By Gene F. Franklin, J. David Powell, Abbas Emami-Naeini
Options handbook for suggestions keep watch over of Dynamic platforms (6th version) through Franklin. came across it on the net someplace, appears lovely sturdy and every little thing seems there.
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Extra resources for Feedback Control of Dynamic Systems Solutions Manual (6th Edition)
Sample text
This can be done as follows, Z 1 e st f (t)dt; F (s) = Z0 1 Z 1 Z 1 e st f (t)dt ds; F (s)ds = 0 s s Interchanging the order of integration, Z 1 Z 1 Z 1 e st ds f (t)dt; F (s)ds = s s 0 Z 1 Z 1 1 1 st F (s)ds = e f (t)dt; t s 0 s Z 1 f (t) st e dt: = t 0 Using this result then, Lfsin tg = L sin t t = = s2 Z 1 1 d ; +1 s tan 1 (1) tan = = 1 ; +1 2 1 tan 2 tan 1 1 s 1 (s); (s); : where a table of integrals was used and the last simpli…cation follows from the related trigonometric identity. 1, entry 7), Lfsin t sin tg = = 1 1 2 +1 s +1 1 : s4 + 2s2 + 1 s2 ; (e) f (t) = Z t cos(t ) sin d : Z t Lff (t)g = L cos(t ) sin d 0 0 = Lfcos(t) sin(t)g: 3010 CHAPTER 3.
52 relies on thermal expansion of actuators under two corners to level the table by raising or lowering their respective corners. The parameters are: Tact = actuator temperature; Tamb = ambient air temperature; Rf = heat ow coe cient between the actuator and the air; C = thermal capacity of the actuator; R = resistance of the heater: Assume that (1) the actuator acts as a pure electric resistance, (2) the heat ‡ow into the actuator is proportional to the electric power input, and (3) the motion d is proportional to the di¤erence between Tact and Tamb due to thermal expansion.
So we will de…ne the total in‡ow to be Win = Wnom + W; so the equations become h_ 1 = h_ 2 = 1 1 1 1 (1 + h1 ) + Wnom + W (30) 20 (100) (100) 1 1 1 1 (1 + h1 ) (1 + h2 ) (30) 20 (30) 20 or, with the nominal in‡ow included, the equations reduce to h_ 1 = h_ 2 = 1 1 h1 + W 600 100 1 1 h1 h2 600 600 Taking the Laplace transform of these two equations, and solving for the desired transfer function (in cc/sec) yields H2 (s) 1 0:01 = : W (s) 600 (s + 1=600)2 which becomes, with the in‡ow in grams/min, H2 (s) 1 (0:01)(60) 0:001 = =: W (s) 600 (s + 1=600)2 (s + 1=600)2 (c) With hole B open and hole A closed, the relevant relations are Win WB h_ 1 = h_ 2 = WB = WB = WC = WC = Ah_ 1 1p g(h1 R Ah_ 2 1p gh2 R h2 ) 1 p 1 g(h1 h2 ) + Win AR A 1 p 1 p g(h1 h2 ) gh2 AR AR 2043 With the same de…nitions for the perturbed quantities as for part (b), we obtain h_ 1 = h_ 2 = p 1 1 (1)(1000)(30 + h1 10 h2 ) + Win (1)(100)(30) (1)(100) p 1 (1)(1000)(30 + h1 10 h2 ) (1)(100)(30) p 1 (1)(1000)(10 + h2 ) (1)(100)(30) which, with the linearization carried out, reduces to p 1 1 1 2 _h1 = (1 + h1 h2 ) + Win 30 40 40 100 p 1 1 1 1 2 h_ 2 = (1 + h1 h2 ) (1 + h2 ) 30 40 40 30 20 and with the nominal ‡ow rate of Win = h_ 1 = h_ 2 = p 10 2 3 removed p 1 2 ( h1 h2 ) + W 1200 100 p p p 1 2 2 2 1 h1 + ( ) h2 + 1200 1200 600 30 However, unlike part (b), holding the nominal ‡ow rate maintains h1 at equilibrium, but h2 will not stay at equilibrium.
Feedback Control of Dynamic Systems Solutions Manual (6th Edition) by Gene F. Franklin, J. David Powell, Abbas Emami-Naeini
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