## Stability of linear and non- linear systems

)2s(. 8. 1. C. 2. 3. C. 3. 3. = ++. +. +. = +. +. = +. +. We want to know what value of Kc causes instability, i.e., at least one root of the above equation is positive.
Stability of linear and nonlinear systems

1. Stability of Linear Systems - Revision In general, a closed loop control system should satisfy: • Closed loop stability • Good disturbance rejection (without excessive control action) • Fast set-point tracking (without excessive control action) • A satisfactory degree of robustness to process variations and model uncertainty • Low sensitivity to measurement noise.

Structure of presentation: • • • • • • •

Stability of linear systems (s domain) – revision Stability of non-linear systems Lyapunov’s first theorem Lyapunov’s direct method Krasovski’s method Questions and Answers Further reading

We are going to consider further the stability of linear and non-linear systems. Starting with linear systems, the stability of a closed loop control system can be checked by • Determining the poles of the characteristic equation • Using the Bode or Nyquist criterion (in the frequency domain) • Using the Routh-Hurwitz criterion.

1

2

Historical note

Linear systems stability – determining poles

• The industrial revolution can be said to have started with the development of a governor for steam engines in 1788. • A governor controls the speed of the engine by regulating the admission of steam (fuel) into the cylinders, so as to maintain a near constant speed whatever the load or fuel supply conditions.

James Watt’s engine, 1788 – Science Museum, London http://www.btinternet.com/~historical.engines/Lap_engine.htm

3

Reference: Marlin, T.E. (2000). Process control, Chapter 10

Reference: http://en.wikipedia.org/wiki/Centrifugal_governor

In 1868, James Clerk Maxwell (1831-1879) linearised the differential equations of the governed steam engine about an equilibrium point and proved that the subsequent closed loop system is stable if the poles 4 have negative real parts.

Example

Example

Root locus best determined using dedicated software e.g. MATLAB

Root locus

5

Example

…. Bode or Nyquist stability criterion

6

Linear systems stability – Bode criterion

7

8

Linear systems stability – Bode criterion

Linear systems stability – Bode criterion

9

Linear systems stability – Bode criterion

11

10

Linear systems stability – Bode criterion

12

Linear systems stability – Routh-Hurwitz criterion

Who was Bode ?

Reference: IEEE Transactions on Automatic Control, Vol. AC-29, No. 3, pp. 193-194, 1984.

Characteristic equation an s n + an −1s n −1 + Λ + a1s + a0 = 0 where an >0 . According to the Routh-Hurwitz criterion, if any of the coefficients a0, a1, …, an-1 are negative or zero, then at least one root of the characteristic equation lies in the right half plane, and thus the closed loop system is unstable. On the other hand, if all of the coefficients are positive, then one must construct the Routh Array shown. For stability, all elements in the first column must be positive.

13

14

Reference: Seborg, D. et al. (2004). Process dynamics and control, Chapter 11.

Linear systems stability – Routh-Hurwitz criterion The first two rows of the Routh Array are comprised of the coefficients in the characteristic equation. The elements in the remaining rows are calculated from coefficients by using the formulas:

b1 =

a n −1a n − 2 − a n a n −3 a n −1

a a − a n a n −5 b 2 = n −1 n − 4 a n −1 . .

b1a n −3 − a n −1b 2 b1 b1a n −5 − a n −1b 3 c2 = b1

Linear systems stability – Routh-Hurwitz criterion 8

Example: G P = (s + 2)3 Characteristic equation is

1+ construct

G V = Gs = 1

GC = KC

1 + G CG V G P G S = 0

8 =0 (s + 2) 3

(s + 2) 3 + 8K C = 0

s 3 + 6s 2 + 12s + 8 + 8K C = 0 We want to know what value of Kc causes instability, i.e., at least one root of the above equation is positive. Using the Routh array,

calculate

c1 =

(n+1 rows must be constructed; n = order of the 15 characteristic eqn.)

1 12 6 8 + 8K C 6(12) − (1)(8 + 8K C ) 0 6 8 + 8K C 0

n =3

Conditions for stability: 72 − (8 + 8K C ) > 0 K C < 8

8 + 8K C > 0

K C > −1

The important constraint is Kc