HW4 Sol

HW4 Sol - Qt S‘f fl(3;2HMH7 634553 d My(3 l g/S wrww “Mm/f 3>.§—HY:9 €“fi9uglfl:failfi(617515 Y g” j Ma 75 yaw 6%hwfl)af3

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Qt S‘f fl (3+;2HMH7) : 634553 d _ My} (3 l g/S ? §+wrww §* “Mm / /f§*+3>.§—HY :9 €“fi9uglfl:failfi (617515 Y g”: / j : Ma): 75 yaw) 6%hwfl)af3 M #54—frw’q§%flmé*“* ) 1. H HG) :gijé @e- K :1 Hwy Jrflr. g? raw/L; St? L. :17 W) =— [LA 359) +4 ( gar) “ire—"Jr 516* 7' r” -—- e Q3.) (9; f} - 3m 3} mad-1:5 4 I “H CH4 "KS"?! , (13) a) 45 «Q mg): #0») : ghbm i“ 1W): 1“ 3135,31 A, O 5?” §L—é$+{9 F 7; JD) {ildl W @1315 “0‘ SR5) 1 5"“ fiéHzo {$1 _ 51 :0 4553-015 5 fl” § Fa) : /O_£J'L 40.0095+m0 J1 13; ML, - -6 £91 “768% :. £6312 +£{WW_ /62.65 S S1_{_%fll 523' V6.13 #31“ V(;) ’75—; (5L+ l-OQLFXJOS x '21- N _ ’7—(f (SL4 [ch )qogiqt» L250 (5+ 510%?)(54r3m1/g # 341.3337 {shz’zflfl +7.??fi4X/o? q )9 c 6‘ :— —-*”“' + If!“ Jr (“ms—y (943%) (Sam) +[52h-Jhx) B) we marl/«74 0‘0 m‘éue’: 0k: 4! .8 J a) #66. n J c : 0.11% M15?” (7“: 0413329 kz’S’V/t Jim-t c VGU : #3:.9 51mm H5518 +. 3.354 (05(1sz MN!) MW in T16 yifgokd/ w grade Volfige marog Hm (wacf'for' K V I [259% Co; (1va fgy-9/°)jm\/ TIM; (M be erHeyx M Phafior nohh‘on at) 0.11641439/‘31/ 5—“— OMIHQAWflWX {4 [1? 44?,9/ “A #4 km a $43 11 S gm 3% x 12m 2 7.12)? (of QM) A C) PhM‘ok andyg‘gj bye com (Age EFMPR voHofle Amigo“ (To (fwd Hag Velmge dryimn +0 4):\AC‘l/\Vol—lfirgf C (1‘ {)mC- (“for (/6245: )( 0,03lfiyl; 4?}. 57V) Vccm : “w : 0,15fl/4f1’ir 2g + 0.0314; 4%?- 9/9 TLW I'S Hm: {7‘2an fade {egfmfé CURL-L e} +L£ fame WM] 0*) man/oer , VCU“) #:d 3: FodVCf'. 3 1).? C05C2K7F gage > LOLNZF ES. +Ifl€ game voIJrl/l amwer Q6) . . . . 5 ' We begm by shorting the 7 cos 41‘ source. and replacing the 5 cos 2? source With 5 s2 + 4 I (3) Define four clockwise mesh currents 1]. lg. 13 and ]x in the top left. top right. bottom left and bottom right meshes. respectively. Then. 55 2 = (12 +1.-'2s)1_s —3 1l —(1.-' 2s)1x [1] 5 +4 0 =-41.+(9.5+s)11—s13—31_. [2] 0 = (3——s+2..-'s)13—s11—31. [3] 0 = (4 —— 3s+ 1_.-"2s) 1x _ 312 —(l_.-"2s) 13 [4] W = (I; —ls) (2s) [5.] Solving all five equations simultaneously using MATLAB. we find that ‘__,,_ 2(ls3[?553 + 100.52 + 1325 + 152] 1 1212s6 + 3311s5 + 2825.54 + 1528053 + 12411143;2 +1111485 +1201) Ts s2 +16- Define four clockwise mesh currents I]. [3. 13 and IM in the bottom left. top left. top right and bottom right meshes. respectit—‘el‘;r (note order Changedfl'om above). Then. Next we short the 5 cos 21‘ source. and replace the '1' cos 41‘ source with U=112+1.525;]ll—T’lg—[l..-"2slllx [I] 0=—41,.+10.5+s112—s13—?11 [2] — 275 =13+s+2..-'s11_.—s13—31. [3] 5 +16 0 = 14+3s+ l_.-"2s] 1x—3 13—11331 1] [4] vi” = 111—1311231 [5] Solving all five equations simultaneously using MATLAB. we find that .3! 51515332215.2 -Ss- 111) 1T1 — {"121235'5 + 331135 +22420s‘1 + 555135;3 + 48'1'3ICIs2 +40590s +4800] The next step is to form the sum \-"[(5) = + V1”. which is accomplished in MATLAB using the function sjy'umdfl J: V 1 = symadclfl" l prime. 1\1-'ldoubleprime): ‘1“) _ lils.3(815»5+110'1"s4 +73133‘3 +1'1"130s2 +21180~s+ 12160) '1. . . . {‘52 + 4111121256 + 3311s5 + 2242054 + 55513s3 + 48230.52 + 405905 + 4800} (b) Using the Hapfaed) routine from MATLAB. we find that 1.410 = [0.21573 501+ 6.903x10-3 cos 21— 2.403 sin 21— 0.116? ism-"1‘ — 0.1943 sot-“5’ cos 0.903: + 0.1611 set-3”“ sin 0.903: — 0.823x10'3 s-“-'-"“‘ + 3.229 cos 41+ 3.626 sin 4!] 31(1) V ...
View Full Document

This note was uploaded on 06/13/2009 for the course EE EE 110 taught by Professor Gupta during the Spring '09 term at UCLA.

Ask a homework question - tutors are online