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$\begingroup$ Thank you very much, Bill. It does, and rather well. Do you think it is possible to do it with RecrrenceTable? I tried, but the result is not correct. $\endgroup$

Pascal77
– Pascal77

2024-04-16 16:39:54 +00:00
Commented Apr 16, 2024 at 16:39
$\begingroup$ Yes you are right. There is a mistake in the recurrence equation. The good one is the following one : y(n)=(a0/b0)*x(n)+(a1/b0)*x(n-1)+(a2/b0)*x(n-2)-(b1/b0)*y(n-1)-(b2/b0)*y(n-2) $\endgroup$

Pascal77
– Pascal77

2024-04-17 07:20:59 +00:00
Commented Apr 17, 2024 at 7:20

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