Control Systems Design in State Space

a2q    aqq V = A11 = Cf1  f2  p  fq  vq + 1  p  vn D a11 a21    aq1 0    0 p p p p p a1q a2q    aqq 0    0 a1q + 1 a2q + 1    aqq + 1 aq + 1q + 1    anq + 1 p p p p p a1n a2n    aqn aq + 1n    ann CAf1  Af2  p  Afq  Avq + 1  p  Avn D Thus, Hence, Next, referring to Equation (10–138), we have (10–139) Referring to Equation (10–136), notice that vector B can be written in terms of q linearly independent column vectors f1 , f2 , p , fq. Thus, we have Consequently, Equation (10–139) may be written as follows: Thus, where A–10–2. Consider a completely state controllable system Define the controllability matrix as M: M = CB  AB  p  An - 1 B D x# = Ax + Bu B11 = F b11b21   bq1 V Bˆ = cB11 0 d b11 f1 + b21 f2 + p + bq1 fq = C f1  f2  p  fq  vq + 1  p  vn D b11 b21    bq1 0    0 B = b11 f1 + b21 f2 + p + bq1 fq B = Cf1  f2  p  fq  vq + 1  p  vn D Bˆ P-1 AP = Aˆ = cA11 0 A12 A22 d AP = P cA11 0 A12 A22 d 820 Chapter 10 / Control Systems Design in State Space Show that where a1, a2, p , an are the coefficients of the characteristic polynomial Solution. Let us consider the case where n=3. We shall show that (10–140) The left-hand side of Equation (10–140) is The right-hand side of Equation (10–140) is (10–141) The Cayley–Hamilton theorem states that matrix A satisfies its own characteristic equation or, in the case of n=3, (10–142) Using Equation (10–142), the third column of the right-hand side of Equation (10–141) becomes Thus, Equation (10–141) becomes Hence, the left-hand side and the right-hand side of Equation (10–140) are the same. We have thus shown that Equation (10–140) is true. Consequently, The preceding derivation can be easily extended to the general case of any positive integer n. A–10–3. Consider a completely state controllable system Define M = CB  AB  p  An - 1 B D x# = Ax + Bu M-1 AM = C01 0 0 0 1 -a3 -a2 -a1 S CB  AB  A2 B D C01 0 0 0 1 -a3 -a2 -a1 S = CAB  A2 B  A3 B D-a3 B - a2 AB - a1 A2 B = A-a3 I - a2 A - a1 A2BB = A3 B A3 + a1 A2 + a2 A + a3 I = 0 CB  AB  A2 B D C01 0 0 0 1 -a3 -a2 -a1 S = CAB  A2 B  -a3 B - a2 AB - a1 A2 B DAM = A CB  AB  A2 B D = CAB  A2 B  A3 B D AM = MC01 0 0 0 1 -a3 -a2 -a1 S∑s I - A∑ = sn + a1 sn - 1 + p + an - 1 s + an M-1 AM = G 010   0 0 0 1    0 p p p p 0 0 0    1 -an -an - 1 -an - 2    -a1 W Example Problems and Solutions 821 822 Chapter 10 / Control Systems Design in State Space and where the ai’s are coefficients of the characteristic polynomial Define also Show that Solution. Let us consider the case where n=3. We shall show that (10–143) Referring to Problem A–10–2, we have Hence, Equation (10–143) can be rewritten as Therefore, we need to show that (10–144) The left-hand side of Equation (10–144) isC01 0 0 0 1 -a3 -a2 -a1 S Ca2a1 1 a1 1 0 1 0 0 S = C-a30 0 0 a1 1 0 1 0 S C01 0 0 0 1 -a3 -a2 -a1 S W = WC 00 -a3 1 0 -a2 0 1 -a1 S W-1C01 0 0 0 1 -a3 -a2 -a1 S W = C 00 -a3 1 0 -a2 0 1 -a1 S M-1 AM = C01 0 0 0 1 -a3 -a2 -a1 S T-1 AT = (MW)-1 A(MW) = W-1(M-1 AM) W = C 00 -a3 1 0 -a2 0 1 -a1 S T-1 AT = G 00  0 -an 1 0    0 -an - 1 0 1    0 -an - 2 p p p p 0 0    1 -a1 W , T-1 B = G 00  0 1 W T = MW ∑s I - A∑ = sn + a1 sn - 1 + p + an - 1 s + an W = Gan - 1an - 2  a1 1 an - 2 an - 3    1 0 p p p p a1 1    0 0 1 0    0 0 W Example Problems and Solutions 823 The right-hand side of Equation (10–144) is Clearly, Equation (10–144) holds true. Thus, we have shown that Next, we shall show that (10–145) Note that Equation (10–145) can be written as Noting that we have The derivation shown here can be easily extended to the general case of any positive integer n. A–10–4. Consider the state equation where The rank of the controllability matrix M, is 2.Thus, the system is completely state controllable.Transform the given state equation into the controllable canonical form. Solution. Since = s2 + 2s + 1 = s2 + a1 s + a2 ∑s I - A∑ = 2 s - 1 4 -1 s + 3 2 = (s - 1)(s + 3) + 4 M = CB  AB D = B0 2 2 -6 RA = B 1-4 1-3R , B = B02R x# = Ax + Bu T-1 B = C00 1 S TC00 1 S = CB  AB  A2 B D Ca2a1 1 a1 1 0 1 0 0 S C00 1 S = CB  AB  A2 B D C10 0 S = B B = TC00 1 S = MWC00 1 S T-1 B = C00 1 S T-1 AT = C 00 -a3 1 0 -a2 0 1 -a1 S Ca2a1 1 a1 1 0 1 0 0 S C 00 -a3 1 0 -a2 0 1 -a1 S = C-a30 0 0 a1 1 0 1 0 S 824 Chapter 10 / Control Systems Design in State Space we have Define where Then and Define Then the state equation becomes Since and we have which is in the controllable canonical form. A–10–5. Consider a system defined by where A = B 0 -2 1 -3 R , B = B0 2 R , C = [1 0] y = Cx x# = Ax + Bu B xˆ# 1 xˆ # 2 R = B 0 -1 1 -2 R B xˆ1 xˆ2 R + B0 1 RuT-1 B = B0.50.5 00.5R B02R = B01R T-1AT = B0.5 0.5 0 0.5 R B 1 -4 1 -3 R B 2 -2 0 2 R = B 0 -1 1 -2 Rxˆ# = T-1 ATxˆ + T-1Bu x = Txˆ T-1 = B0.5 0.5 0 0.5 RT = B 0 2 2 -6 R B2 1 1 0 R = B 2 -2 0 2 RM = B02 2-6R , W = B21 10R T = MW a1 = 2, a2 = 1 Example Problems and Solutions 825 The characteristic equation of the system is The eigenvalues of matrix A are –1 and –2. It is desired to have eigenvalues at –3 and –5 by using a state-feedback control u=–Kx. Determine the necessary feedback gain matrix K and the control signal u. Solution. The given system is completely state controllable, since the rank of is 2. Hence, arbitrary pole placement is possible. Since the characteristic equation of the original system is we have The desired characteristic equation is Hence, It is important to point out that the original state equation is not in the controllable canonical form, because matrix B is not Hence, the transformation matrix T must be determined. Hence, Referring to Equation (10–13), the necessary feedback gain matrix is given by Thus, the control signal u becomes u = -Kx = -[6.5 2.5]Bx1 x2 R = C15 - 2  8 - 3 D B0.50 00.5R = [6.5 2.5] K = Ca2 - a2  a1 - a1 D T-1 T-1 = B0.5 0 0 0.5 RT = MW = CB  AB D B a1 1 1 0 R = B0 2 2 -6 R B3 1 1 0 R = B2 0 0 2 RB 0 1 Ra1 = 8, a2 = 15 (s + 3)(s + 5) = s2 + 8s + 15 = s2 + a1 s + a2 = 0 a1 = 3, a2 = 2 s2 + 3s + 2 = s2 + a1 s + a2 = 0 M = CB  AB D = B0 2 2 -6 R ∑s I - A∑ = 2 s 2 -1 s + 3 2 = s2 + 3s + 2 = (s + 1)(s + 2) = 0 826 Chapter 10 / Control Systems Design in State Space A–10–6. A regulator system has a plant Define state variables as By use of the state-feedback control u=–Kx, it is desired to place the closed-loop poles at Obtain the necessary state-feedback gain matrix K with MATLAB. Solution. The state-space equations for the system become Hence, (Note that, for the pole placement, matrices C and D do not affect the state-feedback gain matrix K.) Two MATLAB programs for obtaining state-feedback gain matrix K are given in MATLAB Programs 10–24 and 10–25. C = [1 0 0], D = [0] A = C 00 -6 1 0 -11 0 1 -6 S , B = C 00 10 S y = [1 0 0]Cx1x2 x3 S + 0u C x # 1 x # 2 x # 3 S = C 00 -6 1 0 -11 0 1 -6 S Cx1x2 x3 S + C 00 10 S u s = -2 + j213 , s = -2 - j213 , s = -10 x3 = x # 2 x2 = x # 1 x1 = y Y(s) U(s) = 10 (s + 1)(s + 2)(s + 3) MATLAB Program 10–24 A = [0 1 0;0 0 1;-6 -11 -6]; B = [0;0;10]; J = [-2+j*2*sqrt(3) -2-j*2*sqrt(3) -10]; K = acker(A,B,J) K = 15.4000 4.5000 0.8000 Example Problems and Solutions 827 A–10–7. Consider a completely observable system Define the observability matrix as N: Show that (10–146) where a1, a2, p , an are the coefficients of the characteristic polynomial Solution. Let us consider the case where n=3. Then Equation (10–146) can be written as (10–147) Equation (10–147) may be rewritten as (10–148) We shall show that Equation (10–148) holds true. The left-hand side of Equation (10–148) is (10–149)N*A = C CCA CA2 S A = C CACA2 CA3 S N* A = C 00 -a3 1 0 -a2 0 1 -a1 S N* N*A(N*)-1 = C 00 -a3 1 0 -a2 0 1 -a1 S∑s I - A∑ = sn + a1 sn - 1 + p + an - 1 s + an N* A(N*)-1 = G 00  0 -an 1 0    0 an - 1 0 1    0 -an - 2 p p p p 0 0    1 -a1 W N = CC* A*C*  p  (A*)n - 1 C* D y = Cx x# = Ax MATLAB Program 10–25 A = [0 1 0;0 0 1; -6 -11 -6]; B = [0;0;10]; J = [-2+j*2*sqrt(3) -2-J*2*Sqrt(3) -10]; K = place(A,B,J) place: ndigits= 15 K = 15.4000 4.5000 0.8000 828 Chapter 10 / Control Systems Design in State Space The right-hand side of Equation (10–148) is (10–150) The Cayley–Hamilton theorem states that matrix A satisfies its own characteristic equation, or Hence, Thus, the right-hand side of Equation (10–150) becomes the same as the right-hand side of Equation (10–149). Consequently, which is Equation (10–148). This last equation can be modified to The derivation presented here can be extended to the general case of any positive integer n. A–10–8. Consider a completely observable system defined by (10–151) (10–152) Define and where the a’s are coefficients of the characteristic polynomial Define also Q = (WN*)-1 ∑s I - A∑ = sn + a1 sn - 1 + p + an - 1 s + an W = Gan - 1an - 2  a1 1 an - 2 an - 3    1 0 p p p p a1 1    0 0 1 0    0 0 W N = CC*  A* C*  p  (A*)n - 1 C* D y = Cx + Du x# = Ax + Bu N*A(N*)-1 = C 00 -a3 1 0 -a2 0 1 -a1 S N*A = C 00 -a3 1 0 -a2 0 1 -a1 S N* -a1 CA2 - a2 CA - a3 C = CA3 A3 + a1 A2 + a2 A + a3 I = 0 = C CACA2 -a3 C - a2 CA - a1 CA2 S C 00-a3 10-a2 01-a1S N* = C 00-a3 10-a2 01-a1S C CCACA2S Example Problems and Solutions 829 Show that where the bk’s (k=0, 1, 2, p , n) are those coefficients appearing in the numerator of the transfer function when C(sI-A)–1B+D is written as follows: where D=b0. Solution. Let us consider the case where n=3. We shall show that (10–153) Note that, by referring to Problem A–10–7, we have Hence, we need to show that or (10–154)WC 00 -a3 1 0 -a2 0 1 -a1 S = C01 0 0 0 1 -a3 -a2 -a1 SW WC 00 -a3 1 0 -a2 0 1 -a1 S W-1 = C01 0 0 0 1 -a3 -a2 -a1 S (WN*) A(WN*)-1 = W CN* A(N*)-1 D W-1 = WC 00 -a3 1 0 -a2 0 1 -a1 S W-1 Q-1 AQ = (WN*) A(WN*)-1 = C01 0 0 0 1 -a3 -a2 -a1 S C(s I - A)-1 B + D = b0 s n + b1 sn - 1 + p + bn - 1 s + bn sn + a1 sn - 1 + p + an - 1 s + an Q-1 B = F bn - an b0bn - 1 - an - 1 b0   b1 - a1 b0 VCQ = [0 0 p 0 1] Q-1 AQ = G 010   0 0 0 1    0 p p p p 0 0 0    1 -an -an - 1 -an - 2    -a1 W 830 Chapter 10 / Control Systems Design in State Space The left-hand side of Equation (10–154) is The right-hand side of Equation (10–154) is Thus, we see that Equation (10–154) holds true. Hence, we have proved Equation (10–153). Next we shall show that or Notice that Hence, we have shown that Next define Then Equation (10–151) becomes (10–155) and Equation (10–152) becomes (10–156) Referring to Equation (10–153), Equation (10–155) becomesC xˆ# 1xˆ# 2 xˆ # 3 S = C01 0 0 0 1 -a3 -a2 -a1 S C xˆ1xˆ2 xˆ3 S + Cg3g2 g1 S uy = CQxˆ + Du xˆ # = Q-1 AQxˆ + Q-1 Bu x = Qxˆ [0 0 1] = C(WN* )-1 = CQ = [1 0 0]C CCA CA2 S = C[0 0 1](WN* ) = [0 0 1]Ca2a11 a110 100S C CCACA2S C(WN*)-1 = [0 0 1] CQ = [0 0 1] = C-a30 0 0 a1 1 0 1 0 SC 0 1 0 0 0 1 -a3 -a2 -a1 S W = C01 0 0 0 1 -a3 -a2 -a1 S Ca2a1 1 a1 1 0 1 0 0 S = C-a30 0 0 a1 1 0 1 0 S W C 0 0 -a3 1 0 -a2 0 1 -a1 S = Ca2a1 1 a1 1 0 1 0 0 S C 00 -a3 1 0 -a2 0 1 -a1 S Example Problems and Solutions 831 where The transfer function G(s) for the system defined by Equations (10–155) and (10–156) is Noting that we have Note that D=b0. Since we have Hence, Thus, we have shown that Note that what we have derived here can be easily extended to the case when n is any positive integer. A–10–9. Consider a system defined by y = Cx x# = Ax + Bu Q-1 B = Cg3g2 g1 S = Cb3 - a3 b0b2 - a2 b0 b1 - a1 b0 Sg1 = b1 - a1 b0 , g2 = b2 - a2 b0 , g3 = b3 - a3 b0 = b0 s 3 + b1 s2 + b2 s + b3 s3 + a1 s2 + a2 s + a3 = b0 s 3 + Ag1 + a1 b0Bs2 + Ag2 + a2 b0Bs + g3 + a3 b0 s3 + a1 s2 + a2 s + a3 = g1 s 2 + g2 s + g3 s3 + a1 s2 + a2 s + a3 + b0 G(s) = 1 s3 + a1 s2 + a2 s + a3 C1 s s2 D Cg3g2 g1 S + D C s-1 0 0 s -1 a3 a2 s + a1 S -1 = 1 s3 + a1 s2 + a2 s + a3 C s2 + a1 s + a2s + a1 1 -a3 s2 + a1 s s -a3 s -a2 s - a3 s2 S G(s) = [0 0 1]C s-1 0 0 s -1 a3 a2 s + a1 S -1Cg3g2 g1 S + DCQ = [0 0 1] G(s) = CQAs I - Q-1 AQB-1 Q-1 B + D Cg3g2 g1 S = Q-1 B 832 Chapter 10 / Control Systems Design in State Space where The rank of the observability matrix N, is 2. Hence, the system is completely observable. Transform the system equations into the ob- servable canonical form. Solution. Since we have Define where Then and Define Then the state equation becomes or (10–157) The output equation becomes y = CQxˆ = B0 1 -1 -2 R B xˆ1 xˆ2 R + B0 2 R uB xˆ# 1xˆ# 2R = B-11 01R B 1-4 1-3R B-11 01R B xˆ1xˆ2R + B-11 01R B02R u xˆ # = Q-1 AQxˆ + Q-1 Bu x = Qxˆ Q-1 = B-1 1 0 1 RQ = b B 2 1 1 0 R B 1 -3 1 -2 R r -1 = B-1 1 0 1 R -1 = B-1 1 0 1 RN = B 1 1 -3 -2 R , W = Ba1 1 1 0 R = B2 1 1 0 RQ = (WN*)-1 a1 = 2, a2 = 1 ∑s I - A∑ = s2 + 2s + 1 = s2 + a1 s + a2 N = CC*  A* C* D = B1 1 -3 -2 RA = B 1-4 1-3R , B = B02R , C = [1 1] Example Problems and Solutions 833 or (10–158) Equations (10–157) and (10–158) are in the observable canonical form. A–10–10. For the system defined by consider the problem of designing a state observer such that the desired eigenvalues for the observer gain matrix are m1 ,m2 , p ,mn . Show that the observer gain matrix given by Equation (10–61), rewritten as (10–159) can be obtained from Equation (10–13) by considering the dual problem. That is, the matrix Ke can be determined by considering the pole-placement problem for the dual system, obtaining the state-feedback gain matrix K, and taking its conjugate transpose, or Ke=K*. Solution. The dual of the given system is (10–160) Using the state-feedback control Equation (10–160) becomes Equation (10–13), which is rewritten here, is (10–161) where For the original system, the observability matrix is Hence, matrix T can also be written as Since we have and (T*)-1 = (WN*)-1 T* = W* N* = WN* W = W*, T = NW CC*  A* C*  p  (A*)n - 1 C* D = N T = MW = CC*  A* C*  p  (A*)n - 1 C* D W K = Can - an  an - 1 - an - 1  p  a2 - a2  a1 - a1 D T-1 z# = (A* - C* K) z v = -Kz n = B* z z# = A* z + C* v Ke = (WN*)-1F an - anan - 1 - an - 1   a1 - a1 V y = Cx x# = Ax + Bu y = [1 1]B-1 1 0 1 R B xˆ1 xˆ2 R = [0 1]B xˆ1 xˆ2 R 834 Chapter 10 / Control Systems Design in State Space Taking the conjugate transpose of both sides of Equation (10–146), we have Since Ke=K*, this last equation is the same as Equation (10–159). Thus, we obtained Equation (10–159) by considering the dual problem. A–10–11. Consider an observed-state feedback control system with a minimum-order observer described by the following equations: (10–162) (10–163) where Axa is the state variable that can be directly measured, and corresponds to the observed state variables. B Show that the closed-loop poles of the system comprise the closed-loop poles due to pole placement Cthe eigenvalues of matrix (A-BK)] and the closed-loop poles due to the minimum- order observer [the eigenvalues of matrix Solution. The error equation for the minimum-order observer may be derived as given by Equation (10–94), rewritten thus: (10–164) where From Equations (10–162) and (10–163), we obtain (10–165) Combining Equations (10–164) and (10–165) and writing we obtain (10–166) Equation (10–166) describes the dynamics of the observed-state feedback control system with a minimum-order observer. The characteristic equation for this system is or @s I - A + BK @ @s I - Abb + Ke Aab @ = 0 2 s I - A + BK 0 -BKb s I - Abb + Ke Aab 2 = 0 Bx# e# R = BA - BK 0 BKb Abb - Ke Aab R Bx e RK = CKa  Kb D = Ax - BK ex - c0 e d f = (A - BK) x + BK c0 e d x# = Ax - BK x = Ax - BK c xa x b d = Ax - BK c xa xb - e d e = xb - x b e# = AAbb - Ke AabB e AAbb - Ke AabB D x b x = cxa xb d , x = c xa xb d u = -Kx y = Cx x# = Ax + Bu K* = AT-1B*F an - anan - 1 - an - 1   a1 - a1 V = (T*)-1F an - anan - 1 - an - 1   a1 - a1 V = (WN*)-1F an - anan - 1 - an - 1   a1 - a1 V Example Problems and Solutions 835 The closed-loop poles of the observed-state feedback control system with a minimum-order observer consist of the closed-loop poles due to pole placement and the closed-loop poles due to the minimum-order observer. (Therefore, the pole-placement design and the design of the minimum-order observer are independent of each other.) A–10–12. Consider a completely state controllable system defined by (10–167) where Suppose that the rank of the following matrix is n+1. Show that the system defined by (10–168) where is completely state controllable. Solution. Define Because the system given by Equation (10–167) is completely state controllable, the rank of matrix M is n. Then the rank of is n+1. Consider the following equation: (10–169) Since matrix is of rank n+1, the left-hand side of Equation (10–169) is of rank n+1. Therefore, the right-hand side of Equation (10–169) is also of rank n+1. Since = CAˆ Bˆ  Aˆ2 Bˆ  p  Aˆn Bˆ  Bˆ D= B AB-CB  A2 B-CAB  p  p  An B-CAn - 1 B  B0 RB AM -CM B 0 R = B A CB  AB  p  An - 1 B D -C CB  AB  p  An - 1 B D B0 R B A -C B 0 RB A-C B0R BM0 01R = B AM-CM B0R BM 0 0 1 RM = CB  AB  p  An - 1 B D Aˆ = B A -C 0 0 R , Bˆ = BB 0 R , ue = u(t) - u(q)e# = Aˆe + Bˆue B A -C B 0 R(n + 1) * (n + 1)C = 1 * n constant matrixB = n * 1 constant matrix A = n * n constant matrix y = output signal (scalar) u = control signal (scalar) x = state vector (n-vector) y = Cx x# = Ax + Bu 836 Chapter 10 / Control Systems Design in State Space we find that the rank of is n+1. Thus, the system defined by Equation (10–168) is completely state controllable. A–10–13. Consider the system shown in Figure 10–49. Using the pole-placement-with-observer approach, design a regulator system such that the system will maintain the zero position Ay1=0 and y2=0 B in the presence of disturbances. Choose the desired closed-loop poles for the pole-placement part to be and the desired poles for the minimum-order observer to be First, determine the state feedback gain matrix K and observer gain matrix Ke . Then, obtain the response of the system to an arbitrary initial condition—for example, where e1 and e2 are defined by Assume that m1=1 kg, m2=2 kg, k=36 Nm, and b=0.6 N-sm. Solution. The equations for the system are By substituting the given numerical values for m1, m2, k, and b and simplifying, we obtain Let us choose the state variables as follows: x4 = y # 2 x3 = y # 1 x2 = y2 x1 = y1 y $ 2 = 18y1 - 18y2 + 0.3y # 1 - 0.3y # 2 y $ 1 = -36y1 + 36y2 - 0.6y # 1 + 0.6y # 2 + u m2 y $ 2 = kAy1 - y2B + bAy# 1 - y# 2B m1 y $ 1 = kAy2 - y1B + bAy# 2 - y# 1B + u e2 = y2 - y2 e1 = y1 - y1 e1(0) = 0.1, e2(0) = 0.05 y1(0) = 0.1, y2(0) = 0, y # 1(0) = 0, y # 2(0) = 0 s = -15, s = -16 s = -2 + j213 , s = -2 - j213 , s = -10, s = -10 CBˆ  Aˆ Bˆ  Aˆ2 Bˆ  p  Aˆn Bˆ D m1 m2 y1 y2 u k b Regulator Figure 10–49 Mechanical system. Example Problems and Solutions 837 Then, the state-space equations become Define The state feedback gain matrix K and observer gain matrix Ke can be obtained easily by use of MATLAB as follows: (See MATLAB Program 10–26.) Ke = B14.40.3 0.615.7RK = [130.4444 -41.5556 23.1000 15.4185] A = E 00 -36 18 0 0 36 -18 1 0 -0.6 0.3 0 1 0.6 -0.3 U = CAaa Aba Aab Abb S , B = E 00 1 0 U = CBa Bb S By1 y2 R = B1 0 0 1 0 0 0 0 R Dx1x2 x3 x4 TD x # 1 x # 2 x # 3 x # 4 T = D 00 -36 18 0 0 36 -18 1 0 -0.6 0.3 0 1 0.6 -0.3 T Dx1x2 x3 x4 T + D00 1 0 T u MATLAB Program 10–26 A = [0 0 1 0;0 0 0 1;-36 36 -0.6 0.6;18 -18 0.3 -0.3]; B = [0;0;1;0]; J = [-2+j*2*sqrt(3) -2-j*2*sqrt(3) -10 -10]; K = acker(A,B,J) K = 130.4444 -41.5556 23.1000 15.4185 Aab = [1 0;0 1]; Abb = [-0.6 0.6;0.3 -0.3]; L = [-15 -16]; Ke = place(Abb',Aab',L)' place: ndigits= 15 Ke = 14.4000 0.6000 0.3000 15.7000 Response to Initial Condition: Next, we obtain the response of the designed system to the given initial condition. Since x = B xa x b R = B y x b Ru = -K xx# = Ax + Bu we have (10–170) Note that where Then, Equation (10–170) can be written as (10–171) Since, from Equation (10–94), we have (10–172) by combining Equations (10–171) and (10–172) into one equation, we have The state matrix here is a 6*6 matrix. The response of the system to the given initial condition can be obtained easily with MATLAB. (See MATLAB Program 10–27.) The resulting response curves are shown in Figure 10–50. The response curves seem to be acceptable. cx# e# d = cA - BK 0 BKF Abb - Ke Aab d cx e d e# = AAbb - Ke AabB e x# = (A - BK) x + BKFe F = c0 I d x - x = c xa xb d - c xa x b d = c 0 xb - x b d = c0 e d = c0 I de = Fe x# = Ax - BK x = (A - BK) x + BKAx - x B 838 Chapter 10 / Control Systems Design in State Space x 1 0 1 2 3 4 t (sec) 0 1 2 3 4 t (sec) 0 1 2 3 4 t (sec) 0 1 2 3 4 t (sec) 0 1 2 3 4 t (sec) −0.05 0 0.1 0.05 0.15 e 1 0 0.05 0.1 0 1 2 3 4 t (sec) e 2 0 0.02 0.04 0.06 x 2 −0.02 0.02 0 0.04 0.06 x 3 −0.6 −0.2 −0.4 0 0.2 x 4 −0.2 0.1 0 −0.1 0.2 Response to initial condition Response to initial condition Figure 10–50 Response curves to initial condition. Example Problems and Solutions 839 MATLAB Program 10–27 % Response to initial condition A = [0 0 1 0;0 0 0 1;-36 36 -0.6 0.6;18 -18 0.3 -0.3]; B = [0;0;1;0]; K = [130.4444 -41.5556 23.1000 15.4185]; Ke = [14.4 0.6;0.3 15.7]; F = [0 0;0 0;1 0;0 1]; Aab = [1 0;0 1]; Abb = [-0.6 0.6;0.3 -0.3]; AA = [A-B*K B*K*F; zeros(2,4) Abb-Ke*Aab]; sys = ss(AA,eye(6),eye(6),eye(6)); t = 0:0.01:4; y = initial(sys,[0.1;0;0;0;0.1;0.05],t); x1 = [1 0 0 0 0 0]*y'; x2 = [0 1 0 0 0 0]*y'; x3 = [0 0 1 0 0 0]*y'; x4 = [0 0 0 1 0 0]*y'; e1 = [0 0 0 0 1 0]*y'; e2 = [0 0 0 0 0 1]*y'; subplot(3,2,1); plot(t,x1); grid; title('Response to initial condition'), xlabel('t (sec)'); ylabel('x1') subplot(3,2,2); plot(t,x2); grid; title('Response to initial condition'), xlabel('t (sec)'); ylabel('x2') subplot(3,2,3); plot(t,x3); grid; xlabel('t (sec)'); ylabel('x3') subplot(3,2,4); plot(t,x4); grid; xlabel('t (sec)'); ylabel('x4') subplot(3,2,5); plot(t,e1); grid; xlabel('t (sec)');ylabel('e1') subplot(3,2,6); plot(t,e2); grid; xlabel('t (sec)'); ylabel('e2') r = 0 yu–y Observer controller + – 4 s(s + 2) PlantFigure 10–51 Regulator system. A–10–14. Consider the system shown in Figure 10–51.Design both the full-order and minimum-order observers for the plant.Assume that the desired closed-loop poles for the pole-placement part are located at Assume also that the desired observer poles are located at (a) s=–8, s=–8 for the full-order observer (b) s=–8 for the minimum-order observer Compare the responses to the initial conditions specified below: (a) for the full-order observer: x1(0) = 1, x2(0) = 0, e1(0) = 1, e2(0) = 0 s = -2 + j213 , s = -2 - j2 13 840 Chapter 10 / Control Systems Design in State Space (b) for the minimum-order observer: Also, compare the bandwidths of both systems. Solution. We first determine the state-space representation of the system. By defining state variables x1 and x2 as we obtain For the pole-placement part, we determine the state feedback gain matrix K. Using MATLAB, we find K to be K=[4 0.5] (See MATLAB Program 10–28.) Next, we determine the observer gain matrix Ke for the full-order observer. Using MATLAB, we find Ke to be (See MATLAB Program 10–28.) Ke = B1436R y = [1 0]Bx1 x2 R Bx# 1x# 2R = B00 1-2R Bx1x2R + B04R u x2 = y # x1 = y x1(0) = 1, x2(0) = 0, e1(0) = 1 MATLAB Program 10–28 % Obtaining matrices K and Ke. A = [0 1;0 -2]; B = [0;4]; C = [1 0]; J = [-2+j*2*sqrt(3) -2-j*2*sqrt(3)]; L = [-8 -8]; K = acker(A,B,J) K = 4.0000 0.5000 Ke = acker(A',C',L)' Ke = 14 36 Now we find the response of this system to the given initial condition. Referring to Equation (10–70), we have This equation defines the dynamics of the designed system using the full-order observer. MATLAB Program 10–29 produces the response to the given initial condition.The resulting response curves are shown in Figure 10–52. Bx# e# R = BA - BK 0 BK A - Ke C R Bx e R Example Problems and Solutions 841 MATLAB Program 10–29 % Response to initial condition ---- full-order observer A = [0 1;0 -2]; B = [0;4]; C = [1 0]; K = [4 0.5]; Ke = [14;36]; AA = [A-B*K B*K; zeros(2,2) A-Ke*C]; sys = ss(AA, eye(4), eye(4), eye(4)); t = 0:0.01:8; x = initial(sys, [1;0;1;0],t); x1 = [1 0 0 0]*x'; x2 = [0 1 0 0]*x'; e1 = [0 0 1 0]*x'; e2 = [0 0 0 1]*x'; subplot(2,2,1); plot(t,x1); grid xlabel('t (sec)'); ylabel('x1') subplot(2,2,2); plot(t,x2); grid xlabel('t (sec)'); ylabel('x2') subplot(2,2,3); plot(t,e1); grid xlabel('t (sec)'); ylabel('e1') subplot(2,2,4); plot(t,e2); grid xlabel('t (sec)'); ylabel('e2') x 1 x 2 e 1 e 2 0.4 0.6 0.8 1 0.2 0 −0.2 0 2 4 6 8 0 2 4 6 8 0 2 4 6 8 −0.4 0.6 0.8 1 1.2 0.4 0.2 0 −0.2 0 1 −1 −2 −0.5 0 −1 −1.5 −3 −2 t (sec) t (sec) 0 2 4 6 8 t (sec) t (sec) Figure 10–52 Response curves to initial condition. 842 Chapter 10 / Control Systems Design in State Space MATLAB Program 10–31 % Obtaining Ke ---- minimum-order observer Aab = [1]; Abb = [-2]; LL = [-8]; Ke = acker(Abb',Aab',LL)' Ke = 6 MATLAB Program 10–30 % Determination of transfer function of observer controller ---- full-order observer A = [0 1;0 -2]; B = [0;4]; C = [1 0]; K = [4 0.5]; Ke = [14;36]; [num,den] = ss2tf(A-Ke*C-B*K, Ke,K,0) num = 0 74.0000 256.0000 den = 1 18 108 To obtain the transfer function of the observer controller, we use MATLAB. MATLAB Program 10–30 produces this transfer function. The result is num den = 74s + 256 s2 + 18s + 108 = 74(s + 3.4595) (s + 9 + j5.1962)(s + 9 - j5.1962) Next, we obtain the observer gain matrix Ke for the minimum-order observer. MATLAB Program 10–31 produces Ke . The result is Ke = 6 The response of the system with minimum-order observer to the initial condition can be ob- tained as follows: By substituting into the plant equation given by Equation (10–79)u = -K x Example Problems and Solutions 843 MATLAB Program 10–32 % Response to intial condition ---- minimum-order observer A = [0 1;0 -2]; B = [0;4]; K = [4 0.5]; Kb = 0.5; Ke = 6; Aab = 1; Abb = -2; AA = [A-B*K B*Kb; zeros(1,2) Abb-Ke*Aab]; sys = ss(AA,eye(3),eye(3),eye(3)); t = 0:0.01:8; x = initial(sys,[1;0;1],t); x1 = [1 0 0]*x'; x2 = [0 1 0]*x'; e = [0 0 1]*x'; subplot(2,2,1); plot(t,x1); grid xlabel('t (sec)'); ylabel('x1') subplot(2,2,2); plot(t,x2); grid xlabel('t (sec)'); ylabel('x2') subplot(2,2,3); plot(t,e); grid xlabel('t (sec)'); ylabel('e') we find or The error equation is Hence the system dynamics are defined by Based on this last equation, MATLAB Program 10–32 produces the response to the given initial condition. The resulting response curves are shown in Figure 10–53. Bx# e # R = BA - BK0 BKbAbb - Ke AabR B xeR e # = AAbb - Ke AabB e x# = (A - BK) x + BKb e = (A - BK) x + B CKa Kb D B0eRx# = Ax - BK x = Ax - BKx + BK(x - x ) 844 Chapter 10 / Control Systems Design in State Space e 0 2 4 6 8 0 0.2 0.4 0.6 0.8 1 t (sec) x 1 x 2 0.6 0.8 1 1.2 0.4 0.2 0 0 2 4 6 8 0 2 4 6 8 −0.2 0 0.5 −0.5 −1 −2 −1.5 −2.5 t (sec) t (sec) Figure 10–53 Response curves to initial condition. The transfer function of the observer controller, when the system uses the minimum-order observer, can be obtained by use of MATLAB Program 10–33. The result is num den = 7s + 32 s + 10 = 7(s + 4.5714) s + 10 MATLAB Program 10–33 % Determination of transfer function of observer controller ---- minimum-order observer A = [0 1;0 -2]; B = [0;4]; Aaa = 0; Aab = 1; Aba = 0; Abb = -2; Ba = 0; Bb = 4; Ka = 4; Kb = 0.5; Ke = 6; Ahat = Abb - Ke*Aab; Bhat = Ahat*Ke + Aba - Ke*Aaa; Fhat = Bb - Ke*Ba; Atilde = Ahat - Fhat*Kb; Btilde = Bhat - Fhat*(Ka + Kb*Ke); Ctilde = -Kb; Dtilde = -(Ka + Kb*Ke); [num,den] = ss2tf(Atilde, Btilde, -Ctilde, -Dtilde) num = 7 32 den = 1 10 Example Problems and Solutions 845 The observer controller is clearly a lead compensator. The Bode diagrams of System 1 (closed-loop system with full-order observer) and of Sys- tem 2 (closed-loop system with minimum-order observer) are shown in Figure 10–54. Clearly, the bandwidth of System 2 is wider than that of System 1. System 1 has a better high-frequency noise- rejection characteristic than System 2. A–10–15. Consider the system where x is a state vector (n-vector) and A is an n*n constant matrix. We assume that A is non- singular. Prove that if the equilibrium state x=0 of the system is asymptotically stable (that is, if A is a stable matrix), then there exists a positive-definite Hermitian matrix P such that where Q is a positive-definite Hermitian matrix. Solution. The matrix differential equation. has the solution Integrating both sides of this matrix differential equation from t=0 to t=q, we obtain X(q) - X(0) = A* a 3q0 X dt b + a 3q0 X dt b A X = eA* t QeAt X # = A* X + XA, X(0) = Q A* P + PA = -Q x# = Ax Frequency (rad/sec) Bode Diagrams of Systems −300 −100 −200 −250 −50 −150 0 −100 −50 P ha se ( de g) ; M ag ni tu de ( dB ) 50 0 10−1 100 101 102 System 1 System 2 System 1 System 2 Figure 10–54 Bode diagrams of System 1 (system with full-order observer) and System 2 (system with minimum- order observer). System 1= (296s+1024) (s4+20s3+144s2 +512s+1024); System 2= (28s+128) (s3+12s2+48s+128). 846 Chapter 10 / Control Systems Design in State Space Noting that A is a stable matrix and, therefore, we obtain Let us put Note that the elements of are finite sums of terms like where the li are the eigenvalues of A and mi is the multiplicity of li . Since the li possess negative real parts, exists. Note that Thus P is Hermitian (or symmetric if P is a real matrix). We have thus shown that for a stable A and for a positive-definite Hermitian matrix Q, there exists a Hermitian matrix P such that We now need to prove that P is positive definite. Consider the following Her- mitian form: Hence, P is positive definite. This completes the proof. A–10–16. Consider the control system described by (10–173) where Assuming the linear control law (10–174) determine the constants k1 and k2 so that the following performance index is minimized: J = 3 q 0 xT x dt u = -Kx = -k1 x1 - k2 x2 A = B0 0 1 0 R , B = B0 1 Rx# = Ax + Bu = 0, for x = 0 = 3 q 0 AeAt xB* QAeAt xB dt 7 0, for x Z 0 x* Px = x* 3 q 0 eA* t QeAt dt x A* P + PA = -Q. P* = 3 q 0 eA* t QeAt dt = P 3 q 0 eA* t QeAt dt teli t p , tmi - 1 eli t,eli t,eAt P = 3 q 0 X dt = 3 q 0 eA* t QeAt dt -X(0) = -Q = A* a 3q0 X dt b + a 3q0 X dt b A X(q) = 0, Example Problems and Solutions 847 Consider only the case where the initial condition is Choose the undamped natural frequency to be 2 radsec. Solution. Substituting Equation (10–174) into Equation (10–173), we obtain or (10–175) Thus, Elimination of x2 from Equation (10–175) yields Since the undamped natural frequency is specified as 2 radsec, we obtain Therefore, is a stable matrix if k2>0. Our problem now is to determine the value of k2 so that the performance index is minimized, where the matrix P is determined from Equation (10–115), rewritten Since in this system Q=I and R=0, this last equation can be simplified to (10–176) Since the system involves only real vectors and real matrices, P becomes a real symmetric matrix. Then Equation (10–176) can be written asB0 1 -4 -k2 R Bp11 p12 p12 p22 R + Bp11 p12 p12 p22 R B 0 -4 1 -k2 R = B-1 0 0 -1 R (A - BK)* P + P(A - BK) = -I (A - BK)* P + P(A - BK) = -(Q + K* RK) J = 3 q 0 xT x dt = xT(0) P(0) x(0) A - BK A - BK = B 0 -4 1 -k2 Rk1 = 4 x $ 1 + k2 x # 1 + k1 x1 = 0 A - BK = B 0 -k1 1 -k2 R = B 0-k1 1-k2R Bx1x2R Bx# 1 x # 2 R = B0 0 1 0 R Bx1 x2 R + B0 1 R C-k1 x1 - k2 x2 Dx# = Ax - BKx x(0) = B c 0 R 848 Chapter 10 / Control Systems Design in State Space Solving for the matrix P, we obtain The performance index is then (10–177) To minimize J, we differentiate J with respect to k2 and set equal to zero as follows: Hence, With this value of k2, we have Thus, the minimum value of J is obtained by substi- tuting into Equation (10–177), or The designed system has the control law The designed system is optimal in that it results in a minimum value for the performance index J under the assumed initial condition. A–10–17. Consider the same inverted-pendulum system as discussed in Example 10–5.The system is shown in Figure 10–8, where M=2 kg, m=0.1 kg, and l=0.5 m. The block diagram for the system is shown in Figure 10–9. The system equations are given by j # = r - y = r - Cx u = -Kx + kI j y = Cx x# = Ax + Bu u = -4x1 - 120x2 Jmin = 15 2 c2 k2 = 120 02J0k22 7 0. k2 = 120 0J 0k2 = a -5 2k22 + 1 8 b c2 = 0 0J0k2 = a 5 2k2 + k2 8 b c2 = [c 0]Bp11 p12 p12 p22 R B c 0 R = p11 c2J = xT(0) Px(0) P = Bp11 p12 p12 p22 R = D 52k2 + k281 8 1 8 5 8k2 T Example Problems and Solutions 849 where Referring to Equation (10–51), the error equation for the system is given by where and the control signal is given by Equation (10–41): where Using MATLAB, determine the state feedback gain matrix such that the following performance index J is minimized: J = 3 q 0 (e* Qe + u* Ru) dt Kˆ x = Dx1x2 x3 x4 T = D uu# x x # T e = Bxe je R = Bx(t) - x(q) j(t) - j(q) RKˆ = CK  -kI D = Ck1 k2 k3 k4  -kI D ue = -Kˆe Aˆ = B A -C 0 0 R = E 020.6010 -0.4905 0 1 0 0 0 0 0 0 0 0 -1 0 0 1 0 0 0 0 0 0 0 U , Bˆ = BB 0 R = E 0-10 0.5 0 U e# = Aˆe + Bˆue A = D 020.601 0 -0.4905 1 0 0 0 0 0 0 0 0 0 1 0 T , B = D 0-1 0 0.5 T , C = [0 0 1 0] 850 Chapter 10 / Control Systems Design in State Space where Obtain the unit-step response of the system designed. Solution. A MATLAB program to determine is given in MATLAB Program 10–34.The result is k1 = -188.0799, k2 = -37.0738, k3 = -26.6767, k4 = -30.5824, kI = -10.0000 Kˆ Q = E10000 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 U , R = 0.01 MATLAB Program 10–34 % Design of quadratic optimal control system A = [0 1 0 0;20.601 0 0 0;0 0 0 1;-0.4905 0 0 0]; B = [0;-1;0;0.5]; C = [0 0 1 0]; D = [0]; Ahat = [A zeros(4,1);-C 0]; Bhat = [B;0]; Q = [100 0 0 0 0;0 1 0 0 0;0 0 1 0 0;0 0 0 1 0;0 0 0 0 1]; R = [0.01]; Khat = lqr(Ahat,Bhat,Q,R) Khat = -188.0799 -37.0738 -26.6767 -30.5824 10.0000 Unit-Step Response. Once we have determined the feedback gain matrix K and the integral gain constant kI, we can determine the unit-step response of the designed system.The system equation is (10–178) [Refer to Equation (10–35).] Since Equation (10–178) can be written as follows: (10–179) The output equation is y = [C 0]Bx j R + [0] rB x# j # R = BA - BK -C BkI 0 R Bx j R + B0 1 R ru = -Kx + kI j Bx# j # R = B A -C 0 0 R Bx j R + BB 0 Ru + B0 1 R r MATLAB Program 10–35 gives the unit-step response of the system given by Equation (10–179). The resulting response curves are presented in Figure 10–55. It shows response curves versus t, versus t, versus t, versus t, and versus t, where the input r(t) to the cart is a unit-step function All initial conditions are set equal to zero. Figure 10–56 is an enlarged version of the cart position versus t. The cart moves backward a very small amount for the first 0.6 sec or so. (Notice that the cart velocity is negative for the first 0.4 sec.) This is due to the fact that the inverted-pendulum-on-the-cart system is a nonminimum-phase system. y C= x3(t) DCr(t) = 1 m D . j C= x5(t) Dy # C= x4(t) Dy C= x3(t) Du# C= x2(t) D u C= x1(t) D Example Problems and Solutions 851 MATLAB Program 10–35 % Unit-step response A = [0 1 0 0;20.601 0 0 0;0 0 0 1;-0.4905 0 0 0]; B = [0;-1;0;0.5]; C = [0 0 1 0]; D = [0]; K = [-188.0799 -37.0738 -26.6767 -30.5824]; kI = -10.0000; AA = [A-B*K B*kI; -C 0]; BB = [0;0;0;0;1]; CC= [C 0]; DD = D; t = 0:0.01:10; [y,x,t] = step(AA,BB,CC,DD,1,t); x1 = [1 0 0 0 0]*x'; x2 = [0 1 0 0 0]*x'; x3 = [0 0 1 0 0]*x'; x4 = [0 0 0 1 0]*x'; x5 = [0 0 0 0 1]*x'; subplot(3,2,1); plot(t,x1); grid; xlabel('t (sec)'); ylabel('x1') subplot(3,2,2); plot(t,x2); grid; xlabel('t (sec)'); ylabel('x2') subplot(3,2,3); plot(t,x3); grid; xlabel('t (sec)'); ylabel('x3') subplot(3,2,4); plot(t,x4); grid; xlabel('t (sec)'); ylabel('x4') subplot(3,2,5); plot(t,x5); grid; xlabel('t (sec)'); ylabel('x5') Comparing the step-response characteristics of this system with those of Example 10–5, we notice that the response of the present system is less oscillatory and exhibits less maximum overshoot in the position response Ax3 versus t B . The system designed by use of the quadratic optimal regulator approach generally gives such characteristics—less oscillatory and well damped. 852 Chapter 10 / Control Systems Design in State Space x 1 0 2 4 6 8 10 t (sec) 0 2 4 6 8 10 t (sec) 0 2 4 6 8 10 t (sec) 0 2 4 6 8 10 t (sec) 0 2 4 6 8 10 t (sec) −0.02 0 0.02 0.04 x 5 0 1 2 3 x 2 −0.05 0.05 0 0.1 0.15 x 3 −0.5 0.5 0 1 1.5 x 4 −0.2 0.2 0 0.4 0.6 Figure 10–55 Response curves to a unit-step input. Cart Position x3 versus t C ar t P os it io n x 3 t (sec) 0 1 2 3 4 5 6 7 8 9 10 1.2 1 0.8 0.6 0.4 0.2 0 −0.2 Figure 10–56 Cart position versus t curve. A–10–18. Consider the stability of a system with unstructured additive uncertainty as shown in Figure 10–57(a). Define true plant dynamics G=model of plant dynamics unstructured additive uncertainty¢a = G  = Example Problems and Solutions 853 G w u K WaI y z − + K (f) (e) w u z y P G K a (d) u z w − + a Ta B A (c) G K a (a) yu A + + – B a K G y u + − A B (b) Figure 10–57 (a) Block diagram of a system with unstructured additive uncertainty; (b)–(d) successive modifications of the block diagram of (a); (e) block diagram showing a generalized plant with unstructured additive uncertainty; (f) generalized plant diagram. 854 Chapter 10 / Control Systems Design in State Space Assume that is stable and its upper bound is known.Assume also that and G are related by =G+ a Obtain the condition that the controller K must satisfy for robust stability. Also, obtain a gener- alized plant diagram for this system. Solution. Let us obtain the transfer function between point A and point B in Figure 10–57(a). Redrawing Figure 10–57(a), we obtain Figure 10–57(b).Then the transfer function between points A and B can be obtained as Define Then Figure 10–57(b) can be redrawn as Figure 10–57(c). By using the small-gain theorem, the con- dition for the robust stability of the closed-loop system can be obtained as (10–180) Since it is impossible to model precisely, we need to find a scalar transfer function such that for all v and use this instead of a. Then, the condition for the robust stability of the closed-loop system can be given by (10–181) If Inequality (10–181) holds true, then it is evident that Inequality (10–180) also holds true. So this is the condition to guarantee the robust stability of the designed system. In Figure 10–57(e), a in Figure 10–57(d) was replaced by . To summarize, if we make the norm of the transfer function from w to z to be less than 1, the controller K that satisfies Inequality (10–181) can be determined. Figure 10–57(e) can be redrawn as that shown in Figure 10–57(f), which is the generalized plant diagram for the system considered. Note that for this problem the matrix that relates the controlled variable z and the exoge- nous disturbance w is given by Noting that u(s)=K(s)y(s) and referring to Equation (10–128), is given by the elements of the P matrix as follows: To make this equal to we may choose P11=0, P12=Wa , P21=I, and P22=G. Then, the P matrix for this problem can be obtained as P = B 0 I Wa -G RWaK(I + GK)-1,£(s) £(s) = P11 + P12K(I - P22K)- 1P21 £(s) z = £(s)w = (WaTa)w = [WaK(I + GK)-1]w £ Hq WaI ¢ 7WaTa 7q 6 1 ¢Wa(jv) s{¢a(jv)} 6 Wa(jv) Wa(jv)¢a 7¢aTa 7q 6 1 K(1 + GK)- 1 = Ta K 1 + GK = K(1 + GK)-1 ¢G G ¢a Problems 855 x = Ax + Bu . y = Cx x2 x3 k2 k1 k3 r u x y = x1 + – + – Figure 10–58 Type 1 servo system. PROBLEMS B–10–1. Consider the system defined by where Transform the system equations into (a) controllable canon- ical form and (b) observable canonical form. B–10–2. Consider the system defined by where Transform the system equations into the observable canon- ical form. B–10–3. Consider the system defined by where By using the state-feedback control it is desired to have the closed-loop poles at Deter- mine the state-feedback gain matrix K. B–10–4. Solve Problem B–10–3 with MATLAB. s = -10.s = -2 ; j4, u = -Kx, A = C 00 -1 1 0 -5 0 1 -6 S , B = C01 1 S x# = Ax + Bu A = C-11 0 0 -2 0 1 0 -3 S , B = C01 1 S , C = [1 1 1] y = Cx x# = Ax + Bu C = [1 1 0]A = C-11 0 0 -2 0 1 0 -3 S , B = C00 1 S , y = Cx x# = Ax + Bu B–10–5. Consider the system defined by Show that this system cannot be stabilized by the state- feedback control whatever matrix K is chosen. B–10–6. A regulator system has a plant Define state variables as By use of the state-feedback control it is desired to place the closed-loop poles at Determine the necessary state-feedback gain matrix K. B–10–7. Solve Problem B–10–6 with MATLAB. B–10–8. Consider the type 1 servo system shown in Figure 10–58. Matrices A, B, and C in Figure 10–58 are given by Determine the feedback gain constants k1, k2, and k3 such that the closed-loop poles are located at Obtain the unit-step response and plot the output y(t)-versus-t curve. s = -2 + j4, s = -2 - j4, s = -10 A = C00 0 1 0 -5 0 1 -6 S , B = C00 1 S , C = [1 0 0] s = -2 + j213 , s = -2 - j213 , s = -10 u = -Kx, x3 = x # 2 x2 = x # 1 x1 = y Y(s) U(s) = 10 (s + 1)(s + 2)(s + 3) u = -Kx, Bx# 1 x # 2 R = B0 0 1 2 R Bx1 x2 R + B1 0 R u 856 Chapter 10 / Control Systems Design in State Space B–10–9. Consider the inverted-pendulum system shown in Figure 10–59. Assume that M=2 kg, m=0.5 kg, l=1 m Define state variables as and output variables as Derive the state-space equations for this system. It is desired to have closed-loop poles at Determine the state-feedback gain matrix K. Using the state-feedback gain matrix K thus determined, examine the performance of the system by computer simu- lation.Write a MATLAB program to obtain the response of the system to an arbitrary initial condition. Obtain the response curves x1(t) versus t, x2(t) versus t, x3(t) versus t, and x4(t) versus t for the following set of initial condition: x1(0) = 0, x2(0) = 0, x3(0) = 0, x4(0) = 1 ms s = -4 + j4, s = -4 - j4, s = -20, s = -20 y1 = u = x1 , y2 = x = x3 x1 = u, x2 = u # , x3 = x, x4 = x # where Design a full-order state observer. The desired observer poles are s=–5 and s=–5. B–10–11. Consider the system defined by where Design a full-order state observer, assuming that the desired poles for the observer are located at s=–10, s=–10, s=–15 B–10–12. Consider the system defined by Given the set of desired poles for the observer to be design a full-order observer. B–10–13. Consider the double integrator system defined by If we choose the state variables as then the state-space representation for the system becomes as follows: y = [1 0]Bx1 x2 R Bx# 1x# 2R = B00 10R Bx1x2R + B01R u x2 = y # x1 = y y $ = u s = -5 + j513 , s = -5 - j513 , s = -10 y = [1 0 0]Cx1x2 x3 S+ C 001.244S u Cx# 1x# 2 x # 3 S = C 00 1.244 1 0 0.3956 0 1 -3.145 S Cx1x2 x3 S A = C 00 -5 1 0 -6 0 1 0 S , B = C00 1 S , C = [1 0 0] y = Cx x # = Ax + Bu A = B-1 1 1 -2 R , C = [1 0] 0 M P z u mg m  sin u x x  cos u  u Figure 10–59 Inverted-pendulum system. B–10–10. Consider the system defined by y = Cx x# = Ax Problems 857 Y(s)R(s) U(s)Observer controller + – s2 + 2s + 50 s(s + 4) (s + 6) Figure 10–60 Control system with observer controller in the feedforward path. It is desired to design a regulator for this system. Using the pole-placement-with-observer approach, design an observer controller. Choose the desired closed-loop poles for the pole- placement part to be s=–0.7071+j0.7071, s=–0.7071-j0.7071 and assuming that we use a minimum-order observer, choose the desired observer pole at s=–5 B–10–14. Consider the system where Design a regulator system by the pole-placement-with- observer approach. Assume that the desired closed-loop poles for pole placement are located at s=–1+j, s=–1-j, s=–5 The desired observer poles are located at s=–6, s=–6, s=–6 Also, obtain the transfer function of the observer controller. B–10–15. Using the pole-placement-with-observer approach, design observer controllers (one with a full-order observer and the other with a minimum-order observer) for the system shown in Figure 10–60. The desired closed-loop poles for the pole-placement part are s=–1+j2, s=–1-j2, s=–5 A = C 00 -6 1 0 -11 0 1 -6 S , B = C00 1 S , C = [1 0 0] y = Cx x# = Ax + Bu The desired observer poles are s=–10, s=–10, s=–10 for the full-order observer s=–10, s=–10 for the minimum-order observer. Compare the unit-step responses of the designed systems. Compare also the bandwidths of both systems. B–10–16. Using the pole-placement-with-observer approach, design the control systems shown in Figures 10–61(a) and (b). Assume that the desired closed-loop poles for the pole place- ment are located at s=–2+j2, s=–2-j2 and the desired observer poles are located at s=–8, s=–8 Obtain the transfer function of the observer controller. Compare the unit-step responses of both systems. [In System (b), determine the constant N so that the steady-state out- put y(q) is unity when the input is a unit-step input.] Y(s)R(s) Observer controller + – 1 s(s + 1) 1 s(s + 1) Y(s) Observer controller + – (b) R(s) N (a) Plant Figure 10–61 Control systems with observer controller: (a) observer controller in the feedforward path; (b) observer controller in the feedback path. B–10–17. Consider the system defined by where a = adjustable parameter 7 0 A = C 00 -1 1 0 -2 0 1 -a Sx# = Ax 858 Chapter 10 / Control Systems Design in State Space Determine the value of the parameter a so as to minimize the following performance index: Assume that the initial state x(0) is given by B–10–18. Consider the system shown in Figure 10–62. Determine the value of the gain K so that the damping ratio z of the closed-loop system is equal to 0.5. Then determine also the undamped natural frequency vn of the closed-loop system. Assuming that e(0)=1 and evaluate 3 q 0 e2(t) dt e # (0) = 0, x(0) = C c10 0 SJ = 3 q 0 xT x dt B–10–21. Consider the inverted-pendulum system shown in Figure 10–59. It is desired to design a regulator system that will maintain the inverted pendulum in a vertical po- sition in the presence of disturbances in terms of angle u and/or angular velocity The regulator system is required to return the cart to its reference position at the end of each control process. (There is no reference input to the cart.) The state-space equation for the system is given by where We shall use the state-feedback control scheme Using MATLAB, determine the state-feedback gain matrix such that the following performance index J is minimized: where Then obtain the system response to the following initial condition: Plot response curves u versus t, versus t, x versus t, and versus t. x # u # Dx1(0)x2(0) x3(0) x4(0) T = D0.10 0 0 T Q = D1000 0 0 0 1 0 0 0 0 1 0 0 0 0 1 T , R = 1 J = 3 q 0 Ax* Qx + u* RuB dt K = Ck1 k2 k3 k4 D u = -Kx B = D 0-1 0 0.5 T , x = D uu# x x # T A = D 020.601 0 -0.4905 1 0 0 0 0 0 0 0 0 0 1 0 Tx # = Ax + Bu u # . + – r = 0 cue K 5 (s + 1) (2s + 1) Figure 10–62 Control system. B–10–19. Determine the optimal control signal u for the system defined by where such that the following performance index is minimized: B–10–20. Consider the system It is desired to find the optimal control signal u such that the performance index is minimized. Determine the optimal signal u(t). J = 3 q 0 AxT Qx + u2B dt, Q = B1 0 0 m R Bx# 1 x # 2 R = B0 0 1 0 R Bx1 x2 R + B0 1 R uJ = 3 q 0 AxT x + u2B dt A = B0 0 1 -1 R , B = B0 1 Rx# = Ax + Bu