How to evaluate stability of a non-causal System?

[rudra]

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

 

[rude man]

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?

 

[rudra]

@rude man,

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

 

[rude man]

@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.

 

[rezeew]

To evaluate the stability of a non-causal system, we need to analyze the poles and zeros of its transfer function. In this case, the transfer function H(s) = 1/(s-α). The stability of a system is determined by the location of its poles in the complex plane.

For a system to be stable, all the poles must lie in the left half of the complex plane (Re(s) < 0). This means that the real part of the pole must be negative. In the given transfer function, the pole is located at s = α. So, for H(s) to be stable, the value of α must be negative.

Now, you have correctly pointed out that while taking the one-sided Laplace transform, the term exp(βt)u(-t) gives 0. This is because the unit step function u(-t) is zero for all negative values of t. So, it does not contribute to the overall Laplace transform.

However, this does not mean that β is not important in determining the stability of the system. In fact, the value of β can affect the stability of the system. For example, if we have a transfer function H(s) = 1/(s-β), where β is positive, then the pole will be located in the right half of the complex plane (Re(s) > 0), and the system will be unstable.

In conclusion, to evaluate the stability of a non-causal system, we need to consider the location of its poles in the complex plane. While taking the one-sided Laplace transform, the unit step function may give a zero contribution, but the value of β can still affect the stability of the system. Therefore, it is important to carefully analyze the poles and zeros of the transfer function to determine the stability of a non-causal system.