HW3 - 78 CHAPTER 2 and similarly for the lower branch Y2 22...

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Unformatted text preview: 78 CHAPTER 2 and similarly for the lower branch, Y2 22 Y ' :2 —0 = — _ 2 1n Y2 Z3 Thus, the total load at the junction is Z1 + Zz YJCT=Y1in+Yzin= 2 20 Therefore, since the common transmission line is also quarter-wave, Z“ = Zg/ZJCT = ZgYJCT = 21+22 = (50+ j50) 9+ (50 — jSO) o = 100 o. Section 2-11: nsients on Transmission Lines .-nerate a bounce diagram for the voltage V(z, t) for a l-m long . - aracterized by Zo— — 50 Q and up— —— 20/ 3 (where c is the velocity of light) if the line 18 fed by a step voltage applied at t— — 0 by a generator circuit with Vg— -— 60 V and RE =100 Q. The line 1s terminated m a load ZL— — 25 9. Use the bounce diagram to plot V(t) at a point midway along the length of the line from t = 0 to t = 25 ns. Solution: ' Rg—Zo_100—50 50 1 Rg+z0 ‘100+50 _ 150 3’ 2L— z0 _25—50_—25 —1 Pg: FL: 2'L+z0 25+50 _ 75 3 ' From Eq. (2.124b), ngo 60 x 50 = — = 20 v. Rg+zo 100+ 50 S + II Also, I l 3 upzzc/3=2x3x108=5“s' The bounce diagram is shown in Fig. P2.50(a) and the plot of V(t) in Fig. P2.50(b). CHAPTER2 79 Volta e F=F=i g 1"=11:-l g 3 z = 0 5 m 3 z=0 ' 2= 1 m 20 ns 4‘ I '-0.08 V 25 “S t t Figure P250: (21) Bounce diagram for Problem 2.50. V(0.5 m, t) 20 V t (ns) 5 10 15 20 25 Figure P250: (b) Time response of voltage. CHAPTER 2 t t Figure P251: (a) Bounce diagram for Problem 2.51. CHAPTER 2 81 Kmmn r (us) 5 10 15 20 25 Figure P251: (b) Time response of current. Problem In response to a step voltage, the voltage waveform shown in Fig. 2—45 (P2. as observed at the sending end of a lossless transmission line with Rg : 50 Q, Z0 2 50 Q, and e, = 2.25. Determine (a) the generator voltage, (b) the length of the line, and (c) the load impedance. vmo 5V 3V 0 6p,s Figure P252: Observed voltage at sending end. Solution: (a) From the figure, V1+ = 5 V. Applying Eq. (2.124b), + VgZO l = ng0 Vg : —2—-', &+%—%+% which gives Vg = 2V1+ : 10 V. 82 CHAPTER 2 __3x108 C (b)“P‘¢.e—‘¢2f2‘s " at At = 6 ,us. But At = Zl/up. Hence, = 2 x 108 m/s. The first change in the waveform occurs I_At,up_6x10‘6 2 (c) Since Rg = Z0, Fg = 0. Hence V2 = 0 and the change in level from 5 V down to 3 V is due to V1_ = —2 V. But x2x108=600m. V— —2 L = —— = —o.4. V;=1"LV1+, or 1"L=V1+ 5 From Problem In response to a step voltage, the voltage waveform shown in Fig. observed at the sending end of a shorted line with Z0 = 50 Q and 2.46 (P2. 3r = 4. Determine Vg, Rg, and the line length. V(0, t) 12 V 3 V 0.75 V Z 0 7 us 14 us Figure P253: Observed voltage at sending end. Solution: c 3 x108 u=———= =l.5x108m/s P 8r t/Z ’ 21 21 = 7 10—6 — —— = . 7’5 x 5 up 1.5 x 108 Hence, I = 525 m. CHAPTER 2 83 From the voltage waveform, V1+ = 12 V. At t = 711$, the voltage at the sending end is V(z = 0,t = 7,us) = V1+ +FLV1+ +rgrLVl+ = ~rgV1+ (because rL = —1). Hence, 3 V: —Fg x 12 V, or Pg = —0.25. From Eq. (2.128), 1+1" 1 — 0.25 g 20(1—rg) (1+O.25) 30 ngo Rg + Z0 ’ Also, which gives Vg = 19.2 V. the sen in of a 50-9 transmission line in response to a step voltage introduced by a generator with Vg = 15 V and an unknown series resistance Rg. The line is 1 km in length, its velocity of propagation is l x 108 m/s, and it is terminated in a load ZL = 100 Q. (a) Determine Rg. (b) Explain why the drop in level of V(O, t) at t = 6 ,us cannot be due to reflection from the load. (c) Determine the shunt resistance Rf and the location of the fault responsible for the observed waveform. Solution: V(0, 1) 5V 3V 0 6us Figure P2.54: Observed voltage at sending end. 84 CHAPTER 2 (a) W = VEZ" . 1 Kg +20 From Fig. 2-45, V1+ = 5 V. Hence, 5 _ 15 x 50 " Rg + 50 ’ which gives Rg = 100 Q and Q = 1/3. (b) Roundtrip time delay of pulse return from the load is _2_l__2x103 2T — up 1 x 108 :20 ,US, which is much longer than 6 gs, the instance at which V(O, t) drops in level. (c) The new level of 3 V is equal to V1+ plus V1— plus V+, V1++Vf+V2+ =5+5rf+5rfrg=3 (V), which yields Ff = —~O.3. But 2 ZLf— Zo = ZLf+ Z0 which gives ZLf = 26.92 0. Since ZLf is equal to Rf and Z0 in parallel, Rf = 58.33 9. Fr —0.3, Proble n a A generator circuit with Vg = 200 V and Rg = 25 Q was used to excite a P lossless line with a rectangular pulse of duration 1: = 0.4 ,us. The line is 200 m long, its up = 2 x 108 m/s, and it is terminated in a load ZL = 125 Q. (a) Synthesize the voltage pulse exciting the line as the sum of two step functions, V31 (t) and ng (t). (b) For each voltage step function, generate a bounce diagram for the voltage on the line. (c) Use the bounce diagrams to plot the total voltage at the sending end of the line. Solution: (a) pulse length = 0.4 gs. VgU) = Vg1(t) + Vg2(t)1 with Vg] (t) = 200U (t) (V), ng (t) = —200 U(t —— 0.4 ,us) (V). CHAPTER 2 85 200 V Z0 = 75 (2 125 (2 Figure P2.55: (a) Circuit for Problem 2.55. V(t) Figure P255: (b) Solution of part (a). w) 1 200 up:2><108= We will divide the problem into two parts, one for Vg1 (t) and another for ng (t) and then we will use superposition to determine the solution for the sum. The solution for ng (t) will mimic the solution for Vg1 (t), except for a reversal in Sign and a delay by 0.4 us. For Vgl (t) = 200 U(t): 1/5. ~ Rg—Zo _ 25—75 _ ’ Rg+zo ’ 25+75 ‘ zL—z0 125—75 FL: 2 : A+A lfi+fi r‘g ' 7 0.25, 86 CHAPTERZ _ V1Zo - 200x75 ‘Rg+zo ‘ 25+75 _ VgZL _ 200x125 ‘Rg+zL _ 25+125 V1+ = 150 V, Vno = 166.67 V. (i) V1 (0, t) at sending end due to Vg1 (t): Vglo) r=rg=— r=rL= i .1. 2 4 z=200m is)? Figure P255: (0) Bounce diagram for voltage in reaction to Vgl (t). CHAPTER 2 (ii) V2(0, t) at sending end due to ng (t): V (t) r rg=— g2 H l 2 z=0 z— r=0.4us W 1.4us 2.4 us 3.4 as 5.4 us 6.4 ”S 0.28V t t Figure P255: (d) Bounce diagram for voltage in reaction to Vg2(t). 87 88 ' CHAPTER 2 (b) .___ '(i) V1(0,t) at sending end due to Vg] (t): mm 1 (HS) 2 4 6 Figure P255: (e) V1 (0, t). (ii) V2(0, t) at sending end: Vzum 1 (HS) Figure P255: (t) V2(0,t). CHAPTER 2 89 (iii) Net voltage V(0,t) = V1(0,t) + V2(O,t): V(0,t) t (HS) Figure P2.55: (g) Net voltage V(O,t), Problem 2.56 For the circuit of Problem 2.55, generate a bounce diagram for the current and plot its time history at the middle of the line. Solution: Using the values for Pg and FL calculated in Problem 2.55, we reverse their signs when using them to construct a bounce diagram for the current. V+ 150 + 1 =_=_=2A 1‘ Z0 75 ’ V+ —150 1+:L:_____=_2 2 20 75 A’ V“, I+=—=1.33A. 9., ZL I CHAPTER 2 Using (263), Z‘ _Z ZL +jZo tanfil m — 0 Zo + jZL tan Bl ~ 50 (75 +j50tan0.81t — W) = (52.38+120.75) :2. Zn is equivalent to a series RL circuit with R Z.n —> L R = 52.38 9 01L = 27th = 20.75 9 or. _ 20.75 _ 2n x 4 x 108 which is a very small inductor. = 8.3 'x 10—9 H, ZL=(50+j]00)§2 The circuit shown in the figure consists of a 100—!) lossless transmission line terminated in a load with ZL = (50 + j 100) Q. If the peak value of the load voltage was measured to be [VLI = 12 V, determine: (a) the time-average power dissipated in the load, (b) the time-average power incident on the line, and (c) the time-average power reflected by the load. 99 100 CHAPTER 2 Solution: (a) Z Z 50 '100 100 50 '100 r: L— 0 :——-+J _ =—— +1 =0.623j82'9°. ZL+ZO 50+]100+100 150+1100 The time average power dissipated in the load is: 1 ~ 2 Pav = EIILi RL 1 I7 2 _ _. J: _ 2 2L RL —1|I7L|2 —lx122x—50 —029W _2|zL|2 L‘z 502+1002" ' (b) . Pav = Paiv“ " [1—12) Hence, P O 29 i — av — - = a a PE”— 1—|1“[2 _1—0.622 047W (C) 10; = —|r|2P;v = —(0.62)2 x 0.47 = —0.18 w. ZL=(75-j50)§2 Use the Smith chart to determine the input impedance Zia of the two-line configuration shown in the figure. CHAPTER 2 101 ,,‘0.625 AW‘NKM /"‘ \ / rm "LL, A}; ..,, /‘/ \ +444" “nix. r "I" *‘fiem. // )EM’" a:N ‘ w 7L ,. K 1: . 'y X "1'; V +2 wt. 1-; "h immnufia’m" J“ 3‘ n A: —"" m1 ‘nu ’/., ' l I _ ii.“ . -:—H 4+ l 1.. L4H) “ ‘ imwd Cmmwcmn In I g mnoc Anr‘imwm‘w w: \t . )7 H1 \ i , 4 \ _‘ ,. JrT H Y _J ' x L. . w, ; An _ 1 L F pr A 4 ’r _ A \ ’ we 5 y - M t L I ’ x V .8} "2 , x \ A» - ‘ \J-X'X "’ \ ' \ "9 «f ‘ , .(X, >9)- . K > ,\ >~ \ >> ’\ N ) \ J , e r_ 14% i \\ t "\ .,/ >_ :00] ~. I, e v . - ly \ \Z’v "13,, e , . Ila A, .- "3‘ / _ 11'\\ ' —_, L‘ » Ac ll). \ m] r. .u . 0433 xx (5»: / ~ ’ is N l . \ IX" 7‘” W" .5 ’ .,.‘ 7;» \ nmew L In I ~4 , a :I‘u \ #:F‘Hfi—H— Smith Chart 1 Solution: Starting at point A, namely at the load, we normalize ZL with respect to 2022 2L 75 — j50 ZL =Z—02——50——= 1.5—j1. (pointA on Smithchart 1) From point A on the Smith chart, we move on the SWR circle a distance of 5X/8 to point 31., which is just to the right of point B (see figure). At Br, the normalized input impedance of line 2 is: Zing = 0.48 — j0.36 (point Br on Smith chart) Next, we unnormalize Zing: ZinZ = Zogzinz = 50 X (0.48 —j0.3()) (241 j l X) S). 102 CHAPTER 2 a“ a: :13 PM u\ “ 37 v “I 039 90 3‘ we M W v m M; e“ ““ a d — = . 4’ y g c o9 \‘9 a“ y 9’ 0’ a" A .93 3 e " 3WR Circle A A‘ a f: 5 2f ‘ A“ i ‘9 g g” «9, *7 , 5‘ g ‘ f a ‘ t a E 5 5 ‘0 73% u \ 5 ~ 2 .5 a; E S 4 {a E = 1 :' ,1 ,+ ;,§‘t;i s 2 :=:n=::::_:- 2-“ - 22” . a . V l: " l ”‘5‘: rC0 .r lZaz, chdu Eco. /\\’u g a 2 Se 8% I ‘ 2 E y ._ v , a a: _ g i g 3E i , u g a \. r 4 “I; s “'5 g 0470}. ‘“ k ’ a) 19159’ fag; D é £967 “a ‘6 a C a? a ’10 9’~ “.5 6‘ ‘a n, 4, e an " :‘Ql‘lgbo 915° ‘2’," filth "9 fin “a my” “7 ‘L 6° an 4’1. ”02% 5 w: v“ 0 . o b “a ‘7'! . q 1 ’ u n‘“ 6 "'7 w m- 'w or “a «‘3 Fa Ha an era "1’ if“ “‘0 PCB 50 ICU 0.345 A. Smith Chart 2 To move along line 1, we need to normalize with respect to Z01. We shall call this 2L1: Zin2 _ 24—j18 = - 4— '0.18 ' B - 201 100 0 2 J (pomt [on Srmth chart 2) After drawing the SWR circle through point B 1, we move 31/ 8 towards the generator, ending up at point C on Smith chart 2. The normalized input impedance of line 1 is: zin = 0.66 —j1.25 which upon unnormalizing becomes: Zin = (66—j125) r2. I. , r F J 104 CHAPTER 2 The normalized load impedance is: 2—5 = 0.33 (point A on Smith chart) 2L _ 75 The Smith chart shows A and the SWR circle. The goal is to have an equivalent impedance of 75 .Q to the lefl of B. That equivalent impedance is the parallel combination of Zn at B (to the right of the shunt impedance Z) and the shunt element Z. Since we need for this to be purely real, it’s best to choose I such that Zin is purely real, thereby choosing Z to be simply a resistor. Adding two resistors in parallel generates a sum smaller in magnitude than either one of them. So we need for Zin to be larger than Zo, not smaller. On the Smith chart, that point is B, at a distance I = k/ 4 from the load. At that point: zin = 3, which corresponds to yin = O..33 Hence, we need y, the normalized admittance corresponding to the shunt impedance Z, to have a value that satisfies: yin+y:1 y=1—y,-u =1.—0.33 =_0.66 1 1 ' =~=——=1.5 Z y 0.66 Z=75x1.5=112.5§2. In summary, Proble . In response to a step voltage, the voltage waveform shown in the figure be w 5 observed at the midpoint of a lossless transmission line with Zo = 50 9 up = 2 x 108 m/s. Determine: (a) the length of the line, (b) ZL, (c) Rg, and (d) Vg. CHAPTER 2 105 12V Solution: (a) Since it takes 3 its to reach the middle of the line, the line length must be 1:2(3x10*6><up)=2x3x10-6x2x103=1200m. (b) From the voltage waveform shown in the figure, the duration of the first rectangle is 6 ,us, representing the time it takes the incident voltage V,+ to travel from the midpoint of the line to the load and back. The fact that the voltage drops to zero at t = 9 ,us implies that the reflected wave is exactly equal to V14" in magnitude, but opposite in polarity. That is, W=-W- This in turn implies that FL = — 1, which means that the load is a short circuit: ZL = O. (c) After V]~ arrives at the generator end, it encounters a reflection coeflicient Fg. The voltage at 15 ,us is composed of: V=W+W+W = (1+FL+FLFg)I/1+ V fia—r—rg From the figure, V V+ = —3 12 = —1 4. Hence, I 1 ngz, 106 CHAPTER 2 which means that 1+r 1+0.25 R = 3 Z = = . . g ( ) 0 (1_0-25)50 3339 (d) VgZO Rg + Z0 _ 12(Rg +Zo) 12(83.3 + 50) V— =——~—= . g Zo 50 32V V1=12= ...
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