Z-Transform Analysis of Discrete-Time Systems

1. Chapter 5 On-Line Computer Control – The z Transform

2. Analysis of Discrete-Time Systems • The sampling process • z-transform • Properties of z-transforms • Analysis of open-loop and closed-loop discrete time systems • Design of discrete-time controllers

3. Continuous signal Disontinuous signal y y* y y* 9 12 t (sec) 1 3 5 7 9 11 t (sec) 1 3 6 y* y* Dt y* Dt y* 9 12 t = nT t t (sec) 6 3 t = nT t t = nT t T = 1 sec • Continuous signal and its discrete-time representation with different sampling rates (a) (b) T = 3 sec (c) (b) (a) (c) • From the response of a real sampler to the response of an ideal impulse sample

4. Impulse Sampler y (t) y* The Sampling Process • At sampling times, strength of impulse is equal to value of input signal. • Between sampling times, it is zero. Laplacing or

5. discrete impulses Continuous output m* (t) m (t) Hold Device 1 m (t) m* (t) T t The Hold Process :From Discrete to Continuous Time • Zero – Order Hold : • Transfer Function : Response of an impulse input : d (t)

6. 2 1 2T T 1 -1 First Order Hold Response to an impulse input Transfer function:

7. m* (nT) m (t) m (t) 0 t 1T 3T 5T 7T 0 t 1T 3T 5T 7T 0 t 1T 3T 5T 7T m* (nT) m (t) m (t) 0 2T 4T 6T 8T 10T t 0 2T 4T 6T 8T 10T t t 10T 0 2T 4T 6T 8T First Order versus Zero Order Hold Comparison of reconstruction with zero-order and first-order holds, for slowly varying signals. (a) (b) (c) (a) (b) (c) Comparison of reconstruction with zero-order and first-order holds, for rapidly changing signals.

8. Sample Z-Transforms y(t) yz(t) Remarks z-transform depends only on the discrete values y(0), y(ז),y(ז)..etc. If two continuous functions have the same sampled values , then z-transform will be the same. It is assumed that the summation exists and is finit. We can also view t in the form Z[ y (s) ] = ŷ(z)

9. Z-Transforms of Basic Functions 1. Unit Step Function 2. Exponential Function

10. Z-Transforms of Basic Functions - Continued 3. Ramp Function 4. Trigonometric Functions

11. Z-Transforms of Basic Functions - Continued 5. Translation

12. Z-transform for Numerical Derivative z-1 is like a back shift operator

13. Properties of z-Transforms 1. Linearity 2. Final Value Theorem

14. Numerical Integration in z-transform Using Trapezoidal Rule or solving

15. Inversion of z-transforms 1.Partial fraction expansion λ1, λ2,… λn are low-order polynomials in z-1 compute c1,c2,…cn. Invert each part separately, we able

16. From Tables of z-transforms y(nT) = -1/2 + 1/2 e11n y:0,1,4,13,0,… 2.Inversion by Long-Division 1z-1+4z-2+13z-3 1-4z-1+3z-2 z-1 z-1-4z-2+3z-3 4z-2+3z-3 4z-2-16z-3+12z-4 13z-3-12z-4 y(0) = 0 y(T) = 1 y(2T) = 4 y(3T) = 13

17. z-transforms of various functions Function Lalpace transform z-transform in time domain unit impluse 1 unit step 1/s ramp: f(t) = at a/s2 f(t) = tn n!/sn+1 f(t) = e-at 1/s+a f(t) =te-at 1/(s+a)2

18. z-transforms of various functions Function Lalpace transform z-transform in time domain f(t) = sinωt f(t) = cosωt f(t) = 1-e-at f(t) = e-at sinωt f(t) = e-at cosωt

19. Discrete-Time Response of systems In computer control: measurements are taken periodically and control actions implemented periodically, This results in a discrete input/discrete output dynamic system. cn en Discrete System

20. Example of Discrete Systems Let a discrete time approximation is Taking z-transform

21. Z-transform for a given continuous system with transfer function G(s) and a ZOH

22. c*(s) y*(s) Example: Pure Integrator with Hold Step response Hence of which impulse a ramp response

23. Example：First order lag system

24. y(t) ＊ ＊ ＊ ＊ ＊ ＊ ＊ ＊ ＊ ＊ ＊ ＊ time Step Response for 1st order lag system From tables, for Note: Compare with discrete approximation to First-order system

25. Generalization or D (z)=Transfer function relating e and c Analogous to Laplace transfer Discrete time input/output model Remark： Note that D(z) is the z-transform of the response of the system to an impulse input

26. Hold H (s) Process Gp (s) c*(s) y (s) y*(s) discrete input discrete output continuous variables Z-transform of a continuous process with Sample and Hold we seek a relationship (Z-transfer function) between c and y. Consider a impulse input c*(z)=1 c*(s)=1 Then HGp(z) called the pulse transfer function (since it represents the z-transform of the pulse response of Gp (s) )

27. Properties of pulse Transfer Function 1. 2.An impulse input is converted into a pulse input by the first order hold element . Hence HG(z) is the pulse response of G(s) sampled at z internals of T. 3.The pulse transfer function of two systems in series can be combined if there is a sample and hold in between. c2 c1 c3 G1(z) G2(z) T

28. disturbance Process Hold T T + D (z) H (s) Gp (s) y(2) set point m (s) ysp (z) - y1(s) sampled output Closed-Loop System or 1. Roots of the Characteristic equation 1+HGp(z)D(z)=0 Determine stability of the closed-loop system 2. Note similarity to continuous system.

29. Example: closed-loop response of a first-order system For proportional control where

30. For a unit step change in set point and

31. The response is very similar to continuous control. The steady state value of y(t) is Hence the offset is

32. Stability of Discrete Systems A system is consider to be stable if output remains bounded for bounded Inputs. Consider a discrete system with transfer function Where P1,P2,…,Pn are n roots of:

34. Im Unstable roots real STABLE REGION Unit circle

36. Example: Stability of closed-loop

37. Example: Stability of closed-loop - Continued

38. Digital Feedback - Control

39. Advantages of velocity Form • No initialization is necessary. [ Cs is not needed ]  Bumpless transfer from manual / automatic • Automatic ‘reset-windup’ protection. • Protection in case of computer failure • Disadvantages: • Since different modes are indistinguishable, on-line tuning methods will not work. • Difficult to put constraints on integral and / or derivative term. • Tuning Digital Controllers: • Ziegler – Nichols • Cohen – Coon settings • Time - integral performance criteria

40. actual ideal time Y(t)

41. Derivation of Deadbeat Controller- Continued

42. Deadbeat

43. Deadbeat 0~10 sec

44. Deadbeat control for (1/(s+1)3) • Sampling time: 2

45. Ringing and Pole-placement Ringing refers to excessive value movement caused by a widely oscillating controller output. Caused by negative poles in D(z). Hence avoid poles near -1. Change controller design such that poles are on the side or near zero on negative side

46. SYS = TF(1,[1 3 3 1]) • Transfer function: • 1 • --------------------- • s^3 + 3 s^2 + 3 s + 1 • >> sysd=c2d(SYS,2) • Transfer function: • 0.3233 z^2 + 0.3073 z + 0.01584 • -------------------------------------- • z^3 - 0.406 z^2 + 0.05495 z - 0.002479 • Sampling time: 2 • 0.3233 0.6306 0.3231 0.0158 • >> p1=[1 -1];p2=[0.3233 0.3073 0.01584] • p2 = • 0.3233 0.3073 0.0158 • >> c=conv(p1,p2) • c = • 0.3233 -0.0160 -0.2915 -0.0158

47. Canceling the ringing pole at z=-0.8958 • ans = • 1.0000 • -0.8958 • -0.0547 • >> p1=[1 0];p2=[1 -0.99];p3=[1 0.0547]; • >> c=conv(p1,p2) • c = • 1.0000 -0.9900 0 • >> c=conv(c,p3) • c = • 1.0000 -0.9353 -0.0542 0 • Warning: Using a default value of 1 for maximum step size. The simulation step size will be limited to be less than this value. • >>

48. Smoothing the Control Action • p=[0.3233 -0.016 -0.2915 -0.01584]; • r=roots(p) • r = • 1.0001 • -0.8959 • -0.0547 • Delete the unstable pole z=-0.8959 • p1=[1 -0.99]; p2=[1 0.0547]; • c=conv(p1,p2) • c = • 1.0000 -0.9353 -0.0542

49. Reconstruct the Control Loop