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

By Norman S. Nise

Teacher strategies guide (ISM) for keep watch over platforms Engineering sixth variation c2011 through N. S. Nise.

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X + 3δx + 2δx =sin (0+ δx) d sinx δx = 0+ cosx ⎮ δx = δx dx ⎮ But, sin (0+ δx) = sin 0 + x =0 .. x =0 .. . Therefore, δx + 3δx + 2δx = δx. C ollecting term s, δx + 3δx + δx = 0 . b. Let x = δx+π. Therefore, .. δx+3δx+2δx =sin (π+δx) But, sin (π+δx) = sin π + .. d sinx ⎮ δx = 0+cosx ⎮ δx = −δx dx x=π x=π . . Therefore, δx+3δx+2δx = -δx. Collecting terms, δx+3δx+3δx = 0 . 51. If x = 0 + δx, ... . -(δx) δx + 10δx + 31δx + 30δx = e -(δx) But e = e-0 + ... -x de dx .. -x ⎮ δx = 1 - e ⎮ δx = 1 - δx x=0 .

9 s 2 The laplace transform of the systems output is £{T (t )} = T ( s ) = Tref s − Tref s+λ + Tref λ 2aπ f 2aπ f = + 2 2 2 s + 4π f s ( s + λ ) s + 4π 2 f 2 2 Dividing by the input one gets T λ 2aπ f s (s) = + 2 U s+λ Tref s + 4π 2 f 2 66. λ (1− e − at ) λ dV (t ) a. By direct differentiation = V0 (αe −αt )e α = λe −αtV (t ) α dt λ b. V (∞) = Lim V (t ) = Lim V0 e α t ⎯⎯→ ∞ t ⎯⎯→ ∞ (1− e − αt ) λ = V0 e α c. Copyright © 2011 by John Wiley & Sons, Inc. 5 0 0 10 20 30 40 d. 6 X 1012 mm3 X 10-3 67.

1), 1 = -ln(1 – x0 ). 6321. 6321 into Eq. 6321 Placing this value into Eq. 718 54. First assume there are n plates without the top plate positioned at a displacement of y2(t) where y2(t) = 0 is the position of the unstretched spring. Assume the system consists of mass M, where M is the mass of the dispensing system and the n plates, viscous friction, fv, where the viscous friction originates where the piston meets the sides of the cylinder, and of course the spring with spring constant, K. Now, draw the freebody diagram shown in Figure (b) where Wn is the total weight of the n dishes and the piston.