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Chapter 5 Analysis of CCS

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Chapter 5 Analysis of CCS

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  1. Chapter 5 Analysis of CCS

  2. Contents 5.1 Introduction 5.2 Stability Analysis 5.3 Steady-state analysis 5.4 Dynamic analysis 5.5 Controllability, reachability, observability, detectability

  3. 5.1 Introduction • To apply a CCS to industry, we must analyze • Stability (z-plane) • Steady-state analysis (steady-state error) • Dynamic analysis (transient propoty) • Analysis of CTS: s-plane • Analysis of CCS: z-plane

  4. 5.2 Stability Analysis 5.2.1 The relationship between s-plane and z-plane • The pole and zero locations of CCS in z-plane are related to the pole and zero locations of CTS in s-plane. • The dynamic specifications of CCS is dependable to T. (1) The left half-s-plane: (2) The jω of s-plane:

  5. 5.2 Stability Analysis (3) The periodic strips A point in z plane →infinite points in s plane A point in s plane → a single point in z plane

  6. 5.2 Stability Analysis (4) Some commonly used contour a. constant-attenuation loci (σ)

  7. 5.2 Stability Analysis b. Constant frequency (ω) • Left half-s-plane, : z-plane 0-(-1) negative real axis • Negative real axis in s-plane : z-plane 0-1 positive real axis • Right half-s-plane, : z-plane 1-∞ positive real axis

  8. 5.2 Stability Analysis

  9. 5.2 Stability Analysis c. Constant damping ratio (ξ)

  10. 5.2 Stability Analysis

  11. 5.2 Stability Analysis

  12. 5.2 Stability Analysis

  13. 5.2 Stability Analysis 5.2.2 Stability definition

  14. 5.2 Stability Analysis (1) Stability and asymptotic stability • Consider the discrete-time state-space equation (possibly nonlinear and time-varying) x (k+1) = f (x(k), k) (4.1) • Let x0(k) and x(k) be solutions of (4.1) when the initial conditions are x0(k0) and x(k0) respectively. Further, let ║·║denote a vector norm. Definitions5.1STABILITY • The solution x0(k) of (4.1) is stable if for a given  > 0, there exists a (, k0)>0 such that all solutions with║x(k0) - x0(k0)║<  such that ║x(k) - x0(k)║<  when kk0.

  15. 5.2 Stability Analysis Definitions 5.2ASYMPTOTIC STABILITY • The solution x0(k) of (4.1) is asymptotically stable if it is stable and if  can be chosen such that║x(k0) - x0(k0)║<  implies that ║x(k) - x0(k)║0 when k. • For linear, time-invariant systems, stability is a property of the system and not of a special solution.

  16. 5.2 Stability Analysis Theorem 5.1 ASYMPTOTIC STABILITY OF LINEAR SYSTEM x(k+1) = x (k) x(0) = a (4.2) • A discrete-time linear time-invariant system (4.2) is asymptotically stable if and only if all eigenvalues of  are strictly inside the unit disc. • Remark: • If a simple pole lie at z = 1, then the system becomes critically stable. Also, the system becomes critically stable if a single pair of conjugate complex poles lies on the unit circle in the z plane. • Any multiple closed-loop poles on the unit circle makes the system unstable.

  17. 5.2 Stability Analysis Example 5.1: • Consider the closed-loop control system shown in Fig. 5.9. Determine the stability of the system when K = 1, T=1. The open-loop transfer function G(s) of the system is

  18. 5.2 Stability Analysis The z transform of G(s) is the closed-loop pulse transfer function for the system is The characteristic equation is which becomes

  19. 5.2 Stability Analysis The roots of the characteristic equation are found to be Since the system is stable. If K > 2.3925 , the system is stable or not?

  20. 5.2 Stability Analysis (2) Input-Output Stability Definition 5.3BOUNDED-INPUT BOUNDED-OUTPUT STABILITY • A linear time-invariant system is defined as bounded-input-bounded-output (BIBO) stability if a bounded input gives a bounded output for every initial value. Theorem 5.2RELATION BETWEEN STABILITY CONCEPTS • Asymptotic stability implies stability and BIBO stability.

  21. 5.2 Stability Analysis 5.2.3 Stability test a. Direct computation of the eigenvalues of  b. Methods based on properties of characteristic polynomials c. The root locus method (assume the open-loop system is known) d. The Nyquist criterion (assume the open-loop system is known) e. Lyapunov’s method (non-linear, time-variant)

  22. 5.2 Stability Analysis (1) The Jury Stability Criterion • Suppose the characteristic equation is (5.4) • Form the table,

  23. 5.2 Stability Analysis • Theorem 5.3 JURY’S STABILITY TEST • If a0 > 0, then (4.4) has all roots inside the unit disc if and only if all a0k, k = 0,1,…,n-1 are positive. If no a0k is zero, then the number of negative a0k is equal to the number of roots outside the unit disc. • Remark • If all a0k are positive for k = 1,2,…,n-1, then the condition a00 > 0 can be shown to be equivalent to the conditions A(1)>0 (-1)nA(-1)>0

  24. 5.2 Stability Analysis Example 5.2: | |

  25. 5.2 Stability Analysis

  26. 5.2 Stability Analysis (2) Routh stability criterion after bilinear transformation The bilinear transformation is defined by

  27. 5.2 Stability Analysis

  28. 5.2 Stability Analysis First substitute (1+w)/(1-w) for z in the characteristic equation as follows Then, clearing the fractions by multiplying both sides of this last equation by (1-w)n, we obtain Once we transform P(z)= 0 into Q(w)= 0, it is possible to apply Routh stability criterion in the same manner as in the continuous-time systems.

  29. 5.2 Stability Analysis Example 5.3 • Discussing the system’s stability (shown as the following Figure) by applying Routh stability criterion.

  30. 5.2 Stability Analysis

  31. 5.2 Stability Analysis

  32. 5.2 Stability Analysis (3) For systems with order 2, Jury’s stability test is simplified to (with the restricted condition that A(z) must be a0 = 1) A(1) > 0 A(-1) > 0 |A(0)| < 1 Proof: Suppose a system with order 2 is: First we will form the Jury table 1 ab b a 1 1-b2a-ab a-ab 1-b2 (1-b2)-{a2(1-b)2/1-b2}={(1-b)(1+b)2/(1+b)}-{a2(1-b)/1+b} =(1-b)/(1+b){ (1+b)2-a2}

  33. 5.2 Stability Analysis • According to the Jury stability test rules, the system is stable if and only if 1-b2>0  |b|<1  -1<b<1  (1-b)/(1+b)>0 and (1-b)/(1+b){(1+b)2-a2}>0  {(1+b)2-a2}>0  |a/(1+b)|<1  -1<a/(1+b)<1 -1-b<a<1+b 1+b+a>0 and 1+b-a>0 • We all know that |b|<1 is equivalent to |A(0)|<1 1+b+a>0 is equivalent to A(1)>0 1+b-a>0 is equivalent to A(-1)>0

  34. 5.2 Stability Analysis Example 5.4: • The system is the same as example 5.3, directly ascertain the K’s value scale in Z-plane.

  35. 5.2 Stability Analysis Example 5.5 Determine the range of K when system is stable where

  36. 5.2 Stability Analysis 5.2.4 Relative Stability Definition 5.4 AMPLITUDE MARGIN • Let the open-loop system have the pulse-transfer function H(z) and let 0 be the smallest frequency such that and such that is decreasing for  = 0 . The amplitude or gain margin is then defined as

  37. 5.2 Stability Analysis Definition 5.5 PHASE MARGIN • Let the open-loop system have the pulse-transfer function H(z) and further let the crossover frequency c be the smallest frequency such that The phase margin marg is then defined as

  38. Exercises (a) (b)

  39. Exercises (c) (d)

  40. 5.3 Steady-state analysis 5.3.1 Steady-state response to input signal • The steady-state performance of a stable control system is generally judged by the steady-state error due to step, ramp, and acceleration inputs. • Suppose the open-loop pulse transfer function is given by the equation where B(z)/A(z) contains neither a pole nor a zero at z =1. Then the system can be classified as a type 0 system, a type 1 system, or a type 2 system according to whether N = 0, N = 1, or N = 2, respectively.

  41. 5.3 Steady-state analysis • Consider the typical discrete-time control system • From the diagram we have the actuating error

  42. 5.3 Steady-state analysis • Consider the steady-state actuating error at the sampling instants, from the final value theorem, we have and where so

  43. 5.3 Steady-state analysis (1) Static Position Error Constant • For a unit-step input r(t) = 1(t), we have • We define the static position error constant Kp as follows • Then the steady-state actuating error in response to a unit-step input can be obtained from the equation

  44. 5.3 Steady-state analysis • If Kp = , which requires that GH(z) have at least one pole at z = 1, the steady-state actuating error in response to a unit-step input will become zero. • For type 0 system: • For type I system: • For type II system:

  45. 5.3 Steady-state analysis (2) Static Velocity Error Constant • For a unit-ramp input r(t) = t, we have • We define the static velocity error constant Kv as follows

  46. 5.3 Steady-state analysis • Then the steady-state actuating error in response to a unit-ramp input can be given by • If Kv = , then the steady-state actuating error in response to a unit-ramp input is zero. This requires GH(z) to possess a double pole z = 1. • For type 0 system: • For type I system: • For type II system:

  47. 5.3 Steady-state analysis (3) Static Acceleration Error Constant • For a unit-acceleration input r(t)= t2/2, we have • We define the static acceleration error constant Ka as follows

  48. 5.3 Steady-state analysis • Then the steady-state actuating error in response to a unit-acceleration input can be obtained from the equation • The steady-state actuating error in response to a unit-acceleration input become zero if Ka = . This requires GH(z) to possess a triple pole z=1. • For type 0 system: • For type I system: • For type II system:

  49. 5.3 Steady-state analysis

  50. 5.3 Steady-state analysis Remark: 1. We should test the stability of the system before we compute the steady-state error of the system. 2. For non-unit input signal, static error constants are not changed, but steady-state error will be different according to the coefficients of the input signal. 3. For multiple input signals, the steady-state error can be computed by superposing multiple steady-state errors. For example, if r(t) = a + bt, then