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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 quarterwave, Z“ = Zg/ZJCT = ZgYJCT = 21+22 = (50+ j50) 9+ (50 — jSO) o = 100 o. Section 211: nsients on Transmission Lines .nerate a bounce diagram for the voltage V(z, t) for a lm 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 509 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. 245, 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 timeaverage power dissipated in the load,
(b) the timeaverage power incident on the line, and (c) the timeaverage 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
—1I7L2 —lx122x—50 —029W
_2zL2 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; = —r2P;v = —(0.62)2 x 0.47 = —0.18 w. ZL=(75j50)§2 Use the Smith chart to determine the input impedance Zia of the twoline
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»  ‘ \JX'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)=2x3x106x2x103=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_025)50 3339 (d)
VgZO
Rg + Z0
_ 12(Rg +Zo) 12(83.3 + 50) V— =——~—= .
g Zo 50 32V V1=12= ...
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