System stability in the s-domain

  • Thread starter Thread starter cocopops12
  • Start date Start date
  • Tags Tags
    Stability System
AI Thread Summary
The stability of a system in the s-domain is determined by the real parts of its poles. When poles have negative real parts, the system exhibits exponential decay, leading to stability. Conversely, poles with positive real parts result in exponential growth, causing instability. If the real parts are zero, the system becomes metastable, characterized by oscillatory behavior. Understanding these concepts can be clarified through partial fraction expansion and inverse Laplace transforms, which reveal the impact of pole positions on system response.
cocopops12
Messages
30
Reaction score
0
Can someone please explain WHY is it when the poles of F(s) have negative real parts, the system is stable.

Why is it when the poles of F(s) have positive real parts the system is unstable?

Why is it when the real parts of the poles of F(s) equal to 0 the system becomes metastable (oscillatory)

Thanks
 
Engineering news on Phys.org
The reasons become clear when you do a partial fraction expansion of the output of the system and find its inverse Laplace transform.

You'll find factors of exp^(p_i*t) in all the terms of the natural response of the system, where p_i is the corresponding pole of the expansion and t is the time.

You can see what happens if the real part of the pole is positive or negative, exponential growth or decay.

You can find a better runthrough here (go down to 'Poles and the Impulse Response'):
http://www-control.eng.cam.ac.uk/gv/p6/Handout3.pdf
 
Last edited by a moderator:
Hi all I have some confusion about piezoelectrical sensors combination. If i have three acoustic piezoelectrical sensors (with same receive sensitivity in dB ref V/1uPa) placed at specific distance, these sensors receive acoustic signal from a sound source placed at far field distance (Plane Wave) and from broadside. I receive output of these sensors through individual preamplifiers, add them through hardware like summer circuit adder or in software after digitization and in this way got an...
I am not an electrical engineering student, but a lowly apprentice electrician. I learn both on the job and also take classes for my apprenticeship. I recently wired my first transformer and I understand that the neutral and ground are bonded together in the transformer or in the service. What I don't understand is, if the neutral is a current carrying conductor, which is then bonded to the ground conductor, why does current only flow back to its source and not on the ground path...
While I was rolling out a shielded cable, a though came to my mind - what happens to the current flow in the cable if there came a short between the wire and the shield in both ends of the cable? For simplicity, lets assume a 1-wire copper wire wrapped in an aluminum shield. The wire and the shield has the same cross section area. There are insulating material between them, and in both ends there is a short between them. My first thought, the total resistance of the cable would be reduced...

Similar threads

Replies
1
Views
3K
Replies
1
Views
3K
Replies
3
Views
1K
Replies
3
Views
2K
Replies
6
Views
4K
Replies
31
Views
11K
Replies
3
Views
2K
Replies
12
Views
3K
Back
Top