5.21,5.23,5.25,5.26,5.49

5.21,5.23,5.25,5.26,5.49 - EEE 480 HW # 6 SOLUTIONS Problem...

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EEE 480 HW # 6 SOLUTIONS Problem 5.21 The root-locus asymptotes for pure-gain compensation are at ¡ 2 : 5. Since the required closed-loop pole locations are ¡ 1 § j , we need to introduce a lag compensator. Observe that asymptotes at ¡ 1 are obtained for poles at 0 ; ¡ 2. Hence, a possible controller would cancel the system pole at ¡ 3 and put a pole at the origin. That is, C ( s )= K s +3 s To compute K , we can form the closed-loop characteristic equation (simple enough), ( s 2 +2 s + K )( s +3). One of its roots is at ¡ 3 due to the cancellation; the other two should be at ¡ 1 § j and have magnitude p 2, hence K =2 . Problem 5.23 Here, the pure-gain asymptotes are at 0 and we need a lead compensator to achieve the desired speci¯cation. At ¡ 2 § j 2, the system double integrator contributes 2 £ 135 o =270 o , so the compensator should contribute 90 o phase lead. Choosing the pole angle as 10 o , the zero should contribute 100 o . Computing the corresponding locations: ² pole: tan10 o =0 : 176 ) p = 2 0 : 176 +2=13 : 3. ² zero: z =2 ¡ (2)(0 : 176) = 1 : 65. ² gain: K = ¯ ¯ ¯ s 2 ( s + p ) s + z ¯ ¯ ¯ ¯ ¯ ¯ s = ¡ 2+ j 2 = (8)( p 11 : 3 2 +2 2 ) p 0 : 35 2 +2 2 =45 : 21. The ¯nal compensator is C ( s )=45 : 21 s +1 : 65 s +13 : 3 Problem 5.25 Here, the pure-gain asymptotes are at 0.5 and we need a lead compensator to achieve the desired speci¯cation. While the procedure used in Problem 5.23 is applicable, let us use a di®erent approach that can yield quick results for simple problems. Guided by the expected root-locus of the compensated system, we select the desired closed-loop characteristic polynomial as ( s 2 +2 s +4)( s +10)= s 3 +12 s 2 +24 s +40 The ¯rst factor is from the speci¯ed dominant pair and the second factor is a reasonable choice for the third closed-loop pole. Next, we write the closed-loop characteristic polynomial in terms of the lead compensator parameters ( s 2 + s )( s + p )+ K ( s + z )= s 3 +(1+ p ) s 2 +( K + p ) s + Kz Equating the coe±cients of the two polynomials we ¯nd: ² p =12 ¡ 1=11 ² K =24 ¡ p =13 ² z =40 =K =3 : 08.
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This note was uploaded on 01/23/2012 for the course EE 4580 taught by Professor Gu during the Fall '10 term at LSU.

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5.21,5.23,5.25,5.26,5.49 - EEE 480 HW # 6 SOLUTIONS Problem...

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