Transfer Functions and System Stability

A good control system must do two things: respond quickly and accurately, and never become unstable. An unstable aircraft autopilot, voltage regulator or chemical reactor controller can oscillate wildly or run away entirely. Transfer functions give engineers a compact way to describe a system, and their poles tell us, before anything is built, whether it will be stable and how it will behave.

What Is a Transfer Function?

For a linear time-invariant system with input \( u(t) \) and output \( y(t) \), the transfer function is the ratio of their Laplace transforms with all initial conditions zero:

$$ G(s) = \frac{Y(s)}{U(s)} $$

For example, a system obeying \( \ddot{y} + 4\dot{y} + 13y = 10u \) has

$$ G(s) = \frac{10}{s^2 + 4s + 13} $$

The transfer function is a property of the system alone, not of any particular input. Once you know \( G(s) \), the output for any input is \( Y(s) = G(s)U(s) \).

Block Diagram Algebra

• Series blocks multiply: \( G_1(s)G_2(s) \).
• Parallel blocks add: \( G_1(s) + G_2(s) \).
• Negative feedback with forward path \( G(s) \) and feedback \( H(s) \) gives the closed-loop transfer function

$$ T(s) = \frac{G(s)}{1 + G(s)H(s)} $$

With unity feedback, \( H(s) = 1 \).

Poles, Zeros and Stability

Write \( G(s) = N(s)/D(s) \). The roots of \( N(s) = 0 \) are the zeros; the roots of \( D(s) = 0 \) are the poles. The denominator \( D(s) = 0 \) is called the characteristic equation.

Each pole \( p \) contributes a term like \( e^{pt} \) to the response. If \( p = \sigma + j\omega \), the term behaves like \( e^{\sigma t}\cos\omega t \). So:

• All poles in the left half-plane (\( \sigma < 0 \)): every term decays, and the system is stable.
• Any pole in the right half-plane (\( \sigma > 0 \)): a term grows without limit, and the system is unstable.
• Non-repeated poles on the imaginary axis: sustained oscillation, called marginally stable.

The imaginary part \( \omega \) sets the oscillation frequency, and the further left a pole lies, the faster its contribution dies away.

The Standard Second-Order System

Many real systems behave approximately like

$$ G(s) = \frac{K\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2} $$

where \( \omega_n \) is the natural frequency and \( \zeta \) is the damping ratio. For \( 0 < \zeta < 1 \) (underdamped), the step response overshoots, and the percentage overshoot is

$$ \%OS = 100\, e^{-\pi\zeta/\sqrt{1 - \zeta^2}} $$

Worked Example 1: Reading Behaviour from the Poles

For \( G(s) = \dfrac{10}{s^2 + 4s + 13} \), find the poles, decide on stability, and find \( \omega_n \), \( \zeta \), the percentage overshoot and the steady-state output for a unit step.

Step 1: Poles. Solve \( s^2 + 4s + 13 = 0 \):

$$ s = \frac{-4 \pm \sqrt{16 - 52}}{2} = \frac{-4 \pm \sqrt{-36}}{2} = -2 \pm j3 $$

Both poles have real part −2, so the system is stable and oscillates at a damped frequency of 3 rad/s.

Step 2: Natural frequency and damping. Comparing with \( s^2 + 2\zeta\omega_n s + \omega_n^2 \):

$$ \omega_n^2 = 13 \Rightarrow \omega_n = \sqrt{13} = 3.606 \ \text{rad/s} $$

$$ 2\zeta\omega_n = 4 \Rightarrow \zeta = \frac{4}{2(3.606)} = 0.555 $$

Step 3: Overshoot.

$$ \frac{\pi\zeta}{\sqrt{1 - \zeta^2}} = \frac{3.1416 \times 0.555}{\sqrt{1 - 0.308}} = \frac{1.743}{0.832} = 2.094 $$

$$ \%OS = 100\, e^{-2.094} = 12.3\% $$

Step 4: Steady state. By the final value theorem with \( U(s) = 1/s \):

$$ y(\infty) = \lim_{s\to0} s \cdot \frac{10}{s^2 + 4s + 13}\cdot\frac{1}{s} = \frac{10}{13} = 0.769 $$

Sanity check. The damped frequency should equal \( \omega_n\sqrt{1 - \zeta^2} = 3.606 \times 0.832 = 3.0 \) rad/s, matching the imaginary part of the poles. A damping ratio near 0.55 is typical of a well-tuned system, and an overshoot of about 12% is consistent with it.

The Routh–Hurwitz Criterion

For high-order characteristic equations, finding roots by hand is impractical. The Routh–Hurwitz criterion tells us how many roots lie in the right half-plane without solving for them. Build the Routh array from the coefficients; the system is stable if and only if every entry in the first column has the same sign (all positive for a positive leading coefficient). Each sign change indicates one right-half-plane pole.

Worked Example 2: Finding the Maximum Stable Gain

A position control system has open-loop transfer function

$$ G(s) = \frac{K}{s(s + 2)(s + 4)} $$

with unity negative feedback. For what range of gain \( K \) is the closed loop stable, and at what frequency does it oscillate at the limit?

Step 1: Characteristic equation. \( 1 + G(s) = 0 \) gives

$$ s(s + 2)(s + 4) + K = 0 \quad \Rightarrow \quad s^3 + 6s^2 + 8s + K = 0 $$

Step 2: Routh array.

• Row \( s^3 \): 1, 8
• Row \( s^2 \): 6, \( K \)
• Row \( s^1 \): \( \dfrac{6(8) - 1(K)}{6} = \dfrac{48 - K}{6} \), 0
• Row \( s^0 \): \( K \)

Step 3: Stability conditions. All first-column entries must be positive:

$$ \frac{48 - K}{6} > 0 \Rightarrow K < 48, \qquad K > 0 $$

So the closed loop is stable for \( 0 < K < 48 \).

Step 4: Oscillation at the limit. At \( K = 48 \) the \( s^1 \) row is zero, and the auxiliary equation from the \( s^2 \) row is

$$ 6s^2 + 48 = 0 \Rightarrow s^2 = -8 \Rightarrow s = \pm j2\sqrt{2} = \pm j2.83 \ \text{rad/s} $$

Sanity check. Substitute \( s = j2.83 \) and \( K = 48 \) into the characteristic equation: \( s^3 = -j22.6 \), \( 6s^2 = -48 \), \( 8s = j22.6 \). Adding: \( -j22.6 - 48 + j22.6 + 48 = 0 \). The poles truly sit on the imaginary axis at \( K = 48 \). It also makes sense that higher gain eventually causes instability: aggressive controllers overcorrect.

Did You Know?

• Edward Routh published his stability test in 1874 and won the Adams Prize for related work in 1877; Adolf Hurwitz independently derived an equivalent test in 1895, prompted by a turbine-governor problem.
• James Clerk Maxwell's 1868 paper "On Governors" is often called the first mathematical analysis of feedback stability.
• Engineers usually keep a safety margin, such as a gain margin of about 6 dB (a factor of 2), below the critical gain found by Routh–Hurwitz.

Key Takeaways

• A transfer function \( G(s) = Y(s)/U(s) \) captures the input–output behaviour of a linear system.
• Poles are the roots of the characteristic equation; all in the left half-plane means stable.
• For second-order systems, \( \omega_n \) and \( \zeta \) predict oscillation frequency and overshoot.
• Feedback changes pole positions; \( T = G/(1 + GH) \) gives the closed loop.
• The Routh–Hurwitz criterion finds stable gain ranges without solving for roots.

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