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Can anybody explain why is this system not causal.

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

How does it depend from future inputs when $n < n_0$.

If $n < n_0$ then $T[x[n]]$ is zero because of summation properties.

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  • $\begingroup$ You are correct about the convention for the sum. Maybe your instructor is using a different definition? $\endgroup$
    – MBaz
    Oct 28 '21 at 22:00
  • $\begingroup$ But why does it depend on the future inputs, for any $n$ the summation goes up to $n$ too. Is there an example that can make it more clear. Thank you. $\endgroup$
    – Derteck pt
    Oct 28 '21 at 22:02
  • $\begingroup$ From what i've searched if $n_0$ was $-\infty$ it would be causal. But since $n_0$ can be anything and summation when $n < n_0$ is possible resulting in zero, I don't know. $\endgroup$
    – Derteck pt
    Oct 28 '21 at 22:22
  • $\begingroup$ As you say, normally the range (for example) 5:3 is empty. However, someone may define it as [5, 4, 3], where $n$ is 3, so it'd be looking into the future. It'd be a completely non-standard definition, though, and probably quite useless too. $\endgroup$
    – MBaz
    Oct 28 '21 at 22:26
  • $\begingroup$ BTW, in case it's not clear: this system is causal, following widely accepted conventions for the $\Sigma$ operation. $\endgroup$
    – MBaz
    Oct 28 '21 at 22:28
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The system is causal (it doesn't depend on future inputs). What it is not is BIBO stable.

Summation shown to be not stable.

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