Homework_7_Solutions

Homework_7_Solutions - has 32% x , 3 x i/ f' g; Kr.» 2 w?...

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Unformatted text preview: has 32% x , 3 x i/ f' g; Kr.» 2 w? “N ’ E k” f g/ 3 g , \Q is"! ”‘ M’ \9/8 N. 3 huge 'E‘hih 333%th himga {h} figihg Zahs ghhciphhiiich pflflfiifigfifi {hc % fiehzi 3i aha @Eigih is given hy E E“ #7 #9 +2 #01 = [fi‘mi-‘xfl—i E 2 E} + $2 : "§27r(2) 2m) 47r Figs £3. Figure fea- hmhiem ssh (a) I] ané I; in the same direction. (b) 11 and I2 in opposite directions. (b) Thc B ficid at point (1,130) can he wriiien as + x -—- ’ --“ = 9 2mm «E "r yszzi «E (c) Simiiafiy, the E ficid at point (2,2,0) can he found as .. - _A M017 M01 _ “A M E ‘ 3‘ +32 ‘ szz) + y27r(2) “[ X + fl 4% (d) When the diE‘CCfiOfl. of the current on ihc :3: axis is switched, ali the 31 hcids also switch direction. Thus, the 3 held ai the origin is now given by ._ __ M017 M01 a Egg fi—El+§2—?2W(Z)+§2fi(2) —§g+§q 4” The B field at point (1,190) is Moi 1 M01 1 m+ ......._ 27mm §2vr(x/§) 2 A #01 31 #03 1 $=B§+Bz=fi Sig;th Ehhgh {if highchhh hi ihig and same cf ihc oihci‘ hmhiems h; Ehig Chapicr, h: i5 hgchii m mic thaw; 325a; {icgiccafi its: Eighi‘c 5.5a and 65h, ihc h—hcfid 3i scmc dismth T hum ihc @éigfi hi“ h 47 , cc ax m Eéga 6.5a figure fee g 6% gggféiii} g fieid a: a fiiefieeee r fiem W “ 27?“ £342 + {2;}2 file Genie? 0f wire of Eefigih 22) Ge this basis, the contributions of eeeh side of file square depicted in Figm‘e 6.56 can simpéy he fomulaied as the supegpesiiion ef the B—fieide ea: various dismmes from edges of wires of Eeegehe L = (1/4 end L : 3a/4. We first note the: {he confirmations due E0 segments 3 and 4 me eqeefie iheee Emma 2 mad Ejezéegg fighiwham mfie, ii fie dear mam 3E3 eegmewée e; fifieéd eeieéi ? fie W2: 3e» that 3:336: fléfiea’ew: eemeébméeee mm} fieki. We Esme engage) 2 “(a/4&2 («z/4P + Gama r:a/4,L=a./4 r=a/4, L=3a/4 = 3) E9]. \/g+ 9E 2 27m 5 end geplsew: (E) %+ “ma/4) 2 “(a/4V 2VFW/492*x/C‘Ba/4>2n—“4753ET4>2 r=3a/4L=a/4 K z (.1. ) w. ( W at?) 2 21m; 15 3 hath it tight: thith eieeelet tithe We eoestdei’ eeeh @éfilfifi ot‘ the leeh seemetehi eeti eeoiv sheegtesttioe The straight pottiotts tie hot etetioce may E heitiet poiiit sitter; along thfitt‘ ems. Assuming the z direction to be out oi” the gage the B fiehi at poiht P totiuceei hy the cement in the shooter—etc pottioe with recites e is 7 p v e; = f“ hm _ _2#oi¢o o 47? 032 47m, Similarly, the 3 field produced by the current iii the circular etc with radius 5 is " 411‘ 52 " Z 47rb E _. ft fltbdthxee “#01qu 0 Theieiore, the totai E produced by the wire loop is 3 =31 +32=2M0f¢0 .1. ~E 47r i) a see iielmheite eoils, Exempie 64:3, we hoew that the Eihehi. at at Elifiifli along the axis of e eitetiiet Qut‘t‘fii‘lt etetggieg loop of tetiies e is hold2 A E 2 2(a2 -t- z2)3/2 E where z is the distance from the loop center, and the z-axis has chosen to be in the direction of the thumb, when the tight-hand fingers move its the direction of the current flow in the loop. (a) Considei a point EKG, O, z) along the z axis at a distance of z from the center of the loop on the left in Figure 6.58 on page 552 of the text. The distance between this point P and the center of the loop on the right is then (d — z). The B—held at this point P is then given as the superpositioh of the contributions from the two lootvs, namely MoNlaz +fi MONIGZ 2(a2 + 22)3/2 2[a2 + (d — z)2j3/2 poNlaz i + i 2 (a2 + z2)3/2 ta? + (d _— z)2]3/2 ep=e 2% where we have taken note of the fact that both of the loops have N turns. (b) We have dB Nlaz 3 - 3 _ d z = “0 2 {jazz + zz) 5/2(22) + inf + (d — z)2l 5/2[2(d ~ m} z zzd/Z zzd/Z " hONIaZ (:33; + 303) W "" 2h;2 + (d/2)2]5/2 2 2 ‘ (c) For d = a we have A MONIaz 1 3 By = Z 2 {a2 + 2:2)3/2 [a2 + (a — z)2]3/2 and 3333 __ taltflieg “a :1 2 5X”, - i 2 5,52”; dz .. 2 L 551a z E ate “fie + to —— z} i 3 iiiei #GIVEQL {wga(a2 + zZ)-‘5/2 +1522(a2 + ZZ)—7/‘2 * BEGZ + (a _ Z)2:§—5f2 dzz Z .r/ +155}; — figs? -%~ ((1, m affirm ' w...“ s: z = efZ we have 2 2 2 ~5/2 2 2 -7/2 2 *5/2 a 2 2 “7/2 deziioNIa [_3<§i> +i5a 4(3):) +i5a 1 422 2 4 4 4 4 42.4 _ .uoNIaz [ 6 <5a2>_5/2 +3 <5a2>“5/Z+3 <5a2>‘5/2j “O '3 2 _ 7353‘ T 7;- ' Eoiiowii‘ig the same procedere as given shove, and with some patience, ii is a Esther straightforwsm and usequ exercise for the student to differentiate the d2 B; /d22 expression given above and shew ihat d3Bz/dz3 : 0 for z = a/2. The natural question i0 ask at this point is whether the fourth derivative is aiso zero. It ean he shows by expanding the expression 32(2) around z = (2/2 that given by 44402 —- a/2)4 BM) 2 i25a4 so that d435,; /dz4 7% 0. The fact that fine first three derivatives are zero provides for an extiemeiy smooih arid unifomi magnetic fielé, which is the basic motivation for using iiie Heimeitz eeii eonflguiation. (d) We have Bz _ While? [ i + 1 j ‘ 2 (a2 + zap/2 [a2 + (d _ @253/2 370? d 2: a, the iieiei at the midpeiiit (z = (1/2) is B —flONja2 Egg. —3/2+ E ~3/2 2‘” 2 4 d 2 -3/2 ~3/2 = ,uoNIaz = so (5) M 2 0.8992 x 10457-1: d d a a {e} At the center of the eoii on the left (2 = O) we have 2 NI BZ z MON“ [CE—3 .5. 24/2653] = 1%; i. 2-Way a 0.8505 x 10—6— 2 2 a a which is qeiie eiose to the value m the mid-point, underscoring the purpose of the Heimeiis eeii eeniigumiions nemeiy io esiehiisii e fiwiieiii wi’iieii is very semis? usiferm. Esme iiisi hf}! symmeiiyg Edieiei iii: iiie eeeiei“ ei iiie eoii es tiie iigiii {2: = a) mess iie iiie same. hetneen tern seieno‘idai eniist The in'nttnai indnctence hi2 hetnieen tern hie-cote r serihed in the ntohiern ean he written as ism N2 _ L~:-—-—=-——~ n «(is l; I} {I 32 1 2 where A32 is the total this linkage through the area 5'2 of coii 2 (with 52 = 3 cm, 2a; = 2.5 cm, and N2 = 3106) due to the B; field produced by the current E in coil 1 (with 51 = 25 cm, 2n; = 2.5 cm and N1 = 1090). Note that the heid 31 can he assumed uniform over the iength of the shorter coil (coii 2) approairnateiy given (from Section 6.2.2) by MOlel 11 since coii 2 is iocated at the center of the ionger coil with 51 > m. Thereforea substituting these vaines into the mntnai inductance expression with dsz = hdsz yields N N 51171 52 ii I” 477 x iG‘7(iGQtl}{ihtl2 ‘.- .J (nag) énitihififiigi 2: eat? “is? «a K» i. 231th Rectangninr teen and other, Since die h—neid produced by the wire is into the pater, its is convenient to choose the area eiement (for the ioop) to also be into the paper, in which case the associated (via the right hand rate) di element is outward at the terminal of the loop farther away from the wire. Based on the discussion on pages 568 to 570 of the text in connection with Figure 7.4 of the text the generai role here is that? in order for {7.1} to he vaiids Wind must he defined to he positive on the terminal at which the di element, the polarity of which is associated with that oi" ds via the right hand rote, points outward. With the loop moving to the right with a veiocity no, the distance between the left edge of the loop and the tong wire varies as fit) = not + m, where m is the initial distance hetween the loop and. the wire. The flux iinked by the loop is given by i i. 2'” Wm 7" 27? M0157 '”°t+“+a d?" MOI!) ( a > The induced voltage is then given hy d‘i’ noib ( vot+n ) [ —voa 1 ,uoIabvo Vina “ —————- ————— = W if > 0 (it 2n vet + r; + a (mt + r02 2?!(?!{}t at whose e It a} .j nth. Extracting newer thorn a newer iinea The constraint doe to the length of the wire in hand can he expressed as 2a+25=2€i0m ~> a=100veb Let the current flowing in the power line he I = Io sin(27rft), where I0 = 4000 A. The magnetic flux linked hy the pick-up loop is a? g b 20+“ at T : neonate:an <20+o\] 2 agroshsawfnmn (12:; mg 23 27W 27? 29 I; an“ I: irreiocesi mirage is their as? p . me w M OVin = ’1}; =3 filo—IQ C05(27l‘]i)b iii < 2% j Power exirsere'i mssimised if ‘i/‘mé is memorized. Reece we heed is fired she raise of h shes maximizes {"éZG m were. This care he sierra»: h}; fiifferehrietieg with respecs re '5} error semi so zero er hf; simpie eiortihg or by rsiai she error. ’E’he resoir is i; 2 62 or. which reesrrs in earn she a. :e 3% he. The choice of she opiimum Eoop dimensions maximizes the induced voitege. Since she wire has nonzero resisrence Rwim, we can View she induced voisage and the wire as a. voltage source wish voltage V; = “Vim and source resistance RS = ngre. To extract maximum amount of power from such a source, we most use a Road marched” to the source, i.e., RL = ES. The wire resistance can he caicuiated simpiy as l l 200 m Rwire = —' = — = W 2 0, 0A O'copperfi'(d/Z)2 (5.8 x 107S_m—i)7r(4’1/10_3/2>2 261g; Thus we choose TL 2 0.2619. We then have 120* 62 (Vim = fflofo 008(271’ ft) [62131 < 62 j] 2 19.9 cos(27rft) The power extracred hy the iced under coeciiriorrs of maximum power transfer (re, RL = BS) is 1 (v.2 2 J (b) Try N = 2. Now our constraint is a + b = 50, or a = 50 — b. The flux iinked is 20+a . MOI ,uofo sm(27rft) 20 — 6 Nb 20 271‘?“ T 27r 2b In 29 Thus, she induced volrage Wind is maximizad if 2b in{{20 — b)/b] is maximized, which occurs for b 2 29.16. {Isirrg this vaiue of b so caicuiate Wind as hefore, we find a smaiier value for PL. holds one for either values of N Z 2. Therefore. we conciuoe that using more sores @0er her show as so misses higher power. Messrs-sigh ershhsiorss. {a} We ieiiow so epprosch sihriisr so Exempie 743. Subsiihriihg the Fife}; phases: iriso she ehesor form of [7. i‘Sc]. we first find the corresponding eiectric fieid phasor 12(9) as i; 1 6H » E =—,—~—~V x 5% z I (3?; AW) W60 jweo fig ~' - 4xi.83x10‘4 . _ mg. 345.83 x 10—46—34?! = ang—g—flry mm 1 33255 meg Weir‘s. ” fresh 7. he; so fine: SEE es i i i x i, r i x 2 ’ 5%. “i A sxfimm 3 5 Ewen 3W9 ‘2 59s x 1 i . 2g _ («374927—483 x 1046?"? jwflfl 9W€0 is . 2 3.83 x 30453411 Am“ W #060 A =3 Note that for this expression to he the same as the Egg) expression given as my) 2 21.83 x io-4e-J'4y A—m‘l the foiiowihg eondition must hoid: 16 4 3 9 ~1 2 :1 —> w: :4x3x10=1.2x10md—s w #060 x/Mofié) With this value of w, Raj) and its counterpart Ety) satisfy both [7.18% and [7.18c]. In addition E(y) satisfies [7.18M (since E(y) = fiEfiy) and therefore V - B = 50V ' E = 0) and E-E(y) satisfies [7.18d] (since Efly) = stew); and so V -B = NOV - H = 0). (b) Substituting w = 1.2 x 109 tad—3‘1 into the My) expression found in part (21) yields E(y) 2 400696141! V—m“ (c) The electric flux density B and the displacement—current density Ed in phases form can he found as o = 60E 2 41:06} 1.3“1'49 pC—m‘z 2:237me 2 «see x tomsss x 16"”)(Ooii >< to"12s*34ys:em2> :3 w sees-MW”) teem—2 7345532. EEecteomegssesée ease its free some. We start with so, 2, t) = 334.9 cos(i.8 x 10% ~ on: -— 250,2) V-m‘l Using Maxwell’s equation {7.1831 and noting that “fix and “(fig are both zero, we have _ es as Vx‘é=fi<~—5f+fifé—f> = —s4.9(—2.5a)] — sin(i.8 >< 109m —— cm: - 2.5%); + fi4.9(—-a)[— sin(i.8 x 10%: — a1: —. 2.5m] 379.3%, z, t) = (4.5M: + a§)(4.9) sin(1.8 x 10%: ~ m: —— 25412:) = - at from which @(m, z, t) can he found by integration over time. We find 393%, z, t) = ~(25ait + e§}(4.9)(i.8 X @9704 cosflfi >< ifiga‘rt - om ~ 2.5oz) We can now use Maxweh’s equatioo 7.18c] to find Fees, 2, t) from @3703, 2, t}. We have §£0%($, z,t) _ iv X a): at - #0 1 2 2 g- A , r, 9 m x 4“ 2‘13. E W §§fi(3.8 K 735 “' {Lfi “w was:ng + e2} W €03€L3 X 3,3937% — as: W KIT—£34 €0I££G\ '7 i I e fleetfis" m m a”; as”; 5 , . ,3 Y.“ “‘1' n * ° fiwfe E30 Ems magma expsessmn E03 @333 z? 3:} gives} as“; gas‘emem séefiemes’g‘é: 9 me}: messimsmg mm fizhe ewes/semi fee gaze, if; we 2233": B772! \ E u ~59; (masses: s L81§fi®sflfi x 3.9% ~ ems w 274%} see-fl E! Eissifises’e gfiegse wees.) {3;} Fee free space; we have _ _wm21rf_2fr EWWVMGE “EM 5: m3; 3 $inee x Q SHE and e '2 3 x m3 m—s‘ , we eeieuéeie ihe gshase censmm as N 27mm x 109 where 27 is the intrinsic impedance of free space which is n = ‘/ yo/eo 1: 120% 2 377g. As a zesuit, the magnesia field phases Efiy) can be feunci as _ ~1 E s. 3 X 308 —- 3207 rad—m and the wavefiength as 27? 271‘ = —~ 2 -—————— 2 0.61 = . A ,8 1207? 67 m 1 67 cm (b) Using {use} my) = -— .——v x my) = ~ . (wag A”) We quo 8y zgw = g Eisejfiy = @éejfly WHO #0 77 my) : sawrieflzm A-m" gang‘s Sefifi‘eées“ gsésessesa (a) Fires, We wsfi’ae the pfieser expressien fer t) z 2100 cos(wt + 4.87m: — 3.6714; + 49) m‘V-m_l Egg, 3;) 2 2543, y) :: fiififlefl'gmewgf’wefl REV—m”); N€K§9 we substimte Em, y) into the Helmhohz wave equation {88¢} 82E; WE; + 8:62 ayz +,62Ez :0 Where ,6 = cue/Moe = w/e = wa/c = ZW/A where c z 3 x 108 m—s'l. Performing the paras} derivatives yields {34,st + {—j3.é7rj§2 a» 53 r. 0 ma» /3 = erg: sad—m"i iter wisieis tine 2: efiéifin‘} m 98% Mtge end; A :2 it fit tn t’esneetisety. exegessien $53}:9 gag? t} with t m {3 anti 3; 2 2n n @722 m; see new: eke, on, on = stoo (308(0487r ~ exm + a) 2 280.9 asymmi see ——>’ «432° e- 6 z "i — 2 36° ms 100 wnieh resuits in 6 2 792° = 0.4477 rad. (e) Using [7.183.], the phasoa‘ magnetic field can be found as i i 6E2 6E2 Hwy) — jWDV >< Remy) — m (it fly - t 533) E’erfonning the patio! derivatives yields fi(~j3.67r) — 3011.870 jwilo E83? = 1006j4.81rxe~j3.6ny€j6 mA_m—1 Substituting jwpzo :: j7106,vthe magnetic field phasot is given by not, y) 2 [—20.159 «a $70.212}ej4‘8me—j3'67ryejg yam-m“i Therefore, the corresponding teal-time expression 53w, y, t) can be written as saw, y, t) 2 {40.159 - @0212] cos(wt + 4.87m“ —— 3.67734 + 0.4417?) mA—m-l where w =_187r x 103 rad—m4. (s) The time—average power carried by this wave can be caleuiated as 3.23 N Wt s _. 21.33 10*5w— *2 2 ’9’; 2 377 X m Aismnet‘i’eet eentflttetee in'tet‘t‘neet We start with em = 6Oe’j4°"z{y — 3's) y—mf1 (a) By competing the given expression with [8.10], it is clear that ,8 = 40%. We thus have 2 2% 2n fl”? “i A=g~4m07T—0.05m c smog 9 f—X— 0°05 —6><10—6GHZ (b) Once again comparing with {8.10}, it is clear that the titeetien of propagation is 3%. Thus, we have _ 1 1 A - ‘ s >< n12: “easement = gamma?) em” E' .’ m 1‘31!) 3!? get: wait i i 3 i 1 z E {e} E‘Eie teiieeteti stave iieitts ate ere) 3 seems — it) = éeett‘tmet + it) 69 em = giee x e = fiefl‘mme we a.“ (d) The total eieettie tieid its air is E1 = e; + E, = 6Ge”j4°"z(§ — 3e) + 606+j40”(—§ + 32) = y 003—32 sin(4~07rx)i + e j60[2j siti(407rm)1 = (—ynzoj sin(407r:c) + (@120 steam”) "he magnitude of each of the two components is equal and varies with distance as 3.20! sie(4t)7rie}1. the sketch 0:? tttis satietieii is itientieai t0 the [Elna/)1 in Figure 8.1% 0f the teat? exeeet tee fact that 3: z i {€25} : W35 m. gwéis fieetesee eeeigtio {a} fat tite eetttet t’teqeeiiey 0f the X-eaiid weather tadat which is 9.4 GEE-i2, the wavelength in the foam material with 627. = 2 is 3 A1: 6 ~———iX——i§——~O.0226m=2.26cm t/e‘zi 72/63 " (9.4 x109)\/§ _ So, the minimum thickness of the tadome (which makes it comeletely invisibie to the radar sigaei at 9.4 GHZ) is d :2 Az/Z 21.133 cm. (b) At 8.5 $st A1 = 3 x 1010/35 x 109 2 3.53 em, A2 = ABA/“52¢ 2 sea/Vi 2 2.50 cm, tee tanwgd) = tanfiZw/Afid} 2 tanKZa/ZSGXLBH 2 41319. Also noting m = m 2 377373 anti the inttinsic impedance of the foam material is 712 = m A /62 2 377/ 1/2 2 2679, we have item [8.52] A2: 773 + 7772 mtflzd) 772 27773 tantflzdi 77 + j26'7(—~0.3i0) 2(267567 + j377(~0.339) 3866-1124" 2918«j23.7° 2353eju‘3052 Using {8.53}, the effective reflection coefficient is Z2('~d) =772 2(267) (347 — 377) — 369.3 gm}; 75.7.371‘114" 2 7276454? 20104575703" Heme, the percentage of the incident power that reflects back at 8.5 GHZ can be eaietiiateti as i (Sax/)1» ! itgaviil At tat eitz, A; = 3 x two/(10.3 x 309) g 2.01 cm. i2 = AIM—‘52? 2 2.91/02 2 2.00 em. and tamfizd) 2 tati[(27r/2.06)(i.i3)] 2 0.310. Using these vaittes with {852}a we have ‘2 377 e j26?(®.3itijt ’2s7 t- $700310} x me = 1176102 x 300 2 (0.3047)2 x 100 2 3.085% thwd} 2 test? i 2 i {iii ere ehesre is the eerrtpiers eeejegete 'riiz. errci ee remit. ei’iieetiee reiieetieii eeeiiiciettt wen. 1 Size W m Vi??? mg if?) C!) m M . S3 W m be at i03 GHZ wiii he the compiex conjugate oi the effective reflection ceefficient at 85 GHz, i.e.. we can write I‘m at 10.3 GE-iz directly from Pee at 8.5 Gi—iz es Fee 2 0. were? 108°. Therefere. the percentage oi the incident power that reflects back at 10.3 Gi-iz is aiso ~1.i)85%. Air Air n 1=377;Q 7114779 . kt w”? -—-—-B* TiM Fig. 8.5. Figure for i‘rehiem 841. The comhihation of mediums 3 and 4 (Radome and air) is equivalent to air, since the impedance n34 = 3779. (c) As shown iii Figure 8.5, the radome—eir comhinatiori on the right behaves just like air, since the radome was designed to eliminate reflections. in ether words, we have 7734 2 3779.. Hence, we can repiece the radome-air combination with just air. resulting in the three layer problem shown in the right-hand panei of Figure 8.5. Using the given properties of the layer medium, namely er 2 4.1 and ten 66 = 0.04, we can calcuiete its intrinsic impedance from {8.22}. We firid = 311/...— mejiian‘VO-O‘i) ~ 186 1ej1~15° ‘72 [1+(o.04)2]1/4 ’“ ' We now have a simple three layer probiem which can be dealt with simply by using {8.50} for the case when the second medium is lossy, nameiy r f. z W e (712 + twins + 772) + (772 ~ "Dim * ’72)e‘2‘120l where . .w .w . ,27r 9.4 x 109} _ 72=9w¢fiaé=yg¢e=2g 64—Jtanécet=J—~———~———(3 X m. 4.1 -J<0-%4)(4oi> Using this value at 72., the eta; found above, and m = 7:13 = 377g, we find ‘3’ a. r . 3» ~ ree e ti.§?% timiii. A riieiectrie eeeteri rei’iecter eeterrrre. This is e three—Eagles" prehierri iii which medium 3 is e perfect eerieecter. with intririsic impeeertce 5:73 2 t}. it cart he solved by rising [8.56] for the case when the second medium is lossy. nameiy e. (it: E mite “r i" it}; e rhiteg ~ age—272d i sit ” , he: i tithes. e mi e mg m 9.373 c. rifle—ETM fire intrinsie eta: can ne toned from {8.22}, transeiy _ no _ 47r X 10—7 m "’ e; — jag! “ 4.5 ~ 30.009 Using this value and m = 377%. we find reg c: seesaw-31° e % Power Reflected = 1m? x 100 = 99.82% in in principle we now have a four layer probiern, two of the layers of which are iossy. “tire general solution is quite difficuit, involving impedance transformation on iossy lines. However; we can observe from part (a) that the coating has very iittie effect, since its ioss tangent is qnire srnail and it is thin (compared to wavelength). Thus, we may ignore the coating and treat the tour layer protoiern as if it is a three layer air-water—metal problem. In that case, we have none again in = O, and m = 377E), whereas in can be found using 6T = 55.4 — 335.45. For a thickness oi" the water iayer of 0.}. inn}, we find almost complete reflection (99.92% of incident power is reflected). However, it the thickness of the water iayer was 1 mm (instead of 0.1 mm) only 46.6% of the incident power is reflected. Thus? condensation and rain may indeed seriously degrade tire eeri’ormance of the reflector antenna. 1.75mm Fig. 3.9, Figure tor Praiaiern 349., ...
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This note was uploaded on 01/17/2011 for the course ECE 305 taught by Professor Staff during the Summer '08 term at Michigan State University.

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Homework_7_Solutions - has 32% x , 3 x i/ f' g; Kr.» 2 w?...

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