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Control Systems Engineering by Nise Solution Manual

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X + 15x + 50x = 2x, or x + 15x +48x = 0 c. x + 15x + 50x = 4 53. The relationship between the nonlinear spring’s displacement, xs(t) and its force, fs(t) is − fs (t) xs (t) = 1 − e Solving for the force, f s (t) = − ln(1 − x s (t)) (1) Writing the differential equation for the system by summing forces, d 2 x(t) dx(t) + − ln(1 − x(t)) = f (t) dt 2 dt (2) 44 Chapter 2: Modeling in the Frequency Domain Letting x(t) = x0 + δx and f(t) = 1 + δf, linearize ln(1 – x(t)). ln(1− x) − ln(1 − x0 ) = d ln(1 − x) δx dx x =x 0 Solving for ln(1 – x), ln(1− x) = ln(1 − x0 ) − 1 1− x x = x0 δx = ln(1− x 0 ) − 1 δx 1− x 0 (3) When f = 1, δx = 0.

Therefore, y = x1. x = y = - 0 Kx - Jx 1 0 1 Dx x + Jx 0 Jω ωz x 18. The equivalent cascade transfer function is as shown below.

Finally, T(s) = 9 4 . s2 + 21 s + 21 41. Writing the equations of motion, (J1s2+K1)θ1(s) - K1θ2(s) = T(s) 2 -K1θ1(s) + (J2s +D3s+K1)θ2(s) +F(s)r -D3sθ3(s) = 0 -D3sθ2(s) + (J2s2+D3s)θ3(s) = 0 where F(s) is the opposing force on J2 due to the translational member and r is the radius of J2. But, for the translational member, F(s) = (Ms2+fvs+K2)X(s) = (Ms2+fvs+K2)rθ(s) Substituting F(s) back into the second equation of motion, (J1s2+K1)θ1(s) - K1θ2(s) = T(s) -K1θ1(s) + [(J2 + Mr2)s2+(D3 + fvr2)s+(K1 + K2r2)]θ2(s) -D3sθ3(s) = 0 -D3sθ2(s) + (J2s2+D3s)θ3(s) = 0 Notice that the translational components were reflected as equivalent rotational components by the square of the radius.

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