EE212_Solutions_Chapter14

EE212_Solutions_Chapter14 - Chapter 14-, Solution 1....

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Unformatted text preview: Chapter 14-, Solution 1. "9 R jE-JRC HCCDJ — , — . - . . \5 R+1/]coC 1—](0RC. '0) (D 1 HCm) = where as" = —. 1 + JCD/Cflu . o (a It (0 H = ‘ Hfio)‘ = :1) = 4H(co) = _—— tan'lf — 1+ (co/coo J " 3 ('30 This is a highpass filter. The frequency response is The same as that for E I except that coE = 1/ RC . Thus. the sketches of H and :1) are shown below. mt} _ R‘ L Chapter 1-1, Solution 3. ¢ A 90° __ (a) The Thevenin impedance across the second capacitor Where "U is taken is R z“ =R—R| 1/56 = n+1 _RC l/sC‘ V " Ym : ‘Vj. : - . 0 R + ifs C. 1 + sRC 0) ZTII Chapter 14, Solution 2. v + 1 __ R 1 1 R ‘ Tn _ res Va HUD) — . — . — , . where (on = — R+J03L 1+J03L/R 1+Jm/coD L H — | H£co)| — 4 43 — AHGD) — Halli/3“; 1(1+ (co/02:0)3 (03 2'- 1/SC Y Chapter 14, Solution 5. i Y _ -x . _ ° zTh+1/sc T: (l—sRC)(1+sCZTh) ' V 1 '03C‘ H<s1—‘° — ‘ _ 1 (a) H103): f =—. h . vi (1+SCZm)(1+ sRcr) (1 + SRC?)(1+ sRC — sRC‘/(1 + 5-ch ‘: R +1033L +1/Jf0€ H( )- 4 1 S _ s=R=C= +3sRC+1 Hfifll = 1 4 jcoRC— mch: (b) RC=(4O»103)(2><10'6)= 80»10'3 : 0.08 1 R (b) .—\ = . . There are no zeros and the poles are at 103(- 1+ JC'JRC - 0.383 4 T8? 51 — — — . H J VG ij j03L(1+J'03RC) (o — — — 2 61‘ ( Vi jcoL + Rf£1+jcoRC) R + jcoL (1+ jcoRC‘) s} = ' ”= _ 32.112 ' RC ij —m1RLc: Hm =—. (J R+j(oL—co‘RLC Chapter 14, Solution 4. L'nnpter 14, 5011111011 E). 1 R ‘1‘” R -—c = W J a ' + J E) (:1) Using current division. _ Lo _ + R H0”) _ Ii _ R —j(oL—l/ij' V0 1 + ijC R . . _ 11(0)) _ , _ _ I I ‘ HG”) _ JC'JRC _ Jo: (20)(0.2:s) ‘ - ij + R R — JC'JLU + J03RC) 1+ chRC 7coJLC 1+ j03[20)(0.25) 7 (0311011025) 1— '(oRC J jaS H03) = -— 11135 — 2.3m R — H(coj = — CD'RLC. + R -5- JCDL (b) EVE apply nodal analysis to the Cil‘C‘lliT below" I £ij 11.5 \ , \ x o " ' / R+ 'coL 'coC R+ '(c-L .- —> (b) H(mj— _ J . — J.( J.) i K K R+Jc-3L+1/Jt-JC 1+](0C(R+JO]L} ‘ :3 L ii R - ij I _ Y,_. + ‘4',“ 43.5?“ Chapter 14H Sulutiun T. *‘ R ij+1fjr0C 055'. _ = .1 But ID = + —> 1." = 21° (J'QL_1},'J-mc} {a} 0.0:? .010ng lmL—UJG’C 2.5 410-3 2105.!“ H I‘ _ H = “31.5.;0-3 =1I‘:”:L_TI.J,T3 Y3 _ R ij+1jjmc {b} _ 0.2 = 201:.ng I i+; _ 0.31:10gmH R god—hfij H —10"“l — 0 4000 215(ij41jij} L_ EEJHL—lfjmt‘) 1 1. ‘ F. ‘ {c} 1042 = 201052” H I 1 _ RC 5.235 =logm H Hm} = —° =+ =% H = 1051‘ =1.210 x10” IE 1—2(101L+],f101C)/'R JmRC+2{1—£0'LC) — jm , . . H - = —. Cha 10r14. 50111th 8. m) jw—lfl—m‘ 0.25} P (a) H: 0.05 jm H.13 = 20105210005 = _ 26.02, c: = 0° H - = —4 — (0) 2+ jm—05m‘ — (0} H: 125 H.13 = 20105210125: 41.94. a = 0° 00 . G (c) H0}: 2+]. = 442220343 H.13 = 2010210 4.422 = 13.01. a = 63.43” 3 6 :1 H1 =—— =3.9—'1.2=4.2541-2355° II} 0 1+]. 2_j .1 H13 = 20mgm 4.254 = 12522. a = — 23.55“ Chapter14‘501ution 9, Chapter 14, solution JD. 1 H(fi)=+ (l—Jm)(1+]m/10) Hum) : I _50_ 2 H10 _ I‘ Hfla =-2010g;c.|l-jml—2fllogml1+jam-"10 JBIIJHUJ) 1er “FL?! 0 = — tan'] ((0) — tan'1 (6-) 1 O) The mannitude and Ehnse plots are shown below. 10:10 Chapter 14: Solution 11. 5(1— jmfl 0) Hm) = jm1+ jun/2) HdB = EGlogm 5 — 2010gL3| 1+ jea/JO | — QOlogmljOJI—2010g1.3|1+ij/3| U = —90°— tan'l mfIO — tan" 03/2 The mannitude and base lots are shown below. Chapter 14, Solution 12. HG'B 0.1(1— m) 40 f(‘1-‘)=m, 2010g0.1=—2O 34 The plots are shown below. 20 - 14 m ‘(dm 0.1 1 " .. G 100 01 JD I 20 40 aIgTA 90‘” CI .. _l 1 - _ 10 100 Chapter 14:501ufiflnl3- Chapter 14= Solution 14. Gm 1+ jm (1/10)(1— jm) _ _ ' (Mamie-a) ' (jwllfl-Jmflo) Hifijzifl , “J” _ 25_ -' jmllifl :jco" '- ofl =20+ZOIOgI,|1—jro|—4010gm jrol—2010g10|1+joJ/10| Jfflll— j; +1?J U=—180°—tan'l£o—tan'lfof10 ' —J I I 'J The mannitude and hose lots are shown below. _ . — . —=—F—F— HdJa _ EDI-ogm 2 — 2Ulog1,3|l+_1{o|— EUlogml Jo| Geo 40 —2ologm|1—jmofi—{jmml| 20 -- I ., CI = 30°- tan'; m—tan'll m1 of: 5 ' . al—mlffi, 0.1 1 1'3 - 100 a] 30 I ‘_ -- The mannitude and Ehase plots are shown below. -40 Ho ‘ 40 -- 26 CI] Chapter 14, Solution 15. 40(l—jm) 2(1—J'm) 11(0) — (3_J'm)(10+jc)) _ (1+je3f2)(1+j0f10) HdB = ZOlogm 2 —2010g1,3| 1+ Jo| — 2010g1,3|1+jm;"2|— Zfllogm 0 = tan'] 0] — tan'] 01/2 — ran'1 (#10 The mannitude and base In t5 are shown below. 1—J'm/10| Chapter 14= Solution 15. Gw= 1“? 1mm + jmj 1 +3 1!— 1:1: Gar; . ED lflloglijml 1 . ' 1:} CI] jm — 40109 ID 2D10g(1.-"1GG} Chapter 14,501uriun17. The mannimde and base lots are shown below. (When (l-J'CDKl-jm/HL 6(0)) 2 GE = —2010gm4 — 2010g13| jc-Jl — 201051;: | 1 — jml — 4010gm|1—jmf2 | CI = -90°- tan'lm— 2 tan'l to}? The maanhude and Ehase Eluts are shown below. 6'ch 1 20 -- __ 0.1 L .- " 10 100 _12 __ .- ' " - (I) -20 —- " -40 I- 45:) Chapter 1-1= Solution 19. H _ + Chapter 14,501utiun 13. (‘5') ‘ 100(1+jmf10—m3,’100} Gian}: 4E1+j<flf2332 HdJa =QDlogm‘jml—Zfllogm100—2010gmll—jmflO—mlfiflfll fiDjmfil+jmf5j{l—jmf10} - 10 _ . . «2 =909—tan-ll “fl Gm = 2fllogm 4pm 4010g1c,|1—_1mf2 |— 2fllogm| _1(:J| aids/100.,- —'2010gm‘1—jmf5‘— EUloglc‘l—jmflfll where EGlogm {fill}: £1.94 CI = —90°— 2 tan'l mffi — tan'; EDIE —tan'1(:y'10 _=—F_F—Il-h‘E maanml‘le and 11359 1"” a” Show" 11910“?- The magnitude and 13111159 Eluts are Shown below. Hdfl 1 NE 40' —— 2E}— CLl _ ' 1 1!}, 100 m 0.1 _9UG _ [I] 4313* — Chapter 14: Solution 20. Chapter 1-4= Solution 11. 10(1—jm—c—J3) m9)=.:1— jm)(1+jm/10) _ (l _ ) JED —_1L'3I N53 = Eo—zologlall—le—zmogmll—Jro/m|—2mogm|1+Jca—m- '1 3' 100(1—Jmflfl){1+lmle—m‘flflfl) .’ \ flztan'll 0]] l—tan'l'OJ-tan—lmflo T _2fl1 |_j ' _j .\1_m , fl _ (nng 103 401031,] 1+_1m _U].nglflfl —2[I'10gm|1—jmflfl|— 201051;: l—jmflfl—mlflflfll f "‘1 m 10 CI = 90°- tan'l m— tau'l {9310— tan-lLl-fjmxl The magnitude and 13111159 Elots are shown below. Tea Chapter 1-4= Solution 12. 2o: 20logm k —> k=10 A zero ofslopo —20 {1133' do: at: = 2 —} l—jcofi A pole of slope - 20 {13:16: at m = 20 —} 1 1+ jmflfl 1 A polo ofslope —20 {113.5 do: at (0:100 — a Hence. m: (1— Jmfzom + Jmfl DID) 1fl*(2+jm} H - =— (“3' [20+jco}[lflfl+jm1 Chapter 1-1= Solution 13. A zero of slope — 20 {IE do: at the origin —} joJ l 1—jcof1 A pole of slope - 20 (1136.12: at to = 1 —> 1 A pole of slope - 40 (13:16: at to = 10 —} Hence, 9 _ {1—jm}(1+jm{1o}3 H _ lflfljm {CO— [l+jo:-}[1l]+jro)2 Chapter 145 Solution 24. The phaae plot is decomposed as shown below. .3. 1 90° — _ - 45¢ _ _ ' I \ arglfl+Jm.-"10] I- I | I | F 0.1 1 10' 100 1000 to 45° — argomn " '- . -'" 1 I '- Mg 7._1+J(o.—"100_.-' l 'u where k' is a constant since argk“ = 0. H2930) = l + jfl.3 kn Henee. Gflm) = where k =1k' is constant If 1 e... _ — H490): R—JL-i‘lfiflL—ilfiflc; 1 "; {4){jxlfl3}{1=<10'5),; Chapter 1-15 Solution 25. - _ a ‘ H4930) = 2009+ J (4):: x19J){49 X10") — 1 1 -.———+—ik d.-" _ - -.r “3 JLC Jfi4flxlfl'5fllx10'5) m 5 ziq‘muj—“Jfl-jkfl H930) = R = 2 kg — Chapter 14. Solution 25. ._ _ 4 2(0E,f—1)—R+J|M 4 L mac-J {a} I ,"5x193 _3 4 "-. 2(ac.f4)=2999-—J|K 4 .49e19 ——(5(103mx10_6); R 100 {b} B =— = _= =m Zita-90H)=2OflO—jfiO—4Dflflf5) L 10x10" limef4J= 1—10.?5kfl (C) Q: 99L 2 1 £2195 19.919-3 214.142 . ' R «.I'ILC.‘ R 59 9.11193 — z “r —R 'flL 2 {QUIJ _ +1 2 _ mac,- 2 ’1 2999+ Emma} 49 19-5 —2 ' _ = _ X _ q _ {QEIf-J' JIM 2 I: j :<-1ll::l_1}|[1lefllrlfi};I 2(9):, f4) = 2999 + 9199— 299935) amp) = 2— 193 kg _." 1 1 H290) = R—lemflL— 1m CI. . — u Chapter 14, Solution 17. At resonance, 1 Z=R=1Dfl, m, =— 3 "~ch Bi d _n_% _L a“ Q‘ E _ F. Hence, R 1.13 SCI Lzflzwzlfifl (on it} C=L-;=25L1F 93L ' {5031306) B—E—E—oefi d"- _L_16_ _ __.ffi :8 Therefore, R=]OQ__ L=16H__ C:25}1F__ B=U.625:'ad.-"s Chapter 14. Solution 13. LetR:1Ufl. L R m 05H _B_2[}_' 1 1 C—————2pF ‘ mfiL ‘ {1000}1(0.5) ‘ Therefore, if R =10fl then L=U.5H, C=2yf__ Q=5 Chapter Lil1 Solution 29. jco 1:"jro r Z —> . l to Z=Jfi—.—+ J . Jm l—Jm z = j Ifm— LH|+ r a (:3; 1+ or m2 + jm Since fit] and ifit} are in phase. 11(2) U 1+ m = =o‘n—— 1 n to 1—m- m‘—oJI—l=fl r —1i«t1+4 m‘=—=t}.613 2 m = 03851 rod '5 Chapter 1-1= Solution 30. Select 11:199. L-i-i-flfl'H-‘i 1H ‘ to3Q ‘ ammo) _ ' ’ _ " fl 1 l GEL {lflflflflflfi} l 1 E = —= —= [[5 RC (town) m 5 Therefore, if R = 10 Q then L=5mH, C=U.2F, B=U.5radts Chapter 14= Solution 31. XL XL =toL —} L:— to B_—_—_ 3 L XL 40x10 _ R _ mR _ Exxllefléxfifixlfli = 3396x105 fads: Chapter 14= Solution 32. Since Q :- lfl , B B {olzmfl—j, m3=mD+: to :5 mi 3 = —” = —X = 511 kradt's Q 12o _ m] = o—flfllfi = 5.9T5xlflfl mars m3 = 5—oo25 = 5.02m 1o‘s mar-g Chapter 14, Solution 33. Q 80 Chapter 14, Solution 34. (a) Qzflj —} =- =—‘ ° 2TEfoR Elxiéxlfifi x40x10-‘ R R 40x103 _, L = = — = 14.21 L1H 00L EKfOQ 231x56x105 x30 m — 1 — 1 — 1 443 load-"3 c _ _ _ — _ _ _ “43 1l8x10_3x60x10_6 B l l = 3.33 [ML-"5 (b) (C) _ RC _ 5x103x60x10_6 o = mom: = 1.44331103 x5x103x60x10‘6 = 432.9 Chapter 1-1= Solution 35. At resonance, 10 or , 1 1 1 "R —} R_Y_25><10'3__4DQ _ _i_L_ Q _ 033“: E C _ (90R _ (200x103}(40)_ _; , L_;_+_ “'3 ' 4E ' mgc ' (4x101'3){1o=<1o-5) ' _E_M_ . B— Q — 80 —2.5k1ad.s B _ 12.)]: (on —:= 200—25 = 191510-3111; B , to]2:90—7:200—2322025101-21 -, Chapter 1-4= Solution 35. 1 EDDIE} d {J 2—: J 1—3 5 l Y{¢Jfl}=fi —} Z(to}=R=2kfl H M 1 111% 4 "a (J ' :—— .— —— = U R J14 (03L 2 - 4 =— {QDH Gfiflflfi—jflfilfifi 1 .J'mfl 2 Y - 1 =—— —C— (an! R J12 tocL; 25 pH = l.42l2+ j53.3 £2 i=fl.5—j?.5k8 1 z ' 2 — — 8.8: I132.-.'r Q 106! j' I F j I F i I 4 I] R J '- El 2 I:I{:_.II I J I I Hit-3,3) = 8.35. — 3132.74 51 1 _ 11419.3): E—J 401E,L— Emma) = 1.4212 — j53.3 Q Chapter 14, Solution 37. 4010121 "1 l-= U.5—le.?i kS .I ij(R+_L_) E—ijR E R—jgmL—L) Z . 1 Jc-JC 1C ,.-1_ LDC =JCIJL-'.-'(R 1 I _) = 1 _ 1 J0”: R +++j01L R2 —(roL——}2 JroC roC (DLR: mL—LCE 1111(2): "- 1 m =0 > 93013452 —LC)=L R‘ —{toL——)‘ roC Thus, 1 m 2—H; «th+ REC‘ Chapter 1-1= Solution 33. 1 R —_1'toL Y—.+ 'mC = SIC—f amt-11. J J 11—- —arL~ At resonance, huff] = {1 , Le. Q = mDRC =l325?:{1032‘125K103x20xlO—5 = 912.3 c mflj‘ 0 m .— — = I] I 2' I R +£3.le Bsz—1l 6=2r:ul RC 25x10-‘20x10‘ — '- ~ - L (b) To increase B b\-' 1.00% means that B‘ = 4 R- + our; = E a C'=1'=+=]UHF f I 1 RB 25x10 x4 — l R‘ l 5D 1|- m : — — ., : .—— —s . ' .1. . '3 LC L‘ {4UX10" )[iflk 1'04) 'x4UXlU'Jg: Since C": CCICC' =10L1F and C1=20 ILLF,‘.X-'€Tl'1&110blau1C3=20t1F. '1 _ '2 m = rad 5 Therefore: to increase the bandwidth, we merely add another 20 HF in a series “ith the first 0119. Chapter 14= Solution 39. ChflF'" 14= 501mm 41- (a) This is a series RLC circuit. a —(:.)j to —'T[(n —_.‘E x — . a is (3' B 1 2 f4 {1) 9(90 86}103 smd' 1 R=2+6=SQ, L=lH._ C204]: mo =;(01+032 = 2K(88)x103 =1?5n — 1 l m =—=—=1.5811‘¢l.-" 1 1 l ,3 ]'l( s 13:E —> C=E=fi=1989nf 811x10 XEXIO - Q:rot,L=1.::811=0I19?6 1 1 1 R 3 (13) me:— —> L: a =q—=164.4H LC m‘oc (176K)‘x19.89x10_9 R B=E=8mdfs (a). me =1?6I=5529krad s (b) This is a parallel RLC circuit. (d) B=8n=25_13krad.-"s (3(6) 1 mo H6“ H 3IuFand6liF —) 3_6_LpF (6} (2:3: St =_2 C=3uFl_ 11:2le. L=2GmH Chapter 1-1-= Solution 40. 1 1 m, — — 1 — Sinai-'5 3 «HLC JfExIU'5H20x10") — (a) L=i—lO=l‘im[-I R 2xlfl3 1 l ., . _ _ , , —2|] ('30 =F=fi= 1.825.: krad:sec Q can]; _ 15x10 x20x10 l l B=—=.—‘=2.'I]k.:l.-" RC {2x10°j(2><10"} ¥ ...
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This note was uploaded on 04/03/2009 for the course EE EE 212 taught by Professor Shaban during the Winter '09 term at Cal Poly.

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EE212_Solutions_Chapter14 - Chapter 14-, Solution 1....

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