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Errata for Digital Control Engineering M. Sami Fadali and Antonio Visioli June 2010 Page No. Area/Element Correction (also see attached pages from book) 13 Example 2.3 {uk } ∞k =0 = {1, 3, 2, 0, 4, 0, 0, 0,...} 49 Example 2.25 ωs = 60 ωb = 600 rad/s. 90 Line above last “unstable poles would.” 96 Example 4.4 (s missing in denominator) G ( s ) = 159 Problem 5.9 Transfer function 183 189 Figure 6.14 Caption Figure 6.17 Caption Third paragraph “The root locus …of 6.51 rad/s” 191 Example 6.10 Sampling period =0.05 s (ωs>25ωd) (could redo the example with T=0.025) 199 “that is almost identical to that of Example 6.8” 197 Above Example 6.13 Figure 6.27 199 Example 6.13 212 Example 6.18 In the solution, change to “Example 6.9” below last equation and in the last sentence. Change “sampling time” to “sampling period” 212 Example 6.18 Solution 212 Example 6.18 Solution Example 6.18 Solution 188 213 216 224 252 Two equations above example Example 6.23 Solution Three eqns. Above (7.42) 0.5848(−0.3549s + 1) 0.1828s 2 + 0.8627 s + 1 K Should be 1 in the denominator (instead of 10) G ( s ) = 2 s + 3s + 1 “with (dashed) and without prewarping (dash-dot) K=46.7. θa is the angle shown (not θc) After the first equation in the solution, delete the phrase “and there are no zeros at infinity.” Change to “Note that there are no poles and zeros outside the unit circle. Using Table 6.5, we choose the desired continuous-time closedloop transfer function as second order with poles at (−4.0530±j 2.3400)/4 for a settling time of 4s and with gain equal to one for zero steady-state error due to a unit step. The continuous-time transfer function is” Last sentence in page should be “By applying (6.43) we have” Change to “We observe that stable pole-zero cancellation has occurred and that an integral term is present in the controller, as expected, since the zero steadystate error condition has been addressed.” Change k to m (k used for time elsewhere) In the sentence below C(z) it should be “Thus, by applying (6.43) we have” The summations are from 0 to ∞ (not 1 to n) Copyright © 2009, Elsevier Inc. All rights reserved. Page No. Area/Element Correction (also see attached pages from book) e At = = ∞ ( At )i i =0 i! ∑ ∞ ∑ i =0 V (Λt ) W i! i { } ⎡ ∞ diag (λ1t )i , (λ2 t )i , L , (λn t )i ⎤ =V⎢ ⎥W i! ⎥⎦ ⎣⎢ i =0 i i ∞ ⎡ ⎧⎪ ∞ (λ1t ) ∞ (λ2 t ) (λn t )i ⎫⎪⎤ W , ,L, = V ⎢diag ⎨ ⎬⎥ i! ⎪⎭⎦⎥ ⎪⎩ i =0 i! i =0 i! i =0 ⎣⎢ ∑ ∑ ∑ ∑ 256 Eqn. (7.48) L⎯→ H (t ) = Ce At B +Dδ (t ) H ( s ) = C[s I n − A] B + D←⎯ 267 Middle of page ⎡0.5⎤ z ⎡1⎤ ⎡− 1⎤ z z X ( z) = ⎢ ⎥ −⎢ ⎥ − 0.5⎢ ⎥ − 0.1 − 0 1 2 z − 1 z − e z − e − 0.2 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ Refer to equation (7.51) (not 7.3) −1 zs 270 Above (7.68) Appendix I Entry 6, continuous time A.9, Page 517518 Appendix III Should be: (not k T) Should be as in the attached pages. Delete the third equation in the section. Froebenius norm: not a title. Move with first two equations to right above Note on page 518. Copyright © 2009, Elsevier Inc. All rights reserved.