6003 where A = jg(j!)j ; = tan Im [G(j!)] Re [G(j!)] = \G(j!) 2. (a) Calculate the magnitude and phae of G() = + 0 by hand for! =, 2, 5, 0, 20, 50, and 00 rad/ec. (b) ketch the aymptote for G() according to the Bode plot rule, and compare thee with your computed reult from part (a). Solution: (a) G() = jg(j!)j = + 0 ; G(j!) = = 0 j! 00 +! 2 0 + j! p ; \G(j!) = tan! 00 +! 2 0! jg(j!)j \G(j!) 2 5 0 20 50 00 0:0995 0:098 0:0894 0:0707 0:0447 0:096 0:00995 5:7 :3 26:6 45:0 63:4 78:7 84:3 (b) The Bode plot i :
Phae (deg) Magnitude 6004 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD 0 0 Bode plot for Problem 6.2 0 0 2 0 3 0 0 0 0 0 2 0 3 20 0 20 40 60 80 00 0 0 0 0 0 2 0 3 3. Sketch the aymptote of the Bode plot magnitude and phae for each of the following open-loop tranfer function. After completing the hand ketche verify your reult uing MATLAB. Turn in your hand ketche and the MATLAB reult on the ame cale. (a) L() = (b) L() = (c) L() = (d) L() = (e) L() = (f) L() = 2000 ( + 200) 00 (0: + )(0:5 + ) ( + )(0:02 + ) ( + ) 2 ( 2 + 2 + 4) 0( + 4) ( + )( 2 + 2 + 5) 000( + 0:) ( + )( 2 + 8 + 64)
Phae (deg) Magnitude 6005 (g) L() = (h) L() = (i) L() = ( + 5)( + 3) ( + )( 2 + + 4) 4( + 0) ( + 00)(4 2 + 5 + 4) ( + )( + 0)( 2 + 2 + 2500) Solution: (a) L() = 0 200 + Bode plot for Prob. 6.3 (a) 0 0 0 5 0 0 2 0 3 0 4 80 00 20 40 60 80 200 0 0 2 0 3 0 4 (b) L() = 00 0 + 2 +
Phae (deg) Magnitude 6006 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode plot for Prob. 6.3 (b) 0 0 0 2 0 0 0 0 0 2 50 00 50 200 250 300 0 2 0 0 0 0 0 2 (c) L() = (+)(0:02+)
Phae (deg) Magnitude 6007 Bode plot for Prob. 6.3 (c) 0 0 0 5 0 2 0 0 0 0 0 2 0 3 50 00 50 200 250 300 0 2 0 0 0 0 0 2 0 3 (d) L() = h ( + ) 2 4 2 2 + 2 + i
Phae (deg) Magnitude 6008 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD 0 0 Bode plot for Prob. 6.3 (d) 0 2 0 4 0 00 200 300 0 0 0 0 0 0 0 0 8 4 (e) L() = + 2 ( + ) p5 + 2 5 +
Phae (deg) Magnitude 6009 0 5 Bode plot for Prob. 6.3 (e) 0 0 0 5 0 2 0 0 0 0 0 2 50 00 50 200 250 300 350 0 2 0 0 0 0 0 2 25 (f) L() = 6 (0 + ) h ( + ) 8 2 + 8 + i
Phae (deg) Magnitude 600 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD 0 5 Bode plot for Prob. 6.3 (f) 0 0 0 3 0 2 0 0 0 0 0 2 0 00 200 300 0 3 0 2 0 0 0 0 0 2 (g) L() = h ( + ) 5 4 5 + 2 3 + i 2 + 4 +
Phae (deg) Magnitude 60 Bode plot for Prob. 6.3 (g) 0 0 0 2 0 0 0 0 0 2 0 00 200 300 400 0 2 0 0 0 0 0 2 (h) L() = 0 0 + 00 + 2 + 5 4 +
Phae (deg) Magnitude 602 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode plot for Prob. 6.3 (h) 0 0 2 0 3 0 2 0 0 0 0 0 2 0 3 00 0 00 0 2 0 0 0 0 0 2 0 3 (i) L() = 25000 ( + ) 0 + h 50 2 + 250 + i
Phae (deg) Magnitude 603 Bode plot for Prob. 6.3 (i) 0 5 0 2 0 0 0 0 0 2 0 3 00 0 00 200 300 0 2 0 0 0 0 0 2 0 3 4. Real pole and zero. Sketch the aymptote of the Bode plot magnitude and phae for each of the following open-loop tranfer function. After completing the hand ketche verify your reult uing MATLAB. Turn in your hand ketche and the MATLAB reult on the ame cale. (a) L() = (b) L() = (c) L() = (d) L() = Solution: (a) L() = ( + )( + 5)( + 0) ( + 2) ( + )( + 5)( + 0) ( + 2)( + 6) ( + )( + 5)( + 0) ( + 2)( + 4) ( + )( + 5)( + 0) ( + ) 50 5 + 0 +
Phae (deg) Magnitude 604 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode plot for Prob. 6.4 (a) 0 0 0 0 0 2 0 0 0 0 0 2 0 3 0 00 200 300 400 0 2 0 0 0 0 0 2 0 3 Frequency (rad/ec) (b) L() = 25 2 + ( + ) 5 + 0 +
Phae (deg) Magnitude 605 Bode plot for Prob. 6.4 (b) 0 0 0 0 0 2 0 0 0 0 0 2 0 3 0 00 200 300 0 2 0 0 0 0 0 2 0 3 (c) L() = 6 25 2 + 6 + ( + ) 5 + 0 +
Phae (deg) Magnitude 606 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode plot for Prob. 6.4 (c) 0 0 0 5 0 0 0 2 0 0 0 0 0 2 0 3 0 50 00 50 200 250 0 2 0 0 0 0 0 2 0 3 (d) L() = 4 25 2 + 4 + ( + ) 5 + 0 +
Phae (deg) Magnitude 607 Bode plot for Prob. 6.4 (d) 0 0 0 5 0 0 0 2 0 0 0 0 0 2 0 3 0 50 00 50 200 250 0 2 0 0 0 0 0 2 0 3 5. Complex pole and zero Sketch the aymptote of the Bode plot magnitude and phae for each of the following open-loop tranfer function and approximate the tranition at the econd order break point baed on the value of the damping ratio. After completing the hand ketche verify your reult uing MATLAB. Turn in your hand ketche and the MATLAB reult on the ame cale. (a) L() = 2 + 3 + 0 (b) L() = ( 2 + 3 + 0) (c) L() = (2 + 2 + 8) ( 2 + 2 + 0) (d) L() = (2 + 2 + 2) ( 2 + 2 + 0) (e) L() = (2 + ) ( 2 + 4)
Phae (deg) Magnitude 608 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (f) L() = (2 + 4) ( 2 + ) Solution: 0 (a) L() = 2 p 0 + 3 0 + 0 0 Bode plot for Prob. 6.5 (a) 0 5 0 0 0 0 0 2 50 0 50 00 50 200 0 0 0 0 0 2 (b) L() = 0 p 0 2 + 3 0 +
Phae (deg) Magnitude 609 Bode plot for Prob. 6.5 (b) 0 0 0 5 0 0 0 0 0 0 0 2 50 00 50 200 250 300 0 0 0 0 0 2 (c) L() = 2 4 5 2 p 2 + 4 + 2 p 0 + 5 +
Phae (deg) Magnitude 6020 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode plot for Prob. 6.5 (c) 0 0 0 0 0 0 0 2 0 50 00 50 0 0 0 0 0 2 (d) L() = 2 6 5 2 p 3 + 6 + 2 p 0 + 5 +
Phae (deg) Magnitude 602 Bode plot for Prob. 6.5 (d) 0 0 0 2 0 0 0 0 0 2 50 00 50 200 0 0 0 0 0 2 4 (e) L() = (2 + ) h i 2 + 2
Phae (deg) Magnitude 6022 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode plot for Prob. 6.5 (e) 0 0 0 2 0 0 0 0 50 00 50 0 50 00 50 0 0 0 0 (f) L() = h 4 i 2 2 + ( 2 + )
Phae (deg) Magnitude 6023 Bode plot for Prob. 6.5 (f) 0 0 50 00 0 0 0 0 50 00 50 0 50 0 0 0 0 6. Multiple pole at the origin Sketch the aymptote of the Bode plot magnitude and phae for each of the following open-loop tranfer function. After completing the hand ketche verify your reult uing MATLAB. Turn in your hand ketche and the MATLAB reult on the ame cale. (a) L() = 2 ( + 8) (b) L() = 3 ( + 8) (c) L() = 4 ( + 8) ( + 3) (d) L() = 2 ( + 8) ( + 3) (e) L() = 3 ( + 4) (f) L() = ( + )2 3 ( + 4)
Phae (deg) Magnitude 6024 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (g) L() = ( + )2 3 ( + 0) 2 Solution: (a) L() = 2 8 8 + Bode plot for Prob. 6.6 (a) 0 0 0 5 0 0 0 0 0 2 50 200 250 300 0 0 0 0 0 2 (b) L() = 3 8 8 +
Phae (deg) Magnitude 6025 Bode plot for Prob. 6.6 (b) 0 0 0 0 0 0 0 0 0 2 250 300 350 400 0 0 0 0 0 2 (c) L() = 4 8 8 +
Phae (deg) Magnitude 6026 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode plot for Prob. 6.6 (c) 0 0 0 0 0 0 0 0 0 2 50 0 50 00 50 0 0 0 0 0 2 (d) L() = 3 8 2 3 + 8 +
Phae (deg) Magnitude 6027 0 5 Bode plot for Prob. 6.6 (d) 0 0 0 5 0 0 0 0 0 2 50 00 50 200 0 0 0 0 0 2 (e) L() = 3 4 3 3 + 4 +
Phae (deg) Magnitude 6028 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD 0 5 Bode plot for Prob. 6.6 (d) 0 0 0 5 0 0 0 0 0 2 50 00 50 200 0 0 0 0 0 2 (f) L() = 4 3 ( + )2 4 +
Phae (deg) Magnitude 6029 Bode plot for Prob. 6.6 (f) 0 5 0 0 0 5 0 2 0 0 0 0 0 2 50 00 50 200 250 300 0 2 0 0 0 0 0 2 (g) L() = 00 3 ( + )2 0 + 2
Phae (deg) Magnitude 6030 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD 0 5 Bode plot for Prob. 6.6 (g) 0 0 0 5 0 2 0 0 0 0 0 2 50 00 50 200 250 300 0 2 0 0 0 0 0 2 7. Mixed real and complex pole Sketch the aymptote of the Bode plot magnitude and phae for each of the following open-loop tranfer function. After completing the hand ketche verify your reult uing Matlab. Turn in your hand ketche and the Matlab reult on the ame cale. (a) L() = (b) L() = (c) L() = ( + 2) ( + 0)( 2 + 2 + 2) ( + 2) 2 ( + 0)( 2 + 6 + 25) ( + 2) 2 2 ( + 0)( 2 + 6 + 25) (d) L() = ( + 2)(2 + 4 + 68) 2 ( + 0)( 2 + 4 + 85) (e) L() = [( + )2 + ] 2 ( + 2)( + 3) Solution:
Phae (deg) Magnitude 603 (a) L() = 0 0 + 2 + p 2 2 + + Bode plot for Prob. 6.7 (a) 0 0 0 5 0 2 0 0 0 0 0 2 50 00 50 200 250 300 0 2 0 0 0 0 0 2 (b) L() = 2 25 h 0 + 5 2 + i 2 + 6 25 +
Phae (deg) Magnitude 6032 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode plot for Prob. 6.7 (b) 0 0 0 5 0 0 0 2 0 0 0 0 0 2 00 200 300 400 0 2 0 0 0 0 0 2 (c) L() = 2 2 25 h 0 + 5 2 + 2 2 + 6 25 + i
Phae (deg) Magnitude 6033 Bode plot for Prob. 6.7 (c) 0 0 0 5 0 2 0 0 0 0 0 2 0 00 200 300 0 2 0 0 0 0 0 2 (d) L() = 4 25 2 2 p 7 0 + 2 + 2 + 7 + 2 p 85 + 4 85 +
Phae (deg) Magnitude 6034 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD 0 5 Bode plot for Prob. 6.7 (d) 0 0 0 5 0 2 0 0 0 0 0 2 00 0 00 200 0 2 0 0 0 0 0 2 (e) L() = 2 p2 3 + + 2 2 + 3 +
Phae (deg) Magnitude 6035 0 5 Bode plot for Prob. 6.7 (e) 0 0 0 5 0 2 0 0 0 0 0 2 0 50 00 50 200 0 2 0 0 0 0 0 2 8. Right half plane pole and zero Sketch the aymptote of the Bode plot magnitude and phae for each of the following open-loop tranfer function. Make ure the phae aymptote properly take the RHP ingularity into account by ketching the complex plane to ee how the \L() change a goe from 0 to +j: After completing the hand ketche verify your reult uing MATLAB. Turn in your hand ketche and the MATLAB reult on the ame cale. (a) L() = + 2 + 0 2 ; The model for a cae of magnetic levitation with lead compenation. (b) L() = + 2 ( + 0) ( 2 ; The magnetic levitation ytem with integral control and lead ) compenation. (c) L() = 2 (d) L() = 2 + 2 + ( + 20) 2 ( 2 2 + 2)
Phae (deg) Magnitude 6036 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (e) L() = (f) L() = ( + 2) ( )( + 6) 2 ( )[( + 2) 2 + 3] Solution: (a) L() = 5 2 + + 0 2 Bode plot for Prob. 6.8 (a) 0 0 0 5 0 0 0 0 0 2 00 50 200 0 0 0 0 0 2 (b) L() = 5 2 + ( + 0) 2
Phae (deg) Magnitude 6037 Bode plot for Prob. 6.8 (b) 0 0 0 5 0 2 0 0 0 0 0 2 50 200 250 300 0 2 0 0 0 0 0 2 (c) L() = 2
Phae (deg) Magnitude 6038 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD 0 5 Bode plot for Prob. 6.8 (c) 0 0 0 2 0 0 0 0 50 0 50 00 50 0 2 0 0 0 0 (d) L() = 20 + 2 p2 2 + 40 (2 + 2 + )
Phae (deg) Magnitude 6039 0 0 Bode plot for Prob. 6.8 (d) 0 5 0 2 0 0 0 0 0 2 300 200 00 0 00 0 2 0 0 0 0 0 2 8 2 (e) L() = + ( ) 6 + 2
Phae (deg) Magnitude 6040 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode plot for Prob. 6.8 (e) 0 0 0 5 0 2 0 0 0 0 0 2 00 50 200 250 300 0 2 0 0 0 0 0 2 (f) L() = 7 ( ) p7 2 + 4 7 +
Phae (deg) Magnitude 604 Figure 6.87: Magnitude portion of Bode plot for Problem 9 Bode plot for Prob. 6.8 (f) 0 0 0 5 0 2 0 0 0 0 0 2 00 50 200 250 300 0 2 0 0 0 0 0 2 9. A certain ytem i repreented by the aymptotic Bode diagram hown in Fig. 6.88. Find and ketch the repone of thi ytem to a unit tep input (auming zero initial condition). Solution: By inpection, the given aymptotic Bode plot i from
y(t) 6042 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Therefore, G() = 0(=0 + ) The repone to a unit tep input i : = + 0 Y () = G()U() = + 0 = + 0 2 y(t) = $ [Y ()] = (t) + 0t (t 0) 25 Prob. 6.9: Unit Step Repone 20 5 0 5 0 0 0.5.5 2 Time (ec) 0. Prove that a magnitude lope of in a Bode plot correpond to 20 db per decade or -6 db per octave. Solution: The de nition of db i db = 20 log jgj () Aume lope = d(logjgj) d(log!) = (2)
6074 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD N = 0; P = 0 =) Z = N + P = 0 The cloed-loop ytem i table for any K > 0: 20. Draw a Nyquit plot for KG() = K( + ) ( + 3) () chooing the contour to be to the right of the ingularity on the j!-axi. and determine the range of K for which the ytem i table uing the Nyquit Criterion. Then redo the Nyquit plot, thi time chooing the contour to be to the left of the ingularity on the imaginary axi and again check the range of K for which the ytem i table uing the Nyquit Criterion. Are the anwer the ame? Should they be? Solution : If you chooe the contour to the right of the ingularity on the origin, the Nyquit plot look like thi : From the Nyquit plot, the range of K for tability i K < 0 (N = 0; P = 0 =) Z = N + P = 0): So the ytem i table for K > 0: Similarly, in the cae with the contour to the left of the ingularity on the origin, the Nyquit plot i:
6075 Figure 6.89: Control ytem for Problem 2 From the Nyquit plot, the range of K for tability i K < 0 (N = ; P = =) Z = N + P = 0): So the ytem i table for K > 0: The way of chooing the contour around ingularity on the j!-axi doe not a ect it tability criterion. The reult hould be the ame in either way. However, it i omewhat le cumberome to pick the contour to the right of a pole on the imaginary axi o that there are no untable pole within the contour, hence P=0. 2. Draw the Nyquit plot for the ytem in Fig. 6.89. Uing the Nyquit tability criterion, determine the range of K for which the ytem i table. Conider both poitive and negative value of K. Solution : The characteritic equation: + K ( 2 + 2 + 2) ( + ) = 0 G() = ( + )( 2 + 2 + 2)
6076 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD For poitive K, note that the magnitude of the Nyquit plot a it croe the negative real axi i 0., hence K < 0 for tability. For negative K, the entire Nyquit plot i eentially ipped about the imaginary axi, thu the magnitude where it croe the negative real axi will be 0.5 and the tability limit i that jkj < 2 Therefore, the range of K for tability i 2 < K < 0: 22. (a) For! = 0: to 00 rad/ec, ketch the phae of the minimum-phae ytem G() = + + 0 =j! and the nonminimum-phae ytem G() = + 0 ; =j! noting that \(j! ) decreae with! rather than increaing. (b) Doe a RHP zero a ect the relationhip between the encirclement on a polar plot and the number of untable cloed-loop root in Eq. (6.28)? (c) Sketch the phae of the following untable ytem for! = 0: to 00 rad/ec: G() = + 0 : =j! (d) Check the tability of the ytem in (a) and (c) uing the Nyquit criterion on KG(). Determine the range of K for which the cloedloop ytem i table, and check your reult qualitatively uing a rough root-locu ketch.
Phae (deg) Magnitude 6077 Solution : (a) Minimum phae ytem, G (j!) = + + 0 j =j! Bode plot for Prob. 6.22 0 0 0 0 0 0 0 0 2 00 80 60 40 20 0 0 0 0 0 0 2 Non-minimum phae ytem, G 2 (j!) = + 0 j =j!
Phae (deg) Magnitude 6078 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode plot for Prob. 6.22 0 0 0 0 0 0 0 0 2 0 50 00 50 200 0 0 0 0 0 2 (b) No, a RHP zero doen t a ect the relationhip between the encirclement on the Nyquit plot and the number of untable cloed-loop root in Eq. (6.28). (c) Untable ytem: G 3 (j!) = + 0 j =j!
Phae (deg) Magnitude 6079 Bode plot for Prob. 6.22 (c) 0 0 0 0 0 0 0 0 2 0 50 00 50 200 0 0 0 0 0 2 i. Minimum phae ytem G (j!): For any K > 0; N = 0; P = 0 =) Z = 0 =) The ytem i table, a veri ed by the root locu being entirely in the LHP.
6080 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD ii. Non-minimum phae ytem G 2 (j!): the be encircled if K < : =K point will not 0 < K < N = 0; P = 0 =) Z = 0 =) Stable < K N = ; P = 0 =) Z = =) Untable Thi i veri ed by the Root Locu hown below where the branch of the locu to the left of the pole i from K < :
iii. Untable ytem G 3 (j!): The K > 0; however, P =, o 608 =K point will be encircled if 0 < K < 0 : N = 0; P = =) Z = =) Untable 0 < K : N = ; P = =) Z = 0 =) Stable Thi i veri ed by the Root Locu hown below right, where the locu croe the imaginary axi when K = 0, and tay in the LHP for K > 0: 23. Nyquit plot and their claical plane curve: Determine the Nyquit plot uing Matlab for the ytem given below with K = and verify that
6082 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD the beginning point and end point for the j! > 0 portion have the correct magnitude and phae: (a) the claical curve called Cayley Sextic, dicovered by Maclaurin in 78 KG() = K ( + ) 3 (b) the claical curve called the Cioid, meaning ivy-haped KG() = K ( + ) (c) the claical curve called the Folium of Kepler, tudied by Kepler in 609. KG() = K ( )( + ) 2 (d) the claical curve called the Folium (not Kepler ) KG() = K ( )( + 2) (e) the claical curve called the Nephroid, meaning kidney-haped. KG() = K 2( + )(2 4 + ) ( ) 3 (f) the claical curve called Nephroid of Freeth, named after the Englih mathematician T. J. Freeth. (g) a hifted Nephroid of Freeth Solution : KG() = K ( + )(2 + 3) 4( ) 3 KG() = K (2 + ) ( ) 3 Thee are all accomplihed by uing Matlab Nyquit function. All intereting hape. To check the magnitude and phae for each, plug in = 0 and = inf and then compare thoe value with the beginning and end point on the Nyquit diagram. (a)
Imaginary Axi Imaginary Axi 6083 Nyquit Diagram 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 Real Axi (b) Nyquit Diagram 20 5 0 5 0 5 0 5 20 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0. 0 Real Axi (c)
Imaginary Axi Imaginary Axi 6084 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD 0.4 Nyquit Diagram 0.3 0.2 0. 0 0. 0.2 0.3 0.4 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0. 0 Real Axi (d) Nyquit Diagram 0.2 0.5 0. 0.05 0 0.05 0. 0.5 0.2 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0. 0 Real Axi (e)
Imaginary Axi Imaginary Axi 6085 Nyquit Diagram 4 3 2 0 2 3 4 3 2 0 2 3 Real Axi (f) Nyquit Diagram 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 0.8 0.6 0.4 0.2 0 0.2 0.4 Real Axi (g)
Imaginary Axi 6086 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD 0.2 Nyquit Diagram 0.5 0. 0.05 0 0.05 0. 0.5 0.2 0.25 0.4 0.3 0.2 0. 0 0. Real Axi
6099 30. If a ytem ha the open-loop tranfer function G() =! 2 n ( + 2! n ) with unity feedback, then the cloed-loop tranfer function i given by! 2 n T () = 2 + 2! n +! 2 : n Verify the value of the PM hown in Fig. 6.36 for = 0:, 0.4, and 0.7. Solution : G() =! 2 n ( + 2! n ) ; T () = G() + G() =! 2 n 2 + 2! n +! 2 n PM from Eq. 6.32 PM from Fig. 6.36 PM from Bode plot 0. 0 0.4 (! = 0:99 rad/ec) 0.4 40 44 43. (! = 0:85 rad/ec) 0.7 70 65 65.2 (! = 0:65 rad/ec) 3. Conider the unity feedback ytem with the open-loop tranfer function G() = K ( + )[( 2 =25) + 0:4(=5) + ] : (a) Ue MATLAB to draw the Bode plot for G(j!) auming K =.
Phae (deg) Magnitude 600 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (b) What gain K i required for a PM of 45? What i the GM for thi value of K? (c) What i K v when the gain K i et for PM = 45? (d) Create a root locu with repect to K, and indicate the root for a PM of 45. Solution : (a) The Bode plot for K = i hown below and we can ee from margin that it reult in a PM = 48 o. 0 5 Bode plot for Prob. 6.3 0 0 0 5 0 0 00 200 300 0 2 0 0 0 0 0 2 0 400 0 2 0 0 0 0 0 2 (b) Although di cult to read the plot above, it i clear that a very light increae in gain will lower the P M to 45 o, o try K = :: The margin routine how that thi yield P M = 45 o and GM = 5 db. (c) K v = lim!0 fkg()g = K = : when K i et for PM=45 K v = :
60 (d) The characteritic equation for PM of 45 : + h ( + ) 5 : 2 + 0:4 i = 0 5 + =) 4 + 3 3 + 27 2 + 25 + 27:88 = 0 =) = :03 j4:78; 0:47 j0:97 32. For the ytem depicted in Fig. 6.94(a), the tranfer-function block are de ned by G() = ( + 2) 2 ( + 4) and H() = + : (a) Uing rlocu and rloc nd, determine the value of K at the tability boundary. (b) Uing rlocu and rloc nd, determine the value of K that will produce root with damping correponding to = 0:707. (c) What i the gain margin of the ytem if the gain i et to the value determined in part (b)? Anwer thi quetion without uing any frequency repone method. (d) Create the Bode plot for the ytem, and determine the gain margin that reult for PM = 65. What damping ratio would you expect for thi PM?
602 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Figure 6.94: Block diagram for Problem 32: (a) unity feedback; (b) H() in feedback (e) Sketch a root locu for the ytem hown in Fig. 6.94(b).. How doe it di er from the one in part (a)? (f) For the ytem in Fig. 6.94(a) and (b), how doe the tranfer function Y 2 ()=R() di er from Y ()=R()? Would you expect the tep repone to r(t) be di erent for the two cae? Solution : (a) The root locu croe j! axi at 0 = j2. K = jh( 0 )G( 0 )j j 0=j2 = jj2 + j jj2 + 4j jj2 + 2j 2 =) K = 80
603 (b) = 0:707 =) 0:707 = in =) = 45 From the root locu given, (c) = 0:9 + j0:9 K = jh( )G( )j j = (d) From the Root Locu : 0:9+j0:9 = j0:0 + j0:9j j3:09 + j0:9j j:09 + j0:9j 2 =) K = 5:9 GM = K a K b = 80 5:9 = 3:5 G()H() = ( + )( + 2) 2 ( + 4) PM=65 when K = 30. Intability occur when K = 80:0. =) GM = 2:67 We approximate the damping ratio by ' P M 00 ' 65 00 = 0:65
Phae (deg) Magnitude 604 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD 0 0 Bode plot for Prob. 6.32 0 5 0 0 0 2 0 0 0 0 0 2 Frequency (rad/ec) 0 00 200 300 400 0 2 0 0 0 0 0 2 (e) The root locu for Fig.6.94(a) i the ame a that of Fig.6.94(b). (f) Y () R() Y 2 () R() = = KG()H() + KG()H() = K ( + )( + 2) 2 ( + 4) + K KG() + KG()H() = K( + ) ( + )( + 2) 2 ( + 4) + K Y () R() Y2() and R() Y2() have the ame cloed-loop pole. However, R() ha doen t have a zero. We would therefore expect a zero, while Y() R() more overhoot from ytem (b). 33. For the ytem hown in Fig. 6.95, ue Bode and root-locu plot to determine the gain and frequency at which intability occur. What gain (or gain) give a PM of 20? What i the gain margin when PM = 20? Solution :
Phae (deg) Magnitude 605 Figure 6.95: Control ytem for Problem 33 0 5 Bode plot for Prob. 6.33 0 0 0 5 0 0 0 2 0 0 0 0 0 2 00 50 200 250 300 0 2 0 0 0 0 0 2
606 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD The ytem with K = give, GM = 52 (! = 5 rad/ec) P M = 0 (! = 0:65 rad/ec) Therefore, intability occur at K 0 = 52 and! = 5 rad/ec. From the Bode plot, a PM of 20 i given by, K = 3:9 (! = 0:33 rad/ec); GM = 52 3:9 = 3 K 2 = 49 (! = 4:6 rad/ec); GM = 52 49 = :06 34. A magnetic tape-drive peed-control ytem i hown in Fig. 6.96. The peed enor i low enough that it dynamic mut be included. The peed-meaurement time contant i m = 0:5 ec; the reel time contant i r = J=b = 4 ec, where b = the output haft damping contant = N m ec; and the motor time contant i = ec. (a) Determine the gain K required to keep the teady-tate peed error to le than 7% of the reference-peed etting. (b) Determine the gain and phae margin of the ytem. I thi a good ytem deign? Solution :
607 Figure 6.96: Magnetic tape-drive peed control (a) From Table 4., the error for thi Type ytem i e = + K j cj (b) Since the teady-tate peed error i to be le than 7% of the reference peed, + K p 0:07 and for the ytem in Fig. 6.96 with the number plugged in, we ee that K p = K: Therefore, K 3: jg()j = 0:79 at \GH = 80 =) GM = jghj = :3 \GH = 73 at jg()j = =) P M = \G + 80 = 7 GM i low =) The ytem i very cloe to intability. P M i low =) The damping ratio i low. =) High overhoot. We ee that to have a more table ytem we have to lower the gain. With mall gain, e will be higher. Therefore, thi i not a good deign, and need compenation. 35. For the ytem in Fig. 6.97, determine the Nyquit plot and apply the Nyquit criterion (a) to determine the range of value of K (poitive and negative) for which the ytem will be table, and (b) to determine the number of root in the RHP for thoe value of K for which the ytem i untable. Check your anwer uing a rough root-locu ketch.
608 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Figure 6.97: Control ytem for Problem 35, 69, and 70 Solution : (a) & b. 3 KG() = K ( + )( + 3) From the Nyquit plot above, we ee that: i. < K < 4 =) 0 < K < 4 There are no RHP open loop root, hence P = 0 for all cae. For 0 < K < 4; no encirclement of - o N = 0, N = 0; P = 0 =) Z = 0 ii. The cloed-loop ytem i table. No root in RHP. 4 < K < 0 =) 4 < K <
609 Two encirclement of the - point, hence N = 2; P = 0 =) Z = 2 Two cloed-loop root in RHP. iii. 0 < K =) K < 0 N = ; P = 0 =) Z = One cloed-loop root in RHP. The root loci below how the ame reult. 36. For the ytem hown in Fig. 6.98, determine the Nyquit plot and apply the Nyquit criterion. (a) to determine the range of value of K (poitive and negative) for which the ytem will be table, and (b) to determine the number of root in the RHP for thoe value of K for which the ytem i untable. Check your anwer uing a rough root-locu ketch. Solution : (a) & b. KG() = K + ( ) 2
60 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Figure 6.98: Control ytem for Problem 36 From the Nyquit plot we ee that: i. < K < 2 =) 0 < K < 2 N = 0; P = 2 =) Z = 2 Two cloed-loop root in RHP. ii. 2 < K < 0 =) 2 < K N = 2; P = 2 =) Z = 0 The cloed-loop ytem i table. iii. 0 < K < =) K < N = ; P = 2 =) Z = One cloed-loop root in RHP.
6 Figure 6.99: Control ytem for Problem 37 iv. < K < =) < K < 0 N = 0; P = 2 =) Z = 2 Two cloed-loop root in RHP. Thee reult are con rmed by looking at the root loci below: 37. For the ytem hown in Fig. 6.99, determine the Nyquit plot and apply the Nyquit criterion. (a) to determine the range of value of K (poitive and negative) for which the ytem will be table, and (b) to determine the number of root in the RHP for thoe value of K for which the ytem i untable. Check your anwer uing a rough root-locu ketch. Solution :
62 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (a) & b. KG() = K ( + ) 2 From the Nyquit plot we ee that: i. < K < =) 0 < K < ii. iii. N = 0; P = 0 =) Z = 0 The cloed-loop ytem i table. < K < 0 =) < K N = ; P = 0 =) Z = One cloed-loop root in RHP. 0 < K < 2 =) K < 2 iv. N = 2; P = 0 =) Z = 2 Two cloed-loop root in RHP. 2 < K =) 2 < K < 0 N = 0; P = 0 =) Z = 0 The cloed-loop ytem i table. Thee reult are con rmed by looking at the root loci below:
63 38. The Nyquit diagram for two table, open-loop ytem are ketched in Fig. 6.00. The propoed operating gain i indicated a K 0, and arrow indicate increaing frequency. In each cae give a rough etimate of the following quantitie for the cloed-loop (unity feedback) ytem: (a) phae margin (b) damping ratio (c) range of gain for tability (if any) (d) ytem type (0,, or 2). Solution :
64 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Figure 6.00: Nyquit plot for Problem 38 For both, with K = K 0 : N = 0; P = 0 =) Z = 0 Therefore, the cloed-loop ytem i table. Fig.6.02(a) Fig.6.02(b) a. PM '7 '45 b. Damping ratio 0.7(' 7 00 ) 45 0.45(' 00 ) c. To determine the range of gain for tability, call the value of K where the plot cro the negative real axi a K : For cae (a), K > K for tability becaue gain lower than thi amount will caue the - point to be encircled. For cae (b), K < K for tability becaue gain greater than thi amount will caue the - point to be encircled. d. For cae (a), the 360 o loop indicate two pole at the origin, hence the ytem i Type 2. For cae (b), the 80 o loop indicate one pole at the origin, hence the ytem i Type. 39. The teering dynamic of a hip are repreented by the tranfer function V () r () = G() = K[ (=0:42) + ] (=0:325 + )(=0:0362) + ) ; where v i the hip lateral velocity in meter per econd, and r i the rudder angle in radian. (a) Ue the MATLAB command bode to plot the log magnitude and phae of G(j!) for K = 0:2 (b) On your plot, indicate the croover frequency, PM, and GM,
Phae (deg) Magnitude 65 (c) I the hip teering ytem table with K = 0:2? (d) What value of K would yield a PM of 30 o and what would the croover frequency be? Solution : (a) The Bode plot for K = 0:2 i : 0 5 Bode plot 0 0 GM=.95 0.0867 rad/ec 0 5 0 4 0 3 0 2 0 0 0 0 00 200 80 deg PM= 23.7 deg 300 400 0 4 0 3 0 2 0 0 0 0 (b) From the Bode plot above :! c = 0:0867 rad/ec P M = 23:7 deg GM = :95 (c) Since P M < 0, the cloed-loop ytem with K = 0:2 i untable. (d) From the Bode plot above, we can get better P M by decreaing the gain K. Then we will nd that K = 0:042 yield P M = 30 at the croover frequency! c = 0:032 rad/ec. The Bode plot with K = 0:042 i :
Phae (deg) Magnitude (db) 66 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD 50 Bode Diagram Gm = 7.68 db (at 0.0576 rad/ec), Pm = 30 deg (at 0.032 rad/ec) 0 50 00 270 80 90 0 0 3 0 2 0 0 0 0 Frequency (rad/ec) 40. For the open-loop ytem KG() = K( + ) 2 ( + 0) 2 : Determine the value for K at the tability boundary and the value of K at the point where P M = 30. Solution : The bode plot of thi ytem with K = i :
Phae (deg) Magnitude (db) 67 Bode Diagram Gm = 64. db (at 8.94 rad/ec), Pm = 4.58 deg (at 0. rad/ec) 00 50 0 50 00 50 35 80 225 270 0 2 0 0 0 0 0 2 Frequency (rad/ec) Since GM = 64: db (' 600), the range of K for tability i : K < 600 From the Bode plot, the magnitude at the frequency with 50 phae i 0:088 ( 34:5 db) at 0:8282 rad/ec and 0:0098 ( 54: db) at 4:44 rad/ec. Therefore, the value of K at the point where P M = 30 i : K = K = 0:088 = 53:2; 0:0098 = 505
68 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Figure 6.0: Magnitude frequency repone for Problem 4 (a) Problem and Solution for Section 6.5 4. The frequency repone of a plant in a unity feedback con guration i ketched in Fig. 6.0. Aume the plant i open-loop table and minimum phae. (a) What i the velocity contant K v for the ytem a drawn? (b) What i the damping ratio of the complex pole at! = 00? (c) What i the PM of the ytem a drawn? (Etimate to within 0 o.) Solution : (a) From Fig. 6.0, (b) Let K v = lim!0 G = jlow frequency aymptote of G(j!)j!= = 00) G () = 2! n + 2! n + For the econd order ytem G (), jg (j!)j!= = 2 ()
69 From Fig. 6.0 : jg (j!)j!=00 = jg(j!)j!=00 jaymptote of G(j!)j!=00 = 0:4 0:2 = 2 (2) From () and (2) we have : 2 = 2 =) = 0:25 (c) Since the plant i a minimum phae ytem, we can apply the Bode approximate gain-phae relationhip. When jgj =, the lope of jgj curve i = -2. =) \G(j!) = 2 90 = 80 P M = \G(j!) + 80 = 0 Note : Actual PM by Matlab calculation i 6.4 ; o thi approximation i within the deired accuracy. 42. For the ytem G() = 00(=a + ) ( + )(=b + ) ; where b = 0a, nd the approximate value of a that will yield the bet PM by ketching only candidate value of the frequency repone magnitude. Solution :
Phae (deg) Magnitude 620 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode Diagram a=3.6 0 0 80 00 20 40 60 0 0 0 0 0 2 0 3 Frequency (rad/ec) a=3.6 80 200 Uncompenated 0 0 0 0 0 2 0 3 a=0 Uncompenated a=0 Without the zero and pole that contain the a & b term, the plot of jgj how a lope of 2 at the jgj = croover at 0 rad/ec. We clearly need to intall the zero and pole with the a & b term omewhere at frequencie greater rad/ec. Thi will increae the lope from -2 to - between the zero and pole. So the problem impli e to electing a o that the - lope region between the zero and pole bracket the croover frequency. That cenario will maximize the PM. Referring to the plot above, we ee that 3:6 < a < 0, make the lope of the aymptote of jgj be at the croover and repreent the two extreme of poibilitie for a - lope. The maximum PM will occur half way between thee extreme on a log cale, or =) a = p 3:6 0 = 5:6
62 Note : Actual PM i a follow : P M = 46:8 for a = 3:6 (! c = 25:0 rad/ec) P M = 58: for a = 5:6 (! c = 7:8 rad/ec) P M = 49:0 for a = 0 (! c = 2:6 rad/ec)
Phae (deg) Magnitude (db) 622 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Problem and Solution for Section 6.6 43. For the open-loop ytem KG() = K( + ) 2 ( + 0) 2 : Determine the value for K that will yield PM 30 and the maximum poible cloed-loop bandwidth. Ue MATLAB to nd the bandwidth. Solution : From the reult of Problem 6.39., the value of K that will yield P M 30 i : 53:2 K 505 The maximum cloed-loop bandwidth will occur with the maximum gain K within the allowable region; therefore, the maximum bandwidth will occur with K = 505: The Bode plot of the cloed loop ytem with K = 505 i : 50 Bode Diagram Gm = 0 db (at 8.94 rad/ec), Pm = 30 deg (at 4.36 rad/ec) 00 50 0 50 00 35 80 225 270 0 2 0 0 0 0 0 2 Frequency (rad/ec)
623 Looking at the point with Magnitude 0.707(-3 db), the maximum poible cloed-loop bandwidth i :! BW; max ' 7:7 rad/ec. Problem and Solution for Section 6.7 44. For the lead compenator where <. D() = T + T + ; (a) Show that the phae of the lead compenator i given by = tan (T!) tan (T!): (b) Show that the frequency where the phae i maximum i given by! max = T p ; and that the maximum phae correpond to in max = + : (c) Rewrite your expreion for! max to how that the maximum-phae frequency occur at the geometric mean of the two corner frequencie on a logarithmic cale: log! max = log 2 T + log : T (d) To derive the ame reult in term of the pole-zero location, rewrite D() a D() = + z + p ; and then how that the phae i given by! = tan jzj uch that! max = p jzjjpj: tan! jpj Hence the frequency at which the phae i maximum i the quare root of the product of the pole and zero location. ;
624 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Solution : (a) The frequency repone i obtained by letting = j!, D(j!) = K T j! + T j! + The phae i given by, = tan (T!) tan (T!) (b) Uing the trigonometric relationhip, then and ince, then tan(a B) = tan(a) tan(b) + tan(a) tan(b) tan() = T! T! T 2! 2 in 2 () = tan2 () + tan 2 ()! in() = 2 T 2 ( ) 2 + 2! 4 T 4 + ( + 2 )! 2 T 2 To determine the frequency at which the phae i a maximum, let u et the derivative with repect to! equal to zero, which lead to d in() d! = 0 2!T 2 ( ) 2 (! 4 T 4 ) = 0 The value! = 0 give the maximum of the function and etting the econd part of the above equation to zero then, or! 4 = 2 T 4! max = p T The maximum phae contribution, that i, the peak of the \D() curve correpond to, or in max = + = in max + in max tan max =! maxt! max T +! 2 maxt 2 = 2 p
625 (c) The maximum frequency occur midway between the two break frequencie on a logarithmic cale, a hown in Fig. 6.53. p log! max = log p T T = log p + log p T T = log 2 T + log T (d) Alternatively, we may tate thee reult in term of the pole-zero location. Rewrite D() a, ( + z) D() = K ( + p) then and or (j! + z) D(j!) = K (j! + p)! = tan jzj! jzj tan = +!! tan jpj! jpj! jzj jpj Setting the derivative of the above equation to zero we nd,! jzj! +!2 jpj jzj jpj 2!! jzj jpj jzj! = 0 jpj and and! max = p jzj jpj log! max = (log jzj + log jpj) 2 Hence the frequency at which the phae i maximum i the quare root of the product of the pole and zero location. 45. For the third-order ervo ytem G() = 50; 000 ( + 0)( + 50) :
Phae (deg) Magnitude (db) 626 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Deign a lead compenator o that PM 50 and! BW 20 rad/ec uing Bode plot ketche, then verify and re ne your deign uing Matlab. Solution : Let deign the lead compenator o that the ytem ha P M 50 &! W B '! c 20 rad/ec. The Bode plot of the given ytem i : 00 Bode Diagram 50 0 50 00 90 35 80 225 270 0 0 0 0 0 2 0 3 Frequency (rad/ec) Start with a lead compenator deign with : D() = T + ( < ) T + Since the open-loop croover frequency! c ('! BW ) i already above 20 rad/ec, we are going to jut add extra phae around! =! c in order to atify P M = 50. Let add phae lead 60. From Fig. 6.54, ' 20 =) chooe = 0:05
Phae (deg) Magnitude (db) 627 To apply maximum phae lead at! = 20 rad/ec;! = p T = 20 =) T = 4:48; T = 89:4 Therefore by applying the lead compenator with ome gain adjutment : D() = 0:2 4:5 + 90 + we get the compenated ytem with : P M = 65 ;! c = 22 rad/ec, o that! BW & 25 rad/ec. The Bode plot with deigned compenator i : 50 Bode Diagram Gm = 5.9 db (at 72.6 rad/ec), Pm = 65.3 deg (at 22 rad/ec) 0 50 00 45 90 35 80 225 270 0 0 0 0 0 2 0 3 Frequency (rad/ec)
628 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Figure 6.02: Control ytem for Problem 46 46. For the ytem hown in Fig. 6.02, uppoe that G() = 5 ( + )(=5 + ) : Deign a lead compenation D() with unity DC gain o that PM 40 uing Bode plot ketche, then verify and re ne your deign uing Matlab. What i the approximate bandwidth of the ytem? Solution : Start with a lead compenator deign with : D() = T + T + which ha unity DC gain with <. The Bode plot of the given ytem i :
Phae (deg) Magnitude (db) 629 Bode Diagram Gm =.58 db (at 2.24 rad/ec), Pm = 3.94 deg (at 2.04 rad/ec) 00 50 0 50 00 90 35 80 225 270 0 2 0 0 0 0 0 2 Frequency (rad/ec) Since P M = 3:9, let add phae lead 60. From Fig. 6.53, ' 20 =) chooe = 0:05 To apply maximum phae lead at! = 0 rad/ec;! = p T = 0 =) T = 2:2; T = 45 Therefore by applying the lead compenator : D() = 2:2 + 45 + we get the compenated ytem with : P M = 40 ;! c = 2:5 The Bode plot with deigned compenator i :
Phae (deg) Magnitude (db) 630 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD 00 Bode Diagram Gm = 24. db (at 2.8 rad/ec), Pm = 40.2 deg (at 2.49 rad/ec) 50 0 50 00 50 90 35 80 225 270 0 2 0 0 0 0 0 2 0 3 Frequency (rad/ec) From Fig. 6.5, we ee that! BW ' 2! c ' 5 rad/ec. 47. Derive the tranfer function from T d to for the ytem in Fig. 6.70. Then apply the Final Value Theorem (auming T d = contant) to determine whether () i nonzero for the following two cae: (a) When D() ha no integral term: lim!0 D() = contant; (b) When D() ha an integral term: where lim!0 D 0 () = contant. Solution : The tranfer function from T d to : D() = D0 () ; 0:9 () T d () = 2 + 0:9 2 2 +2 D()
63 where T d () = jt d j =. (a) Uing the nal value theorem : (b) () = lim (t) = lim(t) = lim t!!0!0 = jt d j lim D() = jt dj contant 6= 0!0 0:9 2 2 (+2)+:8D() 2 (+2) jt d j Therefore, there will be a teady tate error in for a contant T d input if there i no integral term in D(). () = lim (t) = lim(t) = lim t!!0!0 = 0 :8lim!0 D 0 () = 0 0:9 2 3 (+2)+:8D 0 () 3 (+2) jt d j So when D() contain an integral term, a contant T d input will reult in a zero teady tate error in. 48. The inverted pendulum ha a tranfer function given by Eq. (2.3), which i imilar to G() = 2 : (a) Deign a lead compenator to achieve a PM of 30 uing Bode plot ketche, then verify and re ne your deign uing MATLAB. (b) Sketch a root locu and correlate it with the Bode plot of the ytem. (c) Could you obtain the frequency repone of thi ytem experimentally? Solution : (a) Deign the lead compenator : D() = K T + T + uch that the compenated ytem ha P M ' 30 &! c ' rad/ec. (Actually, the bandwidth or peed of repone wa not peci ed, o any croover frequency would atify the problem tatement.) = in(30 ) + in(30 ) = 0:32
Phae (deg) Magnitude (db) 632 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD To apply maximum phae lead at! = rad/ec;! = p T = =) T = 0:57; T = :77 Therefore by applying the lead compenator : D() = K 0:57 + :77 + By adjuting the gain K o that the croover frequency i around rad/ec, K = :3 reult in : P M = 30:8 The Bode plot of compenated ytem i : 20 Bode Diagram Gm =.06 db (at 0 rad/ec), Pm = 30.8 deg (at 0.986 rad/ec) 0 20 40 60 80 40 50 60 70 80 0 2 0 0 0 0 0 2 Frequency (rad/ec) (b) Root Locu of the compenated ytem i :
633 and con rm that the ytem yield all table root with reaonable damping. However, it would be a better deign if the gain wa raied ome in order to increae the peed of repone of the low real root. A mall decreae in the damping of the complex root will reult. (c) No, becaue the inuoid input will caue the ytem to blow up becaue the open loop ytem i untable. In fact, the ytem will blow up even without the inuoid applied. Or, a better decription would be that the pendulum will fall over until it hit the table. 49. The open-loop tranfer function of a unity feedback ytem i G() = K (=5 + )(=50 + ) : (a) Deign a lag compenator for G() uing Bode plot ketche o that the cloed-loop ytem ati e the following peci cation: i. The teady-tate error to a unit ramp reference input i le than 0.0. ii. PM = 40 (b) Verify and re ne your deign uing Matlab. Solution : Let deign the lag compenator : D() = T + T + ; >
Phae (deg) Magnitude (db) 634 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD From the rt peci cation, Steady-tate error to unit ramp = lim!0 =) K < 0:0 D()G() + D()G() 2 2 < 0:0 =) Chooe K = 50 Uncompenated, the croover frequency with K = 50 i too high for a good P M. With ome trial and error, we nd that the lag compenator, D() = 0:2 + 0:0 + will lower the croover frequency to! c ' 4:46 rad/ec where the P M = 40:7. 50 00 Bode Diagram Gm = 8.9 db (at 5.5 rad/ec), Pm = 40.7 deg (at 4.46 rad/ec) 50 0 50 00 50 90 35 80 225 270 0 4 0 3 0 2 0 0 0 0 0 2 0 3 Frequency (rad/ec)
635 50. The open-loop tranfer function of a unity feedback ytem i G() = K (=5 + )(=200 + ) : (a) Deign a lead compenator for G() uing Bode plot ketche o that the cloed-loop ytem ati e the following peci cation: i. The teady-tate error to a unit ramp reference input i le than 0.0. ii. For the dominant cloed-loop pole the damping ratio 0:4. (b) Verify and re ne your deign uing Matlab including a direct computation of the damping of the dominant cloed-loop pole. Solution : Let deign the lead compenator : D() = T + T + ; < From the rt peci cation, Steady-tate error to unit ramp = lim!0 D()G() + D()G() 2 2 < 0:0 =) K < 0:0 =) Chooe K = 50 From the approximation ' P M, the econd peci cation implie P M 00 40. After trial and error, we nd that the compenator, D() = 0 + 00 + reult in a P M = 42:5 and a croover frequency! c ' 5:2 rad/ec a hown by the margin output:
Phae (deg) Magnitude (db) 636 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD 00 Bode Diagram Gm = 3.3 db (at 36 rad/ec), Pm = 42.5 deg (at 52.2 rad/ec) 50 0 50 00 50 90 35 80 225 270 0 0 0 0 0 2 0 3 0 4 Frequency (rad/ec) and the ue of damp veri e the damping to be = 0:42 for the complex cloed-loop root which exceed the requirement. 5. A DC motor with negligible armature inductance i to be ued in a poition control ytem. It open-loop tranfer function i given by G() = 50 (=5 + ) : (a) Deign a compenator for the motor uing Bode plot ketche o that the cloed-loop ytem ati e the following peci cation: i. The teady-tate error to a unit ramp input i le than /200. ii. The unit tep repone ha an overhoot of le than 20%. iii. The bandwidth of the compenated ytem i no le than that of the uncompenated ytem. (b) Verify and/or re ne your deign uing Matlab including a direct computation of the tep repone overhoot.
637 Solution : The rt peci cation implie that a loop gain greater than 200 i required. Since the open loop gain of the plant i 50, a gain from the compenator, K; i required where K > 4 =) o chooe K = 5 From Figure 3.23, we ee that the econd peci cation implie that : Overhoot < 20% =) > 0:5 =) P M > 50 A ketch of the Bode aymptote of the open loop ytem with the required loop gain how a croover frequency of about 30 rad/ec at a lope of -2; hence, the PM will be quite low. To add phae with no decreae in the croover frequency, a lead compenator i required. Figure 6.53 how that a lead ratio of 0: will provide about 55 o of phae increae and the aymptote ketch how that thi increae will be centered at the croover frequency if we elect the break point at D() = 20 + 200 + : Ue of Matlab margin routine how that thi compenation reult in a P M = 59 and a croover frequency! c ' 60 rad/ec.
Phae (deg) Magnitude (db) 638 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD 00 Bode Diagram Gm = Inf db (at Inf rad/ec), Pm = 59.5 deg (at 62.4 rad/ec) 50 0 50 00 90 35 80 0 0 0 0 0 2 0 3 0 4 Frequency (rad/ec) and uing the tep routine on the cloed loop ytem how the tep repone to be le than the maximum allowed 20%. 52. The open-loop tranfer function of a unity feedback ytem i G() = K ( + =5)( + =20) : (a) Sketch the ytem block diagram including input reference command and enor noie. (b) Deign a compenator for G() uing Bode plot ketche o that the cloed-loop ytem ati e the following peci cation: i. The teady-tate error to a unit ramp input i le than 0.0. ii. PM 45 iii. The teady-tate error for inuoidal input with! < 0:2 rad/ec i le than /250. iv. Noie component introduced with the enor ignal at frequencie greater than 200 rad/ec are to be attenuated at the output by at leat a factor of 00,.
Phae (deg) Magnitude 639 (c) Verify and/or re ne your deign uing Matlab including a computation of the cloed-loop frequency repone to verify (iv). Solution : a. The block diagram how the noie, v, entering where the enor would be: b. The rt peci cation implie K v 00 and thu K 00: The bode plot with K = and D = below how that there i a negative PM but all the other pec are met. The eaiet way to ee thi i to hand plot the aymptote and mark the contraint that the gain mut be 250 at! 0:2 rad/ec and the gain mut be 0:0 for! 200 rad/ec. 0 5 Uncompenated Bode Plot 0 0 0 5 50 00 0 0 0 0 0 2 0 3 50 200 250 300 0 0 0 0 0 2 0 3
640 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD In fact, the pec are exceeded at the low frequency ide, and lightly exceeded on the high frequency ide. But it will be di cult to increae the phae at croover without violating the pec. From a hand plot of the aymptote, we ee that a combination of lead and lag will do the trick. Placing the lag according to D lag () = (=2 + ) (=0:2 + ) will lower the gain curve at frequencie jut prior to croover o that a - lope i more eaily achieved at croover without violating the high frequency contraint. In addition, in order to obtain a much phae at croover a poible, a lead according to D lead () = (=5 + ) (=50 + ) will preerve the - lope from! = 5 rad/ec to! = 20 rad/ec which will bracket the croover frequency and hould reult in a healthy P M: A look at the Bode plot how that all pec are met except the P M = 44: Perhap cloe enough, but a light increae in lead hould do the trick. So our nal compenation i (=2 + ) (=4 + ) D() = (=0:2 + ) (=50 + ) with K = 00: Thi doe meet all pec with P M = 45 o exactly, a can be een by examining the Bode plot below.
Phae (deg) Magnitude 64 0 5 Compenated Bode Plot 0 0 0 5 50 00 0 0 0 0 0 2 0 3 50 200 250 300 0 0 0 0 0 2 0 3 53. Conider a type I unity feedback ytem with G() = K ( + ) : Deign a lead compenator uing Bode plot ketche o that K v = 20 ec and PM > 40. Ue Matlab to verify and/or re ne your deign o that it meet the peci cation. Solution : Ue a lead compenation : D() = T + T + ; > From the peci cation, K v = 20 ec ; =) K v = limd()g() = K = 20!0 =) K = 20
Phae (deg) Magnitude (db) 642 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD From a hand ketch of the uncompenated Bode plot aymptote, we ee that the lope at croover i -2, hence the PM will be poor. In fact, an exact computation how that Adding a lead compenation P M = 2:75 (at! c = 4:42 rad/ec) D() = 3 + 30 + will provide a - lope in the vicinity of croover and hould provide plenty of PM. The Bode plot below veri e that indeed it did and how that the P M = 62 at a croover frequency = 7 rad/ec thu meeting all pec. 00 Bode Diagram Gm = Inf db (at Inf rad/ec), Pm = 6.8 deg (at 6.99 rad/ec) 50 0 50 00 90 35 80 0 2 0 0 0 0 0 2 0 3 Frequency (rad/ec) 54. Conider a atellite-attitude control ytem with the tranfer function G() = 0:05( + 25) 2 ( 2 + 0: + 4) :
643 Amplitude-tabilize the ytem uing lead compenation o that GM 2 (6 db), and PM 45, keeping the bandwidth a high a poible with a ingle lead. Solution : The ketch of the uncompenated Bode plot aymptote how that the lope at croover i -2; therefore, a lead compenator will be required in order to have a hope of meeting the PM requirement. Furthermore, the reonant peak need to be kept below magnitude o that it ha no chance of cauing an intability (thi i amplitude tabilization). Thi latter requirement mean we mut lower gain at the reonance. Uing the ingle lead compenator, D() = ( + 0:06) ( + 6) will lower the low frequency gain by a factor of 00, provide a - lope at croover, and will lower the gain ome at the reonance. Thu it i a good rt cut at a compenation. The Matlab Bode plot how the uncompenated and compenated and veri e our intent. Note epecially that the reonant peak never croe magnitude for the compenated (dahed) cae.
Phae (deg) Magnitude 644 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD 0 5 Bode Diagram 0 0 0 5 0 2 0 0 0 0 0 2 00 200 300 400 0 2 0 0 0 0 0 2 The Matlab margin routine how a GM = 6:3 db and P M = 48 thu meeting all pec. 55. In one mode of operation the autopilot of a jet tranport i ued to control altitude. For the purpoe of deigning the altitude portion of the autopilot loop, only the long-period airplane dynamic are important. The linearized relationhip between altitude and elevator angle for the longperiod dynamic i G() = h() () = 20( + 0:0) ( 2 + 0:0 + 0:0025) ft deg : The autopilot receive from the altimeter an electrical ignal proportional to altitude. Thi ignal i compared with a command ignal (proportional to the altitude elected by the pilot), and the di erence provide an error ignal. The error ignal i proceed through compenation, and the reult i ued to command the elevator actuator. A block diagram of thi ytem i hown in Fig. 6.03. You have been given the tak of deigning the compenation. Begin by conidering a proportional control law D() = K.
645 Figure 6.03: Control ytem for Problem 55 (a) Ue Matlab to draw a Bode plot of the open-loop ytem for D() = K =. (b) What value of K would provide a croover frequency (i.e., where jgj = ) of 0.6 rad/ec? (c) For thi value of K, would the ytem be table if the loop were cloed? (d) What i the PM for thi value of K? (e) Sketch the Nyquit plot of the ytem, and locate carefully any point where the phae angle i 80 or the magnitude i unity. (f) Ue Matlab to plot the root locu with repect to K, and locate the root for your value of K from part (b). (g) What teady-tate error would reult if the command wa a tep change in altitude of 000 ft? For part (h)and (i), aume a compenator of the form D() = K T + T + : (h) Chooe the parameter K, T, and o that the croover frequency i 0.6 rad/ec and the PM i greater that 50. Verify your deign by uperimpoing a Bode plot of D()G()=K on top of the Bode plot you obtained for part (a), and meaure the PM directly. (i) Ue Matlab to plot the root locu with repect to K for the ytem including the compenator you deigned in part (h). Locate the root for your value of K from part (h). (j) Altitude autopilot alo have a mode where the rate of climb i ened directly and commanded by the pilot. i. Sketch the block diagram for thi mode, ii. de ne the pertinent G(); iii. deign D() o that the ytem ha the ame croover frequency a the altitude hold mode and the PM i greater than 50 Solution :
Phae (deg) Magnitude (db) 646 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD The plant tranfer function : (a) See the Bode plot : h() 80 () = 0:0 + 2 0: + 2 0:05 0:05 + 80 60 Bode Diagram Gm = Inf db (at Inf rad/ec), Pm = 0.386 deg (at 0.6 rad/ec) 40 20 0 20 40 0 45 90 35 80 0 4 0 3 0 2 0 0 0 Frequency (rad/ec) (b) Since jgj = 865 at! = 0:6, K = jgj j!=0:6 = 0:002 (c) The ytem would be table, but poorly damped. (d) P M = 0:39 (e) The Nyquit plot for D(j!)G(j!) :
647 The phae angle never quite reache 80. (f) See the Root locu : The cloed-loop root for K = 0:002 are : = 0:009; 0:005 j0:6
648 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (g) The teady-tate error e : e = lim = 0!0 a it hould be for thi Type ytem. (h) Phae margin of the plant : + K h() () 000 P M = 0:39 (! c = 0:6 rad/ec) Neceary phae lead and : neceary phae lead = 50 0:39 ' 50 From Fig. 6.54 : =) = 8 Set the maximum phae lead frequency at! c :! = p T =! c = 0:6 =) T = 8 o the compenation i D() = K 8 + 2:2 + For a gain K, we want jd(j! c )G(j! c )j = at! =! c = 0:6: So evaluate via Matlab D(j! c )G(j! c ) and nd it = 2:5 0 3 K!c=0:6 Therefore the compenation i : =) K = D() = 4:0 0 4 8 + 2:2 + which reult in the Phae margin : P M = 52 (! c = 0:6 rad/ec) 4 = 4:0 0 2:5 03