1. Chapter 8:
System Stability

2. Stability of discrete-time systems
Suppose we have the following transfer function of a closed-loop system:
The stability of the system depends on the location of the poles of the closed-loop transfer function, or the roots of the characteristic equation D(z) = 0.
It was shown in Chapter 7 that the left-hand side of the s-plane, where a continuous system is stable, maps into the interior of the unit circle in the z-plane.
Thus, we can conlude that a system in the z-plane will be stable if all the roots of the characteristic equation, D(z) = 0, lie inside the unit circle.

3. Stability of discrete-time systems
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 .

4. 8.1 Factorizing the characteristic equation
The direct method to check the stability of a system is to factorize the characteristic equation, determine its roots, and check if their magnitudes are all less than 1.
It is not usually easy to factorize the characteristic equation by hand. However, using Matlab command “roots”, this is very easy.
This type of test tell us whether a system is stable or not. It does not tell us how the stability is affected if the gain or some other parameter is changed in the system.

5. Example 8.1
The block diagram of a closed-loop system is shown in Figure 8.1. Determine whether or not the system is stable. Assume that T = 1 s.

6. Solution:
The transfer function of the closed-loop system is
Where
For T = 1 sec,
The characteristic equation is
The root of the characteristic equation z = which is outside the unit circle, i.e. the system is not stable.

7. Example 8.2
In the previous example, find the value of T for which the system is stable.

8. Solution:
From the previous example, we got
The characteristic equation is
Hence, the pole is
For stability, the condition |z|<1 must be satisfied;
Thus the system is stable as long as T <

9. 8.2 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
We now form the following:

10. 8.2 Jury’s stability test
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:

11. 8.2 Jury’s stability test
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.

12. Jury test for 2nd order polynomial
For 2nd 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. Jury test for 3rd order polynomial
For 3rd 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

14. Example 8.3
The closed-loop transfer function of a system is given by Where Determine the stability of this system using Jury’s test.

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

16. Example 8.5
Determine the stability of the system having the following characteristic equation:

17. Example 8.7
The block diagram of a sampled data system is shown in Figure 8.2. Use Jury’s test to determine the value of K for which the system is stable. Assume that K > 0 and T = 1 s.

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

20. Example System is unstable!
Determine the stability of the system having the following characteristic equation:
|b0|=0.75< |b3|=1.5
|c0|=1.6875> |c2|=0.75
System is unstable!

21. 8.3 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 left-hand s-plane into the interior of the unit circle in the z-plane. For this transformation, z is replaced by
giving the characteristic equation in w ,

22. 8.3 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.

23. Example 8.6
The characteristic equation of a sampled data system is given by
Determine the stability of the system using the Routh–Hurwitz criterion.

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