Modern Control Engineering

730 Chapter 10 / Control Systems Design in State Space Since both sides of this characteristic equation are polynomials in s,by ...

Section 10–2 / Pole Placement 731 Multiplying the preceding equations in order by a 3 ,a 2 ,a 1 , and a 0 (wherea 0 =1), respect ...

732 Chapter 10 / Control Systems Design in State Space u x A B K ++  Figure 10–2 Regulator system. Equation (10–18) is known ...

Section 10–2 / Pole Placement 733 First, we need to check the controllability matrix of the system. Since the controllability ma ...

734 Chapter 10 / Control Systems Design in State Space Thus, from which we obtain or Method 3: The third method is to use Ackerm ...

Section 10–3 / Solving Pole-placement Problems with MATLAB 735 Comments. It is important to note that matrix Kis not unique for ...

736 Chapter 10 / Control Systems Design in State Space Then we enter K = acker(A,B,J) or K = place(A,B,J) It is noted that the c ...

Section 10–3 / Solving Pole-placement Problems with MATLAB 737 EXAMPLE 10–3 Consider the same system as discussed in Example 10– ...

738 Chapter 10 / Control Systems Design in State Space Response to Initial Condition state variable x^1 −0.5 0 0.5 1 1.5 2 2.5 3 ...

Section 10–4 / Design of Servo Systems 739 x.=Ax+Bu y=Cx x 2 x 3 xn k 2 k 1 k 3 kn r u x y=x 1 +– +– Figure 10–4 Type 1 servo sy ...

740 Chapter 10 / Control Systems Design in State Space the reference input ris a step function. In this system we use the follow ...

Section 10–4 / Design of Servo Systems 741 Since the desired eigenvalues of A-BKare all in the left-half splane, the inverse of ...

742 Chapter 10 / Control Systems Design in State Space The state feedback gain matrix Kis thus K = [160 54 11] Unit-Step Respons ...

Section 10–4 / Design of Servo Systems 743 Unit-Step Response Output y 0 0.6 1.2 0.8 0.4 0.2 1 t Sec 0 0.5 1 1.5 2 2.5 3 3.5 4 4 ...

744 Chapter 10 / Control Systems Design in State Space We assume that the plant given by Equation (10–31) is completely state co ...

Section 10–4 / Design of Servo Systems 745 Define Then Equation (10–37) can be written as (10–38) where (10–39) Define a new (n+ ...

746 Chapter 10 / Control Systems Design in State Space 0 M P z u mg m  sin u x x  cos u  u Figure 10–8 Inverted-pendulum cont ...

Section 10–4 / Design of Servo Systems 747 x.=Ax+Bu y=Cx k 1  kI k 2 k 3 k 4 ru x j j y . +– +– Figure 10–9 Inverted-pendulum c ...

748 Chapter 10 / Control Systems Design in State Space where For the type 1 servo system, we have the state error equation as gi ...

Section 10–4 / Design of Servo Systems 749 MATLAB Program 10–6 A = [0 1 0 0; 20.601 0 0 0; 0 0 0 1; -0.4905 0 0 0]; B = [0;-1;0; ...