is left side of the s-plane stable side?

[truva]

Laplace transform includes exp(-st) and s=σ+jω. σ is negative in the left side and hence exp(-st) goes to infinity. It is not stable. Where am I wrong?

[omkar13]

σ is negative.So,exp(-σt) will tend to 0 as t tends to ∞.Also according to the function for which you want to find LT,limits are imposed on s for stabilty.You cannot directly think of the equation without function.

[truva]

σ is negative.So,exp(-σt) will tend to 0 as t tends to ∞.Also according to the function for which you want to find LT,limits are imposed on s for stabilty.You cannot directly think of the equation without function.

If σ is negative -σ is positive. Hence exp(-σt) goes to ∞ not to 0 as t tends to ∞. !???

[omkar13]

Sorry, I'm also a bit confused in previous reply.Firstly,you shouldnot consider the only integral part without function.
2.σ is not a fixed value.It is a variable which varies according to requirement.
consider LT{tu(t)}.
LT{tu(t)}=∫(-∞ to ∞)t exp(-st)u(t)dt
=∫(0to ∞)t exp(-st)dt
=[texp(-st)/s](0 to ∞)-∫(0 to ∞)1*exp(-st)/s
=1/s^2 IF AND ONLY IF FIRST TERM=0.
now consider first term.
[texp(-st)/s](0 to ∞)=(texp(-s*∞)-0)/s.
HERE WE IMPOSE A CONDITION ON S THAT S IS POSITIVE.HENCE FIRST TERM WILL BE 0 AND WE GET LT AS 1/S^2.IF S IS POSITIVE IT MEANS THAT σ IS ALSO POSITIVE.So what I mean to say is that σ is not a fixed quantity but a variable.If anything above anyone feel irrelevant to topic, please tell me.Thankyou

[es1]

Truva, The question is a bit confusing because the Laplace transform is an operation, not a system. An operation cannot be stable or unstable. You might want to read this introduction: http://www.roymech.co.uk/Related/Control/Laplace_Transforms.html

[truva]

IF S IS POSITIVE IT MEANS THAT σ IS ALSO POSITIVE.

Agree, if σ is positive it is on the right hand side not in the left hand side. This is what I am trying to say.

But most of the DSP books, regarding the IIR filters, say that if σ is in left side the system will be stable and if it is in the right side the system MAY BE unstable. However σ is positive in the right side and the right side is unconditionally stable side. I am really confused.

[omkar13]

Whatever you said is correct.But the stability aspect is for a system.I quoted an example to show that σ is a variable and it has limits which is called REGION OF CONVERGENCE and the limits vary according to the function.For the system to be stable all roots of transfer function of respective system must belong to left half of s-plane.Consider the previous example .
LT{tu(t)}=1/s^2.The roots of this eq. are 0,0 which lies on jw axis hence this system,if designed is marginally stable.You may have ROC extending from some negative value to +∞ but the system is stable only when σ lies in left half of s-plane.
or you can understand like this. σ is only used to make equation integrable.After a transfer function is obtained, we should plot the poles of that T.F.Here,what you h've said is to be checked.Am I clear?

[truva]

I understand now. We look at the value of the VARIABLE s, and see the behavior of the SYSTEM. Thank you very much for your time.