560 hw4 - Uncompensated system: Search along the...

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Unformatted text preview: Uncompensated system: Search along the g": 0.5 1ine and find the operating point is at —1.5356 :: -3} -r- 4 j2.6598 with K = 73.09. Hence, [3/005 = (5' *" 1 * K100 =16.3%; T5 = l 5356 = 2.6 seconds: .8}; 73 09 30 =2.44. A higher—order pole is located at 40.9285. Compensated: Add a pole at the origin and a zero at —0.1 to form a PI controller. Search along the (I: 0.5 line and find the operating point is at -1 .5072 :j2.6106 with K = 72.23. Hence, the estimated . . , —-:.: 41—." . performance specifications for the compensated system are: {31:05 = e ' ' X100 = 16.3%; T; = 1 50_ 7 = 2.65 seconds; K}; = 00. Higher—order po1es are located at 0.0728 and 40.9125. The compensated system should be simulated to ensure effective pole-"zero cancellation. Uncompensated: Searching along the 1350 1ine (t; = 0.707). find the operating point at _ _ 4.604516 —2.32 +J2.32 With K = 4.6045. Hence. Kp = T = 0.921; T5 = } g2 = 1.724 seconds: T}; = ,7 J5. = 1.354 seconds; 9’605 = E 1' 2.32 (on = \I' 2.322 + 2.322 = 3.28 rad-"s; higher-order pole at 6.366. Compensated: To reduce the settling time by a factor of 2, the closed—loop poles should be — 4.64 : ‘1': x100 = 4.33%; 14.64. The summation of angles to this point is 119° . Hence. the contribution of the compensating zero shou1cl be 180° 419° =61° . Using the geometryr shown below. 4. 64 —=tan (610). Or. 2 = 7.21. :c 4.64 ° jun j4.4s s— ol ane -3c —!.64 After adding the compensator zero, the gain at -4.64+j4.64 is K = 4.?7. Hence, _ 4.?7I6ITJI 4 F! _ KP = 6.83. T,=—=0.86 second; Ty: = 0.6?! second; 2x315 " 4.64 4.64 94:03 = a?“ 1‘5 x100 =4.33%; (a, = d4.643 +4.54J = 6.56 rad-"s; higher—order pole at 6.49. The problem with the design is that there is steady-state error, and no effective pole-"zero cancellation. The design should be simulated to be sure the transient requirements are met. 26. 2 2 «. (on =T=W=4954Thus Iln=(9“1,‘1—;2 =4.954 1—0.4037' =4.5324.I—1ence ‘.- -' ’ the design point is —2 +j4.5324. Now, add a pole at the origin to increase system type and drive error to zero for step inputs. Now design a PD controller. The angular contribution to the design point of the system poles and pole at the origin is 101.9%. Thus, the compensator zero must contribute 180° — 101.95 =T8.l'3. Using the geometry below, jco 3 -pl an e j4.5324 4.5324 — = l311(73.10} . Hence. :f = ' .955. The compensated open—loop transfer function with PD Z _3 C a K(s+2.955) compensation is —. Adding the compensator zero to the system and 5(5 — 4H5.- + 6H5 +10} evaluating the gain for this at the point —2 + j4.5324 yields K. = 294.51 with a higher—ordE-I pole at -2.66 and $13.34. (3 — 0.01} PI design: Use GPICS] = — . Hence, the equivalent open—loop Transfer fimdion is 5. _ K(5 +2.955)(5 —0.01) 1 with K = 294.?5. 3' (3 + 4)(3 — 6)(.$ — 10) Ge (3) ...
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This note was uploaded on 09/03/2008 for the course ECE 560 taught by Professor Naber during the Spring '08 term at University of Louisville.

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560 hw4 - Uncompensated system: Search along the...

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