Figure 5-1 Root locus for Problem 5.2.

Transcription

1 K K( +) 5.3 () i KG() = (ii) KG() = ( + )( + 5) ( + 3)( + 5)

2 6 4 Imag Axi - -4 Imag Axi Real Axi Figure 5- Root locu for Problem 5.3 (i) Real Axi Figure 5-3 Root locu for Problem 5.3 (ii)

3 5.7 Deign a controller for the tranfer function G ( ) = ( + )( + 5) to obtain (i) zero teady-tate error due to tep, (ii) a ettling time of le than, and (iii) an undamped natural frequency of 5 rad/. Obtain the repone due to a unit tep and find the percentage overhoot, the time to the firt peak and teady-tate error percent due to a ramp input. The plant tranfer function i G () = ( + )( + 5) For zero teady-tate error due to tep, a PI controller i needed. The implet deign i by cancellation of the pole at with the controller zero. Thi leave the loop gain L () = CG () () = ( + 5) The correponding root locu include a vertical line at.5 i.e. for ufficiently high gain we have an underdamped ytem with ζ ω n =.5 and T = 4/.5 <. For ζ =.5, ω n =.5/.5 = 5 rad/. From the ymmetry of the root locu, the gain K at the deired cloed-loop pole i equal to the quare of the ditance from the location to the origin i.e. K = 5 = 5. The ame reult can be obtained uing a CAD tool or analytically, ince the compenated ytem i imple. The deign yield a econd order ytem and the remaining deign criteria can be computed analytically a follow πζ ζ PO = e % = 6. 3% π π Tp = = =. 76 ω d 5 (. 5) 5 Kv = C() G() = = 5 e( )% = = % + 5 K v 5 Root Locu 4 3 Sytem: l Gain: 4.9 Pole: i Damping:.5 Overhoot (%): 6. Frequency (rad/ec): 4.99 Imaginary Axi Real Axi Figure 5-9 Root locu for problem 5.7

4 5.9 Conider the oven temperature control ytem of Example 3.5 with tranfer function G ( ) = K a) Deign a proportional controller for the ytem to obtain a percentage overhoot le than 5 % The percentage overhoot pecification yield ζ ln(.5) ln(.5) + π =.69 We elect a damping ratio of.7. We can olve the problem analytically ince the ytem i econd order. It i more convenient to ue MATLAB to obtain the gain value. We obtain the root locu plot of Figure 5-4 for the ytem and oberve that for a gain of about 3.5 we have the deired damping ratio.

5 .5 Root Locu Sytem: g Gain: 3.55 Pole: i Damping:.74 Overhoot (%): 4.46 Frequency (rad/ec):.3 Imaginary Axi Real Axi Figure 5-3 Root locu for the oven temperature control ytem. b) Deign a controller for the ytem to reduce the teady-tate error due to tep to zero without ignificant deterioration in the tranient repone. We need a PI controller to reduce the teady-tate error to zero. The proportional control deign yield a pole with real part equal to.5. The PI controller zero i at one tenth thi value, which i approximately. and the controller tranfer function i +. C( ) = K For a gain of 3.5, we have a dominant pair of pole with about the ame damping ratio a for proportional control. However, the tep repone for the ytem i luggih and the ettling time i very large. We increae the gain to K = 7 and obtain the tep repone of Figure 5-5. Although the % ettling time i large (over 7), the 5 % ettling time i about half thi value and the repone i acceptable with an overhoot le than 4 %.

6 .4 Step Repone. Sytem: gcl Peak amplitude:.4 Overhoot (%): 3.73 At time (ec):.46 Sytem: gcl Settling Time (ec): 7.6 Amplitude Time (ec) Figure 5-4 Step repone for the ytem with PI control and a gain of For the inertial ytem governed by the differential equation & θ = τ Deign a feedback controller to tabilize the ytem and reduce the percentage overhoot below % with a ettling time of le that 4. The tranfer function of the ytem i given by G ( ) = We ue PD control to tabilize the ytem. The feedback control doe not reult in a cloed-loop zero. The cloed-loop ytem i econd-order with the damping ratio contraint ζ ln(.) ln(.) + π =.59 We chooe ζ =.6 and calculate the ettling time contraint 4 4 ωn = =.667 T ζ 4.6

7 We chooe ω n = rad/. The correponding deired cloed-loop pole i at cl =. + j.6 The angle of the compenator i ( ) =.87 rad θ C = π L cl / The compenator zero i calculate from ω d a = + ζω n tan ( θ ) C =.6667 The cloed-loop tranfer function with additional precompenator gain i G ( ) = For thi ytem, analytical deign i alo imple ince the cloed-loop chacacteritic equation i ( + a) = + ζω + ω = K n n The tep repone of the ytem hown in Figure 5-5 meet the deign pecification. Figure 5-5 Step repone for the compenated ytem of Problem 5.9

8 5. Conider the oven temperature control ytem of Example 3.5 with tranfer function G ( ) = K c) Obtain the tep repone of the ytem with the a PD cacade controller with gain 8 and a zero at Sytem: gc Time (ec):.36 Amplitude:.984 Step Repone Sytem: gc Final Value:.889 Sytem: gc Settling Time (ec): Amplitude Time (ec) Figure 5-6 Step repone of cacade compenated ytem. d) Obtain the tep repone of the ytem with the a PD feedback controller with a zero at 5 and unity gain and a forward gain of 8.

9 .9.8 Sytem: gf Step Repone Settling Time (ec):.67 Sytem: gf Final Value: Amplitude Time (ec) Figure 5-7 Step repone of feedback compenated ytem. e) Why i the root locu identical for both ytem? For both configuration, the loop gain i given by L ( ) = C( ) G( ) = Hence, the root locu i identical for the two configuration. f) Why are the time repone different although the ytem have the ame loop gain? The time repone are different becaue the cacade configuration ha a cloed loop zero at 5 while the feedback configuration ha no cloed-loop zero. g) Complete a comparion table uing the repone of (a) and (b) including the percentage overhoot, the time to firt peak, the ettling time, and the teady-tate error. Comment on the reult and explain the reaon for the difference in the repone. The tep repone give the value hown in Table P5.. The percentage overhoot for cacade compenation i calculated uing the equation y peak y final % OS = % = % =.7% y.889 final The ytem with cacade compenation ha a fater repone due to the cloed-lop zero at 5 but the repone i more ocillatory. The two ytem have the ame teady-tate error of %.

10 Table P5. Summary of imulation reult for cacade and feedback compenation. Configuration %OS T p T e( )% Cacade.7% % Feedback No overhoot Not defined.67 %

11 5.3 Conider the ytem G ( ) = ( + ) 4 and apply the Ziegler-Nichol procedure to deign a PID controller. Obtain the repone due to a unit tep input a well a a unit tep diturbance ignal. The application of an open-loop unit tep input give the repone hown in Figure 5-. It can be een that K= and L=.3. Since the value of τ+l i the time interval between the application of the tep input and the time when the proce output reache 63.% of it final value, we have τ =3. The Ziegler-Nichol rule given in Table 5. provide the following PID parameter: K p =.77, T i =.6, and T d =.65. The reulting proce output when a tep i applied to the et-point ignal at time t= and a load diturbance ignal i applied at time t=35 i plotted in Figure 5-. Converely, if we compute the um τ +L a the time interval between the application of the tep input and the interection of the tangent line with the traight repreenting the final teady-tate value of the proce output, we obtain τ +L=6, and therefore τ =3. Thu, we can determine the following PID parameter: K p =4.34, T i =.6, and T d =.65. The reulting proce output when a tep i applied to the et-point ignal at time t= and to the load diturbance ignal at time t=6 i hown in Figure 5-. It appear that in thi cae the output i more ocillatory than in the previou cae. Thi actually occur in general becaue the etimate of the dominant time contant i uually higher

12 when etimated uing the interection of the tangent line with the teady-tate value of the proce output. Thi implie that the proportional gain etimate i higher than the etimate obtained by conidering the time when the proce output reache 63.% of it final value proce output time Figure 5- Open-loop tep repone.5 proce output time Figure 5- Set-point and load diturbance tep repone for K p =.77, T i =.6, and T d =.65

13 proce output time Figure 5- Set-point and load diturbance tep repone for K p =4.34, T i =.6, and T d =.65