Timeline for When inverting a transfer function, solving for the input using the output does the causality status change
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Dec 13, 2017 at 13:02 | comment | added | Matt L. | @Michelle: I suggest you formulate a new question. It's getting a bit too cumbersome discussing it in the comments of this answer which is not directly related to these questions. | |
| Dec 13, 2017 at 12:58 | comment | added | Michelle | As you can tell this is not my field of study but I am using the theory to explain my findings in another field involving difference equations and literature doesn't cover ''in theory'' type situations. If (1/G(z)) has all poles strictly inside the unit circle.. what can be said about the input/output relationship ? (G(z) only has poles at z=0) | |
| Dec 13, 2017 at 12:49 | comment | added | Matt L. | @Michelle: No, BIBO stability just means that the output is bounded for any bounded input signal. For a rational transfer function that means that there must be no poles on the unit circle. If you further require causality, all poles must be inside the unit circle. BIBO stability says nothing about the zeros of the transfer function, and, consequently, nothing about the poles of the inverse system. | |
| Dec 13, 2017 at 12:30 | comment | added | Michelle | Is it correct to say that a system is BIBO stable if (and only if) G(z) 's poles are all contained within the unit circle (not on) and all 1/G(z) 's poles are outside the unit circle? | |
| Dec 13, 2017 at 11:00 | vote | accept | Michelle | ||
| Dec 13, 2017 at 7:58 | comment | added | Michelle | Relief! The world is at piece again! Thanx very much. | |
| Dec 13, 2017 at 7:54 | comment | added | Matt L. | @Michelle: If $1/G(z)$ has all its poles outside the unit circle, you get a stable anti-causal filter. | |
| Dec 13, 2017 at 7:43 | comment | added | Michelle | I see, let me explain further...G(z) has only one pole at z=0 and the magnitudes of all the zeros lie outside the unit circle (nothing on) - when I interpret the poles of (1/G(z)) in closed form - is it the anti-causal interpretation (a)^n u(-n-1)? Since y(n) is calculated in terms of past x(n) - but now x(n) is in terms of past/future y(n)? | |
| Dec 12, 2017 at 13:00 | history | answered | Matt L. | CC BY-SA 3.0 |