20112ee141_1_ee141_hw5_sol

# 20112ee141_1_ee141_hw5_sol -...

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EE141 Principles of Feedback Control (Winter 2010) Solutions to Homework 5 Problem 5.2 Only estimation is required. I had to make some numbers to do it in Matlab. The idea here is just get the basic rules right. (b) Assume L ( s ) = s +1 s 2 +0 . 2 s +1 . Angle of departure: ± 138 Break in point: - 2 . 34 -4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 -1.5 -1 -0.5 0 0.5 1 1.5 Root Locus Real Axis Imaginary Axis Figure 1: Root locus for 5.2(b) (f) Assume L ( s ) = s +1 ( s - 1)( s 2 +4 s +5)( s 2 + . 25) . Center of asymptotes: - 1 Angles of asymptotes: ± 45 , ± 135 Break in point: 0 . 057 Break out point: 0 . 623 1

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-10 -8 -6 -4 -2 0 2 4 6 8 -8 -6 -4 -2 0 2 4 6 8 Root Locus Real Axis Imaginary Axis Figure 2: Root locus for 5.2(f) Problem 5.5 (c) Angle of departure: ± 161 . 6 Angle of arrival: ± 159 . 3 ω 0 : none -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 -4 -3 -2 -1 0 1 2 3 4 Root Locus Real Axis Figure 3: Root locus for 5.5(c) (e) Angle of departure: ± 180 Angle of arrival: ± 180 ω 0 : none

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Unformatted text preview: 2-1.8-1.6-1.4-1.2-1-0.8-0.6-0.4-0.2 0.2-2.5-2-1.5-1-0.5 0.5 1 1.5 2 2.5 Root Locus Real Axis Imaginary Axis Figure 4: Root locus for 5.5(e) Problem 5.7 (c) Center of asymptotes:-4 Angles of asymptotes: ± 60 , 180 Angle of departure: ± 14 . 6 , ± 90 Angle of arrival: ± 90 ω : ± 6 . 4-25-20-15-10-5 5-15-10-5 5 10 15 Root Locus Real Axis Figure 5: Root locus for 5.7(c) (e) Center of asymptotes:-1 . 5 Angles of asymptotes: ± 90 Angle of departure: ± 90 Angle of arrival: ± 71 . 6 3 Break out point:-2 . 48 ω : none-3.5-3-2.5-2-1.5-1-0.5 0.5-10-8-6-4-2 2 4 6 8 10 Root Locus Real Axis Imaginary Axis Figure 6: Root locus for 5.7(e) 4...
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20112ee141_1_ee141_hw5_sol -...

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