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HW1solutions - $0 /l¢-/-I‘o n ECE604 Homework 1 Out:...

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Unformatted text preview: $0 /l¢-/-I‘o n ECE604 Homework 1 Out: Tuesday, January 13, 2004 (Session 1) Due: Tuesday, January 20, 2004 (Session 3) [Students with tape delays: due according to session number] These problems are from Chapter 1 of the text by Ramo, Whinnery and Van Duzer (3rd edition, ‘ 1994). 1) 1.2f 2) l.3c 3) 1.4b (Hint: The total beam current is the charge density times the velocity, integrated over the electron beam cross-section) 4) 1.4d 5) 1.7b 6) 1.8a 7) 1.8d 8) 1.8h (Hint: Follow an approach similar to the electric dipole problem, but retain terms up to second order in 8/r) 9) 1.9 10) 1.10d 1,2-(1 A T‘s; (0)0l?) .- Qs \‘ _ 4 Ur Q1 r V . ‘ .: ‘ E _ Ra Sm " 93/ rm Cram Codka E—c—mtmnw : a . (“Cum TL; ML‘S’W‘ 7'445 . 'l w ' h o . -’%—Qs2 SQ wt» ages} a )4 E ’ ‘26. D Qm+rm)’3/L " 2-1*f1 o e — 2.6:. ( \ ~ * 5 (an 70 26: W Y’ é ] _——-—-——._______. «.45 APPIIOMS 6041555 Law to calindrical surFace at arbitrarg r-adius. r a. 4- :mre- EU) = enclosed charge 23 J’OPU'mttr'dr' = JRP.(%— " 131-55 ) Total beam currcht I. = r30) 2Tcrdr and 30') “fivz- “yo 0 - 8: (5)3.“3 + Sig-7);} => I. = 21y). rfl - $10330 + 8.083 dr = 2Wy.‘("i *s‘ih‘lé‘)” =>mraEm = m (gr-£5.53). :EC‘dlrso. =-1t(2,+5‘-5,_-—3L-1-)a?€. r-%°£;) ' r ('L-S E(")'r>a = t++zs.~a.6z—I§I$1rree ’ M‘s? sum 41h "Que! 4mm «<35. Fgmd- gab“, ,r¢4:v9 Q ‘,oka~3..w~[€, Po, WM b”.‘3“"‘_; o I E7 sv'wgmtfi. GEM-":9 is de‘ngRJ 'léglrmé, _’-‘(—.L.;)(I -axisg F” Ms -' we are = 31% r3 - g ; '3 -- - F0 Stash I I . ‘bprJfi, ' g '55“ '2”; E x = ’96 0* scan (10 TWA {23 L91; I .34». $55“ Nu» walk SP M‘CJ'I'Qéad-J‘Rév “QM'fi" W’I"; “9° ) cam-{'12)qu _ a* K: 60.) I"'&J»iuxlb. AMM “W. “was. m #344 iéti‘éiww" is ' Ex ;“_90 X943 §V\x—-d\z4b 36 ' 5K . 1".‘Qob3 ’56: Ewen?” Brow” I. 5 ~\-u<~4l=»l.. . ‘ w_1re-i"'€%;¢ {$4 ' . (A‘s/2g ‘Qide. L-M ’thm'h‘ifi MM km . v: (@033? .J’ q i I . ', o; 0 waé'. ' L210] " ¢1Cr=°»%‘)=.(‘?‘ c— I 1. 1. u | ‘ 26 ’ ' . ’1 a +2 + 2.] Lyk - 1""sz Us SSMAM @mwfwx 4.5 FQ-w Judw‘c alien/e, . “(apple h? 3’ “M'f-"HNJ‘ at 2'25 or!“ > ' ' ' HM?6W\5;W\ IS-Mmoa/u, Lu». W TQM "Mpmdww Q‘W) .‘lf‘ 1;.an #S‘Mbdx._ Cleanse r‘m'pw. , cu ‘ r .SeomJ new "‘4 *. “9M5 " f” ’ l M" ; $12.5 bv+ Wmfi 25"?" 1 4" _ _ . V a . . . I t 3 b -X» /-é('1§rm>° *§?*)+?(rm )1 + I—‘afgw “*3 Jr? (*2? we) 5‘34) I‘L’“ 2" . Na: ¢ :0 Mi aor‘d9 = 7:), MA”); 45¢ch a“ 9 = Co~$~IC//3). $591794”); 'l/rm} I»; &?U;'Foyé.)é‘~e . r'J'hA—e ‘ ‘ a MOd 56.5) = érawa’ ‘sfizb’fi‘ where Thus * d h- ’ 3211.5) = «Ti; h" E‘fiL—x—n—f-Vfifi—J—LML—M’i~w = e (x+.§)‘+53 _ P“ 49‘“ '5”): ' a = flat 1 {13:3 SimpW-ai cation shes I A - _.. .Azx 3" fit‘amwiflw 49 51w 5} \ E. >L(k+_¢é)a-+$a]¥ . {a = [(x -§)‘+ «3‘1‘1 ...
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HW1solutions - $0 /l¢-/-I‘o n ECE604 Homework 1 Out:...

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