Control ENG HW_Part_64

Control ENG HW_Part_64 - 1 0.5t L−1 = e (Cos 0.5t + Sin...

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Unformatted text preview: 1 0.5t L−1 = e (Cos 0.5t + Sin 0.5t ) s − (0.5 + i0.5) 14.8 For the output C to be stable, we analyze the characteristic equation of the system 1+ 1 1 × (τ 3 s + 1) = 0 τ I s (τ 1 s + 1)(τ 2 s + 1) τ I s(τ 1 s 2τ 2 + τ 1 s + τ 2 s + 1) + τ 3 s + 1 = 0 τ I τ 1τ 2 s 3 + τ I (τ 1 + τ 2 ) s 2 + (τ I + τ 3 ) s + 1 = 0 Routh Array s3 τ I τ 1τ 2 s2 τ I (τ 1 + τ 2 ) α s s0 α= τ I + τ3 1 0 1 τ I (τ 1 + τ 2 )(τ I + τ 3 ) − τ I τ 1τ 2 τ I (τ 1 + τ 2 ) Now (1) τ I τ 2τ 1 > 0 Since τ 1 & τ 2 τ 1 > 0;τ 2 > 0 are process time constant they are definitely +ve (2) τ I (τ 1 + τ 2 ) > 0 (3) α > 0 ⇒ τ I (τ 1 + τ 2 )(τ I + τ 3 ) > τ I τ 1τ 2 τ 1τ I + τ 1τ 3 + τ 2τ I + τ 2τ 3 − τ 1τ 2 > 0 τ I (τ 1 + τ 2 ) > τ 1τ 2 − τ 3 (τ 1 + τ 2 ) τI > τ 1τ 2 −τ 3 τ1 +τ 2 also τ I > 0 14.9 In the control system shown in fig find the value of Kc for which the system is on the verge of the instability. The controller is replaced by a PD controller, for which the transfer function is Kc(1+s). if Kc = 10, determine the range for which the system is stable. Characteristics equation 1+ 6 Kc =0 ( s + 1)( s + 2)( s + 3) or ( s + 1)(s + 2)(s + 3) + 6 Kc = 0 ( s 2 + 3s + 2)(s + 3) + 6 Kc = 0 s 3 + 6 s 2 + 11s + (6 + 6 Kc) = 0 Routh array s3 13 s2 3 1 + Kc s 1 + Kc 3− ) 3 ...
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This note was uploaded on 11/13/2011 for the course COP 4355 taught by Professor Koslov during the Spring '10 term at University of Florida.

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Control ENG HW_Part_64 - 1 0.5t L−1 = e (Cos 0.5t + Sin...

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