1. Lecture 6: Stability of discrete-time systems 1

2. Suppose that we have the following transfer function of a closed-loop discrete-time system: The system is stable if all the poles (roots of the characteristic equation D(z) = 0) lie inside the unit circle in the z-plane. 2 Stability of discrete-time systems

3. There are several methods to check the stability of a discrete-time system such as: Factorizing D(z) = 0 and finding its roots. Jury’s test. Routh–Hurwitz criterion. 3 Stability of discrete-time systems

4. Factorizing the characteristic equation The direct method to check system stability is to factorize the characteristic equation, determine its roots, and check if their magnitudes are all less than 1. Although it is not usually easy to factorize the characteristic equation by hand, this is easy using MATLAB command “ roots ”. 4

5. Example 1 Check the stability of the following closed-loop discrete-time system. Assume that T = 1 s. Solution: The transfer function of the closed-loop system is 5

6. Where The characteristic equation is thus 6

7. Example 2 In the previous example, find the value of T for which the system is stable. Solution: From the previous example, we found 7

8. The characteristic equation is For stability, the condition |z|<1 must be satisfied; Thus the system is stable as long as T < 0.549. 8

9. Jury’s stability test Jury’s stability test is similar to the Routh–Hurwitz stability criterion used for continuous systems. To describe Jury’s test, express the characteristic equation of a discrete-time system of order n as Then, we form the following: 9

10. The elements of this array are defined as follows:  The elements of each even-numbered row are the elements of the preceding row, in reverse order.  The elements of the odd-numbered rows are defined as: 10 Jury’s stability test

11. The necessary and sufficient conditions for the characteristic equation to have all roots inside the unit circle are given as Jury’s test is applied as follows: Check the three conditions (I) and stop if any of them is not satisfied. Construct Jury’s array and check the conditions (II). Stop if any condition is not satisfied. 11 Jury’s stability test

12. Jury test for 2 nd order polynomial For 2 nd characteristic equation: Jury’s test reduces to the following simple rules: no roots of the system characteristic equation will be on or outside the unit circle provided that 12

13. Jury test for 3 rd order polynomial For 3 rd order characteristic equation: Jury’s test reduces to the following simple rules: no roots of the system characteristic equation will be on or outside the unit circle provided that 13

14. Example 3 The closed-loop transfer function of a system is given by Where Determine the stability of this system using Jury’s test. 14

15. Solution The characteristic equation is Or Applying Jury’s test All conditions are satisfied and the system is stable. 15

16. Determine the stability of the system having the following characteristic equation: 16 Example 4

17. Example 5 The block diagram of a sampled data system is shown below. Use Jury’s test to determine the value of K for which the system is stable. Assume that K > 0 and T = 1 s. 17

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19. Apply Jury’s test: The third condition is Combining all inequalities together, the system is stable for K < 2.4 19

20. Determine the stability of the system having the following characteristic equation: Example 6 |b 0 |=0.75< |b 3 |=1.5 |c 0 |=1.6875> |c 2 |=0.75 System is unstable! 20

21. Routh–Hurwitz Criterion The stability of a sampled data system can be analyzed by transforming the system characteristic equation into the s- plane and then applying the well-known Routh–Hurwitz criterion. A bilinear transformation is usually used to transform the interior of the unit circle in the z-plane into the left-hand s- plane (or w-plane). For this transformation, z is replaced by giving the characteristic equation in w, 21

22. Routh–Hurwitz Criterion The Routh-Hurwitz array is formed as shown. The first two rows are obtained from the equation directly and the other rows are calculated as shown. The Routh–Hurwitz criterion states that the number of roots of the characteristic equation in the right hand s-plane is equal to the number of sign changes of the coefficients in the first column of the array. Thus, for a stable system all coefficients in the first column must have the same sign. 22

23. Example 7 The characteristic equation of a sampled data system is given by Determine the stability of the system using the Routh–Hurwitz criterion. 23

24. Now, we form Routh array: To check the answer, the roots of the characteristic equation, are found using Matlab command roots([ 2 1 1 1]) to be 0.1195 + 0.8138i 0.1195 - 0.8138i -0.7390 As we are interested their magnitudes, we can write directly abs(roots([ 2 1 1 1])). This gives 0.8226, 0.8226, 0.7390 which are all less than one, i.e. the roots lie inside the unit circle. Hence, we can conclude that the system is stable. 24 No sign change in the first column, so the system is stable.