001 x 100k 1meg 100k 100 meg fp 1 2a1 12100m159pf fp

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Unformatted text preview: 100 Meg fp = 1/ 2Πa1 = 1/2Π.100M.159pF fp = 10 Hz Sol 8: Sol 13) T = ( A o1 ´b f ) T = 0.01 ´ 1000 = 10 A) W o= 1k ´ 10k ´ 11 = 10488 Q= Wo 10488 = = 0.953 W 1+W 2 11000 K= 1 2Q K= 1 = 0.52 2 ´ 0.953 B) Underdamped ! C) 5500 ± 5500[1 - 3.63] 2 ± j8924 1 d) E) Q = 0.5 W o= 5500 1 + T = 3.025 T = 2.025 1k ´ 10k (1 + T ) = 5500 2 Bf = T 2.025 = = 2.02 AOL 1000 Sol 16) Sol 17) a) T = 99 Q = 2.87 b) S1,2 c) f0 = (10k.100k.100)1/2 = 316K K = 0.17 fR = 311K 55k + j311k ; 55k – j311k Af = 10,000/100 = 100 d) e)...
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This document was uploaded on 02/18/2014 for the course ECE 463 at Rutgers.

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