Errata and Comments

Okuyama, Yoshifumi

doi:10.1007/978-1-4471-5667-3_8

Yoshifumi Okuyama²

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Chapter 1: Mathematical Descriptions and Models

1.
Page 5: Eq. (1.10),
2.
Page 6: Eq. (1.18),
3.
Fig. 1.5: Time sequence of the solution for Example 1.6
4.
Below Fig. 1.5: Note that the response is delayed by one step as shown in Fig. 1.5 if (i.e., y(k)=x ₁(k−1)) is applied to the computer program for (1.55). This response corresponds to the result of Exercise (7).
5.
Page 7: Eq. (1.21),
6.
Page 13: Eq. (1.35),
7.
Page 13:
8.
Page 13: For simplicity, the initial conditions are assumed to be zero (i.e., y(0)=y(1)=⋯=y(n−1)=0 and also u(0)=u(1)=⋯=u(n−1)=0).

Comments: There would be no contradiction, because y(κ) and u(κ) in (1.35) defined for κ=k+n (κ≥n). However, in a computer simulation, backward expressions (1.19), (1.21), and (1.40) should be used.
9.
Page 13: Eq. (1.36),
10.
Page 14: Eq. (1.38), The is given as
and
11.
Page 18:
12.
Page 18:
Tips: The z-transforms of the above equations are given as:
$$ \begin{bmatrix} z-1 & -1\\ 0.5 & z \end{bmatrix} \begin{bmatrix} \hat{x}_1(z)\\ \hat{x}_2(z) \end{bmatrix} = \begin{bmatrix} \hat{u}(z)\\ \hat{u}(z) \end{bmatrix} $$
Then,
$$\begin{bmatrix} \hat{x}_1(z)\\ \hat{x}_2(z) \end{bmatrix} = \frac{1}{z^2-z+0.5} \begin{bmatrix} z & 1\\ 0.5 & z-1 \end{bmatrix} \begin{bmatrix} z/(z-1)\\ z/(z-1) \end{bmatrix}. $$
Thus,
$$\hat{y}(z)=\hat{x}_1(z) =\frac{z^2+z}{(z^2)-z+0.5)(z-1)} =\frac{z^2+z}{z^3-2z^2+1.5z-0.5}. $$
13.
Page 19: The caption in Fig. 1.6, Fig. 1.6 Block diagram for Example 1.6, where a ₁=1, , and b ₁=b ₂=1
14.
Page 23: Eq. (1.68),
15.
Page 24: In Table 1.2, The fourth line in ‘Discrete time’,
16.
Page 25: Eq. (1.76),
where
$$\boldsymbol{I}= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} $$
17.
Page 26: Eqs. (1.78), (1.79), and (1.80), My first manuscript was written as follows:
$$\begin{cases} \boldsymbol{x}(k+1)=\boldsymbol{\Phi}(h)\boldsymbol{x}(k)+\boldsymbol{\Gamma}(h)u(k),\quad \boldsymbol{\Gamma}(h) =\int_0^h\boldsymbol{\Phi}(\tau)\boldsymbol{B}\mathrm{d}\tau\\ y(k)=\boldsymbol{Cx}(k). \end{cases} $$

$$\begin{cases} z\hat{\boldsymbol{x}}(z)=\boldsymbol{\Phi}\hat{\boldsymbol{x}}(z) +\boldsymbol{\Gamma}\hat{u}(z)\\ \hat{y}(z)=\boldsymbol{C}\hat{\boldsymbol{x}}(z). \end{cases} $$

$$\hat{y}(z)=\boldsymbol{C}[\boldsymbol{I}-\boldsymbol{\Phi}z^{-1}]^{-1}\boldsymbol{\Gamma}z^{-1}\hat{u}(z). $$
These expressions might be preferable to (1.78), (1.79), and (1.80).

Chapter 2: Discretized Feedback Systems

1.
Page 50: Eq. (2.7),
2.
Page 56: In Eq. (2.25),
3.
Page 69: Eq. (2.62),
4.
Page 71: In the last line ⋯and the problem is proved.

Chapter 3: Robust Stability Analysis

1.
Page 75: Eq. (3.8),
2.
Page 94: Eq. (3.46) in Theorem 3.3,
3.
Page 94: The verification of robust stability using the above modified Hall diagram (off-axis -circles) is based on the following theorem.

Chapter 4: Model Reference Feedback and PID Control

1.
Page 114: In Eqs. (4.16), (4.17), (4.20), and (4.21),
2.
Page 117: In Eq. (4.22),
3.
Page 117: In the first line under Fig. 12, ⋯ , characteristic equation is given as
4.
Page 132:
5.
Page 140: In Exercise (3), determine the characteristic equation , $\mathcal{F}(z)=0$, for Example 4.1 (A) ⋯ .
6.
Page 140: (4) Show that the approximate PID control system in Fig. 4.18 is obtained from the model-reference feedback system in Fig. 4.17, when $\mathcal{D}_{m}(\cdot)$ and $\mathcal{D}_{f}(\cdot)$, are .

Chapter 5: Multi-Loop Feedback Systems

1.
Page 153:
2.
Page 153: ⋯, where $y_{j}^{(j)}$ ⇒ ⋯, where $\tilde{y}_{j}^{(j)}$
3.
Page 153: ⋯all principal minors of matrix $\mathcal{A}$ ⇒ ⋯all principal minors of matrix
4.
Page 163:
5.
Page 163: ⋯, where $y_{j}^{(j)}$ ⇒ ⋯, where $\tilde{y}_{j}^{(j)}$
6.
Page 163: ⋯all principal minors of matrix $\mathcal{A}$ ⇒ ⋯all principal minors of matrix
7.
Page 177:

Chapter 6: Interval Polynomials and Robust Performance

1.
Page 186: Equation (6.17) should be written as follows:
2.
Page 191: Fig. 6.3 ⋯ for discrete control system ⇒ Fig. 6.3 ⋯ for discrete control
3.
Page 192:
The proof of Lemma 6.1 is given as follows:
$$\begin{aligned} & \qquad x^2+(y-\gamma)^2= \frac{[1-(1+\gamma^2)\theta^2]^2+[-\gamma+2\theta(1+\gamma^2) -\gamma(1+\gamma^2)\theta^2]^2}{ [1-2\gamma\theta+(1+\gamma^2)\theta^2]^2} \\ & {=}\,\frac{(1+\gamma^2)[1+(1+\gamma^2)\theta^4-2\theta^2 +4(1+\gamma^2)\theta^2+\gamma^2(1+\gamma^2)\theta^4 -4\gamma\theta-4\gamma(1+\gamma^2)\theta^3+2\gamma^2\theta^2]}{ [1-2\gamma\theta+(1+\gamma^2)\theta^2]^2} \\ & {=}\,\frac{(1+\gamma^2)[1+4\gamma^2\theta^2 +(1+\gamma^2)^2\theta^4 -4\gamma\theta-4\gamma(1+\gamma^2)\theta^3 +2(1+\gamma^2)\theta^2]}{ [1-2\gamma\theta+(1+\gamma^2)\theta^2]^2} =1+\gamma^2. \end{aligned}$$
Thus, Lemma 6.1 has been proved. □
4.
Page 204:

Chapter 7: Relation to Discrete Event Systems

1.
Page 228: Fig. 7.6 Petri net systems ⇒ Fig. 7.6 Petri net systems
2.
Page 232: In (7.21),
3.
Page 234: Their notation is ⋯ ⇒
4.
Page 234:
and
5.
Page 235: In (7.28) and (7.33),
6.
Page 236:
7.
Page 237: ⋯ and three events. ⇒ ⋯ and events.
8.
Page 239:
9.
Page 239: The equations for continuous-time systems should be corrected as follows:
and
furthermore,