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## Chapter 5 Analysis of CCS

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**Chapter 5**Analysis of CCS**Contents**5.1 Introduction 5.2 Stability Analysis 5.3 Steady-state analysis 5.4 Dynamic analysis 5.5 Controllability, reachability, observability, detectability**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**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:**5.2 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**5.2 Stability Analysis**(4) Some commonly used contour a. constant-attenuation loci (σ)**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**5.2 Stability Analysis**c. Constant damping ratio (ξ)**5.2 Stability Analysis**5.2.2 Stability definition**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 kk0.**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.**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.**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**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**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?**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.**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)**5.2 Stability Analysis**(1) The Jury Stability Criterion • Suppose the characteristic equation is (5.4) • Form the table,**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**5.2 Stability Analysis**Example 5.2: | |**5.2 Stability Analysis**(2) Routh stability criterion after bilinear transformation The bilinear transformation is defined by**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.**5.2 Stability Analysis**Example 5.3 • Discussing the system’s stability (shown as the following Figure) by applying Routh stability criterion.**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}**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**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.**5.2 Stability Analysis**Example 5.5 Determine the range of K when system is stable where**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**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**Exercises**(a) (b)**Exercises**(c) (d)**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.**5.3 Steady-state analysis**• Consider the typical discrete-time control system • From the diagram we have the actuating error**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**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**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:**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**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:**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**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:**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