I thought you needed the poles and zeroes to be at the right hand side to make the system unstable. Therefore here is the question in two parts:
- Why are repeated poles unstable at the origin?
- What difference does it make to stability if the repeated poles are at the origin or a non-zero -ve number.
- Also let's consider 1/s^3 where the poles are repeated at the imaginary axis, when we convert it back to Time domain what does 0.5*t^2 mean? What does time factor stand for ?
Also please try to explain it in layman's term as well.
Best Answer
This is a good question.
Case 1:
If there is any pole of a linear control system on the open RHS of the complex plane, i.e. the presence of any pole with a positive real part, then the system is unstable.
Case 2:
On the other hand if all the poles of the linear control system lie on the open LHS of the complex plane, i.e. if all poles have negative real parts, then the system is asymptotically stable.
Case 3:
The special case when all the poles belong to the closed LHP [$\mbox{Re} \ z \leq 0$] of the complex plane is addressed next. If the system has no pole on the imaginary axis, then we can deduce stability of the system from the first two cases. So we suppose that there are some poles on the imaginary axis.
We look at the multiplicity of the poles on the imaginary axis including the multiplicity of the origin (as a pole). If the poles on the imaginary axis are found to be simple (multiplicity = 1), then the linear system is Lyapunov stable or marginally stable. If there is any pole on the imaginary axis which is repeated (multiplicity > 1), then the linear system is unstable. This can be established by expressing the system in a partial fraction expansion and calculating the inverse Laplace transform. The repeated poles on the imaginary axis cause a stability problem.
I shall give some examples to illustrate Case 3.
Consider the linear system with the transfer function $$ G_1(s) = {4 \over s^2 + 4} $$
It has two purely imaginary poles : $s = \pm 2 j$
The natural response for the system is: $$ y_1(t) = \sin 2 t $$ which oscillates indefinitely. Such a linear system is called marginally stable.
Let us consider another linear system with the transfer function $$ G_2(s) = {4 s \over s^4 + 8 s^2 + 16} = {4 s \over (s^2 + 4)^2} $$
This system has repeated poles on the imaginary axis: $$ s_{1,2} = \pm 4 j \ \ \mbox{and} \ \ s_{3, 4} = \pm 4 j $$
The natural response for this linear system is: $$ y_2(t) = t \sin 2 t $$ which is unbounded. Hence, this linear system is unstable.
(I have enclosed a MATLAB plot for the step response of the linear system with system transfer function $G_2(s)$.
A similar argument can be used to establish the fact that a linear system is unstable whenever it has repeated poles on the imaginary axis with multiplicity $k \geq 2$.