1 Errata of ”Process Control - Theory and applications” Do not hesitate to contact the author for any comment or information:
[email protected]
Page 43: δ(t) = 0 ∀t 6= 0 δ(0) = +∞ R +∞ δ(t)dt = 1 −∞ Z
+∞
f (t)δ(t − t0 )dt = f (t0 ) −∞
Page 215: the black circles were lightly displaced as shows the corrected figure . 2
1/GM
-
1.5
1
ωφ @ −1@@ Rr
Im
0.5
0
6 -0.5
r
PM
ωg
-1
-1.5
-2 -2
-1.5
-1
-0.5
0 Re
0.5
1
1.5
Equation (8.48) ∂Yi ∂Uj U =0,k6=j k λij = ∂Yi ∂Uj Y =0,l6=i
l
Equation (13.78) S(q)
= q −h 1
s R(q) = K q − 1 + T τI T (q) = R(q)
i
Equation (14.158) d(tc +
p p 2|A| − t2c ) = 0 =⇒ tc = |A| dtc
2
2 Equation (14.295) S = F T [S − S G (GT S G + R)−1 GT S] F + M Q M Equation (15.30) yˆ(t + 1) = 1.97y(t) − 0.97y(t − 1) + 1.2∆u(t) + 0.58∆u(t − 1) yˆ(t + 2) = 2.9109y(t) − 1.9109y(t − 1) +1.2∆u(t + 1) + 2.944∆u(t) + 1.1426∆u(t − 1) yˆ(t + 3) = 3.8236y(t) − 2.8236y(t − 1) +1.2∆u(t + 2) + 2.944∆u(t + 1) + 4.6357∆u(t) + 1.6883∆u(t − 1)
Equation (15.32) yˆ(t + 1|t) = 0.58∆u(t − 1) + 1.97y(t) − 0.97y(t − 1) yˆ(t + 2|t) = 1.1426∆u(t − 1) + 2.9109y(t) − 1.9109y(t − 1) yˆ(t + 3|t) = 1.6883∆u(t − 1) + 3.8236y(t) − 2.8236y(t − 1) Equation (15.35) ∆u(t)
= 0.5181[r(t + 1) − yˆ(t + 1|t)] + 0.1823[r(t + 2) − yˆ(t + 2|t)] −0.0435[r(t + 3) − yˆ(t + 3|t)] = −0.4354∆u(t − 1) − 1.385y(t) + 0.7281y(t − 1) +0.5181r(t + 1) + 0.1823r(t + 2) − 0.0435r(t + 3)
Equation (17.147) u=
v − Lrf h(x) − β1 Lr−1 h(x) − . . . − βr−1 Lf h(x) − βr h(x) f Lg Lr−1 h(x) f
