By Norman S. Nise

ISBN-10: 0470547561

ISBN-13: 9780470547564

Teacher options guide (ISM) for keep an eye on platforms Engineering sixth variation c2011 by way of N. S. Nise.

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**Extra resources for Control Systems Engineering - Instructor Solutions Manual**

**Example text**

I Solutions to Problems 2-41 2kI 0 mH 02 δH ( s) =− δI ( s ) 2kI 02 s2 − mH 03 59. The two differential equations for this system are: M b &x&s + K a ( x s − x w ) + C a ( x& s − x& w ) = 0 M us &x&w + K a ( x w − x s ) + C a ( x& w − x& s ) + K t ( x w − r ) = 0 Obtaining Laplace transform on both sides gives ( M b s 2 + Ca s + K a ) X s − ( K a + Ca s ) X w = 0 ( M us s 2 + Ca s + ( K a + K t )) X w − ( K a + Ca s ) X s = RK t Solving the first equation for X s and substituting into the second one gets Xw K t ( M b s 2 + Ca s + K a ) ( s) = R (M us s 2 + Ca s + ( K a + Kt ))(M b s 2 + Ca s + K a ) − ( K a + Ca s)2 60.

3 2 F ( s ) s (8s + 12s + 26s + 18) 28. a. (4s 2 + 8s + 5)X1 (s) − 8sX 2 (s) − 5X3 (s) = F(s) −8sX1 (s) + (4s 2 + 16s)X 2 (s) − 4sX 3 (s) = 0 −5X1 (s) − 4sX 2 (s) + (4s + 5)X 3 (s) = 0 Solving for X3(s), Copyright © 2011 by John Wiley & Sons, Inc. 3 Chapter 2: Modeling in the Frequency Domain X 3 (s) = 2-22 (4s 2 + 8s + 5) -8s F(s) 2 −8s (4s + 16s) 0 -4s 0 −5 Δ F(s) = −8s (4s 2 + 16s) −5 −4s Δ or, X 3 (s) 13s + 20 = 3 F(s) 4s(4s + 25s 2 + 43s + 15) b. (8s 2 + 4s + 16) X1 (s) − (4s+ 1) X 2 (s) − 15 X 3 (s) = 0 −(4s + 1) X1 (s) + (3s 2 + 20s+ 1) X 2 (s) − 16sX3 (s) = F(s) −15 X1 (s) − 16sX 2 (s) + (16s+ 15) X3 (s) = 0 Solving for X3(s), X 3 (s) = (8s 2 + 4s + 16) -(4s+1) 0 2 (3s + 20s+1) F(s) −(4s+1) −15 -16s 0 Δ (8s 2 + 4s + 16) -(4s+1) -F(s) −15 −16s = Δ or X3(s) 128s 3 + 64s 2 + 316s + 15 = F(s) 384s 5 + 1064s 4 + 3476s 3 + 165s 2 29.

Solutions to Problems 2-19 21. a. 1 2 x10−6 s 1 Z 2 ( s ) = 105 + 2 x10−6 s Z1 ( s ) = 5 x105 + Therefore, − Z 2 (s) 1 ( s + 5) =− Z1 ( s ) 5 ( s + 1) b. ( s + 5) ⎛5 ⎞ Z1 ( s) = 105 ⎜ + 1⎟ = 105 s ⎝s ⎠ 5 ⎞ ⎛ 5 ( s + 10) Z 2 ( s ) = 105 ⎜1 + ⎟ = 10 ( s + 5) ⎝ s+5⎠ Therefore, − s ( s + 10 ) Z 2 ( s) =− 2 Z1 ( s ) ( s + 5) 22. a. 625) b. 25 x106 110 x103 + s Therefore, Z1 ( s) + Z 2 ( s ) 2640s 2 + 8420s + 4275 = Z1 ( s ) 1056s 2 + 3500s + 2500 Copyright © 2011 by John Wiley & Sons, Inc. Chapter 2: Modeling in the Frequency Domain 2-20 23.

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