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CHAPTER 4:

Generalized Predictive Control

1

S.K. Hosseini Sani

Assistant Professor

Electrical Engineering

Ferdowsi University

Mashhad, Iran1392

Generalized Predictive Control

2

4.1 Introduction

1987سالدرclarkeتوسطGPCکنترلطرح•

تحقیقوپژوهشدروصنعتدرمحبوبروش•

فرآیندتبدیلتابعمدلبرمتکی•

ناپایداروفازمینیممغیرسیستمهایبهاعمالقابلیت•

3

4.2 Formulation of Generalized Predictive Control

کارنقطهحولخطیغیرهایسیستماکثرسازیخطی•

𝐴(𝑧−1 )𝑦(𝑡) = 𝑧−𝑑𝐵(𝑧−1)𝑢(𝑡 − 1) + 𝐶(𝑧−1)𝑒(𝑡)

Controller Auto-Regressive Moving-Average (CARMA) model.

•𝑒(𝑡)صفرمیانگینباسفیدنویز

•Dسیستمتاخیر

4


𝐴(𝑧−1 )𝑦(𝑡) = 𝑧−𝑑𝐵(𝑧−1)𝑢(𝑡 − 1) + 𝐶(𝑧−1)𝑒(𝑡)

•𝐴,𝐵 ,𝐶درجهازایهاییجملهچند

𝐴 𝑧−1 = 1 + 𝑎1𝑧−1 + 𝑎2𝑧−2 + … + 𝑎𝑛𝑎𝑧−𝑛𝑎

𝐵(𝑧−1) = 𝑏0 + 𝑏1𝑧−1 + 𝑏2𝑧−2 + … + 𝑏𝑛𝑏𝑧−𝑛𝑏

𝐶(𝑧−1) = 1 + 𝑐1𝑧−1 + 𝑐2𝑧−2 + … + 𝑐𝑛𝑐𝑧

−𝑛𝑐

5


• CARMA:

𝐴(𝑧−1 )𝑦(𝑡) = 𝑧−𝑑𝐵(𝑧−1)𝑢(𝑡 − 1) + 𝐶(𝑧−1)𝑒(𝑡)

بودن(non-stationary)ایستاغیربهتوجهبا(CARIMA)مدلازاستفاده•صنعتیفرآیندهایاکثردرنویز

• CARIMA:

𝐴 𝑧−1 𝑦 𝑡

= 𝐵 𝑧−1 𝑧−𝑑 𝑢 𝑡 − 1 + 𝐶 𝑧−1𝑒 𝑡

∆(4.1)

𝑤𝑖𝑡ℎ ∆ = 1 − 𝑧−1

6


𝐴 𝑧−1 𝑦 𝑡 = 𝐵 𝑧−1 𝑧−𝑑 𝑢 𝑡 − 1 + 𝐶 𝑧−1𝑒 𝑡

∆(4.1)

𝑤𝑖𝑡ℎ ∆ = 1 − 𝑧−1

.میگیریمیکراCمحاسباتسادگیبرای•

.شوددادهبرشکهصورتیدرBوAدر𝐶−1گرفتننظردر•

رنگینویزگرفتننظردر:ترکلیحالت•

7


هزینهتابعکردنمینیمایزازناشیکنترلسیگنالاعمالGPCکنترلهدف•

𝑢(𝑡),𝑢(𝑡محاسبهنتیجهدرو(multistage)ایمرحلهچند + 1),...

𝐽 𝑁1, 𝑁2, 𝑁𝑢 =

𝑗=𝑁1

𝑁2

𝛿 𝑗 𝑦(𝑡 + 𝑗 | 𝑡) − 𝑤(𝑡 + 𝑗) 2

+

𝑗=1

𝑁𝑢

𝜆 𝑗 ∆𝑢 𝑡 + 𝑗 − 1 2

(4.2)

8


دیوفانتینمعادلهگرفتننظردربا•

1 = 𝐸𝑗 𝑧−1 𝐴 𝑧−1 + 𝑧−𝑗 𝐹𝑗 𝑧−1

𝑤𝑖𝑡ℎ

𝐴 𝑧−1 = ∆𝐴 𝑧−1 4.3

𝑗درجاتاز𝐹𝑗و𝐸𝑗ایهایجملهچندیکتایتعیین − 𝑛𝑎و1

𝑧−𝑗𝐹𝑗(𝑧باقیماندهبهرسیدنتا𝐴(𝑧−1) بر1تقسیمروشبکارگیری−1)

9


𝐸𝑗∆درزیرمعادلهضرببا• 𝑧−1 𝑧𝑗

• 𝐴 𝑧−1 𝑦 𝑡 = 𝐵 𝑧−1 𝑧−𝑑 𝑢 𝑡 − 1 + 𝐶 𝑧−1 𝑒 𝑡

∆

داریم• 𝐴 𝑧−1 𝐸𝑗 𝑧−1 𝑦 𝑡 + 𝑗

= 𝐸𝑗(𝑧−1)𝐵(𝑧−1)∆𝑢(𝑡 + 𝑗 − 𝑑 − 1) + 𝐸𝑗(𝑧

−1)𝑒(𝑡 + 𝑗) (4.4)

𝐴 اعمالبا• 𝑧−1 = ∆𝐴 𝑧−1

1 − 𝑧−𝑗𝐹𝑗 𝑧−1 𝑦 𝑡 + 𝑗

= 𝐸𝑗 𝑧−1 𝐵 𝑧−1 ∆𝑢 𝑡 + 𝑗 − 𝑑 − 1 + 𝐸𝑗 𝑧−1 𝑒 𝑡 + 𝑗

نتیجهدر

𝑦(𝑡 + 𝑗)= 𝐹𝑗(𝑧

−1)𝑦(𝑡) + 𝐸𝑗(𝑧−1)𝐵(𝑧−1)∆𝑢(𝑡 + 𝑗 − 𝑑 − 1)

10


• The best prediction of 𝑦(𝑡 + 𝑗) is therefore

𝑦(𝑡 + 𝑗 | 𝑡)= 𝐺𝑗(𝑧

−1) ∆𝑢(𝑡 + 𝑗 − 𝑑 − 1) + 𝐹𝑗(𝑧−1)𝑦(𝑡)

𝑤ℎ𝑒𝑟𝑒

𝐺𝑗(𝑧−1) = 𝐸𝑗(𝑧

−1)𝐵(𝑧−1).

• It is very simple to show that the polynomials 𝐸𝑗 and 𝐹𝑗 can

be obtained recursively (study at book).

11


• To solve the GPC problem the set of control signals

𝑢(𝑡), 𝑢(𝑡 + 1), … , 𝑢(𝑡 + 𝑁) has to be obtained in order to

optimize Expression (4.2).

𝐽 𝑁1, 𝑁2, 𝑁𝑢 =

𝑗=𝑁1

𝑁2

𝛿 𝑗 𝑦(𝑡 + 𝑗 | 𝑡) − 𝑤(𝑡 + 𝑗) 2

+

𝑗=1

𝑁𝑢

𝜆 𝑗 ∆𝑢 𝑡 + 𝑗 − 1 2

(4.2)

12


13


• Now consider the following set of 𝑗 ahead optimal

predictions:

𝑦(𝑡 + 𝑑 + 1 | 𝑡) = 𝐺𝑑+1∆ 𝑢(𝑡) + 𝐹𝑑+1𝑦(𝑡) 𝑦(𝑡 + 𝑑 + 2 | 𝑡) = 𝐺𝑑+2∆ 𝑢(𝑡) + 𝐹𝑑+2𝑦(𝑡)

… 𝑦(𝑡 + 𝑑 + 𝑁 | 𝑡) = 𝐺𝑑+𝑁∆ 𝑢(𝑡) + 𝐹𝑑+𝑁𝑦(𝑡)

• which can be written as:

𝐲 = 𝐆𝐮 + 𝐅 z−1 𝐲 t + 𝐆′ z−1 ∆𝑢 𝑡 − 1 (4.6)

14


• Where:

𝐲 =

𝑦(t+d+1|t) 𝑦(t+d+2|t)

⋮ 𝑦(t+d+N|t)

,

𝐮 =

∆𝑢 (t)∆𝑢 (t+1)

⋮∆𝑢 (t+N−1)

,

𝐆 =

𝑔0

𝑔1

⋮𝑔𝑁−1

0𝑔0

⋮𝑔𝑁−2

……⋮…

00⋮

𝑔015


𝐺′ 𝑧−1 =

𝐺𝑑+1 𝑧−1 − 𝑔0 𝑧

𝐺𝑑+2 𝑧−1 − 𝑔0 − 𝑔1𝑧−1 𝑧2

⋮𝐺𝑑+1 𝑧−1 − 𝑔0 − 𝑔1𝑧

−1 − ⋯ − 𝑔𝑁−1𝑧−(𝑁−1) 𝑧𝑁

𝐹 𝑧−1 =

𝐹𝑑+1(𝑧−1)

𝐹𝑑+2 𝑧−1

⋮𝐹𝑑+𝑁 𝑧−1

16


• Notice that the last two terms in Equation (4.6)

• 𝐲 = 𝐆𝐮 + 𝐅 z−1 𝐲 t + 𝐆′ z−1 ∆𝑢 𝑡 − 1 (4.6)

• only depend on the past and can be grouped into 𝐟 leading

to:

𝐲 = 𝐆𝐮 + 𝐟

• Notice that if all initial conditions are zero, the free

response 𝐟 is also zero.

17


• If a unit step is applied to the input at time 𝑡; that is,

∆𝑢(𝑡) = 1, ∆𝑢(𝑡 + 1) = 0, . . . , ∆𝑢(𝑡 + 𝑁 − 1) = 0

• the expected output sequence:

𝑦 𝑡 + 1 , 𝑦 𝑡 + 2 , . . . , 𝑦 𝑡 + 𝑁 𝑇

• is equal to the first column of matrix 𝐺.

• That is, the first column of matrix 𝐺 can be calculated asthe step response of the plant when a unit step is applied tothe manipulated variable.

18


• The free response term can be calculated recursively by:

𝑓𝑗+1 = 𝑧 1 − 𝐴 𝑧−1 𝐟𝑗 + 𝐵(𝑧−1)∆𝑢(𝑡 − 𝑑 + 𝑗)

• with 𝐟0 = 𝑦(𝑡) and ∆𝑢(𝑡 + 𝑗) = 0 𝑓𝑜𝑟 𝑗 ≥ 0.

19


• Expression (4.2) can be written as:

𝐉 = 𝐆𝐮 + 𝐟 − 𝐰 T (𝐆𝐮 + 𝐟 − 𝐰) + λ𝐮T𝐮 (4.7)

• Where:

𝐰 = 𝑤 𝑡 + 𝑑 + 1 𝑤 𝑡 + 𝑑 + 2 · · · 𝑤 𝑡 + 𝑑 + 𝑁 𝑇

20


• Equation (4.7) can be written as:

𝑱 =1

2𝐮T𝐇𝐮 + 𝐛T𝐮 + 𝑓0 (4.8)

• Where:

𝑯 = 𝟐(𝑮𝑇𝑮 + 𝜆𝑰)

𝐛T = 2 𝐟 − 𝐰 T𝐆

𝑓0 = 𝐟 − 𝐰 T (𝐟 − 𝐰)

21


• The minimum of 𝑱, assuming there are no constraints on the

control signals, can be found by making the gradient of 𝑱equal to zero, which leads to:

𝐮 = −𝐇−1𝐛 = 𝐆T𝐆 + λ𝐈−1

𝐆T (𝐰 − 𝐟 ) (4.9)

• Notice that the control signal that is actually sent to the

process is the first element of vector 𝐮, given by:

∆𝑢 𝑡 = 𝐊 𝐰 − 𝐟 (4.10)

• Where 𝐾 is the first row of matrix 𝐆T𝐆 + λ𝐈−1

𝐆T .22


• This has a clear meaning that can easily be derived from

Figure 4.1:

23


• if there are no future predicted errors, that is, if 𝐰 − 𝐟= 𝟎, then there is no control move, since the objective will

be fulfilled with the free evolution of the process.

24


• However, in the other case, there will be an increment in

the control action proportional (with a factor K) to that

future error.

25


• Notice that the action is taken with respect to future errors,

not past errors, as is the case in conventional feedback

controllers.

26


• Notice that only the first element of 𝐮is applied and the

procedure is repeated at the next sampling time.

• The solution to the GPC given involves the inversion (or

triangularization) of an 𝑁 × 𝑁 matrix which requires a

substantial amount of computation.

27


• In [58] the concept of control horizon is used to reduce the

amount of computation needed, assuming that the projected

control signals are going to be constant after 𝑁𝑢 < 𝑁.

• This leads to the inversion of an 𝑁𝑢 × 𝑁𝑢 matrix which

reduces the amount of computation (in particular, if 𝑁𝑢

= 1 it is reduced to a scalar computation, as in EPSAC),

but restricts the optimality of the GPC.

28

4.4 An Example

29

4.4 An Example

• In order to show how a Generalized Predictive Controllercan be implemented, a simple example is presented.

• The controller will be designed for a first-order system forthe sake of clarity.

• The following discrete equivalence can be obtained when afirst-order continuous plant is discretized:

1 + 𝑎𝑧−1 𝑦 𝑡 = 𝑏0 + 𝑏1𝑧−1 𝑢 𝑡 − 1 +𝑒 𝑡

∆

• In this example the delay 𝑑 is equal to 0 and the noisepolynomial 𝐶(𝑧−1)is considered to be equal to 1.

30

4.4 An Example

• The algorithm to obtain the control law described in the

previous section will be used on the preceding system,

obtaining numerical results for the parameter values 𝑎= −0.8, 𝑏0 = 0.4 and 𝑏1 = 0.6, the horizons being 𝑁1

= 1and 𝑁2 = 𝑁𝑢 = 3.

• As has been shown, predicted values of the process output

over the horizon are first calculated and rewritten in the

form of Equation (4.6), and then the control law is

computed using Expression (4.9).

31

4.4 An Example

• Predictor polynomials 𝐸𝑗(𝑧−1), 𝐹𝑗(𝑧

−1) from 𝑗 = 1 𝑡𝑜 𝑗

= 3 will be calculated solving the Diophantine Equation

(4.3), with:

𝐴(𝑧−1) = 𝐴(𝑧−1)(1 − 𝑧−1) = 1 − 1.8𝑧−1 + 0.8𝑧−2

• In this simple case where the horizon is not too long, the

polynomials can be directly obtained by dividing 1 by 𝐴(𝑧−1) with simple calculations.

32

4.4 An Example

• With these values and the polynomial:

𝐵(𝑧−1) = 0.4 + 0.6𝑧−1,

• the values of 𝐺𝑖(𝑧−1) are:

𝐺1 = 0.4 + 0.6𝑧−1

𝐺2 = 0.4 + 1.32𝑧−1 + 1.08𝑧−2

𝐺3 = 0.4 + 1.32𝑧−1 + 2.056𝑧−2 + 1.464𝑧−3

34

4.4 An Example

• and so the predicted outputs can be written as:

𝑦(𝑡 + 1|𝑡)

𝑦 ̂(𝑡 + 2|𝑡)𝑦 ̂(𝑡 + 3|𝑡)

=0.4 0 01.32 0.4 02.056 1.32 0.4

∆𝑢(𝑡)

∆𝑢(𝑡 + 1)∆𝑢(𝑡 + 2)

+

0.6∆𝑢 𝑡 − 1 + 1.8𝑦 𝑡 − 0.8𝑦(𝑡 − 1)

1.08∆𝑢 𝑡 − 1 + 2.44𝑦 𝑡 − 1.44𝑦(𝑡 − 1)

1.464∆𝑢 𝑡 − 1 + 2.952𝑦 𝑡 − 1.952𝑦(𝑡 − 1)

f

35

4.4 An Example

• The following step is to calculate 𝐻−1𝑏.

• If 𝜆 is taken as equal to 0.8

(𝐺𝑇𝐺 + 𝜆𝐼) − 1𝐺𝑇 =0.133 0.286 0.147

−0.154 −0.165 0.286−0.029 −0.154 0.1334

• As only ∆𝑢(𝑡) is needed for the calculations, only the first row

of the matrix is used, obtaining the following expression for the

control law:

∆𝑢(𝑡) = −0.604 ∆𝑢(𝑡 − 1) − 1.371𝑦(𝑡) + 0.805𝑦(𝑡 − 1)+ 0.133𝑤(𝑡 + 1) + 0.286𝑤(𝑡 + 2) + 0.147𝑤(𝑡 + 3)

36

4.4 An Example

• where 𝑤(𝑡 + 𝑖) is the reference trajectory which can be

considered constant and equal to the current setpoint or a

first-order approach to the desired value.

• Then the control signal is a function of this desired

reference and of past inputs and outputs and is given by:

𝑢(𝑡)= 0.396𝑢(𝑡 − 1) + 0.604𝑢(𝑡 − 2) − 1.371𝑦(𝑡)+ 0.805𝑦(𝑡 − 1) + 0.133𝑤(𝑡 + 1) + 0.286𝑤(𝑡 + 2)+ 0.147𝑤(𝑡 + 3) (4.14)

• Simulation results show the behaviour of the closed-loop

system.37

4.4 An Example

• In the first graph of Figure 4.2 the reference is constant and

equal to 1, and in the second one there is a smooth

approach to the same value, obtaining a slightly different

response, slower but without overshoot.

38

4.4 An Example

39

4.4 An Example

• The GPC control law can also be calculated without the use

of the Diophantine equation.

• To obtain the control law it is necessary to know matrix 𝐆

and the free response 𝐟 , to compute 𝐮 = 𝐆T𝐆

40

4.4 An Example

• Matrix 𝐺 is composed of the plant step response

coefficients, so that the elements of the first column of this

matrix are the first 𝑁 coefficients, that can be computed as:

𝑔𝑗 = −

𝑖=1

𝑗

𝑎𝑖𝑔𝑗−𝑖 +

𝑖=0

𝑗−1

𝑏𝑖 𝑤𝑖𝑡ℎ 𝑔𝑘 = 0 ∀𝑘 < 0

• where 𝑏𝑖 and 𝑎𝑖 are the parameters of the numerator and

denominator of the transfer function.

41

4.4 An Example

• Therefore, as the prediction horizon is 3, 𝐴 = 1− 0.8𝑧−1 𝑎𝑛𝑑 𝐵 = 0.4 + 0.6𝑧−1

𝑔0 = 𝑏0 = 0.4

𝑔1 = −𝑎1𝑔0 + 𝑏0 + 𝑏1 = 1.32

𝑔2 = −𝑎1𝑔1 − 𝑎2𝑔0 − 𝑎3𝑔0 + 𝑏0 + 𝑏1 = 2.056

42

4.4 An Example

• and the matrix is given by:

𝐺 =0.4 0 01.32 0.4 02.056 1.32 0.4

• which logically coincides with the one obtained by the

previous method.

43

4.4 An Example

• The free response can also be calculated without the use of theDiophantine equation, just noting that it is the response of theplant assuming that future controls equal the previous control𝑢(𝑡 − 1) and that the disturbance is constant.

• Thus, using the transfer function:

𝑦(𝑡) = 0.8𝑦(𝑡 − 1) + 0.4𝑢(𝑡 − 1) + 0.6𝑢(𝑡 − 2)

𝑦(𝑡 + 1) = 0.8𝑦(𝑡) + 0.4𝑢(𝑡) + 0.6𝑢(𝑡 − 1)

• If both equations are added and 𝑦(𝑡 + 1) is extracted:

𝑦(𝑡 + 1)= 1.8𝑦(𝑡) − 0.8𝑦(𝑡 − 1) + 0.4 ∆𝑢(𝑡) + 0.6 ∆𝑢(𝑡 − 1)

44

4.4 An Example

• Now, considering that in the free response only the controlincrements before instant t appear:

𝑓(𝑡 + 1) = 1.8𝑦(𝑡) − 0.8𝑦(𝑡 − 1) + 0.6 ∆𝑢(𝑡 − 1)

𝑓(𝑡 + 2) = 1.8𝑓(𝑡 + 1) − 0.8𝑦(𝑡)= 2.44𝑦(𝑡) − 1.44𝑦(𝑡 − 1) + 1.08 ∆𝑢(𝑡 − 1)

𝑓(𝑡 + 3) = 1.8𝑓(𝑡 + 2) − 0.8𝑓(𝑡 + 1)= 2.952𝑦(𝑡) − 1.952𝑦(𝑡 − 1) + 1.464 ∆𝑢(𝑡 − 1)

• Vector f obtained this way is the same as the one previouslyobtained, so the control law is the one given by Equation(4.14).

45

4.3 The Coloured Noise Case

46


𝐶(𝑧−1)نبودن1صورتدرخروجیبینیپیش•

𝐴 𝑧−1 𝑦 𝑡 = 𝐵 𝑧−1 𝑧−𝑑 𝑢 𝑡 − 1 +𝐶 𝑧−1 𝑒 𝑡

∆ 𝑧−1

:جدیددیوفانتینمعادله•

𝐶 𝑧−1 = 𝐸𝑗 𝑧−1 𝐴 𝑧−1 + 𝑧−𝑗𝐹𝑗 𝑧−1

:رتبه چندجمله ای خارج قسمت و باقیمانده

𝛿(𝐸𝑗(𝑧−1)) = 𝑗 − 1

𝛿(𝐹𝑗(𝑧−1)) = 𝛿( 𝐴(𝑧−1)) − 1.

𝐸𝑗∆ضرب معادله خروجی در • 𝑧−1 𝑧𝑗و استفاده از معادله دیوفانتین:

𝐶(𝑧−1)(𝑦(𝑡 + 𝑗) − 𝐸𝑗(𝑧−1)𝑒(𝑡 + 𝑗))

= 𝐸𝑗(𝑧−1)𝐵(𝑧−1)∆𝑢(𝑡 + 𝑗 − 1) + 𝐹𝑗(𝑧

−1)𝑦(𝑡)

47


امید ریاضی صفرنویز آینده غیرقابل تخمین

:امید ریاضی قسمت چپ معادله پیش بینی𝐸[𝐶(𝑧−1)(𝑦(𝑡 + 𝑗) − 𝐸𝑗(𝑧

−1)𝑒(𝑡 + 𝑗))] = 𝐶(𝑧−1) 𝑦(𝑡 + 𝑗|𝑡)

:پیش بینی خروجی•

𝐶 𝑧−1 𝑦 𝑡 + 𝑗 𝑡= 𝐸𝑗 𝑧−1 𝐵 𝑧−1 ∆𝑢 𝑡 + 𝑗 − 1 + 𝐹𝑗 𝑧−1 𝑦 𝑡

𝐶برایدیوفانتینمعادله• 𝑧−1قسمتخارجو𝑀𝑗

1 = 𝐶 𝑧−1 𝑀𝑗 𝑧−1 + 𝑧−𝑘𝑁𝑗 𝑧−1

:رتبه چندجمله ای خارج قسمت و باقیمانده

𝛿(𝑀𝑗(𝑧−1)) = 𝑗 − 1

𝛿(𝑁𝑗(𝑧−1)) = 𝛿(𝐶(𝑧−1)) − 1.

48


𝑀𝑗(𝑧دربینیپیشمعادلهضرب•:دیوفانتینمعادلهازاستفادهو(1−

𝑦 𝑡 + 𝑗 𝑡= 𝑀𝑗𝐸𝑗 𝑧−1 𝐵 𝑧−1 ∆𝑢 𝑡 + 𝑗 − 1 + 𝑀𝑗 𝑧−1 𝐹𝑗(𝑧

−1)𝑦(𝑡)

+ 𝑁𝑗(𝑧−1)𝑦(𝑡)

:بازنویسی•

𝑦(𝑡 + 𝑗|𝑡)= 𝐺(𝑧−1)∆𝑢(𝑡 + 𝑗 − 1) + 𝐺𝑝(𝑧−1)∆𝑢(𝑡 + 𝑗 − 1)

+ (𝑀𝑗 𝑧−1 𝐹𝑗(𝑧−1) + 𝑁𝑗(𝑧

−1))𝑦(𝑡)

:𝐺(𝑧−1)رتبه

• 𝛿(𝐺(𝑧−1)) < 𝑗 →𝐺(𝑧−1)∆𝑢(𝑡:آیندهخروجیدرآیندهورودیهایاثر + 𝑗 − 1)𝐺𝑝(𝑧−1)∆𝑢(𝑡:آیندهخروجیدرگذشتهورودیهایاثر + 𝑗 − 1)

49هزینهتابعوخروجیبینیپیشاساسبرآیندهبهینهورودیمحاسبه•


وورودیهایسیگنالدرنظرگرفتن:بینیپیشمحاسبهدیگرروش•

:فیلترشدهخروجی

𝑦𝑓 𝑡 =1

𝐶 𝑧−1𝑦 𝑡 ,

𝑢𝑓(𝑡) =1

𝐶 𝑧−1𝑢(𝑡)

مدل نهایی مانند نویز سفید:

𝐴 𝑧−1 𝑦𝑓 𝑡 = 𝐵 𝑧−1 𝑢𝑓 𝑡 +𝑒 𝑡

∆

:استفاده از روش نویز سفید•

𝑦𝑓 :سیگنال پیش بینی (𝑡 + 𝑗|𝑡)

𝑦𝑓 سیگنال خروجی اصلی (𝑡 + 𝑗|𝑡) فیلترشده𝐶(𝑧−1)است50

chapter 4: generalized predictive controlfumblog.um.ac.ir/gallery/839/mpc_lecture_92_4_1.pdf ·...

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