# hw4 - E105 Homework 4 Solutions Stanford University Winter...

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Unformatted text preview: E105 - Homework 4 Solutions Stanford University - Winter 2006 Stephen Russell February 8, 2006 Problem 1 - FPE 4.6 Consider a system with the plant transfer function G ( s ) = 1 s ( s +1) . You wish to add a dynamic controller so that ω n = 2 rad/sec and ζ ≥ . 5. Several dynamic controllers have been proposed: 1. D ( s ) = s +2 2 2. D ( s ) = 2 s +2 s +4 3. D ( s ) = 5 s +2 s +10 4. D ( s ) = 5 ( s +2)( s +0 . 1) ( s +10)( s +0 . 01) (a) Using MATLAB, compare the resulting transient and steady-state responses to reference step inputs for each controller choice. Which controller is best for the smallest rise time and smallest overshoot? Our transfer function takes the form: Y ( s ) R ( s ) = D ( s ) G ( s ) 1 + D ( s ) G ( s ) Controller 1) D ( s ) G ( s ) = s + 2 2 ¶ 1 s ( s + 1) ¶ = s + 2 2 s 2 + 2 s Y ( s ) R ( s ) = s +2 2 s 2 +2 s 1 + s +2 2 s 2 +2 s = s + 2 2 s 2 + 3 s + 2 1 Controller 2) D ( s ) G ( s ) = 2( s + 2) s ( s + 1)( s + 4) = 2 s + 4 s 3 + 5 s 2 + 4 s Y ( s ) R ( s ) = 2 s +4 s 3 +5 s 2 +4 s 1 + 2 s +4 s 3 +5 s 2 +4 s = 2 s + 4 s 3 + 5 s 2 + 6 s + 4 Controller 3) D ( s ) G ( s ) = 5( s + 2) s ( s + 1)( s + 10) = 5 s + 10 s 3 + 11 s 2 + 10 s Y ( s ) R ( s ) = 5 s +10 s 3 +11 s 2 +10 s 1 + 5 s +10 s 3 +11 s 2 +10 s = 5 s + 10 s 3 + 11 s 2 + 15 s + 10 Controller 4) D ( s ) G ( s ) = 5( s + 2)( s + 0 . 1) s ( s + 1)( s + 10)( s + 0 . 01) = 5 s 2 + 10 . 5 s + 1 s 4 + 11 . 01 s 3 + 10 . 11 s 2 + 0 . 1 s Y ( s ) R ( s ) = 5 s 2 +10 . 5 s +1 s 4 +11 . 01 s 3 +10 . 11 s 2 +0 . 1 s 1 + 5 s 2 +10 . 5 s +1 s 4 +11 . 01 s 3 +10 . 11 s 2 +0 . 1 s = 5 s 2 + 10 . 5 s + 1 s 4 + 11 . 01 s 3 + 15 . 11 s 2 + 10 . 6 s + 1 Step responses for each controller were plotted using MATLAB, see Figure 1. System 1 has the smallest percentage overshoot, M p , while System 4 has the fastest rise time. System 3 seems to have a nice compromise between the two characteristics. (b) Which system would have the smallest steady-state error to a ramp reference input? e ss = lim s → sE ( s ) E ( s ) = E ( s ) R ( s ) R ( s ) = 1 1 + D ( s ) G ( s ) 1 s 2 ¶ Thus, our steady-state error expression for a ramp input is e ss,ramp = lim s → 1 s (1 + D ( s ) G ( s )) 2 System 1 Time (sec) System 2 Time (sec) Amplitude System 3 Time (sec) System 4 Time (sec) Amplitude 1 2 3 4 5 6 7 8 System: sys Peak amplitude: 1.03 Overshoot (%): 3.48 At time (sec): 4.05 System: sys Rise Time (sec): 1.98 1 2 3 4 5 6 7 8 9 0.2 0.4 0.6 0.8 1 1.2 1.4 System: sys Rise Time (sec): 1.68 System: sys Peak amplitude: 1.08 Overshoot (%): 8.4 At time (sec): 3.54 1 2 3 4 5 6 7 8 System: sys Peak amplitude: 1.05 Overshoot (%): 4.95 At time (sec): 3.77 System: sys Rise Time (sec): 1.82 5 10 15 20 25 30 0.2 0.4 0.6 0.8 1 1.2 1.4 System: sys Rise Time (sec): 1.63 System: sys Peak amplitude: 1.14 Overshoot (%): 13.9 At time (sec): 3.73 Figure 1: 4.6(a) - Step responses Controller 1) e ss = lim s → 1 s ‡ 1 + s +2 2 s 2 +2 s · = lim s → 2 ¢ s ( s + 1) ¢ s (2 s 2 + 3 s + 2) = 2 2 = 1 Controller 2)...
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hw4 - E105 Homework 4 Solutions Stanford University Winter...

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