NPTEL Online Course: Control Engineering Ramkrishna Pasumarthy Assignment-11 : s 1. Consider a system described by state space model [ ] [ 0 1 1 x + u 5 1 2] y = [ 1 2 ] x What is the transfer function of the system? (a) 5s+13 s 2 +s+5 (b) 5s+13 s 2 +s 5 [Correct] (c) 5s s 2 +s+5 (d) 5s+13 s 2 s 5 Given [ ] 0 1 A = 5 1 [ 1 B = 2] C = [ 1 2 ] The transfer function is given by C(sI A) 1 B. Substituting A,B and C, we get: 5s + 13 G(s) = s 2 + s 5 2. In Problem No.1, is the system controllable? (a) Yes [Correct] (b) No (c) Cannot be determined with given information 1
We find the controllability matrix as follows: C = [BAB] [ ] 1 2 C = 2 3 The rank of C is 2. Therefore the system is controllable. 3. In Problem No. 1, is the system observable? (a) Yes [Correct] (b) No (c) Cannot be determined with given information We find the observability matrix as follows: [ ] C O = CA [ ] 1 2 O = 10 1 The rank of O is 2. Therefore the system is observable. 4. In Problem No.1, if the closed-loop characteristic equation is determined for a feedback of the form u = [K 1 K 2 ]x, what is the coefficient of s given K 1 = 2; K 2 = 2, rounded to the nearest integer? Using state feedback u = Kx, we get: (A BK)x ( [ ] [ ] 0 1 1 [K1 ] ) K 5 1 2 2 x [ ] K1 1 K 2 x 5 2K 1 1 2K 2 The characteristic equation becomes s 2 + (1 + K 1 + 2K 2 )s + (3K 1 + 5K 2 5) = 0 Substituting k 1 = 2; k 2 = 2, we get the coefficient of s to be 7. 5. For K 1 = 1; K 2 = 1, the closed loop system is: 2
(a) Stable (b) Unstable [Correct] (c) Marginally Stable By substituting K 1 = 1; K 2 = 1 in the characteristic equation obtained in the previous problem, we get s 2 7 = 0. Therefore the system will have poles on the right half plane and it will be unstable. 6. Given the system ] [ ] [ ] [ [ẋ1 4 1 x1 0 = + u(t) ẋ 2 2 1 x 2 1] For u(t) being a step input, what is the steady state value of x 1 rounded to two decimal places? At steady state, the values of x 1 and x 2 do not change i,e,. ẋ 1 = ẋ 2 = 0 We get the following equations by substituting ẋ 1 = ẋ 2 = 0 in the model: 4x 1 + x 2 = 0 2x 1 x 2 + u = 0 As u(t) is a step input, we can substitute u(t) = 0 and solve the above equations to get steady state values of x 1 & x 2. We get x 1 = 0.17 x 2 = 0.67 7. In Problem No.6, what is the steady state value of x 2 rounded to two decimal places? x 2 = 0.67 8. Consider a system with state space equations 5 1 1 0 3 2 x + 0 1 u 0 0 4 1 y = [ 1 0 0 ] x What is the transfer function? 3
2s+5 (a) s 3 +2s 2 23s 60 (b) 2s+5 s 3 +12s 2 +47s+60 [Correct] (c) 2s+10 (d) s 3 +4s 2 23s 60 2s 5 s 3 +2s 2 +23s+60 The transfer function is given by C(sI A) 1 B. Substituting A,B and C as per the given model, we get: G(s) = 2s + 5 s 3 + 12s 2 + 47s + 60 9. In Problem No.8, what is the DC gain of the system rounded to two decimal places? DC gain is given by: K = lim s 0 G(s) K = lim s 0 2s + 5 s 3 + 12s 2 + 47s + 60 K = 5 60 K = 0.08 10. In Problem No. 8, is the system observable? (a) Yes [Correct] (b) No (c) Cannot be determined with given information We find the observability matrix as follows: C O = CA CA 2 1 0 0 O = 5 1 1 25 8 11 The rank of O is 3. Therefore the system is observable. 4
11. For the state space system [ ] [ 0 1 0 x + u 6 5 1] y = [ 1 0 ] x We design a state feedback controller u = Kx, K = [k 1 k 2 ], such that the closed-loop poles have a damping coefficient ξ = 0.707 and step response peak time t p = 3.14. What should the natural frequency ω n of the system with desired closed loop poles rounded to two decimal points? [Ans: 1.41] The step response peak time is given as: π t p = ω n 1 ζ 2 3.14 3.14 = ω n 1 0.707 2 ω n = 2 = 1.41 12. In Problem No.10, what is the value of k 1 in the designed state feedback controller rounded to nearest integer? [Ans: -4] Using the values of omega n and ζ in the previous problem, the desired characteristic equation is s 2 + 2s + 2 = 0. Using state feedback u = Kx, we get: (A BK)x ( [ ] 0 1 6 5 [ 0 1 6 k 1 5 k 2 [ ] 0 [k1 ] ) k 1 2 x ] The characteristic equation becomes s 2 + (5 + k 2 )s + (6 + k 1 ) = 0 Comparing this with the desired characteristic equation, we get: 6 + k 1 = 2 k 1 = 4 13. In Problem No.10, what is the value of k 2 in the designed state feedback controller rounded to nearest integer? [Ans: -3] x 5
Continuing from above problem, 5 + k 2 = 2 k 2 = 3 14. Consider the non-linear system x 2 4. How many equilibrium points does the system have? Since the model equation is of degree 2, the number of equilibrium points is equal to 2. They can be calculated by setting 0 x 2 4 = 0 x 2 = 4 x = ±2 15. What the value of stable equilibrium point of the system? To determine the stability of equilibrium points, we find ẍ. ẍ = 2x Now ẍ x = 2 = 4 < 0 Therefore, the stable equilibrium point is -2 16. What the value of unstable equilibrium point of the system. ẍ x =2 = 4 > 0 Therefore, the unstable equilibrium point is 2 17. Consider the non-linear system represented by the following set of equations: ẋ 1 = u 1 cos x 3 ẋ 2 = u 1 sin x 3 ẋ 3 = u 2 Find the linearised state space model of the system with equilibrium conditions [x 1 x 2 x 3] T and u 1 = 0; u 2 = 0. What is the matrix A in the state space model? 6
(a) A = 0 0 0 0 0 0 [Correct] 0 0 0 (b) A = 1 0 0 0 1 0 0 0 1 1 0 0 (c) A = 0 1 0 0 0 1 0 0 0 (d) A = 0 0 0 0 0 1 Let us take : f 1 (x, u) = ẋ 1 = u 1 cos x 3 f 2 (x, u) = ẋ 2 = u 1 sin x 3 f 3 (x, u) = ẋ 3 = u 2 Following Jacobi Linearization, A = f x x,u Therefore, A = x 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 0 0 0 A = 0 0 0 0 0 0 A becomes a zero matrix because every f has an u in it and it is given that at equilibrium, u 1 = 0; u 2 = 0 18. In Problem No.17, what is the dimension of matrix B? (a) 3 2 [Correct] (b) 2 3 (c) 3 3 (d) 3 1 7
Following Jacobi Linearization, B = f u x,u Therefore, B = f 1 It is a matrix of dimension 3 2 19. In Problem No.17, what is the first row first column element of matrix B? (a) cos(x 3) [Correct] (b) sin(x 3) (c) 0 (d) 1 = (u 1 cos(x 3 )) = cos(x 3) x,u 20. In Problem No.17, what is the last row last column element of matrix B? (a) cos(x 3) (b) sin(x 3) (c) 0 (d) 1 [Correct] = = 1 (u 2 ) x,u 8