Control Systems Engineering - Instructor Solutions Manual by Norman S. Nise

By Norman S. Nise

Teacher recommendations handbook (ISM) for keep watch over platforms Engineering sixth version c2011 via N. S. Nise.

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At equilibrium I2 k d 2H mg = k = 0 . From which we get that or H 0 = I 0 2 2 mg H dt b. Following the ‘hint’ procedure: 2( I 0 + δI ) 2 ( H 0 + δH ) d 2δH m =k dt 2 ( H 0 + δH ) 4 2( I 0 + δI ) ( H 0 + δH ) 2 δH = 0 ,δI = 0 δH − k ( H 0 + δH ) 4 δH = 0 ,δI = 0 After some algebraic manipulations this becomes: 2kI 0 d 2 δH 2kI 02 = m δH − δI 2 3 dt mH 0 mH 02 Obtaining Laplace transform on both sides of the equation one obtains the transfer function: Copyright © 2011 by John Wiley & Sons, Inc. δI Solutions to Problems 2-41 2kI 0 mH 02 δH ( s) =− δI ( s ) 2kI 02 s2 − mH 03 59.

38 Ea (s) s( s + ) 21 43. 75) 3 4 1 3 Since θ2(s) = 1 θm(s), 4 1 θ 2 (s) 12 = . 75) 44. 33 1000 Kt Tstall 100 Ea 12 1 2 = = ; Kb = = . 33 12 6 3. 92 = = . Since θL(s) = θm(s), = . 92 Copyright © 2011 by John Wiley & Sons, Inc. Solutions to Problems 2-31 45. From Eqs. 46), RaIa(s) + Kbsθ(s) = Ea(s) (1) Also, Tm(s) = KtIa(s) = (Jms2+Dms)θ(s). Solving for θ(s) and substituting into Eq. (1), and simplifying yields Dm ) Ia (s) 1 Jm = Ea (s) Ra s + Ra Dm + K b Kt Ra J m (s + (2) Using Tm(s) = KtIa(s) in Eq.

Solving for θ(s) and substituting into Eq. (1), and simplifying yields Dm ) Ia (s) 1 Jm = Ea (s) Ra s + Ra Dm + K b Kt Ra J m (s + (2) Using Tm(s) = KtIa(s) in Eq. (2), Dm ) Tm (s) Kt Jm = Ea (s) Ra s + Ra Dm + K b Kt Ra J m (s + 46. For the rotating load, assuming all inertia and damping has been reflected to the load, (JeqLs2+DeqLs)θL(s) + F(s)r = Teq(s), where F(s) is the force from the translational system, r=2 is the radius of the rotational member, JeqL is the equivalent inertia at the load of the rotational load and the armature, and DeqL is the equivalent damping at the load of the rotational load and the armature.

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