- #1
shawrix
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Is exponential function e to the power x[t] stable? My book uses BIBO and says its stable but for -x[t] it says its not stable. Is my book wrong?
Is BIBO applicable for non-linear functions?
- Thread starter shawrix
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- Exponential Signal Stable
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Discussion Overview
The discussion revolves around the applicability of the Bounded Input Bounded Output (BIBO) stability criterion to non-linear functions, specifically focusing on the exponential function and its behavior under different inputs. Participants explore the stability of functions like e^x[t] and e^(-x[t]) in both continuous and discrete contexts.
Discussion Character
- Debate/contested
- Technical explanation
- Mathematical reasoning
Main Points Raised
- One participant questions the stability of the exponential function e^x[t] and contrasts it with e^(-x[t]), suggesting a potential inconsistency in their textbook.
- Another participant argues that both e^(±x(t)) are bounded if x(t) is bounded, implying that the BIBO criterion is satisfied.
- A participant raises the question of how the BIBO criterion applies to discrete inputs, specifically x[n].
- One participant asserts that the BIBO criterion holds for e^(-x[t]) by demonstrating that it remains bounded under certain conditions.
- Another participant references two books that claim e^(-x[n]) is unstable, providing a proof based on impulse response that suggests the system does not converge.
- A later reply claims to have proven that the output remains bounded as time approaches infinity, suggesting that the previous references may be incorrect.
- One participant expresses confusion about the BIBO criterion, noting that it applies similarly to both discrete and continuous systems, emphasizing the distinction in the input domain.
- A participant questions whether BIBO can be applied to non-linear functions, indicating uncertainty about the generalizability of the criterion.
Areas of Agreement / Disagreement
Participants express differing views on the stability of e^(-x[n]) and the applicability of the BIBO criterion to non-linear functions. There is no consensus on the correctness of the references cited, and the discussion remains unresolved regarding the overall applicability of BIBO to non-linear functions.
Contextual Notes
Participants reference different interpretations of the BIBO criterion and its implications for both continuous and discrete systems. There are unresolved assumptions regarding the definitions of stability and the conditions under which BIBO applies.
- #2
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Yes, it does seem that exp(+-x(t)) is bounded whenever x(t) is bounded.
- #3
shawrix
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what about x[n]?
- #4
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There's no difference between the two.
If $$|x[t]| \leq M$$ then $$0<e^{-x[t]}< e^M$$ thus $$|e^{-x[t]}|<e^M< \infty$$
So you still have BIBO criterion satisified.
If $$|x[t]| \leq M$$ then $$0<e^{-x[t]}< e^M$$ thus $$|e^{-x[t]}|<e^M< \infty$$
So you still have BIBO criterion satisified.
- #5
shawrix
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I have two books as reference and they both say e to the power -x[n] is unstable. They take x[n] and y[n] in terms of impulse and impulse response and prove that for n=0 we have output e^-1 and for n/=0 it is 1. Then it uses the bibo stability condition for causal and stable system and prove that the system never converges and hence it is unstable.
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- #6
shawrix
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Yes 
i finally proved it myself, as t-> infinite the output will also become bounded ie 1. Both books are wrong...
i finally proved it myself, as t-> infinite the output will also become bounded ie 1. Both books are wrong...
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- #7
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I am not sure I understand. BIBO criterion says that is you have bounded input then your output will also be bounded, as you've seen what I wrote, from bounded input also the output of e^x(t) is bounded.
BTW in this case it doesn't matter if your system is discrete or in the continuum, either way the same BIBO condition is satisified. The difference is that the domain of the input in one is the natural numbers and on the other is real numbers.
BTW in this case it doesn't matter if your system is discrete or in the continuum, either way the same BIBO condition is satisified. The difference is that the domain of the input in one is the natural numbers and on the other is real numbers.
- #8
shawrix
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Wait, Can we apply BIBO for non-linear functions like this one?
Ps i have edited my prev post, i mentioned incorrectly what i had found.
Ps i have edited my prev post, i mentioned incorrectly what i had found.
Last edited:
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