How to evaluate stability of a non-causal System?

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Discussion Overview

The discussion revolves around evaluating the stability of a non-causal system using the Laplace transform. Participants explore the implications of the one-sided Laplace transform and the behavior of the system's response as time approaches infinity, particularly focusing on the roles of parameters α and β.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states that for stability, α must be negative, based on their calculation of H(s) = 1/(s-α).
  • Another participant questions the physical meaning of the problem, suggesting it deals with a nonexistent situation.
  • Concerns are raised about the behavior of the term exp(βt)u(-t) in the context of the one-sided Laplace transform, particularly how it approaches zero.
  • Some participants propose that β should be positive based on the logic of the non-causal part of the system.
  • There is a suggestion to consult a professor regarding the physical meaning of the problem, indicating uncertainty about its relevance.

Areas of Agreement / Disagreement

Participants express differing views on the physical significance of the problem and the implications of the parameters α and β. There is no consensus on the overall stability evaluation or the relevance of the non-causal system.

Contextual Notes

Participants note that the problem may involve assumptions about the physical context of non-causal systems, which remain unresolved. The implications of the one-sided Laplace transform and the behavior of the exp(βt)u(-t) term are also points of contention.

rudra
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For this problem I have taken laplace(one-sided) transform of h(t) which gives me
H(s)=1/(s-α). From this I can state that α must be -ve for G(s) to be stable.
But my problem is while taking one-sided Laplace Transform the exp(βt)u(-t) part gives 0.
So in H(s) according to my calculation, β doesn't appear. I don't know what I am doing wrong. Please help
 
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rudra said:
Imp.jpg




For this problem I have taken laplace(one-sided) transform of h(t) which gives me
H(s)=1/(s-α). From this I can state that α must be -ve for G(s) to be stable.
But my problem is while taking one-sided Laplace Transform the exp(βt)u(-t) part gives 0.
So in H(s) according to my calculation, β doesn't appear. I don't know what I am doing wrong. Please help

As I said, this problem deals with a nonexistent situation (anyone diosagree?).

But: the u(t) part is easy: what does exp(αt) do as t → ∞?

Now for the noncausal part: u(-t) = 1 for t < 0 and = 0 for t => 0. So for any negative value of t, what does exp(βt) do as t gets more and more negative, approaching t → -∞, with β positive or negative?
 
@rude man,

I think your approach gives the proper soln. α should -ve and β by your logic should be positive.
 
rudra said:
@rude man,

I think your approach gives the proper soln. α should -ve and β by your logic should be positive.

That's where I would put my chips. t → -t in the u(-t) term.

I still think you should ask your prof if that problem has any physical meaning. Unless you're a math purist I see no reason to worry about it.
 

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