Exam2F10_sol - Name: _______________________________...

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Unformatted text preview: Name: _______________________________ Section: 9:30_Ariyur 11:30_Savran 1:30_Chiu ME 365 EXAM 2 Thursday November 4, 2010 6:30‐7:30 p.m. EE 129 PROBLEM 1 2 3 4 TOTAL POINT DISTRIBUTION SCORE 35 points 35 points 40 points 40 points 150 points Percentage: 2/3 Total = Don’t forget your name and please circle your section instructor. Exam is 6 pages including this cover page: make sure you’re not missing any pages. If you use extra pages, indicate on the problem page that you’re continuing onto an extra page. Pay attention to units. Explain your reasoning. Correct answer with wrong explanation = no credit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ode Diagram! 0! Magnitude 20! (dB)! 40! 60! 80! 1000! ! Phase (deg)! 45! 90! 135! 1801 ! 0! 1 ! 1 0! 2 ! 3 ! 1 0 Frequency ! (rad/sec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j ! ) = G F (j ! ) " G S (j ! ) " L 1,2 = G F (j ! ) " G S (j ! ) " Z Z in 2 +Z out 1 in 2 (5 + 50, 000j ! ) 0.02 10 0 1+ 0.01 ! j = " " (5 + 50, 000j ! ) 1+ 0.001 ! 1+ 0.01 ! j j 300 + 1+ 0.01 ! j = = ! ! ! ! ! ! ! ! ! ! ! 2 (5 + 50, 000j ! ) " (1+ 0.001 ! )(1+ 0.01 ! ) (5 + 50, 000j ! ) + 300(1+ 0.01 ! ) j j j = ! 2 " (1+ 0.001 ! )(1+ 0.01 ! ) j j 1 300(1+ 0.01 ! ) j 1+ (5 + 50, 000j ! ) 2 (5 + 50, 000j ! ) " (1+ 0.001 ! )(1+ 0.01 ! ) (305 + 50, 003j ! ) j j ! !"# $%&'()*+,-.*,/%00%1(&2,34-(5*,/(0-*+,&*-1%+6, , 7*-*+8(&*,-.*,/+*9:*&4;,+*'<%&'*,/:&4-(%&,%/,-.*,3=%5*,%<>38<,4(+4:(-?/(0-*+@, , Z F1 " Z F2 " = R, Z I1 V1 Z =# Vin Z = = F I 1 C1 ! j + R1 = C1 ! j RC1 ! j RC1 ! j R =# =# =# 1 R1 1 ! + Cj 1 R1 1 ! + Cj 1 j! + C1 ! j 1 , ( R 2 ) + ( 2j ! ) 1 C Z Vo =# V1 Z F2 1 R1 1 ! + Cj =# Z I2 = R2 1 ( R 2 ) + ( 2j ! ) 1 C R2 I2 , =# 1 1 =# 1 R 2C 2j ! + 1 0. j ! + 1 V1 Vo RC1 ! j RC1 ! j V $ o= % = = Vin Vin V1 ( + R 1 1 ! )( + R 2C 2j ! ) 1 Cj 1 ( + j ! )( + 0. j ! ) 1 1 1 , !$# A:30(-3-(5*0;,)*'4+(=*,-.*,/:&4-(%&,%/,-.*,3=%5*,B<>C8<,4(+4:(-,!!"#!$%&''()*+,$%&'')+-) .&/0)%&''#@,,"3'*),%&,-.*,<+*5(%:',3&'1*+D,)*-*+8(&*,-.*,<3'',=3&),!123$+44) 4-5625/1"5'#,/%+,-.*,3=%5*,%<>38<,4(+4:(-?/(0-*+@, , T his is a band- pass f ilter. 1 1 = = 1 [rad / sec] 6 , R 1 1 0.1! 1 " 1 ! 1 #6 C 0 0 0 1 1 L P cut- of f f requency is = = 1 [rad / sec] 0 6 R 2C 2 1! 1 " 0.1! 1 #6 0 0 HP cut- of f f requency is $ Pass band: 1to 1 [ ad/sec] 0r , !7# 7*-*+8(&*,-.*,530:*,%/,-.*,+*'('-3&4*,7,&**)*),-%,2(5*,3,-%-30,23(&,%/,EF,/%+,-.*,3=%5*, B<>C8<,4(+4:(-?/(0-*+@, , 1 has gain of 1in the pass- band 1+ 0.1 " j In order f or the combined f ilter has gain of 1 in the pass- band, the high- pass f ilter 0 , need to have a gain of 1 in the pass- band. 0 L ow - pass f ilter ! R C 1j " is R C 1 1+ j " R = 1 6 = 1[M '] 0 T he pass- band gain of the high- pass f ilter ! # R C 1 = R $ 1 % 1 !6 = 1 00 0 & ...
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This note was uploaded on 12/26/2011 for the course ME 365 taught by Professor Merkle during the Fall '07 term at Purdue University-West Lafayette.

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