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**Additional info for Feedback Control of Dynamic Systems Solutions Manual (6th Edition)**

**Example text**

55. Assume that Eq. 74) describes ‡ow through the equal-sized holes at points A, B, or C. (a) With holes at A and C but none at B, write the equations of motion for this system in terms of h1 and h2 . Assume that h3 = 20 cm, h1 > 20 cm; and h2 < 20 cm. When h2 = 10 cm, the out‡ow is 200 g/min. (b) At h1 = 30 cm and h2 = 10 cm, compute a linearized model and the transfer function from pump ‡ow (in cubic centimeters per minute) to h2 . (c) Repeat parts (a) and (b) assuming hole A is closed and hole B is open.

DYNAMIC MODELS (a) Relation between torque constant and electric constant. Torque constant: 1 ounce 1 inch 0:2778 N = 1 Ampere 2:540 1A 10 2 m = 7:056 10 3 N m= A Electric constant: 1V 1 J=(A sec) = = 9:549 1000 RPM 1000 30 rad/ s 10 3 N m= A So, 1 oz in= A = = 7:056 10 3 V=1000 RPM 9:549 10 3 (0:739) V=1000 RPM (b) 25 V=1000 RPM = 25 1 oz in= A = 33:872 oz in= A 0:739 (c) 25 V=1000 RPM = 25 9:549 10 3 N m= A = 0:239 N m= A 19. The electromechanical system shown in Fig. 50 represents a simpli…ed model of a capacitor microphone.

18, and assuming the area of both tanks is A; the values given for the heights ensure that the water will ‡ow according to 1 1 WA = [ g (h1 h3 )] 2 R 1 1 WC = [ gh2 ] 2 R WA WC = Ah_ 2 Win WA = Ah_ 1 From the out‡ow information given, we can compute the ori…ce resistance, R; noting that for water, = 1 gram/cc and g = 981 cm/sec2 ' 1000 cm/sec2 : 1p 1p WC = 200 g= mn = gh2 = g 10 cm R p R p 1 g= cm3 1000 cm= s2 10 cm g 10 cm R = = 200 g= mn 200 g=60 s s 1 1 g cm2 s2 100 = 60 = 30 g 2 cm 2 3 2 2 200 cm s g (b) The nonlinear equations from above are h_ 1 = h_ 2 = 1 1 p g (h1 h3 ) + Win AR A 1 p 1 p g (h1 h3 ) gh2 AR AR The square root functions need to be linearized about the nominal heights.