Why are repeated poles at the origin regarded as unstable?

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:

1. Why are repeated poles unstable at the origin?
2. What difference does it make to stability if the repeated poles are at the origin or a non-zero -ve number.
3. 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.

• Where did you get these statements from? Repeated poles don't make a system unstable. Consider $1/(s + 1)^n$ which has $n$ "repeated" poles at $-1$ but the system is stable. Also zeros don't need to be in the right half plane in order to make a system unstable. Consider $(s + 1)/(s - 1)$ which is unstable although it has a zero at $-1$. Mar 29, 2022 at 8:30
• I changed my question. My bad I missed out the origin Mar 29, 2022 at 8:35

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 $$k \geq 2$$.

$G_2(s)$" />

• Why is it tho? Does it converge diverge like what's going on. I am having difficulty picture it in my mind. lets say for 1/s+1 when we convert it to time domain it is e^-t. We can picture it going to zero as it goes to infiniti. But i Cant do the same with the multiplicity of poles at the origin. Mar 29, 2022 at 9:49
• I added two examples for Case 3. Kindly check the details, thanks! Mar 29, 2022 at 12:29
• Thanks! Makes much more sense. I need one more clarification about the tsin(t) graph. What transformation occured here from sint. Does the amplitude correspond to the increasing x values? Mar 30, 2022 at 2:09
• Shouldn't y_1(t) = 2 sin(2t) and y_2(t) = 2tsint(2t)? It doesn't change anything but just for the purpose of rigour :-). Mar 16 at 4:11

A system is

1. unstable if and only if its impulse response grows unboundedly with time,
2. marginally stable (or "stable") if and only if its impulse response is bounded, and
3. asymptotically stable if and only if its impulse response is bounded and converges asymptotically to zero.

Let us consider the following transfer function

$$H(s)=\dfrac{1}{(s-\alpha)^n},\ n\ge1,\ \alpha\in\mathbb{R}.$$

The impulse response is given by

$$h(t)=\dfrac{t^{n-1}}{(n-1)!}e^{\alpha t}H(t),$$ where $$H(t)$$ is the Heaviside function.

We have the following cases:

1. If $$\alpha>0$$ and $$n\ge1$$, then $$h(t)\to\infty$$ as $$t\to\infty$$ (unstable).
2. If $$\alpha<0$$ and $$n\ge1$$, then $$h(t)\to0$$ as $$t\to\infty$$ (asymptotically stable).
3. If $$\alpha=0$$ and $$n=1$$ then $$h(t)=H(t)\to1$$ as $$t\to\infty$$ (stable).
4. If $$\alpha=0$$ and $$n>1$$ then $$h(t)=\dfrac{t^{n-1}}{(n-1)!}H(t)\to\infty$$ as $$t\to\infty$$ (unstable).

Analogous results hold true when $$\alpha$$ is a complex number with non-zero imaginary part.

As a rule of thumb, if you have a transfer function with repeated poles on the imaginary axis, then the system is unstable.

This may not be the case in the state-space framework where eigenvalues on the imaginary axis can be repeated and the system is still stable. This is the case, when the eigenvalues on the imaginary axis are semi-simple.