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Stability of the below function was asked.

$$ T(x[n]) = \sum_{k=n_0}^n x[n] $$

I understand that $x[n]$ shouldn't exceed $M$ value. $x[n]\le M$. I mean, we are looking for if the input is bounded, what will the output be.

However, i didn't understand from where $|n-n_0|M$ came from in the below function.

$$ \left |T(x[n]) \right| \le \sum_{k=n_0}^n \left| x[n] \right| \le |n-n_0|M $$

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  • $\begingroup$ um, your question is basic math: if all elements of an $l$-element sum are below some value $M$, then what is the maximum value of that sum? $\endgroup$
    – Marcus Müller
    May 25, 2016 at 14:35
  • $\begingroup$ M * l . i see now where it came from. $\endgroup$
    – Uygar Uçar
    May 25, 2016 at 14:50

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Well, we are summing over $k$ in the range $n_0$ to $n$. The number of elements in the sum is therefore $n-n_0$. We also know that the sum must be positive and that all terms in the sum ($x[n]$) have absolute value less than $M$.

To account for $n-n_0$ being negative, but our sum MUST be positive, we take its absolute value.

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  • $\begingroup$ Just to be a bit pedantic: the number of elements in the sum is $|n-n_0|+1$, so the bound is actually $(|n-n_0|+1)M$. $\endgroup$
    – Matt L.
    May 25, 2016 at 17:26

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