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Aug 11, 2020 at 14:55 vote accept helloworld922
Aug 11, 2020 at 14:55 comment added helloworld922 Ah, I think I know what my problem is. For my actual problem, y[n] is actually the exponent for the actual variable I'm interested in, so when I saw this growing to overflow somewhere in the simulation, I couldn't figure out what was wrong, even though y[n] itself will eventually stabilize.
Aug 11, 2020 at 13:17 answer added Matt L. timeline score: 2
Aug 11, 2020 at 9:25 comment added AJN I couldn't reproduce unbounded output with this Octave code ? Is it same as or different from your question ? How did you simulate the system ? I have deleted my two earlier comments.
Aug 11, 2020 at 8:10 history edited helloworld922 CC BY-SA 4.0
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Aug 11, 2020 at 7:48 comment added helloworld922 @AJN I added the z-transform for the oscillatory input, and it does indeed have a repeat pole at \$z=-1\$, though I'm not entirely sure why that matters?
Aug 11, 2020 at 7:46 history edited helloworld922 CC BY-SA 4.0
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Aug 11, 2020 at 7:41 comment added helloworld922 @aconcernedcitizen I'm a bit surprised that negative inputs are not consider "bounded". Most definitions I've found for BIBO stability only require that \$\|x\|_{\infty}\$ be "finite".
Aug 11, 2020 at 7:37 comment added AJN @OP can you add the z transform of the input signal also into your question ? Does it form a double pole (not a complex pair, but 2 poles at same location) on the unit circle ? That may give a clue.
Aug 11, 2020 at 7:24 comment added a concerned citizen This may be beter suited for dsp.ee. That said, \$-x_0\$ falls outside BIBO, so you can't expect it to behave. Which leaves you with two choices: either change the input, or change the transfer function.
Aug 11, 2020 at 7:17 history asked helloworld922 CC BY-SA 4.0