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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 n = 2 rad/sec and . 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)...
View Full Document

Page1 / 14

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online