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Unformatted text preview: g , E J“ Page 1 Copy «Title» ECE 202 Summer 2009 Name Q Q (.Qfl ECE 202 Summer 2009 Exam 3 7/24/09 Problem Score 1 /25 2 /25 3 /25 4 /25 Total /100 Page 2 Copy «Title» ECE 202 Summer 2009 Name Question 1: (25 Points) For the circuit below, find: 1) The filter classification (High—pass, Low—pass, Band—pass, or Band—reject) Why? 2) Find Hmax 3) Find the cutoff frequency/frequencies R1 L F \//\ /\ f‘f‘vfifl\ + + Vin Vout R“? i r\ 1 _; Lou/Opotgfi Elia, The, gqivmfiw ‘3 ,0 S‘Mflv 6d; 10w gap/mam" aim M O)?“ Q) [4’9" Qty/cyan. TIM/‘43)”; be“ g’Bfl/érzcib" MU (9g: [saga/A +0 OVJYPVJH amok Mk aqjcfilc/ma‘cg‘ (/4 U [9C s‘vfay‘gggayc a. “(WAX OCLUKSQ‘) S30] [mi rH‘ {yea/afar, (is (A 6},M,+ 600 V AIV H ,. i: l k i [WM {2’ TR?) Page 3 Copy «Title» Ra M” ‘r; r 06.1 ( R: H23) ECE 202 Summer 2009 ' 7Q» “293% “a L“ ( WMWMT" \3 02 \thay’vwafi' Name Page 4 Copy «Title» ECE 202 Summer 2009 Name Question 2: (25 Points) For the circuit below, 1) Find the BW 2) Find Hmax Hint: Using Om” formulas (located on back of exam), transform R2 and L as done in class, then combine the three resistances into Req to get H(s) for a parallel RLC circuit. l T * é R2 l 10 mOhms iin(t) "\ Re C R vout(t) 4cos(10t) 2 300 Ohms ,/~\ 100th E 50 Ohms J I : 0,1H we 1 l l E l o - Lu 1- we 0,} m WC) v I .W, mm a ‘7 6 / infirm/z». Q»; L: 0.1 it a 1/2): 493%: We 12%: {Hiram :— 30;; n 9 ‘ in 42m“ f QL fl New Circuit: Page 5 Copy «Title» ECE 202 Summer 2009 Name Page 6 Copy «Title» ECE 202 Summer 2009 Name Question 3: (25 Points) For the two—pole Sallen-Key Low—Pass Normalized Butterworth—type filter below,prove that this is a Normalized Butterworth low-pass filter find new component values to make (Dc = 1M rad/s and C1=C2 = 1nF. Remember, A Normalized Butterworth filter has a cutoff frequency of 1 rad/s, and its two-pole form is $2 + x/ES + 1. The transfer function of the circuit is: K _ R1R2C1C2 H(S)‘ 2+( 1 + 1 +1—K) +_1___ 3 R161 R261 R262 5 R1R26162 Where: K : Ra + Rb Ra Given: 1 1 R1=1O\/2; R2=20\/§; 61 2561:; C2 =fiF; 12(1sz Hint: In other words, use the given formulas and the Normalized Butterworth equation to prove that it is a Butterworth function. Then, use frequency and magnitude scaling to get the correct new component values. Rb Page 7 Copy «Title» ECE 202 Summer 2009 Name e , I“) ‘ ' [AK ,3 V” M .Q‘ W .f— W) g H375“ “"9 + (M 1 Rm (2969 S + Ragaééa (7%; ,r ,L W») i,“ ML“ ‘1 % Rté‘ 9%} 123 C2 m RIR'AGC’A - + ( l/ I: ' I V : . M ‘ a ( * ‘ 4/1223. :- / W" 105$ '7 «713 mega 3:2 / 9‘“ P FY54] (/96, “gray/0x57 éwda £597“ {I‘M-k g" +0 (W‘flk/f Kg: :: MO 5 Q :Cane I; .3 5:3)”0 (Ml/1!“) new W I New N 3‘ , LuQ u 4?, M5,5W'-hwuh. 350» (Eng ~ {Tim WM“ :: lmf— :7 Km“ 30 / I» ’CQV m 2 Km M; Page 8 Copy «Titie» ECE 202 Summer 2009 Name Question 4: (25 Points) Find the resonant frequency of Zn CP—h—JV } < /”\J[-‘V\,r‘\___ R1 0 10uH j 200 Ohms 5UP Zin r R2 10 Ohms ~ 1 o . _ M ’ “1“ M“ J (2. (/JL 2 -ng lama K‘+3W¢ "Lme; : R‘ILDZ’f' J13. a JWQJ, ‘. ' 1221 [3011, deL'HQR Q93.de l, " ‘ Z? . 1 :1 M- ,J 'RewL'HZ‘wL Zme) - ‘2‘ A)"; + O I 9; flaw, ‘23 1“ will? ‘ WWW ‘ \3 WWW W0 2%: Re, «, _« :1, 9:) WERQD‘LL’ Ra “(er . m... r; 0):? [Rach " L1] ' Ra ' » W1 ._ r’“ “9 a: WK“ Raffles—L1 ” ' woes’e — e’ .. LL ’0 m 57. '1’ 0 Ma mk/ ) ~—— " r10 2; 9(55 ”‘” g6 “” 50/ § 1 Ha ‘ [<90 .100 M -' 4}) "5— {’0 (76?, ‘0')@ C’ Page 9 Copy «Title» ECE 202 Summer 2009 Name Page 10 Copy «Title» ECE 202 Summer 2009 Name Page 11 Copy «Title» ECE 202 Summer 2009 Name Laplace Transform Pairs Item t L t =F 5 Number f( ) {f( )} ( ) 1 MW) K K 2 Ku(t)or K ‘3- 1 3 W) 5—2‘ 71! 4 (3711105) Sn+1 1 5 —at t e u( ) S J; a —at 6 t5 (S + a)2 t n! n —(1 7 t 6 MO (5+a)n+1 w 8 sm(wt) u(t) 52 i: wz 9 cos(wt) u(t) 32 ng - _—__—.___ 10 e sm(wt)u(t) (S + 602 + a)2 11 mat s + a e cos(wt) u(t) (S + 002 + 602 12 I Zws tsm(a)t) u(t) ($2 + wz)2 52 _ (U2 13 tcos(wt)u(t) #62 + a)2)2 s ' + a) cos 14 sin(wt + @1105) #81“: 2 (p S + a) _ 5 cos g0 — (1) sm q) 15 cos wt+ u t ——-————— < «2) < > 52 + (92 16 e"a‘[sin(wt) wt cos(wt)]u(t) 2mg [(5 + a)2 + (02]2 s + a _at . 2 _.__..—____ 17 te sm(wt)u(t) w[(s+a)2 Jr(1)2]2 C — C a C15 + C2 1 _at 2 1 ) . ] _________ 8 9 [C1 cos(a)t) + <—-———w sm(wt) MO (5 + (1)2 + a)2 19 2 A2 -13 fli+fl +39 cos[wt—tan (EHMO s+a+jw 3+a—jw 20 2 A2 32 "at ‘1 B t AHB + A_jB + te cos[wt — tan u() (S + a +jw)2 (S + a _jw)2 Page 12 Copy «Title» ECE 202 Summer 2009 Name Laplace Properties Property fit) £{f(i)}=F(S) Linearity alffit) «E— {szgle} (£1515) + agFg (5) Time Shift f0: — T)u_(£ — T) e“5TF(3) Muitipfication _ . d , f t a: f m_ ' ‘ W H M} (Isms) Muitipiicatéon V 7 . HHS) ‘ byt“ inf“) “1}” d3" iirequency Shift fat/Kt) F(s + a) Time {If V w Differentiation Era) SFCS) W I6) } Second-order dzflr) 2 e e ,(l. " m " mm x J ” Differentiation 05:2 S {Nb} 5’“) f {0 > nth-order d’lffit) , .. __ u ,_. g x M m M Differentiation din S"F(S}—s” 1m} )-5” 2WW )—’~~f“ W 3 , e, be" " (aiii fflfldq “3) Law} (@014 Tame we 3 +—S*— Integration 0 1 {7(3) {13)}; ; (snag T Time or 1 3 Frequency fCaf} -F (_) Scaiing a. a EE-202,Ex 3 Sp 08 page 12 Original Circuit Exact Approximate Equivalent Circuit at Equivalent circuit, for high Q, (00 (QL > 6 and QC > 6) and (0 within (1 i 0.05) (00 ...
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