20112ee141_1_EE141_hw6_sol

20112ee141_1_EE141_hw6_sol - = 0 . 5. We can see that the...

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EE141 Principles of Feedback Control (Winter 2010) Solutions to Homework 6 Problem 5.8 (b) Center of asymptotes: - 8 3 Angles of asymptotes: ± 60 , 180 Break out point: 0 . 543 Break out angle: ± 90 ω 0 : none -15 -10 -5 0 5 -20 -15 -10 -5 0 5 10 15 20 Root Locus Real Axis Imaginary Axis Figure 1: Root locus for 5.8(b) (e) Center of asymptotes: - 3 Angles of asymptotes: ± 60 , 180 Break out point: 0 . 488 Break out angle: ± 90 ω 0 : none 1
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-15 -10 -5 0 5 10 -25 -20 -15 -10 -5 0 5 10 15 20 25 Root Locus Real Axis Imaginary Axis Figure 2: Root locus for 5.8(e) Problem 5.13 (a) Center of asymptotes: - 6 Angles of asymptotes: ± 90 Angle of departure: ± 135 . 8 , ± 90 Angle of arrival: ± 139 Break out point: - 5 . 32 Break out angle: ± 90 Break in point: - 2 . 31 Break in angle: ± 90 ω 0 : none -14 -12 -10 -8 -6 -4 -2 0 2 -50 -40 -30 -20 -10 0 10 20 30 40 50 Root Locus Real Axis Figure 3: Root locus for 5.13(a) 2
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(b) The two dashed lines in the above figure are corresponding to
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Unformatted text preview: = 0 . 5. We can see that the two root loci from the imaginary axis poles will not touch these dashed lines. Thus no K will cause all roots to have a damping ratio greater than 0 . 5 (c) When K is about 35 , 90, we can have some closed-loop poles with = 0 . 707. You can use Matlab to nd these values for K . (d) K = 35 is used for this following plot. The step response shows the basic form of a well damped response with the vibration of the response element added. 2 4 6 8 10 12 14 0.2 0.4 0.6 0.8 1 1.2 1.4 Step Response Time (sec) Amplitude Figure 4: Root locus for 5.13(d) 3...
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This note was uploaded on 07/02/2011 for the course EE 141 taught by Professor Balakrishnan during the Spring '07 term at UCLA.

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20112ee141_1_EE141_hw6_sol - = 0 . 5. We can see that the...

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