key_le2_41_sum11

key_le2_41_sum11 - E CE 3 O4L L ecture E xam N o 2...

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ECE 3O4L Lecture Exam No. 2 FRIoay Jur,y 8, 12-l:L0 p. M. GTID No. 9 Summer ?OLL V.(t): E[ | i-t*) r fte -e fltr^il vs t: R c- t(t) 1= 1'753wttt E: Eo K$n:zo [ = Rr r ltrll ( ttr t R' llfts) s a+ le ll( 3 + Vaa 3.8 f t-e'frtJ* 3'oel ic*tr 3& t,T teS i,f **ia.s O'L?Iat V3s V.1 is: f 3 f;?n3"?14d4& Io vt i V.. t a* Rs # 3'eos r' ", a ?, a, ?^ *ryt I.< Name | | e!> CI i - (i) l." n '' 5 $.?tqv ll ltl/(gnd) l6 zt + 1ll(r16) 3+16 e'tll t1)* 6i'35>Jl Cr-,h 3 O,8?Zau *t e [tr. s- ?. a B , Instructions. Totally Closed Book and Note. Calculator Permitted. Four Equally Weighted Problems. All Work Must Be Shown for Credit. 1. What is the current a5(t) in the circuit shown below when f :2 ms? The source is e(t) : 20 Y u(t) where u(t) is the unit step function. The component ralues are: R1 ){ kQ, R2:3 kf), Rs:2 k0, -Ra:16 kf,), Rr:11 k0, and C:0.15 p,F. 5 I -S 3s c*) iz V1 R 3 aq is(t) : O.3_7 I Nvt A
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2. Theexcitation, e(t), in the circuit shown below is e(t) : 10 cos(ut) v' Determine the resonant frequency, quality factor, and half power bandwidth for the current i(t)' The component values u'""t"i, --R2: Rs: ftn: t5Ct' L:3 mH' and C :0'22p'F' The frequencies
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This note was uploaded on 08/30/2011 for the course ECE 3041 taught by Professor Brewer during the Spring '08 term at Georgia Tech.

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key_le2_41_sum11 - E CE 3 O4L L ecture E xam N o 2...

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