**Introduction**

Transfer function stability is solely determined by its denominator. The roots of a denominator are called **poles**.

Poles located in the left half-plane are stable while poles located in the right half-plane are not stable.

The reasoning is very simple: the Laplace operator **"s"**, which is location in the Laplace domain, can be also written as:

Left half-plane has negative *sigma*.

For a first-order system, the following transfer function corresponds to this time domain differential equation:

Consider a step change as the input:

Hence:

The output converges to 0.2, a steady-state and stable value.

**Stability of a Generic Polynomial**

Generally, consider the following generic denominator:

If such a polynomial is expanded and terms are collected, we obtain:

**Claim: If all polynomial coefficients are positive, all roots are negative and the system is stable**

In short:

- If all polynomial coefficients are positive, then all poles are negative.
- If all roots are negative, then all polynomial coefficients are positive.

If all three terms **a, b, and c** are positive, then all three poles are located in the left-half plane and the system is stable. Also, notice that all coefficient of the expanded polynomial are positive.

E.g., a = 7, b = 2, c = 4, (poles are located at -7, -2, and -4).

For a polynomial containing **complex poles** as shown below, the imaginary part sign does not matter since there needs to be the complex pole counterpart.

Thus, it suffices that **a** and **b** are positive:

How do we know that the second polynomial (s^{2}+2bs+b^{2}+c^{2}) has negative roots?

The **b ^{2}** and

Therefore, if all coefficients are positive, all roots are negative and the system is stable.

**Claim: If any polynomial coefficient is negative, at least one root is positive and the system is unstable**

Without a proof here.

**Claim: If a polynomial coefficient is zero, at least one root is positive and the system is unstable**

See the following polynomial:

Case 1: s^{2}(a+b+c) = 0

This means that at least one term (from a, b, c) is negative, otherwise the sum would be positive.

Case 2: s(ab+bc+ca) = 0

From the way the coefficient is structured, at least one term has to be negative or all three terms need to be zero otherwise the coefficient would be positive.

Case 3: abc = 0

This term can be zero. This also means that at least one root is zero as well.

e.g. a = 0

*Major takeaway: based on this analysis (by no means a complete proof) when a coefficient of a power of s is zero, then the system is unstable.*

Consider the following system. For which gain Kp does the closed-loop system becomes stable?

The plant transfer function G(s) is shown below:

The complete closed-loop transfer function, including the proportional controller Kp, is:

**Answer: **No proportional controller gain can make the system stable since the closed-loop transfer function has zero "s" and "s^{2}" term coefficients. All coefficients of "s^{n}" must be positive.

Root locus follows (notice the two right half-plane poles).

Further Reading

In MatLab, DSPs, and FPGAs.

.

The fundamentals of signal flow.

Introduction to dynamic systems.

It is all sine waves.

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