific subject.

For an impulse response $h[n]$ to be BIBO stable, equation $(1)$ need to be satisfied $$ \sum\limits_{n=-\infty}^{\infty}\big\lvert h[n]\big\rvert = B < \infty\tag{1} $$

  • So, for $h_1[n]$, $$ \sum\limits_{n=-\infty}^{\infty} \big\lvert 2^{-n} u[n]\big\rvert = \sum\limits_{n=-\infty}^{\infty} \big\lvert 2^{-n}\big\rvert\cdot\big\lvert u[n]\big\rvert = \sum\limits_{n=0}^{\infty} 2^{-n} $$ according to Wolfram this sum is equal to $2$, and therefore, system $h_1[n]$ is BIBO stable.

  • For $h_2[n]$, $$ \sum\limits_{n=-\infty}^{\infty} \big\lvert (0.1)^{n}\big\rvert =\sum\limits_{n=-\infty}^{\infty} (0.1)^{n} $$ since $-\infty\le n\le\infty$, the system is unstable

  • $\begingroup$ Both are BIBO stable. You see, second one actually converges if n is increased. $\endgroup$ – Parth Apr 10 '18 at 3:24
  • $\begingroup$ if $h_2[n]$ was also multiplied by the unit step $u[n]$, then $(0.1)^n u[n]$ is BIBO stable also, i believe. $\endgroup$ – robert bristow-johnson Apr 10 '18 at 3:58
  • $\begingroup$ @robertbristow-johnson Yes, that's what I thought too, but in the problem doesn't have that $u[n]$. Thanks for helping! $\endgroup$ – rmendes Apr 10 '18 at 4:03

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