HW11solution

HW11solution - ' \ So /‘4.‘ Z 6 fl ECE604 Homework 11...

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Unformatted text preview: ' \ So /‘4.‘ Z 6 fl ECE604 Homework 11 Out: Tuesday, April 13, 2004 (Session 25) Due: Tuesday, April 20 (Session 27) [Students with tape delays: due according to session number] Problem numbers are from the text by Ramo, Whinnery and Van Duzer (3rd edition, 1994). 1. 8.7a 2) 8.7g 3) 8.8d 4) Start with the gaussian function f(t) given by Eq. 8.16(13). Using this in Eq. 8.16(10), show that the envleope broadens with 2 as given by Eq. 8.16(14). Hint: Use the following Fourier transform identity: _ 1/2 J exp (—I‘t2 ) e_Jwtdt = exp (—0)2 /4I‘) 5) 10.3d Hint: Review section 6.4 and note equivalence between 0 and 038”. 6) 10.43 7) A simple air—filled transmission line resonantor may be constructed by shorting both ends of a transmission line, as pictured below left, for ideal (lossless) resonator. transmission line: char. impedance Zo, velocity c _ I Resonator With loss Ideal (lossless) resonator The ideal resonator supports resonant modes such that m half—wavelengths fit within the m?» resonator, i.e., T = L, where m is a positive integer. a) What is the frequency of the mth resonance? b) Now consider a resonator with loss, which we model by replacing one of the shorted ends with a load RL, where RL << 20. Work out an approximate expression for the energy storage time and the Q of the mth resonance. You may assume that the resonant frequencies are approximately the same as for part (a). c) Give numbers for the resonant frequencies and the Q’s for L = 3 meters, Z0 = 509., and RL = 19. 8) In this problem we consider data transmission in two different metal waveguide structures: (I) the parallel-plate waveguide and (II) the square waveguide. Both waveguides are taken to be air-filled and lossless. (H) (I) _ , ,. , ., a=3.c'r_nr—_>‘|> parallel metalplateS-v .sgqgammeialgui‘de (Separation a = .6 cm); (Si-flea, = (a) We wish to transmit information using a baseband transmission scheme. Our data rate is R (units of bits/second), and each bit has durationT = 1/R. For each data bit equal to ‘1’, we simply transmit a constant DC voltage during time T. For each data bit equal to ‘0’, the transmitted voltage is zero during the duration T of that bit. (See sketch below, showing - transmitted waveform for data ‘1011’). "time voltage We desire to operate at a data rate R = 100 Mbit/sec (108 bits/sec). For very fundamental reasons, one of the two waveguide structures is expected to give much better performance. Which one, and why? (A short answer is sufficient if properly justified). (b) Now we wish to transmit data via binary amplitude modulation of a 6—GHz sinusoid. For each data bit equal to ‘1’, we transmit a 6-GHz sinusoidal burst of duration T. For each data bit equal to ‘0’, the transmitted voltage is zero during the duration T of that bit. (See sketch below, showing transmitted waveform for data ‘1011’). '1 , _ .. . . I6=GH2 T {—T v -' 'si'newave g0 0 time D -> Again we desire to operate at a data rate R = 100 Mbit/sec. Comment again on the relative merits of the two waveguide structures for this data transmission scheme. Is there a reason to prefer one waveguide structure over the other? 6-1 (1—8) owho 'bo boUA TE. cu'nd. TM. Use. a =' 3m, -b=l-5c,w\.¥‘= met-ta. 8-‘1a.‘. 5% 6‘5 {lor g‘ass =4, km.“ =' [ME-r wag-W 5% Rs $46“ Wm ‘gfi‘gb'g‘ '“M’é' m1. Pam 233-33) ‘0‘“ *4” ' 313" ‘fi% M.k¢*°*\a¢‘3 , {I ..__/I H V: —&06A - I I I i A l/ a l/ II II, :: —3\o):: . I} - m ‘1’, I II I aka/x It To» show “ML Wposvfion ogvm' '. M «AM» we» { » - _ 525 am seawmag gym my 4%) r _ = “if: wapzymkamm mm are: way 53 ohm, (5 7' A . _ I ,E'tt kr- 0W 18g; WT: Bios; 8mm m) aw: ‘ = 4%69. am, 377 _ ' ‘_‘c°5'/.xw)z é - w W1” {axmbfz N03: Mpg—mo \N_—_-.\‘U5 M 4W féuu . . .. . a 2.11%)... 74¢ (aft 227? _."””“’”e"f'*)“ M a M w _. ..mm7‘w’<4e? 7%.» %M%“_v’w_., .(e {’7‘ ’7) a a. A. Ample! 6' a. — A .0,“ “03%;; = w |. WI. But wg' 32L 0‘4va = J; L(ea')‘—UL.dv =- %f, E‘Av . Q :- Wog‘jvE‘dV - w,e' - i; . ' ‘95 JV‘EHN ”' °' Emplet \s rene, Q= Howe“: (93 comflred with ~ID‘ -(-’or cube with @123 Via-“5- mm. -Not¢ that wave vanishes -l’or TEMP (E‘s. "Io-51.0345); Hi pao or a? both In and n=o. For TM"...P modes : Nav; vanishes 1-? owner m or nae. Resonant {are m -?ro , to. L1) '5, 'I leub' cau‘ I I ti cg .m E}; 4 1 Per uc W3 00. = 3% [m‘+n‘-+ ‘1‘ Below we. tabulate nanvanishinfl modes, with nwmalieed resonant. and Aeaenei-acg . There are. ‘l difierent Muendcs. 7) , a) erh MSWMM lzragugno? W‘H be 7%: ; m%L 53 {ESWW WOW mm [055] ref/6cm Wefi p=i§5% m“ 71/23 Hoax/w and, Mid «/ {$3} «L M 7%: daffy/7%. C: The 3mm anan M 4/ ngeq/L me in Me rgsmm‘w M9 tame. (4:359 is W Wand—W 17%, HEB-2192), mm A" is grand Wflmflx Wm WM Jazz 79mg, Me 37%? $456 péa‘ cmée y H16 Mpmmfvh/ flew? @prm‘fimfim U; (ETA): UECtH ggfwéw A: Ugfi],p? t Cm log Yea/Wan a; Ugém 9%.A4épr Uaflfle A :ug‘c%:§9i- (91% 8W4?” 7’7;ch 'Z /\S 4974/73/45 75/3 /7}77€ L; W31 72/” U509 “50 fa //€ 015% M/‘fie/Mg U562): @g6%3? :UZflZv _> Z: A 3L ' I701 c0799 9 W0 U; ' .. ,_-. ‘. 6? ” ‘Wf 7%?" “WW 75” » W=Afi~§WW€rmvflU H C m Iv K, TEzo {PT-Em; . milk. _ I I I. ' ._ ..C%:~fé%/f3r)w..._ _ I ‘ . i. T991. ‘ “C 7‘. . 5‘ HTEROFTAMM _ .3. V gm “64*: 6,? _ .6:ng _ _ 15.11» 2-2 I I l 2‘: ._ ., ~_-_-_‘._.fl-..._-__ I §_..—-.T_ ' : I .. I .. .‘ V _ .. _ m, _. ._ ,_ _ M F ‘ ._ J. ‘ . ._ .. ., . . .. . ,. .,. 5 . _ 7? i3: :3 fl. . .. .. .. . , i ,_ = ~_______._I____~*'_-__.___.wr. ...,.___._..___ yr. :3 3 WVF’fli../aw, Mew/W “An/'5 Mfifviéf . . . W@. “(339; 9’77??? . ' "5:! a»; flu V. . _ . 7 _ .. .. “C.” ‘ l .97 _ _ W _ .A I ...
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This note was uploaded on 12/12/2010 for the course ECE 604 taught by Professor Staff during the Spring '08 term at Purdue.

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HW11solution - ' \ So /‘4.‘ Z 6 fl ECE604 Homework 11...

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