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By Wan E.A.

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Example text

3 The "Stick on a Cart" example O m u After normalization, we have: ; =u 49 (87) ) s2 (s) ; (s) = U (s) (88) (s) 1 U (s) = s2 ; 1 (89) This results in the following model of the system: The root locus diagram is shown below. The u θ 1/(s^2 - 1) Figure 10: X -1 X 1 Figure 11: above gure indicates an unstable system, irrespective of the gain. With lead compensation. u + e 1 2 s -1 s+α s+β θ - X O X 50 X The root loci of the lead-compensated system are now in the LHP, which indicates a stable system.

Ko + - k s+1 28 (35) k sk+1 0 1 + skk+10 = s + 1kk+0 kk (36) 0 0 Final values = 1 +kkkk 0 (37) Set k2 = 1 +kkkk0 0 ko r + - k2 k s+1 Sensitivity for k0 ! k0 + k0 Final value = k2 1 +k (kk(0k+ +k0k) ) 0 0 2 k2 4 3 5 1 1 + k(k +1 k ) 1 1+x 1;x k2 1 ; k(k +1 k ) = 1 +kkkk0 k2 k(kk(0k+ +k0k) ;) 1 0 0 0 0 0 = 1 + 1 +1kk 0 for k large ! reduced sensitivity. 0 (38) (39) 0 (40) (41) (42) Example: Integral Control ko r + - k/s kk0 s (s+1) k k0 + 1 s s+1 s+1 = s (s +k1)k0+ k k 0 Final value 29 (43) lim 0 s !

This adds a zero at s = ;2. The modi ed locus is hence the circle. Figure 7: The e ect of the zero is to move the root locus to the left, improving stability. Also, by adding the zero, we can move the locus to a position having closed-loop roots and damping ratio 0:5 We have "compensated" then dynamics by using D(s) = s + 2. The trouble with choosing D(s) based on only a zero is that the physical realization would contain a di erentiator that would greatly amplify the high-frequency noise present from the sensor signal.

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