solution hw5

# solution hw5 - Problem 1 ySP (s) + k = 1.6 5 (s + 1)(2s +...

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Problem 1 The closed loop transfer function can be written as follows: CL 22 SP 8 y(s) 8 8 9 (s 1)(2s 1) G( s ) 8 y( s ) 2 s 3 s9 29 s 1 3 s1 1 (s 1)(2s 1) ++ == = = + + + (S1.1) Thus, comparing (S1.1) to the standard form of 2 nd order systems, one obtains: 2 2 9 2 3 0.471 τ= ⇒ τ= = (S1.2) 2 1 2 2 4 0.354 ξτ = ξ = = (S1.3) k 8 9 0.889 (S1.4) Since ξ is smaller than 1, the system is underdamped. a) From textbook, we know that: max 2 yy ( ) max dev final value OS exp 0.305 final value y( ) 1 ⎛⎞ −πξ ⎜⎟ = = −ξ ⎝⎠ (S1.5) The output of the system for a step change of magnitude 0.1 in the set point is: 2 0.1 8 9 y(s) s29 s 13 = (S1.6) Applying the final value theorem, yields: 2 s0 0.8 9 0.8 y( ) limsy(s) lim 0.0889 29s 13s 1 9 →→ ∞= = = = (S1.7) Hence, the maximum value of the response is: max y y( )(1 OS) 0.116 =∞+ = (S1.8) k1 . 6 = 5 (s 1)(2s 1) + + SP s ) y(s) +

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b) For a servo problem, the offset is given by: offset new set point y( ) 0.1 0.0899 0.0111 =− = = (S1.9) c) From the textbook, the period of the oscillation is: 2 2 where 1 Τ= π ω ω= −ξ τ (S1.10) Therefore, the value of the period of the oscillation Τ is: 3.17min Τ= (S1.11) Problem 2 The resistance of a liquid to a hydrostatic pressure can be defined as the rate of changing of the liquid level due to the change of the output low rate. Thus: dh R dq = (S2.1) Assuming linear resistances, R 1 and R 2 can be directly evaluated by plotting h (ft) versus
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## This note was uploaded on 10/04/2009 for the course CHE 470 taught by Professor Smith during the Spring '09 term at Rice.

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solution hw5 - Problem 1 ySP (s) + k = 1.6 5 (s + 1)(2s +...

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