Hw5-S05_Solution - HW 5 - SOLUTION ME 304 CONTROL SYSTEMS...

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HW 5 - SOLUTION ME 304 Transient Response CONTROL SYSTEMS (01) April 20, 2005 PROBLEM 1: Step Response Time (sec) Amplitude 0 5 10 15 20 25 30 0 2 4 6 8 10 12 14 16 18 20 First order system, K G(s) s1 = τ+ (K: gain, τ : time constant) Step input with magnitude 10, 10 R(s) s = Steady state value of the response = 10K = 20 => K=2 Initial slope of the response = 10K 20 5 = τ => τ =5 PROBLEM 2: g=10 m/s 2 R h a) First obtain governing differential equation of the system: Tank: dP QC dt −= C Q Valve: PR Q = Combine the above equations => dQ CR Q 0 dt + = Take Laplace transform remembering initial conditions are not zero, 0 CR(sQ(s) Q ) Q(s) 0 −+ = => 0 CRQ Q(s) (CRs 1) = + Take the inverse Laplace transform to obtain time response tt 0 CR CR 0 CRQ Q(t) e Q e CR −− == Prof. Dr. Y. Samim Ünlüsoy, Res. Asst. Kemal Çal ı ş kan 1
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HW 5 - SOLUTION ME 304 Transient Response CONTROL SYSTEMS (01) April 20, 2005 Initial flow rate can be expressed as, 0 Q = 0 gh R ρ Then, tt 0 CR 0.130 3 gh 10 1.5 Q(t) e e 0.5e R3 0 −− ρ == = t b) 3 1 3 Q(3) 0.5e 0.5e 0.184 === [m 3 /s] 1 RQ(3) h(3) 1.5e 0.5518 [m] g ρ PROBLEM 3: Response to unit step input
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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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Hw5-S05_Solution - HW 5 - SOLUTION ME 304 CONTROL SYSTEMS...

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