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 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

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.
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.

15. 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.

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 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 > , the system is stable or not?

20. 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.

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
a b
b a
1-b a-ab
a-ab 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|<  <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 b+a>0 and 1+b-a>0
We all know that
|b|< is equivalent to |A(0)|<1
1+b+a> is equivalent to A(1)>0
1+b-a> 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

51. 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.

52. 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.

53. 5.3 Steady-state analysis
Solution:
Judge the stability of the system:

54. 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

55. 5.3 Steady-state analysis
Table 5.2 Static error constants for typical closed-loop configurations of discrete-time control systems

56. 5.3 Steady-state analysis

57. 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

58. 5.3 Steady-state analysis

59. 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.

60. 5.3 Steady-state analysis
where ĜD(z) does not involve any zeros at z = 1.

61. 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.

62. 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

63. 5.3 Steady-state analysis

64. 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.

65. 5.4 Dynamic analysis
where n > m. When there is no multiple roots of the characteristic equation, Y(z) can be rewritten as

66. 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.

67. 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.

68. 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.

69. 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

70. 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.

71. 5.4 Dynamic analysis (5) The poles are a pair conjugate complex number
where
and the similar transient response is

72. 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).

73. 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).

74. 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.

75. 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.

76. 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.

77. 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.

78. 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.

79. 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.

80. 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.

81. 5.5 Controllability, reachability, observability, detectability
Example: Determine the system is controllable or not.
(1)
The system is controllable.

82. 5.5 Controllability, reachability, observability, detectability
(2)
The system is not controllable.

83. 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.

84. 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.

85. 5.5 Controllability, reachability, observability, detectability
Example: A system with unobservable state

86. 5.5 Controllability, reachability, observability, detectability
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.

87. 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):

88. 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.

89. 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)

90. 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

91. 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 :

92. 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.

93. 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.

94. Homework

95. Homework

96. Homework

97. Homework

98. Homework