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# filled13 - 1 2 1 2 2 2 p p z z r r(6.3(8.5 4K K 6.0 r...

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ME 475 Handouts - Spring 2011 1 Slide 1 ME 475 Session 13: Root Locus Compensation Galen King Purdue University PID Controller Design 3 ( Complex zeros ) s ) z s )( z s ( K ) s ( G 2 1 c - - = where ϖ - σ - = ϖ + σ - = j z j z 2 1 Slide 2 ME 475 Session 13: Root Locus Compensation Galen King Purdue University PID Controller Design 3 Im s Re s j4 j3 j2 j1 - j4 - j3 - j2 - j1 - 2 - 4 - 6 j5 j6 - j5 - j6 - 8

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ME 475 Handouts - Spring 2011 2 Slide 3 ME 475 Session 13: Root Locus Compensation Galen King Purdue University Apply Angle Criterion 1 2 d z z 198.4 φ = = φ + φ r 1 2 z 1 z 2 1 2 180 , 180 161.6 φ = - α φ = - α α + α = r r r choose 5.5: σ = 1 2 1 2 1 2 1 2 6 6 tan 2(6 ) 6 0.5 6 6 tan 2(6 ) 6 0.5 tan tan 2(6 ) 2(6 ) tan( ) 1 tan tan 1 2(6 )2(6 ) - ϖ - ϖ α = = = - ϖ - σ + ϖ + ϖ α = = = + ϖ - σ α + α - ϖ + + ϖ α + α = = - α α - - ϖ + ϖ Slide 4 ME 475 Session 13: Root Locus Compensation Galen King Purdue University Apply Magnitude Criterion
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Unformatted text preview: 1 2 1 2 2 2 p p z z r r (6.3)(8.5) 4K K 6.0 r r (1.86)(10.22) = = ⇒ = c 6(s 5.5 j4.2)(s 5.5 j4.2) G (s) s + + +-= 2 c 6(s 11s 47.9) G (s) s + + = ME 475 Handouts - Spring 2011 3 Slide 5 ME 475 Session 13: Root Locus Compensation Galen King Purdue University Root Locus with PID Controller 3-8-6-4-2 2-8-6-4-2 2 4 6 8 Real Axis Imag Axis Root Locus with PID Control (Complex Zeros) Slide 6 ME 475 Session 13: Root Locus Compensation Galen King Purdue University PID 3 Step Response 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 1.2 1.4 Time (sec.) Amplitude Step Responses Complex zeros Repeated real zeros...
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filled13 - 1 2 1 2 2 2 p p z z r r(6.3(8.5 4K K 6.0 r...

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