# ch02 - T W O Modeling in the Frequency Domain SOLUTIONS TO...

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T W O Modeling in the Frequency Domain SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Transfer Functions Finding each transfer function: Pot: V i (s) θ i (s) = 10 π ; Pre-Amp: V p (s) V i (s) = K; Power Amp: E a (s) V p (s) = 150 s+150 Motor: J m = 0.05 + 5 ( 50 250 ) 2 = 0.25 D m =0.01 + 3 ( 50 250 ) 2 = 0.13 K t R a = 1 5 K t K b R a = 1 5 Therefore: θ m (s) E a (s) = K t R a J m s(s+ 1 J m (D m + K t K b R a )) = 0.8 s(s+1.32) And: θ o (s) E a (s) = 1 5 θ m (s) E a (s) = 0.16 s(s+1.32) Transfer Function of a Nonlinear Electrical Network Writing the differential equation, d(i 0 + δ i) dt + 2(i 0 + i) 2 5 = v(t) . Linearizing i 2 about i 0 , (i 0 + δ i) 2 -i 0 2 =2 i i=i 0 δ i=2i 0 δ i. . Thus, (i 0 + δ i) 2 =i 0 2 +2i 0 δ i.

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Solutions to Problems 13 Substituting into the differential equation yields, d δ i dt + 2i 0 2 + 4i 0 δ i - 5 = v(t). But, the resistor voltage equals the battery voltage at equilibrium when the supply voltage is zero since the voltage across the inductor is zero at dc. Hence, 2i 0 2 = 5, or i 0 = 1.58. Substituting into the linearized differential equation, d δ i dt + 6.32 δ i = v(t). Converting to a transfer function, δ i(s) V(s) = 1 s+6.32 . Using the linearized i about i 0 , and the fact that v r (t) is 5 volts at equilibrium, the linearized v r (t) is v r (t) = 2i 2 = 2(i 0 + δ i) 2 = 2(i 0 2 +2i 0 δ i) = 5+6.32 δ i. For excursions away from equilibrium, v r (t) - 5 = 6.32 δ i = δ v r (t). Therefore, multiplying the transfer function by 6.32, yields, δ V r (s) V(s) = 6.32 s+6.32 as the transfer function about v(t) = 0. ANSWERS TO REVIEW QUESTIONS 1. Transfer function 2. Linear time-invariant 3. Laplace 4. G(s) = C(s)/R(s), where c(t) is the output and r(t) is the input. 5. Initial conditions are zero 6. Equations of motion 7. Free body diagram 8. There are direct analogies between the electrical variables and components and the mechanical variables and components. 9. Mechanical advantage for rotating systems 10. Armature inertia, armature damping, load inertia, load damping 11. Multiply the transfer function by the gear ratio relating armature position to load position. 12. (1) Recognize the nonlinear component, (2) Write the nonlinear differential equation, (3) Select the equilibrium solution, (4) Linearize the nonlinear differential equation, (5) Take the Laplace transform of the linearized differential equation, (6) Find the transfer function. SOLUTIONS TO PROBLEMS 1. a. F ( s ) = e st dt 0 =− 1 s e st 0 = 1 s b. F ( s ) = te st dt 0 = e st s 2 ( st 1) 0 = ( st + 1) s 2 e st 0
14 Chapter 2: Modeling in the Frequency Domain Using L'Hopital's Rule F ( s ) t →∞ = s s 3 e st t →∞ = 0. Therefore, F ( s ) = 1 s 2 . c. F ( s ) = sin ω t e st dt 0 = e st s 2 + 2 ( s sin t cos t ) 0 = s 2 + 2 d. F ( s ) = cos t e st dt 0 = e st s 2 + 2 ( s cos t + sin t ) 0 = s s 2 + 2 2. a. Using the frequency shift theorem and the Laplace transform of sin ω t, F(s) = ω (s+a) 2 + ω 2 . b. Using the frequency shift theorem and the Laplace transform of cos ω t, F(s) = (s+a) (s+a) 2 + ω 2 .

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ch02 - T W O Modeling in the Frequency Domain SOLUTIONS TO...

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