ECE 320 Exam2-solns

ECE 320 Exam2-solns - NAME Solu Ham)” ECE 320 EXAM II...

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Unformatted text preview: NAME Solu Ham)” ECE 320 EXAM II Nov. 15, 2007 One two—sided note sheet allowed. No calculators allowed. 1. (15 pts) Convolution. Compute the convolution y(t) : x(r) * Mt) for the Signals given below. 0.8.L 0.6L Ml l 0.2} O O 1 2 3 4 5 6 t X{A~t) tos {*3 i A £3 0 A no Overhw> z? 50w::x&)*htd =0 6m t743 9.0 2. (’Ffipts) Bode Approximation. Find the Bode plot straight—line approximation corresponding to 108(s+i0) the transfer function H(s) = —-~—--~—-~2~ (s +100)(s +1000) . You can do your work on this page and plot your result on the: next page. H63): ’03,. (0(T§“+: [00(1900 + 6) {000L(I:oo +1): MnfiiLLw :: ([500 +;)(M:é +02, {FHij 3 10 {ajwflfl + 10 ‘53”) J 1+ (“J/10):, " 10 (03M {[+(W/100)L w 40h)?” W 4:53» fails: w 410 : 20mm ()0) ¢ 20 AG 0 (4,) '> lo : 10 (03”) w/(o (new; H“; one, abo‘JQ) “w“m%00%+mwuwMJ {one (fie): km W! MONO-J LA) iHQ’m) ; dB _3OF ‘ _ :i L ._'* 4O 1 }L_1.| l _.L_;i;a;i‘i ) 1o0 101 102 103 10“ IS" 3. (Q pts) Steady~State Response‘far a Sinusoidal Input. W's (s +100)(5 +1000) x0) : lOsin(IOOt), find the steady-state output 32(5) . (No need to simplify too much; just set up the expressions to the point where it can be simplified by a calculator.) Consider the system fer which H (S) = . Given the input 3%) : tO'(H(J'0°J]5{m(Ioot + GHUIMU W mat-too L ‘i [cola-[00” ‘ i looz+ i000 imam] =~ R H 4. (20 pts) Passive BPF. (a) Design a passive series RLC bandpass filter with B = 100 and (no 2 1000. Let C = 0.1 uF and then select appropriate vaiues for R and L. Sketch your circuit. (b) Give numerical values for me 2 H0030) and Q for your circuit. (C) ’ ‘ . “ 1 , :---:'-,-:" ""“'"i‘ "" . "Tm"i"'n”“"i I I I u/‘l :3 45-0 d What is the transfer function H s of our circuit? ( ) ( ) y watt {org 5. (30 pts) Broadband Butterworth BPF. Design an active broadband Butterworth BPF with the specifications below. Sketch the entire circuit. (REMINDER: Iogm(2) = 0.3, longO) = 1.3) c passband gain K = 20 dB ' CDC] 2 ' @CZ : o the frequency response is 50 dB down from the passband gain at cg: 5 0 the frequency response is 15 dB down from the passband gain at 0%,:— 2000 o Use 1 k9 resistors for the LPF and 1 ii}: capacitors for the HPF. I ’S’ LPFnAeV,‘ n: M -{ wzpi 10 log”) (7400/1000) 10 [03,0 (Z) 4’ wfz Atsz HFF (WARP: m: {0 __,( 57" z—{ffnlsl w(ojw(/M/;} 20 (03,3(20) 16 f W w" 1.3 L?F “francsfe/r {an 3 HL(5') =2 WA (srwaMSzv‘wuf + We: ’FCV‘ J), HH(€) :: 5 1 gz+fiwgs+ we, » LPF lsfevralm' 5413.2. I -—~ :2 l w... A y - F t: can , 1000 mac ~—> c .w 0000);: ,1»: (R 1k) ’ LP? Zugk“ caer stavaeh “- A l C t 2’ '2 l’up [MC-L * (000 W R .Ci :2; I ((600); L t w 3(0002": ' -Wzoy F a game; 5”" <1 - Rumor "‘ m r HP? 1AA 17er {5132: w .a. J; «pic =1» M T 1 '/2‘I00‘C [0-4 loo ...
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This note was uploaded on 04/02/2008 for the course ECE 320 taught by Professor Strickland during the Spring '07 term at Arizona.

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ECE 320 Exam2-solns - NAME Solu Ham)” ECE 320 EXAM II...

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