Solution+to+2009+2nd+Midterm

Solution+to+2009+2nd+Midterm - EECS‘l45 Electrical...

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Unformatted text preview: EECS‘l45 Electrical Engineering Analysis Fall 2009 Second Midterm Examination Chin C. Lee 1:00-1:50pm, Nov. 6, Friday Each problem has 20 points. 1. Present the three most important second-order partial differential ‘equations in engineering. For each equation, you need to write down the name (title) of the equation, the equation itself, and the type of problems that it can solve. 2. The electric field component of an electromagnetic plane wave is given by: E(x, y, z, t) = (3, 4, DEC exp [j(kz —— (Um, where (3, 4, 0) = 3i+4j+0k, a vector. Using one of Maxwell's equations, derive the magnetic field component of this piane wave H(x, y, z, t). You are required to present your derivation procedure step by step. Guessing alone will not give your any credit. 3. f(z) = z2 + 11(22). Find the points on z-plane at which f(z) is not a conformal mapping function. 4. w(z) = u(x,y) + j v(x,y) = sinh(z). Find u(x,y) and v(x,y), and prove that grad u(x,y)-grad v(x,y) = O 5. w(z) = e'z. A domain on the z—plane is chosen as a rectangle defined by: 0 s x s 1 and -1 s y s 1. Find and sketch the range (image) on the w—plane. You must label your range clearly with equations. JM Jud Heth 1%.sz (WOW? eqthah) Won/e Ylfo‘vjem WC—FMSRM . (a), I’llQr/‘f (OmeC'hcA Ea _ ) DPPLSFM FEM?!” ; 6D ' S a t WU = 1:9” D'i’Pt/UQA 5) D: C / @ : D’Iflp’fi/‘SW'+X V Fay 1/,me Camclvfl'hon I) D: I 'E—\—- L+h9kmmi I ojl‘lc‘lfwhb’fy) Poisson} L“ at C QM“ 913mm“) E/“Thfimfic FYoLk’M : {QUVé IL CI’IWQQ ‘H’Pe éfapg 5? €:0 9 Vivcfi):o (Lame 5%) Pal/(Addy; Law '7 VX E: : ~M UK E: ; I a 6 5 3% ?f\ a BQJI‘Z “CZ . J)c% : Ltd: 3/ O) )k E3 4 _\ WE szfl yaIA ZSLiZXT'at A , 5k \-—-\ I: {—4, 3/ Q) _ k kzzMwlv (—4, 5/0) kid/we vO/v\ ': *4 . E; H L 5 O) )7; E0 a: a1 ESL __}u(_/t R g _ zit/m: «DD-1:38 'U‘Z’W‘i} Q’ S“ _.1 ~ [FOE , ;Lkedm¢) Lljo'e jLF€~w€ J ic) W H L ._._ H E + z; K 4.’ cli'Fn/w\ N0+ CJW'FOYVHM’ TD -,-__ O Jy- “‘4 kc. d? dTC(Z) ] Ol—g : “'2 23 :0 __' :> E— 23):?) l :7 Z: Z; :7 Z4 : \ _ C05 2mm+ 3914279” CWIO,-'L4+€jer) :> Z : cos‘flg + J (my? n10 => 2, : coso +5gm0:{ 4 V111 :3 E72 : CO§g+ngmg :3 4 W12 :7 23:Cogm+ngfi/:_l Ar _ - 5_+r g “Fin 9 i- mg?) :3 1' COSTZ‘IV J l ,2 N. ‘F(‘£) wnolJeFmeJ => E20 4. I _X _.\ Jr‘leILJ/J 39, [mt /) J ] t : QX.CLOSY+‘)g|m\/) —- 1 Z .2: cosy Cex— Q’X) + Sirfl/ LQX‘I’ 9/) ; Coax 9th + 'ng\/-Coslax K S m: COSY.smFX1 V: smx/ Casi/xx _) <3 -._) V Ux ' {i— {@6775th )f‘ + C #27 C0871-glthJJ+CfiCOSy-srmt~x)}< " ax cosy-cad“? Jr (dé'N/‘é‘Vfl’WJJJF OT; 3 a J a . ~> L w @5th v v : C37 €na“‘/'CO§1“X)F+CFT 5“”7' “3wa +52 / k / __\ .3 1 Sim/ swim“? + cosy-Cmth + QR g' V“ "V V 3 C05 7» COSI’W' SHM/ 9th —— Shm/ - g‘hquACOS\/.cogqu : O ...
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Solution+to+2009+2nd+Midterm - EECS‘l45 Electrical...

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