# Chapter 15 - Solved Problems

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1 Chapter 5 - Solved Problems Solved Problem 5.. Contributed by - Alvaro Liendo, Universidad Tecnica Federico Santa Maria, Consider a plant having a nominal model given by G o (s) = s + 2 The aim of the control design is to follow a constant reference. Also, an input disturbance has significant energy in the frequency band [0, 6][rad/s]. Find a proper controller using the affine parameterization. Solutions to Solved Problem 5. Solved Problem 5.2. Contributed by - Alvaro Liendo, Universidad Tecnica Federico Santa Maria, () G o (s) = s 2 (s + )(s + 3) (2) Assume that the bandwidth of the closed loop has to be limited to 8[rad/s] in order to have good noise immunity. We further assume that the reference is constant. Design a controller using the affine parameterization. Solutions to Solved Problem 5.2 Solved Problem 5.3. Contributed by - Alvaro Liendo, Universidad Tecnica Federico Santa Maria, G o (s) = s 2 +.4s + Find a PID controller that stabilizes the plant and provides zero steady state error at zero frequency. Solutions to Solved Problem 5.3 Solved Problem 5.4. Contributed by - Alvaro Liendo, Universidad Tecnica Federico Santa Maria, (3) G o (s) = e s (s + )(s + 2) (4) using the Smith controller of Figure 5.5 of the book, find a controller that stabilizes the plant and allows good tracking (although delayed!!) of a reference with energy in the band [ 0, 2] [rad/s]. Solutions to Solved Problem 5.4

2 Solved Problem 5.5. Contributed by - Alvaro Liendo, Universidad Tecnica Federico Santa Maria, G o (s) = s Find a controller that stabilizes the plant, ensures perfect reference tracking at zero frequency and provides good compensation of disturbances with energy in the frequency band [ 0, ] [rad/s]. Solutions to Solved Problem 5.5 Solved Problem 5.6. Contributed by - Alvaro Liendo, Universidad Tecnica Federico Santa Maria, (5) Find all stabilizing controllers for this plant. Solutions to Solved Problem 5.6 G o (s) = s (6) Solved Problem 5.7. Contributed by - Alvaro Liendo, Universidad Tecnica Federico Santa Maria, Show that the controller found in Problem 5.5 is consistent with the solution to Problem 5.6 Solutions to Solved Problem 5.7 Solved Problem 5.8. Contributed by - Alvaro Liendo, Universidad Tecnica Federico Santa Maria, Consider a discrete time plant having nominal transfer function given by G oq (z) = z + 2 (z 0.9)(z 0.) Find a controller that stabilizes the plant, ensure perfect tracking of a constant reference and exhibits a closed-loop step response faster than (0.7) k. Solutions to Solved Problem 5.8 Solved Problem 5.9. Contributed by - Alvaro Liendo, Universidad Tecnica Federico Santa Maria, Consider a discrete time plant having nominal transfer function given by G oq (z) = (z + 2) Find all controllers that stabilize the plant and ensure zero steady state error when tracking a constant reference. Solutions to Solved Problem 5.9 (7) (8) 2

3 Chapter 5 - Solutions to Solved Problems Solution 5.. To achieve perfect tracking, T o (0) = is needed. Also, we have that S io (s) = [ T o (s)]g o (s) = (s + 2) [ T o(s)] (9) Then, to achieve good disturbance rejection, we need the bandwidth of T o (s) to be at least 6[rad/s] Also [ T o (s)] should cancel the factor (s + 2) in S io. This leads to T o ( 2) =. We choose 36(τs + )(s + 2) s 2 + 8s + 36 (0) The condition T o ( 2) = Q( 2)G o ( 2) = implies τ = 6. Thus 6(s + 6)(s + 2) s 2 + 8s + 36 The quality of the disturbance compensation can be evaluated by plotting the magnitude of the input sensitivity function. This is shown in Figure. () 0 20 Magnitude [db] S io (jω) db Frequency [rad/s] Figure : Input sensitivity frequency response From Figure we see that, in the disturbance frequency band, [0, 6] [rad/s], the gain is at most 7 [db]. Solution 5.2. We know that (s 2) must be a zero of T o (s) to ensure an internally stable closed loop. Also, T o (0) must be zero to ensure perfect steady state reference tracking. Also, to achieve the fastest possible loop, we set the poles in T o (s) close to 8[rad/s]. Say we choose T o (s) = 32(s 2) (s 2 + 2s + 64) (2) This leads to 3

4 32(s + )(s + 3) (s 2 + 2s + 64) (3) Solution 5.3. We use the theory developed in subsection of the book. To achieve a proper inverse we select F Q (s)[g o (s)] = F Q (s)(s 2 +.4s + ) (4) We choose a natural frequency of 2.5 and a damping factor of Hence And we then have F Q (s) = 0.6s s + (5) s2 +.4s + 0.6s s + (6) Using equations (5.4.6) to (5.4.9) in the book we finally have K P = 2.0 (7) K I =.92 (8) K D = 0.3 (9) τ D = 0.3 (20) The closed-loop step response using this controller is shown in Figure Closed loop response Time [s] Figure 2: Closed-loop step response 4

5 Solution 5.4. Using the structure of Figure 5.5 in the book, we find a controller for the rational part of G o (s), denoted by G o (s). An appropriate Q(s) (achieving a closed-loop bandwidth larger than 2 [rad/s]) is leading to 20(s + )(s + 2) (s s + 4)(s + 0) (2) e s T o (s) = 40 (s s + 4)(s + 0) Note that T o (0) =, i.e., the controller provides integral action. (22) Solution 5.5. To ensure internal stability of the closed loop it is necessary and sufficient that Q(s) be stable and that (s ) be a zero of Q(s) and of ( G o (s)q(s)). We choose where (s )Q(s) (23) P (s) with P (s) a polynomial satisfying P (0) = 6, to achieve zero steady state error for constant references. (This forces Q(0) =.) Furthermore, (24) G o (s) s2 + 5s + 6 P (s) must have a zero at s = and one other zero. This leads to (25) To satisfy P (0) = 6, we choose α = 0. This leads to s 2 + 5s + 6 P (s) = (s )(s + α) (26) P (s) = (6 α)s + (6 + α) (27) With this choice we have that 6s + 6 6(s + )(s ) (28) (29) T o (s) = Q(s)G o (s) = 6(s + ) Note that T o (s) satisfies the interpolation constraint T o () =. (30) 5

6 Solution 5.6. We know that all controllers that stabilize this plant have the form (see equation (5.7.) in the book) s [ ] P (s) F (s) E(s) + Q (s ) u(s) E(s) where Q u (s) is any stable transfer function; F (s), E(s) are stable polynomials and P (s), F (s), E(s) satisfy the pole assignment equation (3) (s )L(s) + P (s) = E(s)F (s) (32) sl(s) + (P (s) L(s)) = E(s)F (s) (33) Note that P (s)/l(s) is any stabilizing controller for this plant. Choosing L(s) =, P (s) = 2 and E(s) = we have F (s) = s + This leads to (s ) (s + ) [2 + (s )Q u(s)] (34) Solution 5.7. We only need to find the Q u (s) in Problem 5.6 that leads to Q(s) in Problem 5.5 and verify that is stable After some algebra, we find (s ) (s + ) [2 + (s )Q 6(s + )(s ) u(s)] = (35) Q u (s) = 4s + 6 This shows consistency because Q u (s) is found to be stable as required in the solution to Problem 5.6. Solution 5.8. To have a stable Q q (z), T oq (z) must contain the unstable zero of G oq (z). Also, to follow a constant reference we need T oq () =. Finally, to ensure that the closed-loop step response has transients faster than (0.7) k, T oq (z) must contain only poles inside the disk with radius 0.7. This requires that Q q (z) cancels the slow plant pole (located at z = 0.9). Considering this, we choose The condition T oq () = leads to K = 0.27 and (36) T oq (z) = K z + 2 (z 0.) 2 (37) This yields z + 2 T oq (z) = 0.27 (z 0.) 2 (38) (z 0.9) Q q (z) = 0.27 (z 0.) (39) 6

7 Solution 5.9. We first find all stabilizing controllers for this plant. They have the form (see equation (5.7.) in the book) Q q (z) = z + 2 [ ] Pq (z) F q (z) E q (z) + Q (z + 2) uq(z) E q (z) (40) where Q uq (z) is any stable transfer function; F q (z), E q (z) are stable polynomials and P q (z), F q (z), E q (z) satisfy the pole assignment equation (z + 2)L q (z) + P q (z) = E q (z)f q (z) (4) zl q (z) + [P q (z) + 2L q (z)] = E q (z)f q (z) (42) Note that P q (z)/l q (z) is any stabilizing controller for this plant. Choosing L q (z) =, P q (z) = 2.5 and E q (z) = we have F q (z) = z 0.5 This leads to Q q (z) = (z + 2) (z 0.5) [ (z + 2)Q uq(z)] (43) This parameterizes all controllers that stabilize the plant. We next need to ensure perfect reference tracking. This is achieved by setting Q q () = [G oq (z)] = 3. Hence, we require Q q () = 6[ Q uq ()] = 3 (44) This leads to Q uq () =. The only other requirement is that Q u q(z) must be stable. 7

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