HW 3 Sol - 3-30. Resistive line impedance. Using [3.32],...

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Unformatted text preview: 3-30. Resistive line impedance. Using [3.32], the line impedance of the 1009 line of length l terminated by a load impedance of 120 —— 3200 S2 is given by — ‘ 0 '100T ' _ Z(z=_£)=(100)(120 320 )+3 “(100)120+3(100T 200) 100 +j(120 —- j200)T W (100+200T) +j120T [120 +j(100T -— 200)][(100 + 200T) -— leOT} = 100 ( ) (100 + 200T)2 + (120T)2 where T = tan(fil). To find the minimum electrical length l/A at which the line impedance is purely real, we set the reactive (imaginary) part of the above expression to zero, i.e., .SBm{Z(z m --l)} = 0 we (100T — 200)(100 + 200T) — (120)2T = O —> SOT2 — 111T u so = 0 —» T 2' 2.604, —O.384 From tanwl) 2: 2.604, we have 21rl/A : 1.204 rad -—r 1/). 2 0.192 and from tan(fl£) 2 “0.384, we have Zed/A 2 w0.367 rad == (—0.367 + 71') rad : 2.77 rad —> l/A 2: 0.442. Therefore, the minimum electrical length for a purely resistive line impedance to occur is l/A : 0.192. To find the value of the line impedance Z(z) at z c: —0.192/\ we substitutel 2 0.192/\ (i.e., T 2 2.604) into the real part of the of Z which yields 120(100 + 200T) + 120T(100T —- 200) Re{Z(z : 4)} z (100) (100 + 200132 + (120152 2 19.39 T316011 3-3;? Wnown termination. (a) Using [3.32] and [3.37], we can write Z + ‘Zt filmin):ZD L 3 o 311(51)::70 a I '— 30 + J ZL Emmi) S "e 30 +J'ZL tanwimin) = SZL + jSZO tanQBlmin) —* ZLIS " j tanCfilminH = Zoll -" 2'3 tfln(fllmin)l Hence, the expression for the load imp edance in terms of the first volta e mini ' ' from the load foilows as g mum posmon away (b) Substituting 20 = 509, S = x/i 1min = 25 mm: 311d /\ 2 = .10 cm into the abo ' the unknown load impedance can be found as / “3 expression, 1 — 33/3 tan[(27r/20)(2.5)] 1 — jfi Z = so —-——.___.____ = L ( ) x/i — j tanner/20x25» (50) \/§ — j 2(50)£1_‘3;@L[3_fl : (x/i ~ gun/5 +j) Using the value of ZL, the load reflection coefficient can be calculated as 50 (I) {06+ \/§)+j(1— 3)] = 250/3T — j) 9 F_ZL—Zo 25(fi—j)~50 L— 2 ZL+ZEJ 250/5“ j)+50 : 0.263e‘j90" 3-20. Two antennas. (a) The input impedance of the A / 2 long, 509 line terminated by ZL; = 40 ——j30 Q is Zin1= Z“ 2 40 — 330 Q. and the input impedance of the 3A/4 long, 509 line terminated by 2L2 = 100 + 3'50 9 is 2 _ . (50) 50 = 50(2 30:21]“.110Q z- =_____,=w "‘2 100+j50 2+5; 22-112 The parallel combination of these two impedances at the junction can be considered as the equivalent load impedance for the main line given by Z )H _j30)(20 “' __ -- “5“ ‘“‘ "‘2 " (40—j30)+(20—j101 ‘ M60 _ 3-40 _25 - 3'50 _ (25 -— j50)(3 +32) m 175 _ leO -372.“ _ MW 13 2: 13.5 - 37.69 a The equivalent load reflection coefficient at the junction can be calculated as ZLeq — Z0 (13.5 —j7.69) u- 50 37‘3e—j168° ” .1610 r a mi“ 2 M 2 .._......_.H___._ z I 3 ml ZLeq + Z0 (13.5 - 37.69) + 50 gage—35.91:: 0 5846 Therefore, the standing—wave ratio on the main line is evaluated as 1 + 111%,] 1 + 0.534 = m— : —— r: .31 S 1—1133“ 1—0534 3 (b) The input impedance of the A / 4 iong, Z0 = 509 main line seen from the source end is given by _ 202 N (50)2 N (50903.5 + j7.69) _ m _ =140+ son am 115- j7.69 (13.5)2+(7.69)2 5' The equivalent circuit seen by the source is as shown. The current phasor at the source end of the line can be found as _ V0 _h 34ejO ‘ Rs +2“in “ 100+(140+j80) Is 2 0.134e"j18‘4° A Therefore, the time-average power dissipated in the source resistance R5 = 1009 and the time- average power input to the transmission line system represented by Zia = 140 + 3‘80 9 at the source end can be calculated as 1 1 pfifi = #142125 2 E(0.13:1)2(100) : 0.903 w 1 1 pm = pLeq 7. 511812113“ : E(0.134)2(140) 2 1.26 w Therefore, the total time—average power supplied by the sinusoidal source is H0111 = P12. + Pin 2 2.17 W e 1 .2511wa >m¢mwrww m xwwww -Wuwm )rwmwmwmym mammmmw mmmwmmww—‘M' NWWWWW R5=100$2 1s vo=34ef°v o I zjn=14o+jsotz Fig. 3.5. Figure for Problem 3-20. The equivalent circuit as seen from the source. (c) Since the lines are lossless, all of the time—average power input to the line at the source end must be absorbed by the two antennas. At the junction position, Zinl and Zing are in parallel and share the same voltage across them, so the powers delivered to Zinl and Zing (which are the powers delivered to Z“ and 2L2) can be expressed in terms of the junction voltage VJ- as : — d = — P1 2 Z“ 5R8{ mi} an P2 2 Zm {Rd m2} Taking the ratio of these powers yield {’1 x “13L; z [21,”!2 = (40)2+(30)2 :04 P2 1:12 9%{Zinz} 20 ' [2M2 (20)2 + (10)2 In addition, we also know that Pin 2 P} + P2 2 1.26 W. So, by-solving these two equations simultaneously, we find the powers delivered to each antenna as PL] 2 0.361 W and PL: 2 0.903 W respectively. 3-25. Matching with series stub. The input impedance of the 750 line of length 1 connected to the ZL = 90 + 3135 .0 load (not including the series stub) is given by (90 + j135) +3751" 90 + j(135 + 7ST) = ————— = 75 W 21 (75)7S+j(90+j135)T ( )(75 —135T)+390T ' 75— 135T ~ '90:“ . =(75)[90+5(135+75T)][( ) 3 1275+” (75 — 135T)2 + (90302 where T = tan(fll) and fl = 21r/A where A = 20 cm. Equating the real parts of the above equation, we have 90(75 -— 135T) + 90T(135 +75T) = 75 (75 — 135132 + (90132 (75) —, [(1332 + (90)2 — 90 x 751T2 — 150 x 135T +7505 * 90) = 0 —+ 7331"2 a 810T «— 45 = o —> T 2 1.09, —0.0529 WWWWM meageranAmwflawlwlywwwaxwmw mammmmmmw inme 2.63 cm and from tanwlg) 2 —0.0529, we have 27rl2/A 2 —0.0528 rad = (--0.0528 + 11“) rad 2 3.09 rad ——> lz/A 2 0.492 —) Z; 2 (0.492)(20 cm) 2 9.33 cm respectively. Next, we have to find out which one of these two lengths require the minimum stub length is for the short-circuited series stub. So, for each case, we evaluate the imaginary (reactive) part of the input impedance Z 1 and use that to determine the stub length ls needed to cancel out that imaginary part. So, equating i From tanQBll) 2 1.09, we have 27rl1/A 2 0.827 rad -—+ ll/A 2 0.132 —> £1 2 (0.132)(20 cm) 2 the imaginary part of Z; to the minus of the short—circuited stub impedance ZS, we have j§m{zl} : “ZS = _jZ0 tan{filsc) _ 2 _’ (muss +75T)(75 — 135T) (90) T ____ 45m In] (75 - 135T)2 + (901")2 20 i Substituting T 2 1.09 yields | ZSE 231249 -—+ £5] 2 3.27 cm and T 2 —0.0529 yields Z52 2 —j124.Q -—+ ls2 2 6.73 cm Therefore, the minimum length of the 750. transmission line which yields minimum length 15 (i.e., lsI 2 3.27 cm) for the series short~circuited stub is ill 2 2.63 cm. Note that this problem could also be solved using {3.46} in Chapter 3. First, we calculate the load reflection coefficient as ZL — Zo _ (90+j135) —75 1" = '7’ = __ ....._________ L Peg ZL+ZO (90+j135)+75 135.4317“ . .. rem N 344-4 11381393., ... 0.637s Substituting ,0 2 0.637 and 2,!) 2 44.4° 2 0.774 rad in [3.46] results in the same two solutions 5 obtained above for the length of the transmission line as __ A —1 A "1 m — 47r(¢ cos p) __ 41f [0.774 cos (0637)] _ 0.132A,0.492A 2:]:038 rad respectively. The corresponding lengths for the short-circuited series stub can be found using X = izo—z-fim = 30 tanwlsc) \/1 —,;:>2 where the minus sign corresponds to 9; = COS—1p in the range 0 S 61 _<_ 7r/2 (i.e., l 2 0.492A) and the plus sign corresponds to 92 = cos'”1 p in the range war/2 S 02 g 0 (i.e., l 2 0.132A) respectively. Substituting p 2 0.637 yields 2(0.637) 1/1 — (0.637)2 which is the same result obtained in the first solution. The minimum length of the short-circuitecl series stub needed for the matching network can be found from X = 75 tanQBlsc) 2 +1249 where ,6 = 21r/(20 cm) as Esq 2 0.163A 2 3.27 cm. X 2 21:75 2 3:124!) 3—28. Quarter-wave transformer. (a) To match a load of RL = 209 to a transmission line of Z0 2 809 at frequency f0 : 1.5 GHz (or wavelength A0 = tip/f0), the length and the char- acteristic impedance of the transformer must be I = MM and ZQ = x/ZOR = 1/(80)(20) = 4052 respectively. With thesc choices, the input impedance of the 4052 quarter-wave transformer terminated with RL = 209 at its end is Zin :2 809, Le, it is perfectly matched to the 809 line. (b) Using the quarter-wave transformer designed in part (a), the input impedance of the transformer at f] = f0 / (1.25) = 1.2 GHz (or A] = 1.25%) can be evaluated using [3.32] as W 40 + j20tan[(27r/x\1)()\o/4)] 20 + 3400.03) 125.930” _____ : (mm—FT 40+ 3206.08) 73.432510 Zia =(40) 2(40) 2 68.03j23'3°9 Next, the reflection coefficient at the input of the transformer can be calculated from the value of Zin found as Zn. — 29 N 68.097233" ._ so =mz... + z. - N(62.2+ 327.4) — 30 N 32.792123" N 0 zzfiejmo "(62.2+j27.4)+80 _ user—10.90 — - I“in Therefore, the standing-wave ratio 8 on the Zn = 809 line at f1 = 1.2 GHz is given by 1+1r- | 1+0.226 S = m 2’ —-—— z . 1 — mil 1 — 0.226 1 58 Following a similar approach at f2 = 1.8 GHz, we have 2. .. WW __ N "j23.79° '“ 40 + 320 tan[(27r/A2)(l .2212) /4)] — 67958 and Zia — 80 1+ [Fin] Zin +80 1_ [rial Irinl z 20.2257 ——+ S: z 1.58 3-46. A lossy high-speed interconnect. Note that this is not a low~loss interconnect. Using the line parameters L a 10.1 nH—(cmr', 0 = 1.1 pF—(cmrl, R = 143.5 Q-(cm)", and G = 0.022 S-(cm)“1, the propagation constant 7 and the characteristic impedance 20 at 5 GHz can be calculated as '1 =W 2W 23.78e3'61-5" 2 1.80 + 33.32 where 1.80 is in np—(cm)_l and 3.32 is in rad-(cmr‘, and Z _ (R+ij) N 143.5 +j317.3 ° ' (G +3100) ‘ 0.022 + 30.0346 34331617" m. _ m1/————~0_04108j57.50 .. 92.23J _ 92.0 H 36.55 :2 wwwaxwmuwm 1m Wmmmilnmmwmmmvarn.mm,m.w..w wWWWWWmmWme1“wamr-mcwwwummw“WW. 3-43. A lossy high—speed interconnect. (a) Using the line parameters R = 7.59~(cm)—1, L =_4._6 nH—(cm)“”1, C = 0.84 pF—(cmr‘i, and G = 0, the propagation constant 7 and the characteristic impedance 20 of the lossy interconnect can be evaiuated as '7 =0: +jfi = x/(R+ij)(ij) =v/(75 +327: x 2 ><10g x 4.6 x 10-50927: x 2 x 109 x 0.84 x 10-12) 2 58.36316” x 1.056 x 10—28j900 z 0.7848j86’30 20.0506 4- 30.783 where a 2 0.0506 np—(crn)‘1 and ,6 2 0.783 rad~(cm)‘1, and R + ij Z =1]— 0 ij “ 7.5 +j27r x 2 x109 x 4.6 x10—9 " j27r X 2 x109 x 0.84 x 10—9 58.3 Jim-6° . o g : 74.36—33'696 2 74.2 — 34.79 9 1.056 x 10" 3390 (b) Using the value of Zn, the load reflection coefficient can be found as respectively. HZ], — 2:0 N 50 — (74.2 — 34.79) ‘” ZL + 20 “ 50 + (74.2 m 34.79) 3169" . 0 :fi'ififi 2: 0.1983317] FL Using the values of FL and 7 caiculated and assuming the load position to be at z = 0, the reflection coefficient at any other position 2 < 0 along the interconnect can be written as = 116272 : 0.19883'171‘380.IOIzej15662: 2 0-19860.10113j(1.5662+2.98) The impedance of the interconnect at any position 2 seen looking toward the load gate can be expressed in terms of 1"(z) as 1 + l"(z) 1 — Hz) So, the input impedance of the interconnect at the source end (Le, 25 = —10 cm) is given by 2(2) = Zg 1+ F(z = “10) 2 743631696. 1+ 0.072157%; 1 — Hz = —10) 1 — 0.0721e—26 _- o 1.072 « 0.00754 _- o _- a 33.696 0 928-570 00754 2 (74.38 33.696 )(L1546 30.368 ) Esme—i456" 2 35.5 ~— 3633 o Zin =Z(z = —10 em) = Z9 274.36 Using the voltage divider principle (see Figure 3.651“) except the 21“ value is as found above), the source-end voltage V; is , 8 . —-j4.56° : Z,n V0 2 5 8e ' ( 1) Z, + Z,“ 50 + 85.5 m 36.83 —j4.56° ‘ o "2% 3 0-6333—31-58 V Vs The voltage at any position along the interconnect is given by V(z) = V+e"Vz[1 + F(z)] from which V+ can be written as i " V(z) V " e‘Tz[1+l"(z)] Using the source-end voltage VS : O.633e'"51"53° V at z = —10 cm, we can calculate ‘V’r as 0.633e"j1‘68° . o w z 2 0.356 1139-“ V 6050593335 {1 +0.072Ie—35°] ‘9 The voltage at the load end of the interconnect can be found as VL = V(z z 0) = V+{1+ 11] g 0.356e—J'39-3°[1 + 0.198e3'17‘] g 0_2378—j87.6° V (c) The time—average power delivered to the interconnect from the source end is 1 135:5 Vs Z. Rm 9: .— m— (85.5) w 2 2.33 mW 1 2 1 0.633 2 2 85.8 The time—average power delivered to the 509 load is 1 1w2 1 (0.287)2 333725 50 W 2 0.822 mW Thus, using the conservation of energy principle, the pOWer dissipated along the line is Host 2 P5 "‘ PL 2 "" g 111W ...
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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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HW 3 Sol - 3-30. Resistive line impedance. Using [3.32],...

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