1 (s + 3)(s + 2)(s + a) G(s) = C(s) = K P + K I

Transcription

1 MAE 43B Linear Control Prof. M. Krstic FINAL June 9, Problem. ( points) Consider a plant in feedback with the PI controller G(s) = (s + 3)(s + )(s + a) C(s) = K P + K I s. (a) (4 points) For a given constant a, find the ranges of the controller gains K P and K I so that the closed-loop system is stable. (The conditions on K P and K I may depend on a.) (b) (6 points) Set K I =, i.e., let C(s) be a P controller. Find the range of K P so that the closed-loop system is stable for all a. The condition on K P must not depend on a. Part (a) represents the scenario where the designer knows the value of the parameter a. Part (b) represents the scenario where the designer does not know the value of the parameter a, only knows that a, and tries to design a controller that works for all a. Solution. (a) The characteristic equation of the closed-loop system is ( + K P + K ) I s (s + 3)(s + )(s + a) =, which we can rewrite as s(s + 5s + 6)(s + a) + K P s + K I = s 4 + (a + 5)s 3 + (5a + 6)s + (6a + K P )s + K I =. The corresponding Routh array is For stability we must have s 4 : 5a + 6 K I s 3 : a + 5 6a + K P s : 5a + 6 6a+K P a+5 K I s : 6a + K P (a+5)k I 5a+6 6a+K P a+5 s : K I. 5a + 6 6a + K P a + 5 > 6a + K P (a + 5)K I 5a + 6 6a+K P a+5 > K I >,

2 which are equivalent to K P < 5(a + 5a + 6) < K I < (6a + K P ) {5(a + 5a + 6) K P } (a + 5). (b) When K I =, the characteristic equation of the closed-loop system is which we can rewrite as + K P (s + 3)(s + )(s + a) =, (s + 5s + 6)(s + a) + K P = s 3 + (a + 5)s + (5a + 6)s + (6a + K P ) =. The corresponding Routh array is For stability we must have which are equivalent to s 3 : 5a + 6 s : a + 5 6a + K P s : 5a + 6 6a+K P a+5 s : 6a + K P. 5a + 6 6a + K P > a + 5 6a + K P >, 6a < K P < 5(a + 5a + 6). () Since a, the lower bound 6a in () is less than or equal to and the upper bound 5(a +5a+6) in () is greater than or equal to 3. Therefore, in order for K P satisfy the condition in () for all a, K P must be in the range < K P < 3. Problem. (9 points) Consider the closed-loop system in Problem with G(s) = (s + 3)(s + )(s + a) and the PI controller C(s) = K P + K I s. Let K I = K P. Find the locus of the closed-loop roots with respect to K P in following cases: (a) (3 points) < a < (b) (3 points) a = (c) (3 points) < a <

3 Solution. (a) The characteristic equation of the closed-loop system is + K P (s + ) s(s + 3)(s + )(s + a) =. There are four branches to the locus, starting at s = 3, s =, s = a, and s =, respectively. The real-axis segments s 3, s, and a s are parts of the locus. There are 3 asymptotes centered at ( 3 a) ( ) = 4 a and at the angles ±6 and Therefore, the locus of the closed-loop roots with respect to K P is illustrated in Figure. 4 Root Locus Figure : The root locus of the closed-loop system with respect to K P when < a < (This plot is drawn with a =.5). (b) The characteristic equation of the closed-loop system is + K P s(s + 3)(s + ) =. There are three branches to the locus, starting at s = 3, s =, and s =, respectively. The real-axis segments s 3 and s are parts of the locus. There are 3 asymptotes centered at 3 = 5 and at the angles 3 3 ±6 and 8. Therefore, the locus of the closed-loop roots with respect to K P is illustrated in Figure. (c) The characteristic equation of the closed-loop system is + K P (s + ) s(s + 3)(s + )(s + a) =. There are four branches to the locus, starting at s = 3, s =, s = a, and s =, respectively. The real-axis segments s 3, s a, and s are parts of the locus. There are 3 asymptotes centered at ( 3 a) ( ) = 4 a and at the angles ±6 and Therefore, the locus of the closed-loop roots with respect to K P is illustrated in Figure 3.

4 Root Locus Figure : The root locus of the closed-loop system with respect to K P when a =. 4 Root Locus Figure 3: The root locus of the closed-loop system with respect to K P when < a < (This plot is drawn with a =.5). Problem 3. (5 points) Consider a closed-loop system with a plant G(s) = e εs s where < ε < is a constant. Sketch (qualitatively) the Nyquist plot of G(s). Then consider G(s) in feedback with a constant gain K and determine the range of K for which the closed-loop system is stable. Supporting material and hint: (i) Denote φ(ω) = εω + arctan ω π. (ii) Note that φ (ω) = ε + + ω.

5 (iii) Observe that φ() = π, denote by ω (ε) the positive solution of φ(ω) = π, and denote by ω n (ε) the solutions of φ(ω) = (n + )π for n =,,... (iv) Do not calculate ω (ε), ω (ε), ω (ε),..., but do use them in your answer. Solution. In order to draw the Nyquist plot of G(s), evaluate G(s) at s = jω for ω so that G(jω) = e jεω jω = ω + e j(π arctan ω) e jεω = ω + ejφ(ω), from which it follows that ) G(j) =, ) G(jω) =, and ω + 3) G(jω) = φ(ω). To investigate φ(ω), we first look into the φ (ω). It is clear that 4) φ () = ε >, 5) φ (ω) is monotonically decreasing as ω increases from to, and 6) lim ω φ (ω) = ε <, from which it follows that 7) there exists a unique positive solution ω > of the equation φ (ω) =, 8) φ(ω) is monotonically increasing for ω ω, 9) φ(ω) is monotonically decreasing for ω > ω. Moreover, we know ) φ() = π, ) lim ω φ(ω) = from the definition of φ(ω), ) φ(ω) < π for all ω from the fact that arctan ω < π for all ω. From the arguments above, we can conclude the followings: 3) The Nyquist plot of G(s) starts off at (, ) : from ) 4) The distance from the origin to the Nyquist plot is monotonically decreasing and the Nyquist plot converges to : from ) 5) The Nyquist plot moves counterclockwise first : from 8) 6) The Nyquist plot turns back before it reaches the negative imaginary axis and, then, moves clockwise endlessly. : from 8), ), 9), and ) Finally, using the properties 3), 4), 5), and 6), we can qualitatively draw the Nyquist plot of G(s) which is illustrated in Figure 4 and 5. From the properties of φ(ω) and Figure 4 and 5, it follows that the Nyquist plot of G(s) intersects the negative real axis infinitely many times at, G(jω (ε)), G(jω (ε)), G(jω (ε)), and < G(jω (ε)) = ω (ε) + < G(jω (ε)) = ω (ε) + < G(jω (ε)) = ω (ε) +. <....

6 Nyquist Diagram Figure 4: The Nyquist plot of G(s) (This plot is drawn with ε =.5). Nyquist Diagram Figure 5: The magnified plot of Figure 4 around the origin. Since G(s) has one unstable pole, the closed-loop system is stable if, and only if, #CCW #CW= where the encirclement is around (, j). And, this only happens when < < K K G(jω (ε)) = ω. (ε) + Therefore, the range of K is < K < ω (ε) +. Problem 4. (4 points) (a) (6 points) Sketch the Nyquist plot of the plant G(s) = s + s(s ). and de- 5(s + ) (b) (8 points) Sketch the Nyquist plot of the plant G(s) = (s + )(s 3s + ) termine the stability of a unity feedback system with this plant.

7 Solution. (a) Evaluated values of G(s) at s = jω for ω > is G(jω) = jω + ω jω (jω + )(ω j) = ω(ω + 4) = 3ω + j( ω + ) ω(ω + 4) = 3 ω j ω + ω(ω + 4), from which it follows that ) lim ω + G(jω) = j(+ ), ) lim ω G(jω) = j( ), 3) lim ω G(jω) =, 4) G(jω) intersects the imaginary axis at ω =, and 5) G(jω) intersects the real axis at ω = with G(j ) =. Therefore, the Nyquist plot of G(s) is given in Figure 6. 3 Nyquist Diagram Figure 6: The Nyquist plot of G(s) in (a). (b) Evaluated values of G(s) at s = jω for ω is G(jω) = from which it follows that ) G(j) = 5, 5(jω + ) (jω + )( ω j3ω + ) = 5(jω + )( jω + )( ω + + j3ω) (ω + ){( ω + ) + 9ω } 5(ω ω ) = (ω + ){( ω + ) + 9ω } + j 5ω(ω 3) (ω + ){( ω + ) + 9ω },

8 ) lim ω G(jω) =, 3) Re[G(jω)] < and Im[G(jω)] < for sufficiently large ω, 4) G(jω) intersects the imaginary axis at ω =.55 with Im[G(j.55)] >, and 3 5) G(jω) intersects the real axis at ω =, with Re[G(j 3/)] = =.485 <. 7 Therefore, the Nyquist plot of G(s) is given in Figure 7. 4 Nyquist Diagram Figure 7: The Nyquist plot of G(s) in (b). Since G(s) has two unstable poles, the closed-loop system is stable if, and only if, #CCW #CW= = where the encirclement is around (, j). And, from the Nyquist plot in Figure 7, we have #CCW #CW=. Therefore, the closed-loop system is stable. Problem 5. ( points) Sketch the Bode plots of the following systems: (a) (6 points) G(s) = (b) (6 points) G(s) = (s + ) 3 (s.) (s + )(s + ) s + s + (s +.)(s +.s + ) Solution. (a) An approximate Ĝ(jω) is obtained in the following: ) For < ω, log Ĝ(jω) = log G(j) = log = : db/dec Ĝ(jω) = G(j) =. ) For < ω, log Ĝ(jω) = log Ĝ(j ) + ( ) (log ω log ) = 4(log ω + ) : 4 db/dec Ĝ(jω) = Ĝ(j ) ( 9 ) = + 8 = 8.

9 3) For < ω, 4) For < ω, 5) For < ω 3, log Ĝ(jω) = log Ĝ(j) + ( + 3) (log ω log ) = 6 + log ω : db/dec Ĝ(jω) = Ĝ(j) = = 45. log Ĝ(jω) = log Ĝ(j) + ( + 3 ) (log ω log ) = 4 : db/dec Ĝ(jω) = Ĝ(j) 9 = 45 9 = 36. log Ĝ(jω) = log Ĝ(j ) + ( + 3 ) (log ω log ) = 4 (log ω ) : db/dec Ĝ(jω) = Ĝ(j ) 9 = 36 9 = 7. The approximate Ĝ(jω) and the Bode plot of G(s) are given in Figure 8. Magnitude (db) Bode Diagram Frequency (rad/sec) Phase (deg) Frequency (rad/sec) Figure 8: The approximate Ĝ(jω) and the Bode plot of G(s) in (a). (b) The system G(s) can be rewritten as G(s) = s + ζ ω n s + ω n (s +.)(s + ζ ω n s + ω n) where ζ =., ω n =, ζ =.5, and ω n =. An approximate Ĝ(jω) is obtained in the following: ) For < ω, log Ĝ(jω) = log G(j) = log = : db/dec Ĝ(jω) = G(j) =.

10 ) For < ω, 3) For < ω, 4) For < ω, log Ĝ(jω) = log Ĝ(j ) + ( ) (log ω log ) = (log ω + ) : db/dec Ĝ(jω) = Ĝ(j ) 9 = 9 = 9. log Ĝ(jω) = log Ĝ(j) + ( ) (log ω log ) = 6 log ω : 6 db/dec Ĝ(jω) = Ĝ(j) 9 = 9 8 = 7. log Ĝ(jω) = log Ĝ(j) + ( + ) (log ω log ) = 6 (log ω ) : db/dec Ĝ(jω) = Ĝ(j) + 9 = = 9. The approximate Ĝ(jω) and the Bode plot of G(s) are given in Figure 9. Magnitude (db) Bode Diagram Frequency (rad/sec) Phase (deg) Frequency (rad/sec) Figure 9: The approximate Ĝ(jω) and the Bode plot of G(s) in (b). In the Bode plot, the peaks in magnitude and the sharp changes in phase are caused by ζ =. and ζ =.5.