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55. 3) Fup (s) T H R E E Modeling in the Time Domain SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: State-Space Representation . Ea(s) 150 For the power amplifier, V (s) = s+150 . Taking the inverse Laplace transform, ea +150ea = p 150vp. Thus, the state equation is • e a = −150e a + 150vp . For the motor and load, define the state variables as x1 = θm and x2 = θ m. Therefore, . x 1 = x2 (1) Using the transfer function of the motor, cross multiplying, taking the inverse Laplace transform, and using the definitions for the state variables, .

222 . 122) θ m (s) 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 ) Jm Ia (s) 1 = Ea (s) Ra s + Ra Dm + K b Kt Ra J m (s + (2) Using Tm(s) = KtIa(s) in Eq. (2), Dm ) Jm Tm (s) Kt = 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.

For the parallel analogy, treating the equations of motion as nodal equations yields In the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries. 42 Chapter 2: Modeling in the Frequency Domain 48. Writing the equations of motion in terms of angular velocity, Ω(s) yields K1 K )Ω1 (s) − (D1 + 1 )Ω2 (s) = T(s) s s K1 (K + K2 ) −(D1 + )Ω 1 (s) + (J 2 s + D1 + 1 )Ω 2 (s) = 0 s s K K − 2 Ω 2 (s) − D2Ω 3 (s) + (D2 + 2 )Ω 4 (s) = 0 s s K3 (J3 s + D2 + )Ω 3 (s) − D2 Ω 4 (s) = 0 s (J1s + D1 + For the series analogy, treating the equations of motion as mesh equations yields In the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries.