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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; higherorder 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 steadystate 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—ordEI 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.
 Spring '08
 Naber

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