Optimal Control. Lecture Notes. February 13, Preliminary Edition

Transcription

1 Lecture Notes February 13, 29 Preliminary Edition Department of Control Engineering, Institute of Electronic Systems Aalborg University, Fredrik Bajers Vej 7, DK-922 Aalborg Ø, Denmark

2 Page II of IV Preamble These notes have been prepared for a basic course in optimal control at the 8th term of studies in Electrionics and Information Technology, Aalborg University. The students are expected to be acquainted with classical feedback control theory. Key Words Optimal control. Riccati equation. Linear Quadratic Gaussianl, Kalman observer, Stability; Affiliation of the Authors The notes are based on original notes in danish by Ole Sørensen. The notes have been translated and adapted by Palle Andersen. Department home page: The of the last author is: pa@es.aau.dk

3 Page III of IV Contents 1 Stationary LQ-Controllers Steady State Values of L(k) and S(k) Example: LQ control of a discrete time 2nd order system Example: LQ control of a continuous time 2nd order system References and disturbances Modelling reference and disturbance Control of System with a Reference Model Example, First Order System with constant reference Example, second order system with constant reference Using a Disturbance Model in the Control Law 21 5 Stochastic LQ Control with Incomplete State Information Incomplete State Information Separation theorem Summary of LQ method for stochastic, discrete time systems with incomplete state information Duality between controller and observer Innovation model Stability of Multivariable Systems Stability for Multivariable System in general Stability for an LQ-conrolled System

4 Chapter 4 Stationary LQ-Controllers 4.1 Steady State Values of L(k) and S(k) Very often - perhaps most often - LQ control will be used with values of L() and S() determined for N. For convenience recursive expressions for controller calculations are repeated here Discrete Time: L(k) Q 2 + Γ T S(k + 1)Γ 1 Γ T S(k + 1)Φ S(k) Q 1 + Φ T S(k + 1)Φ Φ T S(k + 1)ΓQ 2 + Γ T S(k + 1)Γ 1 Γ T S(k + 1)Φ Continuous time: L(t) Q 1 2 BT S(t) ds(t) dt Q 1 + A T S(t) + S(t)A S(t)BQ 1 2 BT S(t) If we seek an optimal controller for a problem where the performance index is a sum or integral over a time interval increasing towards infinity, the controller will become independent of time L() L(k) L(k+1) L. The steady state value of L and the corresponding steady state value of S() S(k) S(k + 1) S are solutions of a set of steady state Riccati equations Discrete time: L Q 2 + Γ T SΓ 1 Γ T SΦ S Q 1 + Φ T SΦ Φ T SΓQ 2 + Γ T SΓ 1 Γ T SΦ Continuous time: 19

5 Page 2 of 36 L Q 1 2 BT S Q 1 + A T S + SA SBQ 1 2 BT S In both cases the equations for S are nonlinear and difficult to solve. The equations are referred to as algebraic Riccati equations, ARE s. If the order of the system is three or larger the equations are almost impossible to solve by hand. Fortunately it is possible to solve the equations iteratively, simply by using the same algorithm as in the case with finite time horizon in the performance sum with the one difference, that you do not stop the iteration loop for S(k) and L(k) after a fixed number (N) of repetitions, but continue the iteration until no more significant (using a suitable criterion) changes of S(k) and L(k) appear. Solutions of the stationary Riccati equations in continous or discrete time may also be found using the MatLab Control Toolbox function lqr. Having found S and L we will use the optimal controller u(k) Lx(k). The obtained value of the performance index will be J (x()) x T ()Sx() In practice there is seldom any use for the value of the performance function. However the performance function can only obtain a stationary value if the state vector x(k) goes to origo in state space. An optimal controller using the steady state value of L to calculate the control signal will therefore always force the state from an initial value x() towards origo. 4.2 Example: LQ control of a discrete time 2nd order system. We consider the servomechanism shown in the block diagram in figure 1.1. A continuous time state space model of this system is 1 ẋ(t) x(t) + u(t) Ax(t) + Bu(t) y(t).13 x(t) Cx(t) u(s) 76.7 x 2 (s) 1 x 1 (s) y(s).13 s s Figure 4.1: Block diagram for servomechanism The model is transformed to discrete time using a sampling time of 1 ms: x(k + 1) y(k).13 x(k) Hx(k).3816 x(k) u(t) Φx(k) + Γu(k)

6 Page 21 of 36 We will use the performance function I (x T (k)q 1 x(k) + u(k)q 2 u(k)) With a diagonal weighting matrix Q 1 1 As it may be seen only the state variable representing position is weighted, while there is no weight on the state variable representing speed. The weight matrix Q 2 is in this case with only one input a scalar q 2. We will now give q 2 several values and calculate the optimal controller in each case using the general recursive equations until L(k) has achieved its steady state value within 1 %. The result is q 2 : L() q 2.1 : L() q 2.1 : L() q 2.1 : L() Using these four controllers we have simulated closed loop progress of the output y(t) and the input u(t) with the initial state x() x1 () x 2 () 1 The results of these simulations are shown in figures 1.2, 1.3, 1.4 and 1.5. Comments on the simulations: q 2 In this case the input signal is not weighted in the performance function. This results in a DEAD-BEAT controller forcing the state to zero as fast as possible. From the output plot alone this seems to be the perfect controller, where the state is driven from the initial state to zero after only one sampling period. But it is important to note, that the system is simulated in discrete time and the signals are shown only at the sampling times. Consequently we know nothing about, how the real continuous time system behaves between sampling times. The plot of the input behaviour shows, that the price for the apparently perfect control is high. The input is highly oscillatory with amplitudes of approximately 5, indicating that the output probably also will oscillate with large amplitudes between sampling times. q 2.1 This gives a very fast progress of the output with a small overshoot, but still the control input is very large with a maximum of approximately 1. q 2.1

7 Page 22 of 36 This gives a reasonably fast progress for the output with a small overshoot, and now the control input is more reasonable with a maximum of 22. q 2.1 This gives a very slow progress for the output, and only a very small control signal with a maximum of Figure 4.2: Simulation with q 2 upper plot u, lower plot y x Figure 4.3: Simulation with q 2.1 upper plot u, lower plot y

8 Page 23 of x Figure 4.4: Simulation with q 2.1 upper plot u, lower plot y Figure 4.5: Simulation with q 2.1 upper plot u, lower plot y

9 Page 24 of Example: LQ control of a continuous time 2nd order system We will consider the continuous time system ÿ(t) + aẏ(t) u(t) A state space model of the system can be given with the state variables x 1 (t) y(t) and x 2 (t) ẏ(t) resulting in the equations ẋ 1 (t) ẏ(t) x 2 (t) ẋ 2 (t) ÿ(t) aẏ(t) + u(t) ax 2 (t) + u(t) These equations combine directly to a full state space model ẋ(t) We choose the performance function I 1 a y(t) 1 x(t) (x T 1 (t) So we have the below constants for our problem A 1 a, B 1 x(t) + 1 u(t) x(t) + u(t)qu(t))dt 1, Q 1 Q, Q 2 q We have chosen only to "punish" the output y(t) x 1 (t), and put no restrictions on the second state variable x 2 (t). We partition the steady state value of the matrix S(t) as below S11 S S 12 S 21 S 22 Now we have for determination of L S11 S 12 S 12 S 22 L Q 1 2 BT S 1 q 1 S11 S 12 S 12 S 22 1 q S 12 S 22 and as we seek the stationary values the derivative of S will be zero:

10 Page 25 of 36 ds(t) dt Q 1 + A T S + SA SBQ 1 2 BT S 1 S11 S + 12 S11 S a S 12 S 22 S 12 S 22 a S11 S 12 1 S11 S 1 12 S 12 S 22 1 q S 12 S 22 1 S11 as S 11 as 12 S 12 as 22 S 12 as 22 S12 1 q S 12 S 22 S 22 This may be written as three scalar equations (the two off diagonal elements give the same equation because S is symmetric): q S S 11 as 12 1 q S 12S S 12 2aS 22 1 q S2 22 From 1. we obtain : S 12 q From 3. we obtain : S 22 aq + a 2 q 2 + 2q q This give us the optimal controller u(t) Lx(t): u(t) 1 q S 12 S 22 x(t) 1 q (S 12x 1 (t) + S 22 x 2 (t)) 1 q ( qx 1 (t) + ( aq + a 2 q 2 + 2q q)x 2 (t)) Using ẋ(t) Ax(t) + Bu(t) and u(t) Lx(t) we achieve: ẋ(t) (A BL)x(t) 1 ( a 1 1 S 12 q a S 22 q 1 q S 12 S 22 )x(t) x(t) The closed loop poles are determined by the characteristic equation for the matrix A BL

11 Page 26 of 36 det(si (A BL)) s 2 + (a + S 22 q )s + S 12 q Comparing this with the characteristic equation for a second order system in standard form you will find 1 characteristic frequency : ω n q 1/4 1 damping ratio : ζ 1 + a2 q 2 2 For some extreme values of a and q we find: q q a ω n 1 q 1/4 ζ These values found with the LQ criterion justify, that classic controller design often uses 1 values of the damping ratio in the proximity of 2 For pure inertia systems, ÿ kst u(t) (i.e. a ) a damping ratio of 1 2 will be optimal independent of the choice of weighting q.

12 Chapter 5 References and disturbances Until now we have considered optimal control of a system described in state space using a quadratic performance function. The purpose was to bring the system from an initial state to origo in state space. The balance between speed and reasonable use of control signal was determined by weighting in the performance index. A practical control problem will naturally most often be to make the output y(t) track a reference signal r(t) and/or keep the output close to the reference without using too large control signals u(t). Furthermore the system will most often have disturbances influencing the output. It is obvious, that it is not possible to optimise a controller for all possible reference signals and disturbances. In classical control it is a common practice to adjust parameters in for example a PID controller to give the "best possible" step response. However this does not imply, that the response of other reference forms is "best possible". If you use optimal control the same conditions are valid. It is possible to find a controller which is optimal to certain classes of reference- and disturbance signals 5.1 Modelling reference and disturbance In this section we will consider models of signals with the intention to use such models to describe the behavior of an external signal as a reference or disturbance, call it r(t). Some simple signal type may be modelled using the general state space description, where the initial condition x() is given x(k + 1) Φ r x(k) r(k) H r x(k) A description like this without any input, just initiated by the value of x() is called an autonomous state space description see figure

13 Page 28 of 36 z -1 x(t) H r r(t) Φ r Figure 5.1: Block diagram of an autonomous state space model a) Constant (step), r(t) K K is any constant. The state space description of this is very simple: x(k + 1) x(k), r(k) x(k) x() K b) Ramp, r(t) K 1 t K 1 is an arbitrary constant. r(t) may be modelled as the solution of the difference equation (T s : sampling time): r(k) r(k 1) r(k 1) r(k 2) T s T s or: r(k) 2r(k 1) r(k 2) Here we introduce x 1 (k) r(k) and x 2 (k) r(k 1), and achieve : x 2 (k + 1) x 1 (k) and x 1 (k + 1) 2x 1 (k) x 2 (k), leading to x(k + 1) r(k) 1 x(k) x(k), x() K 1

14 Page 29 of 36 c) Acceleration, r(t) K 2 t 2 K 2 may be an arbitrary constant. r(t) may be found as a solution to the difference equation (T s : sampling time): or: r(k) r(k 1) T s r(k 1) r(k 2) T s T s r(k 1) r(k 2) T s r(k) 3r(k 1) 3r(k 2) + r(k 3) r(k 2) r(k 3) T s T s Here we introduce the state variables x 1 (k) r(k), x 2 (k) r(k 1) and x 3 (k) r(k 2), leading to: x 3 (k + 1) x 2 (k) x 2 (k + 1) x 1 (k) x 1 (k + 1) 3x 1 (k) 3x 2 (k) + x 3 (k) and achieve x(k + 1) r(k) 1 x(k) x(k), x() K 2 4K 2 Similar descriptions may be found for continuous time signals. As an example we show the model of a cosine d) Cosine, r(t) K 3 cos(at) K 3 is an arbitrary constant. r(t) is the solution of the differential equation : r(t) a 2 r(t) We introduce the state variables x 1 (t) r(t) and x 2 (t) ṙ(t), to achieve: ẋ 1 (t) x 2 (t) and ẋ 2 (t) a 2 x 1 (t), or as a full state space description ẋ(t) 1 a 2 r(t) 1 x(t) K3 x(t), x()

15 Page 3 of 36 e) System with more than one reference If a system has m outputs, the reference vector must also have m elements. We will show an example of this: A system with two outputs should also have two references. We want to model one reference as a constant of size K. The other reference is modelled as a ramp with slope K 1. The constant and the ramp may be modelled as shown below in an autonomous state space model with two outputs x r (k + 1) x r (k) x r () K K 1 The first state variable represents the constant and the second state variable represents the ramp. The third state variable is not present in the reference vector which is taken as output in 2 3 output matrix with zeros in the third column. The full reference model is thus r(k) x r (k + 1) Φ r x r (k), where Φ r r(k) H r x r (k), where H r x r (k) and x r () K K 1

16 Chapter 6 Control of System with a Reference Model System model : x s (k + 1) Φ s x s (k) + Γ s u s (k) y(k) H s x s (k) Reference model : x r (k + 1) Φ r x r (k) r(k) H r x r (k) For this system we will introduce the control error e(k) r(k) y(k) H r x r (k) H s x s (k) We want to minimize the performance function: I (e T (k)q 1e e(k) + u T (k)q 2 u(k)) + e T (N)Q Ne e(n) k The system state x s (k) is augmented by the reference state x r (k), giving the augmented state vector x(k) x T s (k) xt r (k)t. The augmented system is described by xs (k + 1) Φs xs (k) Γs + x r (k + 1) Φ r x r (k) xs (k) e(k) H s H r x r (k) We will write this with more the compact notation u(k) where x(k + 1) Φx(k) + Γu(k) e(k) Hx(k) 31

17 Page 32 of 36 Φs Φ Φ r Γs, Γ, H H s H r The performance function may be rewritten: I (e T (k)q 1e e(k) + u T (k)q 2 u(k)) + e T (N)Q Ne e(n) k ((Hx(k)) T Q 1e (Hx(k)) + u T (k)q 2 u(k)) + (Hx(N)) T Q Ne (Hx(N)) k (x T (k)h T Q 1e Hx(k) + u T (k)q 2 u(k)) + x T (N)H T Q Ne Hx(N) k (x T (k)q 1 x(k) + u T (k)q 2 u(k)) + x T (N)Q N x(n) k Note that this performance function has the structure we have used earlier. Therefore we can use the recursive expressions derived earlier to find L(k) and S(k), using the weighting matrices Q 1 H T Q 1e H Q N H T Q Ne H H T s H T r Q 1e H s H r H T s Q 1e H s H T s Q 1eH r H T r Q 1eH s H T r Q 1eH r H T s Q Ne H s H T s Q NeH r H T r Q NeH s H T r Q NeH r These values of Φ, Γ, Q 1, Q N and Q 2 results in a state feedback matrix L() representing an optimal controller: xs (k) u(k) L()x(k) L s () L r () x r (k) L s ()x s (k) L r ()x r (k) A block diagram of this control system is shown in figure 3.1 Most often the reference can be modelled as a step with the autonomous model x r (k + 1) Ix r (k) r(k) Ix r (k) This results in the structure shown in figure 3.2:

18 Page 33 of 36 x(t) z -1 -L r () u(k) Γs -1 z x (k) H s y(k) Φ r Φ s H r -L s () r(k) Figure 6.1: Block diagram of optimal control system with reference r(k) -L () r u(k) Γs -1 z x (k) y(k) H s Φ s -L s () Figure 6.2: Block diagram of optimal control system with step-reference We will now show that the feedback from the states of the original system L s () x s (k), will be equal to feedback we would get if the reference had been zero as it was earlier. For the augmented system we have the equations L(k) Q 2 + Γ T S(k + 1)Γ 1 Γ T S(k + 1)Φ S(k) Q 1 + Φ T S(k + 1)Φ ΓL(k) We will partition the matrix according to the partition of the state vector in system states and reference states S(k) S11 (k) S 12 (k) S 21 (k) S 22 (k) L(k) L s (k) L r (k) This gives us the following partition of the Riccati equation

19 Page 34 of 36 S(k) S11 (k) S 12 (k) S 21 (k) S 22 (k) H T s Q 1e H s H T s Q 1e H r Φ T + s Φ T r H T r Q 1e H s H T r Q 1e H r Φs Γs Φ r H T s Q 1e H s + Φ T s S 11 (k + 1)(Φ s Γ s L s (k)) H T r Q 1e H s + Φ T r S 21 (k + 1)(Φ s Γ s L s (k)) S11 (k + 1) S 12 (k + 1) S 21 (k + 1) S 22 (k + 1) L s (k) L r (k) H T s Q 1e H r + Φ T s S 11 (k + 1)Γ s L r (k) + Φ T s S 12 (k + 1)Φ r H T r Q 1e H r Φ T r S 21 (k + 1)Γ s L r (k) + Φ T r S 22 (k + 1)Φ r For the feedback matrix we find: L(k) L s (k) L r (k) 1 Q 2 + Γ T S11 (k + 1) S s 12 (k + 1) Γs S 21 (k + 1) S 22 (k + 1) Γ T S11 (k + 1) S s 12 (k + 1) Φs S 21 (k + 1) S 22 (k + 1) Φ r 1 Q 2 + Γ T s S 11(k + 1) Γ T s S Γs 12(k + 1) Γ T s S 11(k + 1) Γ T s S Φs 12(k + 1) Φ r Q 2 + Γ T s S 11(k + 1)Γ s 1 Γ T s S 11(k + 1)Φ s Γ T s S 12(k + 1)Φ r From this we can pull out the sub matrix corresponding to the original system: L s (k) Q 2 + Γ T s S 11(k + 1)Γ s 1 Γ T s S 11(k + 1)Φ s S 11 (k) H T s Q 1eH s + Φ T s S 11(k + 1)(Φ s Γ s L s (k)) These matrices give an optimal controller for the system alone i.e. with the reference equal to zero and with the performance index I (y T (k)q 1e y(k) + u T (k)q 2 u(k)) + y T (N)Q Ne y(n) k (x T (k)h T s Q 1e H s x(k) + u T (k)q 2 u(k)) + x T (N)H T s Q Ne H s x(n) k We have now shown the interesting result that L s (), the feedback proportionality matrix from the system state x s (k) is independent of the presence of a reference vector and can be obtained using the weight matrix

20 Page 35 of 36 Q 1 H T s Q 1yH s H T s Q 1eH s in other words L s () is independent of the reference model 6.1 Example, First Order System with constant reference System : x s (k + 1) a s x s (k) + bu(k) y(k) cx s (k) Reference : x r (k + 1) x r (k), x r () K r(k) x r (k) With e(t) r(t) y(t), we seek a minimum for the performance index I (e T (k)q 1e e(k) + u T (k)q 2 u(k)) + e T (N)q Ne e(n) k (q 1e e 2 (k) + q 2 u 2 (k)) + q Ne e 2 (N) k Because the weighting matrices are scalars we can fix q 1e 1 and consider q 2 q in the interval ;. For the augmented system the two dimensional state vector x T (k) x s (k) x r (k) is introduced giving the model: x(k + 1) The performance index is: a 1 b x(k) + e(k) c 1x(k) Hx(k) u(k) Φx(k) + Γu(k) I (x T (k)h T q 1e Hx(k) + u T (k)q 2 u(k)) + x T (N)H T q Ne Hx(N) k with the weighting matrices Q 1 H T q 1e H Q 2 q 2 q Q Ne H T q Ne H c 1 c 1 c 2 c 1 c 1 c 1 c 2 c 1 c 1 q Ne c 1 Combining this with the augmented matrices Φ, Γ and H you may find L() and S() using the usual recursive expressions.

21 Page 36 of 36 This leads to the controller u(k) L()x(k) L s () L r () xs (k) x r (k) L s ()x s (k) L r ()x r (k) corresponding to the structure shown in figure 3.3. r(k) 1 x (k) u(k) b x (k) -L r () + z - a c -L s () Figure 6.3: Structure of first order system with constant reference. We will now detail the expressions for L(k) and S(k). L(k) Q 2 + Γ T S(k + 1)Γ 1 Γ T S(k + 1)Φ q + b 2 S 11 (k + 1) 1 abs 11 (k + 1) bs 12 (k + 1) abs 11 (k + 1) bs 12 (k + 1) q + b 2 S 11 (k + 1) q + b 2 S 11 (k + 1) L s() L r () S(k) Q 1 + Φ T S(k + 1)Φ ΓL(k) c 2 c as11 (k + 1) as + 12 (k + 1) a bls (k) bl r (k) c 1 S 21 (k + 1) S 22 (k + 1) 1 c 2 + as 11 (k + 1)(a bl s (k)) c as 11 (k + 1)bL r (k) + as 12 (k + 1) c + S 21 (k + 1)(a bl s (k)) 1 S 21 (k + 1)bL r (k) + S 22 (k + 1) S11 (k + 1) S 12 (k + 1) S 21 (k + 1) S 22 (k + 1) From this calculation it may be seen, that the feedback from the system state will be determined from the coupled pair of recursive equations L s (k) abs 11(k + 1) q + b 2 S 11 (k + 1), S 11(k + 1) c 2 + as 11 (k + 1)(a bl s (k)) that is, precisely the expressions you would achieve if the reference was zero and you chose to weight the output y(t) using the weighting matrix Q 1y Q 1e.

22 Page 37 of Example, second order system with constant reference Once again we consider the discrete time system x(k + 1) y(k).13 x(k).3816 x(k) u(t) We used Q 1 1 I.e. the state x 1 (k) was punished with factor 1, corresponding to a weight on the output y(t) of 1/(,13) , while the state x 2 was weighted with a factor. Simulations showed that a suitable value of Q 2 q 2 would be,1. For this example we will augment the state description with a reference state as a third state variable: x(k + 1) e(k).13 1x(k) x(k) u(t) we will use the performance index I (e T (k)q 1e e(k) + u T (k)q 2 u(k)) Again we choose q 2.1 and q 1e 5917, i.e. the same weight on the output as we used when the reference was zero. We achieve: Q Use if the above weighting matrices lead to the optimal controller L() A simulation of the closed loop is shown in figure 3.4 showing input and output, with the initial state

23 Page 38 of 36 x T () 1 The 3rd state variable which is the reference stateis initialized to 1 introducing a unity step on the reference with the system state initially being zero. It can be seen that L() is unchanged from the earlier calculation resulting in the same closed loop dynamics. This is confirmed by the simulated step response. Note that the steady state error is zero only because the system itself has an integration. If this had not been the case you would have seen a steady state error because the controller in principle is a proportional controller Figure 6.4: Closed loop simulation of step response, upper plot u(t). Lower plot y(t)

24 Chapter 7 Using a Disturbance Model in the Control Law Model of system : x s (k + 1) Φ s x s (k) + Γ s u s (k) + Γ d d(k) y(k) H s x s (k) Model of disturbance : x d (k + 1) Φ d x d (k) d(k) H d x d (k) We want to minimize the performance function: I (y T (k)q 1y y(k) + u T (k)q 2 u(k)) + y T (N)Q Ny y(n) k The system state vector x s (k) is augmented with the disturbance state vector x d (k), thus the augmented state vector includes x(k) x T s (k) x T d (k)t. The equations describing the augmented states are: xs (k + 1) Φs Γ d H d xs (k) x d (k + 1) Φ d x d (k) xs (k) y(k) H s x d (k) Γs + u(k) In short notation this can be written with x(k + 1) Φx(k) + Γu(k) y(k) Hx(k) Φs Γ Φ d H d Φ d Γs, Γ, H H s The performance function may be rewritten: 39

25 Page 4 of 36 I (y T (k)q 1y y(k) + u T (k)q 2 u(k)) + y T (N)Q Ny y(n) k (x T (k)h T Q 1y Hx(k) + u T (k)q 2 u(k)) + x T (N)H T Q Ny Hx(N) k (x T (k)q 1 x(k) + u T (k)q 2 u(k)) + x T (N)Q N x(n) k We observe that the performance index has the same form as we have used in general derivations, therefore we can use the earlier derived general recursive expressions to determine L(k) og S(k) using Q 1 H T H T Q 1y H s Q N H T H T Q Ny H s H T Q 1y H s s Q 1y H s Q Ny H s H T s Q Ny H s This vil give us a value of L() for the optimal controller: xs (k) u(k) L()x(k) L s () L d x d (k) L s ()x s (k) L d ()x d (k) Also in this case it can be shown that L s (), the feedback from the system state vector is precisely the same as if there was no disturbance. L s () is in other words independent of the disturbance model. The structure of the controlled system is shown in figure 4.1.

26 Page 41 of 36 z 1 Φ d x (k) d Ld () Γd H d u(k) x (k) y(k) s + Γs + z 1 H s Φ s L s () Figure 7.1: Structure of controlled system with disturbance model

27 Bibliography AM89 ÅW84 Brian D. O. Anderson and John B. Moore., Linear Quadratic Methods. Prentice-Hall Information and System Sciences Series. Prentice-Hall International, Englewood Cliffs, NJ, USA, K.J. Åström and B. Wittenmark. Conputer Controlled Systems: Theory and Design. Prentice-Hall Information and System Sciences Series. Prentice-Hall Inc., Englewood Cliffs, NJ, USA, FPW97 Gene F. Franklin, J. David Powel, and Michael L. Workman. Digital Control of Dynamic Systems. Addison Wesley, California,USA, KS72 H. Kwakernak and R. Sivan. Linear Systems. Wiley Interscience, New York, USA,