Hw6-S05_Solution - HW 6 - SOLUTION ME 304 Transient...

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HW 6 - SOLUTION ME 304 Transient Response Specifications & Stability CONTROL SYSTEMS (01) Due on April 27,2005 PROBLEM 1: 2 16 C(s) 16 s(s 2) 16 R(s) s 5.6s 16 1 (1 0.225s) s(s 2) + == + + ++ + compare with the standard form: 2 n 2 nn K G(s)= s2 s ω 2 + ξω + ω => K=1 , 2 n 16 ω = , n 25 ξω = . 6 then, n 4 ω= 0.7 ξ= a) Rise time, 2 2 1 1 22 1 10 . 7 tan tan 0.7 0.821 s 14 1 0 . 7 ξ π = −− r n t w b) Maximum overshoot, ξ 0.7 - π () - π 2 2 1- ξ - π 0.98 1-0.7 M = e = e = e = 0.046 (4.6%) p c) %2 settling time, 44 t 1.43 s s ξ w0 . 7 4 n == = PROBLEM 2: 32 23 2 10s 31s 21s G(s) (s 3s 2)(s s 2s)(s 50) = + a) 2 10s(s 3.1s 2.1) 10s(s 2.1)(s 1) (s 2)(s 1)s(s 0.5 1.32j)(s 0.5 1.32j)(s 50) + + + + + + + − + (s 2)s(s s 2)(s 50) + 1 0.225s + + - 16 s(s 2) + C(s) R(s) ¾ s and (s+1) terms in the numerator and denominator cancel each other. ¾ Zero at s=-2.1 is close to pole at s=-2, thus these can also be canceled ¾ Pole at s=-50 is too far from the imaginary axis (more than 5 times) compared to poles at s=-0.5±1.32j. So, the poles at s=-0.5±1.32j will be dominant and effect of the pole at s=-50 can be neglected. Prof. Dr. Y. Samim Ünlüsoy, Res. Asst. Kemal Çal ı ş kan 1
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HW 6 - SOLUTION ME 304 Transient Response Specifications & Stability CONTROL SYSTEMS (01) Due on April 27,2005 The response of the original system can be approximated with the simplified transfer function 2 10 10 G(s) (s 0.5 1.32j)(s 0.5 1.32j) (s s 2) == ++ +− + + Compare with the standart form: 2 n 2 nn K G(s)= s2 s ω 2 + ξω + ω => K=5 , 2 n 2 ω = , n 21 ξω = then, n 2 ω= 0.354 ξ= Rise time, 2 2 1 1 22 1 1 0.354 tan tan 0.354 1.461 s 1 2 1 0.354 ξ π
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This note was uploaded on 05/16/2011 for the course ME 304 taught by Professor Samim during the Spring '11 term at Middle East Technical University.

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Hw6-S05_Solution - HW 6 - SOLUTION ME 304 Transient...

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