Ch12s - Chapter 12 Problem Solutions 12.1(a Af = A 1 A 5...

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Chapter 12 Problem Solutions 12.1 (a) 5 5 5 5 1 5 10 120 1 (5 10 ) 120 (120)(5 10 ) 5 10 0.008331 f A A A β β β β = + × = + × + × = × = (b) 3 3 3 3 5 10 120 1 (5 10 ) 120 (120)(5 10 ) 5 10 0.008133 β β β × = + × + × = × = 12.2 (a) 0.15 1 f A A A β β = = + T A β = (i) T = ∞ (ii) 4 3 80 dB 10 1.5 10 A A T = = = × (iii) 15 T = (i) 1 6.667 f A β = = (ii) 6.662 f A = (iii) 6.25 f A = (b) (i) T = ∞ (ii) 3 2.5 10 T = × (iii) 25 T = (i) 1 4.00 f A β = = (ii) 3.9984 f A = (iii) 3.846 f A = 12.3 (a) 1 125 1 f A A A β β = = + 0.0080 β = (b)

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[ ] [ ] (125)(0.9975) 124.6875 124.6875 1 (0.008) 124.6875 1 (0.008) A 124.6875 1 0.9975 49,875 f A A A A A A = = = + + = = = 12.4 (a) 4 4 3 10 100 1 (10 ) 9.9 10 β β = + = × (b) 4 100 ( 0.10) 0.001 0.10% 10 f f f f f dA A dA A A A dA A = = = − = − 12.5 4 4 4 3 0.001 ( 0.10) 5 10 500 5 10 500 1 (5 10 ) 1.98 10 f f f f f dA A dA A A A A A β β = = × = × = + × = × 12.6 (a) For Fig. P12.6(a) 1 1 1 1 200 10 1 200 1 10 o o i o v v v v β β = = + + 1 1 200 10 40 1 200 1 10 f A β β ⎞⎛ = = ⎟⎜ + + ⎠⎝ 2 1 1 1 1 50 (1 200 )(1 10 ) 1 210 2000 β β β β = + + = + + 2 1 1 2000 210 49 0 β β + = 1 1 210 44100 4(2000)(49) 2(2000) 0.1126 β β ± + = = For Fig P12.6(b) 2 2 (200)(10) 40 1 (200)(10) 0.0245 β β = + = Fig. P12.6(a) (b)