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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: tan 10 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. The ¯nal compensator is C ( s ) = 13 s + 3 : 08 s + 11 This approach is based on the so-called \pole-placement" design which computes the parameters of a general (high-order) compensator to assign the closed-loop poles to desired locations. The advantage of pole placement

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

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