# Page 1 Errata for Digital Control Engineering M. Sami Fadali and

```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&gt;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&plusmn;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)
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.