Stability criteria - System Analysis and Control

Suppose a system has transfer function

G(s) = \frac{b(s)}{a(s)},

(1)

where $b(s)$ and $a(s)$ are the numerator and denominator polynomials, respectively. The denominator $a(s)$ is also called the characteristic polynomial of the system, and its roots are the poles of the system.

Remember the definition of stability: a system is stable if all of its poles have negative real part, i.e., they lie in the left-half plane (LHP). It is always possible to check stability by:

Computing all roots of $a(s)$ , and
Checking that each root has negative real part.

But if we only care whether the system is stable (not the exact pole locations), computing all of the roots is often impractical and unnecessary. We will learn a faster way!

Normalize the characteristic polynomial¶

Many stability tests assume the characteristic polynomial is monic (leading coefficient equals 1). So first, normalize $a(s)$ by dividing through by its leading coefficient.)

For example,

\frac{3s+7}{-2s^2+5s+1} \;=\; \frac{-\tfrac{3}{2}s-\tfrac{7}{2}}{s^2-\tfrac{5}{2}s-\tfrac{1}{2}},

(2)

and the normalized characteristic polynomial is

s^2-\frac{5}{2}s-\frac{1}{2}.

(3)

Positive coefficients¶

Fact (necessary condition)

If the system is stable, then all coefficients of the normalized characteristic polynomial are positive. Mathematically, we can write this as:

\textsf{(stable system)} \;\implies\; \textsf{(all coefficients positive)}

(4)

An equivalent way of saying this^[1]: if any coefficient of the normalized characteristic polynomial is negative (or zero), then the system is not stable (it is unstable or marginally stable). We can write this alternative statement as:

\textsf{(any coefficient negative or zero)} \;\implies\; \textsf{(not stable)}.

(5)

Necessary vs sufficient conditions

A necessary condition for a statement $P$ is a condition $Q$ such that if $P$ is true, then $Q$ must also be true. In other words, $P$ implies $Q$ . However, $Q$ being true does not necessarily imply that $P$ is true.
A sufficient condition for a statement $P$ is a condition $R$ such that if $R$ is true, then $P$ must also be true. In other words, $R$ implies $P$ . However, $P$ being true does not necessarily imply that $R$ is true.

For example, let $P$ , $Q$ , and $R$ be the following statements:

$P$ : “You are on the swim team.”
$Q$ : “You can swim.”
$R$ : “You are wearing swim team gear (sweater, backpack, baseball cap, etc.).”

In this case, $P \implies Q$ (if you are on the swim team, then you can swim), so $Q$ is a necessary condition for $P$ . However, $Q$ does not imply $P$ (just because you can swim does not mean you are on the swim team).

On the other hand, $R \implies P$ (if you are wearing swim team gear, then you must be on the swim team), so $R$ is a sufficient condition for $P$ . However, $P$ does not imply $R$ (just because you are on the swim team does not mean you are wearing swim team gear on a given day).

In the example above, the condition of having all positive coefficients is a necessary condition for stability, but it is not a sufficient condition. In other words, if any coefficient is negative or zero, we can conclude that the system is not stable. However, if all coefficients are positive, we cannot conclude that the system is stable; it could still be unstable.

Why? If all roots are in the left-half plane, then the polynomial factors into terms like

real roots: $(s+r_i)$ with $r_i>0$ , and
complex conjugate roots: $(s+p_i)^2 + q_i^2$ with $p_i>0$ ,

so the characteristic polynomial factors as

a(s)= \underbrace{(s+r_1)\cdots(s+r_k)}_{\textsf{real roots}}\ \underbrace{\Bigl((s+p_1)^2+q_1^2\Bigr)\cdots\Bigl((s+p_\ell)^2+q_\ell^2\Bigr)}_{\textsf{complex conjugate roots}}

(6)

and multiplying these factors cannot produce a negative coefficient because there are no negative signs in the factors! This means that if any coefficient is negative, its factored form cannot look like the above, so it must have at least one root in the right-half plane, and therefore the system is unstable.

However, the converse is not true: positive coefficients do not guarantee stability. For example,

s^3+s^2+4s+30=(s+3)(s^2-2s+10),

(7)

which has roots at $-3$ and $1\pm 3j$ , so it is unstable (two poles in the right-half plane), even though all coefficients are positive.

Exact conditions for stability¶

There are stronger tests that are necessary and sufficient (they tell you exactly when the system is stable). For low-order polynomials, the conditions are simple.

Let the characteristic polynomial be written in the form

a(s)=s^n+a_{n-1}s^{n-1}+\cdots+a_1 s + a_0.

(8)

First or second order¶

If $a(s)=s+a_0$ or $a(s)=s^2+a_1 s + a_0$ , then the system is stable if and only if all coefficients are positive. In other words, the condition we saw earlier for positive coefficients is all you need to check.

Third order¶

If $a(s)=s^3+a_2 s^2 + a_1 s + a_0$ then the system is stable if and only if all coefficients are positive and the following inequality holds:

a_2 a_1 > a_0.

(9)

Fourth order¶

If $a(s)=s^4+a_3 s^3 + a_2 s^2 + a_1 s + a_0$ then the system is stable if and only if all coefficients are positive and the following inequality holds:

a_3 a_2 a_1 > a_1^2 + a_3^2 a_0

(10)

Fifth order and above¶

For order 5 and higher, there are stability conditions, but they become significantly more complicated. A standard tool is the Routh–Hurwitz stability criterion, which provides a systematic way to check stability without computing the roots, but we will not cover it here because it is rarely needed.

Examples¶

Example 1: second-order check¶

Is the following system stable?

G(s)=\frac{s-2}{s^2+7s+34}

(11)

Answer. Yes. The characteristic polynomial is $s^2+7s+34$ , and for second order we only need $7>0$ and $34>0$ . The numerator is irrelevant for determining stability.

Example 2: positive counterexample¶

Recall the counterexample we saw earlier in Eq. (7):

a(s)=s^3+s^2+4s+30

(12)

Is the system with this characteristic polynomial stable or unstable?

Answer. Unstable! Here $a_2=1>0$ , $a_1=4>0$ , $a_0=30>0$ , but the third-order condition

a_2 a_1 > a_0

(13)

fails because $1\cdot 4 \not> 30$ . Therefore the polynomial has at least one root in the right-half plane and the system is unstable. This confirms our previous result when we calculated the roots explicitly. Note that we did not have to calculate the roots to determine stability; we just had to check the coefficients and the inequality.

Example 3: integral control¶

Consider the unity-feedback loop with plant and controller given by

G(s)=\frac{1}{(s+1)(s+2)} \quad\textsf{and}\quad C(s)=\frac{k}{s}

(14)

What values of $k$ produce a stable closed-loop transfer function?

Answer. The closed-loop transfer function is

\frac{C(s)G(s)}{1+C(s)G(s)} = \frac{\dfrac{k}{s}\,\dfrac{1}{(s+1)(s+2)}}{1+\dfrac{k}{s}\,\dfrac{1}{(s+1)(s+2)}} = \frac{k}{s(s+1)(s+2)+k}.

(15)

Expanding the denominator gives the characteristic polynomial

a(s)=s^3+3s^2+2s+k.

(16)

This is third order with $a_2=3$ , $a_1=2$ , $a_0=k$ . Stability requires

k>0 \quad\text{and}\quad a_2 a_1 > a_0 \;\;\Longleftrightarrow\;\; 3\cdot 2 > k.

(17)

So the closed-loop system is stable for

0<k<6.

(18)

This confirms our intuition that integral control can cause instability if the gain is too high, and it also gives us an exact range of gains that guarantee stability.

Test your knowledge¶

Solution to Exercise 1 #

The closed-loop transfer function is

\frac{C(s)G(s)}{1+C(s)G(s)} = \frac{\dfrac{k}{(s+1)^4}}{1+\dfrac{k}{(s+1)^4}} = \frac{k}{(s+1)^4+k}.

(19)

The characteristic polynomial is

a(s)=(s+1)^4+k=s^4+4s^3+6s^2+4s+(1+k).

(20)

All coefficients must be positive, so this implies $1+k>0$ , which means $k>-1$ . The fourth-order condition is

\begin{aligned} &a_3 a_2 a_1 > a_1^2 + a_3^2 a_0 \\ &\implies\; 4\cdot 6 \cdot 4 > 4^2 + 4^2 (1+k) \\ &\implies\; 96 > 16 + 16(1+k) \\ &\implies\; 80 > 16(1+k) \\ &\implies\; 5 > 1+k \\ &\implies\; k < 4. \end{aligned}

(21)

Combining these two conditions, we find that the closed-loop system is stable for

\boxed{-1 < k < 4}

(22)

Footnotes¶

This is known as the contrapositive of the original statement, and it is logically equivalent to the original statement. Formally, if $P$ implies $Q$ , then not $Q$ implies not $P$ .
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