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 Analysis of CTS: s-plane Analysis of CCS: z-plane 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

5.2 Stability Analysis c. Constant damping ratio (ξ)

5.2 Stability Analysis

5.2 Stability Analysis

5.2 Stability Analysis

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. Definitions 5.1 STABILITY 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.

5.2 Stability Analysis Definitions 5.2 ASYMPTOTIC 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 5.3 BOUNDED-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 5.2 RELATION 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

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

5.2 Stability Analysis

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

5.2 Stability Analysis

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 a b b a 1 1-b2 a-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

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

5.3 Steady-state analysis 4. For a different closed-loop configuration, it is noted that if the closed-loop discrete-time control has a closed-loop pulse transfer function, then the static error constants can be determined by an analysis similar to the one just presented. If the closed-loop discrete-time control system does not have a closed-loop pulse transfer function, however, the static error constants cannot be defined, because the input signal cannot be separated from the system dynamics.

5.3 Steady-state analysis Example 5.6 where a. T = 0.5, k = 6 and 10, compute the ess respectively; b. T = 0.5, determine the range of the k for satisfying ess  0.05.

5.3 Steady-state analysis Solution: Judge the stability of the system:

5.3 Steady-state analysis a. when T = 0.5, k should satisfy 0 < k < 8.15, so we should only need to compute ess for k = 6. b. with the stability condition 0 < k < 8.15, we conclude that

5.3 Steady-state analysis Table 5.2 Static error constants for typical closed-loop configurations of discrete-time control systems

5.3 Steady-state analysis

5.3 Steady-state analysis 5.3.2 Steady-state response to disturbances Let us assume that the reference input is zero, or R(z) = 0 in the system shown in Fig. 5.11(a), but the system is subjected to disturbance N(z). For this case the block diagram of the system can be redrawn as shown in Figure 5-11(b). Then the response C(z) to disturbance N(z) can be found from the closed-loop transfer function If |GD(z)G(z)|>>1, then we find that

5.3 Steady-state analysis

5.3 Steady-state analysis Since the system error is We find the error E(z) due to the disturbance N(z) to be Thus, the larger the gain of GD(z) is, the smaller the error E(z). If GD(z) includes an integrator [which means that GD(z) has a pole at z = 1], then the steady-state error due to a constant disturbance is zero.

5.3 Steady-state analysis where ĜD(z) does not involve any zeros at z = 1.

5.3 Steady-state analysis Remark: 1. If a linear system is subjected to both the reference input and a disturbance input, then the resulting error is the sum of the errors due to the reference input and the disturbance input. The total error must be kept within acceptable limits. (1) Suppose N(z)=0, essr can be obtained using the above method. (2) Suppose R(z)=0, essn can be obtained using the above method. (3) So at last ess=essn+essr.

5.3 Steady-state analysis 2. The point where the disturbance enters the system is very important in adjusting the gain of GD(z)G(z). For example, consider the system shown in Fig. 5.12(a) and (b), their closed-loop pulse transfer function for the disturbance are

5.3 Steady-state analysis

5.4 Dynamic analysis Considering the Fig. 5.9 The z transform of the output of the system is where (z) is the z pulse transfer function of the closed-loop control system.

5.4 Dynamic analysis where n > m. When there is no multiple roots of the characteristic equation, Y(z) can be rewritten as

5.4 Dynamic analysis (1) pi is the positive real number The transient response for the poles to be positive real number is Let then the transient response component of the system is given by Since |pi| < 1, and therefore  < 0. The transient response is the exponential attenuation curve and is monotonic. If |pi| is smaller, and the || is larger, so that the result is that the pole is nearer to the original and the transient response is faster.

5.4 Dynamic analysis (2) pi is the negative real number The transient response for the poles to be negative real number is The transient response is the exponential positive and negative attenuation curve respectively. So the negative roots correspond to high- frequency oscillation with frequency ωs/2, which is also called ringing. Ringing period is 2T, frequency is 1/2T, radian frequency is 2π/2T= ωs/2.

5.4 Dynamic analysis (3) pi is equal to zero For single pole case, it is a one step attenuation. When pi is zero, the transient response is the fastest, which is also called deadbeat control.

5.4 Dynamic analysis (4) Multiple poles case: suppose there are m poles p ( remark: ) For |p|<1 p 0, 如果F(z)/z在z=p处有三重极点,那么部分分工必包含须有(z+p)/(z-p)3项. F(z)/z=c1/(z-p)^2+c2/(z-p) C1=[(z-p)^2F(z)/z]z=p C2={d/dz[(z-p)^2F(z)/z]}z=p

5.4 Dynamic analysis For 0<p<1，the response is monotonic exponential attenuation curve. For -1<p<0 , the response is a high-frequency oscillation attenuation curve with frequency s/2. For p=0, e.g., two poles system, M multiple poles imply a m steps attenuation.

5.4 Dynamic analysis (5) The poles are a pair conjugate complex number where and the similar transient response is

5.4 Dynamic analysis The transient response is periodic damping oscillation form. The more closed to the origin the pole is, the faster attenuation is. The larger i is (T: 0  /2  ), the more intensely oscillation does (oscillation frequency 0 s/4  s/2).

5.4 Dynamic analysis Summary 1. System with poles in the unit circle corresponds to attenuation curve (stable). 2. Positive real pole corresponds to monotonic response Negative real pole corresponds to high-frequency oscillation with frequency s/2. 3. The more the pole closed to the origin is, the faster attenuation is, when poles are on the origin, the transient response is the fastest, which is also called deadbeat control. 4. The larger i is (T: 0  /2  ), the more intensely oscillation does (oscillation frequency : 0 s/4  s/2).

5.4 Dynamic analysis So we have the Fig. 5.13. If we desire to get the better transient response (better performance), the poles of the closed-loop pulse- transfer-function should locate in the right half of unit circle on z plane, and near the real axis and origin.

5.5 Controllability, reachability, observability, detectability 1. Whether it is possible to steer a system from a given initial state to any other state. 2. How to determine the state of a dynamic system from observations of inputs and outputs. 5.5.1 Controllability and Reachability Consider the system (5.17) Assume that the initial state x(0) is given.

5.5 Controllability, reachability, observability, detectability The state at time n, where n is the order of the system, is given by (5.17) If Wc has rank n, then it is possible to find n equations from which the control signals can be found such that the initial state is transferred to the desired final state x(n). Remark: The solution is not unique if there is more than one input signal.

5.5 Controllability, reachability, observability, detectability The system (5.17) is controllable if it is possible to find a control sequence can be found such that the origin can be reached from any initial state in finite time. Reachability: The system (5.17) is reachable if it is possible to find a control sequence such that an arbitrary state can be reached from any initial state in finite time.

5.5 Controllability, reachability, observability, detectability Theorem 5.4 REACHABILITY The system in (5.17) is reachable if and only if the matrix Wc has rank n.

5.5 Controllability, reachability, observability, detectability Remark: 1.The matrix Wc is usually referred to as the controllability matrix because of its analogy with continuous-time system. 2.The reachability of a system is independent of the coordinates.

5.5 Controllability, reachability, observability, detectability 3. Controllability and reachability are equivalent if  is invertible. Otherwise, controllability does not imply reachability. Example: The system (5.17), where The system is reachable because has rank 2 , that is it has full rank. But when We know the system is not reachable, but is controllable, because , the origin is reached in two steps for any initial condition by using u(0) = u(1) = 0.

5.5 Controllability, reachability, observability, detectability Example: Determine the system is controllable or not. (1) The system is controllable.

5.5 Controllability, reachability, observability, detectability (2) The system is not controllable.

5.5 Controllability, reachability, observability, detectability 5.5.2 Observability and Detectability UNOBSERVABLE STATES The states x0≠0 is unobservable if there exists a finite k1≥n-1 such that y(k) = 0 and u(k)=0 for 0 ≤ k ≤ k1 when x(0) = x0. Observable: The system in (5.17) is observable if there is a finite k such that knowledge of the input u(0), …, u(k-1) and the outputs y(0), …, y(k-1) is sufficient to determine the initial state of the system. Detectability: A system is detectable if the only unobservable states are such that decay to the origin. That is, the corresponding eigenvalues are stable. The observability matrix is independent of the coordinates in the same way as in the controllability matrix.

5.5 Controllability, reachability, observability, detectability Theorem 5.5 OBSERVABILITY The system (5.17) is observable if and only if Wo has rank n. The state x(0) is unobservable if x(0) is in the null space of Wo.

5.5 Controllability, reachability, observability, detectability Example: A system with unobservable state

5.5 Controllability, reachability, observability, detectability 5.5.3 Kalman’s Decomposition The state space is partitioned into four parts : 1. reachable and observable, 2. not reachable but observable, 3. reachable and not observable, 4. neither reachable nor observable.

5.5 Controllability, reachability, observability, detectability The pulse-transfer operator of the system is given by That is, the pulse-transfer operator is only determined by the reachable and observable part of the system. Theorem 5.6 Kalman’s decomposition A linear system can be partitioned into four subsystems with the following properties (shown in Fig. 5.14):

5.5 Controllability, reachability, observability, detectability Further, the pulse-transfer function of the system is uniquely determined by the subsystem that is observable and reachable.

Summarization Stability Analysis The relationship between s-plane and z-plane -The left half-s-plane -The jw of s-plane -Periodic strips -Constant attenuation line -Constant frequency line -Constant damp ratio Stability definition -Stability -Asymptotic stability -BIBO stability Stability test a. The eigenvalues of  b. The properties of characteristic polynomials (Jury, Routh)

Summarization Steady-state analysis Steady-state response to input signal Static Position Error Constant Static Velocity Error Constant Static Acceleration Error Constant Steady-state response to disturbances Dynamic analysis pi is the negative real number pi is the positive real number pi is zero The poles are a pair conjugate complex number Controllability, reachability, observability, detectability

Homework 1. Determine the range of K when system is stable where 2. Use Jury stability criterion and Modified Routh criterion to analyze the stability of the system :

Homework 3. a. T = 0.5, k = 6 and 10, compute the ess respectively; b. T = 0.5, determine the range of the k for satisfying ess  0.05.

Homework 4. Determine steady-state error coefficients and steady-state error when the input is t and t2 respectively, where k =1. When , determine the interval of the k which should make ess  0.05.

Homework

Homework

Homework

Homework

Homework