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Unformatted text preview: 13113334019me 1_ Texas State University—San Marcos
School of Engineering EE 3340 Spring 2010 Exam 1  Feb. 11, 2010 Closed book and notes. You may use only pen or pencil, calculator, and the formula
sheet provided. Express all numerical answers as a number between 1 and 999999. followed by the appropriate engineering unit (mA, A, mV, etc.) as needed. _
For 1this exam you may also use the attached Smith chart and linear scale by folding the
linear scale as described in the formula—sheet section at the back. NM:_§M£_~IJ‘_§_~ .Multigle Choice: Put the letter corresponding to the ﬁst answer in the blank provided. 1. (4 pts.) In the time and spacedependent'wave expression 
V(x, t)  V0 (205611  ﬁz + it), the phase constant B has the dimensions of (A) radians (B) meters (C) Crucible)“1 (D) meters per second Answer: C 2. (4 pts) If the reference characteristic impedance 20 = 50 Q, the reﬂection coefﬁcient
I‘ of a short circuit (Z = O) is (A) +1 @ ‘1 (C) 0 (D) Cannotbedetermined from information given I” —» Z 2
—~ 1... Answer: '5 ' a "‘ BE 3340 Exam 1 2 3. (4 pts.) Two coaxial cables are fabricated with identical dimensions, except that in Cable #1, the dielectric between the inner and outer conductors is foam plastic (3, =13), _ L i
and in Cable #2, the dielectric is solid PTFE plastic (3, =22). Which of the following Z;
statements is true? C I (A) #1 has a h'
(meanest Answer: B , r 2 3259+ 5 o 3 4. (4 pts.) A quarter—wave mat%hing section is designed with a 3114 section of 5042
transmission line so that a 25—9 resistive load is transformed to Zn 2 100 +j0 Q at f z 1 GHz. Suppose everything remains the same except that the frequency of measurement is
raised to j‘ = 1.05 GHz. What is true about the new input impedance 2m"? (A) The real part remains about the same but the imaginary part is now inductive. kl t (B) The imaginary part remains about the same but the real part rises above 100 Q. ‘lWN‘M
. ... 333 v a? at Mew—3‘ 1st??? asses: 333333 .3...” s .....3 Eg/P‘leﬁl l‘ (D) The imaginary isaiimmmains about the same but the real part Answer: ______C_r____ Problems: When working the following problems, show as much work g m cm! I will
try to give you partial credit for partially correct answers, but if all you write down is an
answer and it's wrong, that is difﬁcult For problems involving the Smith chart, please
use the attached Smith charts at the end of this exam, folding the radiallyscaled parameter scales as instructed. Note in the problem that your work is shown on the Smith
chart and leave it attached to the exam. 5. (20 points) Design of 75—9 coaxial cable. Suppose that a certain type of plastic—foam
dielectric has a relative dielectric cons _ __=l.3. Its relative permeability ,u, = l. The outer conductor gshield) has an‘ « fan‘s . . ‘ . memenpmsg
Standard AWG wire gauges in mm are given in the following table for 12—gauge to 20—
gauge wire: gr: (:3 EE3340Exam 1%,; [ ZO'QOE: Q‘s}; ‘ 3
Ci M 3 Choose a wire gauge for the center conductor so that the characteristic impedance of e
resulting coaxial cable is as close as possible to 20 = 75 9. Once you have selected the
wire gauge. calculate the actual impedance of the cable and the wave velocity u in m s". .. ‘ ‘M
at; to {2: lug) :79» 75“50%.MC%:)
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_, Me 45 :m {.5 cuatedimpedance: “ALA Wave velocity: 1  ‘3 '19 g M/S 6. (24 points) VSWIR to impedance. A long, low—loss transmission line whose impedance
is 30 = 50 S2 is connected between a 50—9 generator on the left and an unknown load on
the right. Before the load is counected, the line is connected to a perfect short circuit (R =
0) and the standing—wave voltage V}(x) is measured. The distance x in meters is
measured from the generator end and goes from 0 to the length L = 7 20 m of the line,
Wen the unknown load is connected, a different standing—
wave pattern V2(x) results. These measurements are shown in the following ﬁgure: .xam 1 Vollaga magnitude (U) _ \ (1.4 = ,"  ’i
i 5 ‘ V2 (unknown)
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(a) (4 points) Calculate the VSWR ofthe unknown load by estimating the aximum and
minimum values of its standing wave. ~ I \/ .ngﬁcéimim/O USMR: Wul" —;( o 90A )
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2&2» *“ B VSWR of unknown load = W (b) (5 points) Estimate by how man. neters the+ioeation of thevoltage minimum shifted
in going from the short to the un own load: _. ‘7 b D V"\
170 A,” ' Minimum shifted N meters (0) (10 points) Using the information you found 1n (a) and (b), calculate the complex
valued reflecnon coefﬁcient Fof the Load. You may use the Smith charts attached to the
end of this exam, but indicate your work on them if you do. Express the angle of the reﬂection _oefficient1n degrees.
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m Angle of F(degrees)= . (31% / Magnitude IN = (d) (5 points) Using your result in (c), calculate the complex—valued impedance of the unknown load in ohms. Express it in rectangular coordinates as a sum of a real part R . and an imaginary part jX. _ M
. 31 49 We " Z L {.0 {5 ”:OSIJJCL X504“
‘355 @2950 :17er ”O U0 Unknown load 2: 2 Eli " $3 2
TH : 7. (22 points) Shuntstub impedance matching. At a certain frequency, a load
impedance has the value ZL z 34.51862 82. Design a shunt—stub matching network with
5082 transmission lines that will match this load impedance so that 23”,, = 50 $2 in the
shuntstub matching circuit below. Use the shortest possible lengths of line, make the
shunt stub shorted at the far end, and express line lengths Lu and LPA in terms of
wavelengths it (e. g. a quarter—wave line has L z 0.25 A). You may use the Smith charts attached to the back of this exam, but label your work on those charts and indicate in this
problem that part of your work is there. Why756.2 I Z
'50 _ " l
.I‘ “r «4 1..v "32. . LFVG‘W [3.2%] EB 3340 Exam 1 6 (a) (3 points). Normalize the actual load impedance ZL to 50 9 so that normalized 7
(dimensionless) impedance 2‘; = 2,150 S2.  so ““J “ {foo—(.121, (b) (4 points) Using either a calculator or the Smith chart, convert the normalized load
impedance to normalized load admittance yL — 1/24,. {pm a J. : 1 : 0.3 35¢+51i°~ﬂ0§+ o. ; I
__.... FL 4"“ FiﬁTrait" )7 j S )4“
or $6: Sm‘eh Laure ﬁrm 31L: D_l+1 6.5 (c) (5 points) Perform the operations on the Smith chart to move the normalized load
admittance to the g z 1 unit circle with L”. mn too». A‘: MEMO a: one: mutel grew" h: 3.01: we GA" YLT— l +j2lﬁ
A;'l$“0°%=lDi\3 L!:L96 LSE‘M (d) (5 points) Then ﬁnd the normalized admittance ym that must be presented by the
shunt stub to move the impedance to 3’0 = 1.7 e ml 7% 2 ~51!” 10(1de +j1~0i EB 3340 Exam 1 Y» : am
y“: “F“ (9)13?) (c) (5 points) Finally, ﬁnd the length of the shunt stub Ln needed to produce the ad h t d “311
requir sun ami‘ TBYM ’ 20M) [F l :0 (“\wa
(“kaukwk . ?\M:2 EsTJEVk‘W\ LL: lo ”fonkéﬂ) (3* {VIM ‘ 3+I~AQQJ A: Y“ _ ______ :Ej1‘0‘f \pkx‘c‘u {5 15?1:[/0?LL1
3“ L”: Mus Sw—\€"h (~th . 5“ #1 A: balm—0.1»: 0.6a“ —§L
__./ 8. (18 points) Transients on transmission lines. The circuit above consists of a +1 VDC ideal voltage source, an ideal switch 31, a 509
resistor, and at 43cm section of 50—9. transmission line shorted at the righthand end.
Before I = O the switch is open. At 2‘ = 0, the switch is closed and remains closed. The
propagation velocity it along the cable is u z: 2 x 103 m s'l. Draw a "bounce diagram" in the ﬁgure above, showing the progress of the voltage wave
along the transmission line in the graph below it, indicating the wave‘s position with a
dashed line. Then, use the information from the bounce diagram to determine the voltage
as a function of time at two fixed points: ' (a), (9 points) Atx = 20 cm (in the center of the line). (b) (9 points) At x = 0 cm (at the lefthand end of the line). l a _ _L___,* ._ “tout
Txmc ti. £°" ﬁtrwbt “f0 MVBU'SK ""13 0 L€ 61“ LA. .— m5:2h
So
l"\"€\ul Vaguu [\e\"'\* +{U ”2/:— U S“
get: 750:» 5° 0% “‘4’ (”I U;+O$v ’
h 0“ (’Lof’f’ {'5 .— Zp‘néo v t—ip . v{ 80 “‘ Awo" as ‘“ EE 3340 Exam 1
V(20 cm)
1 .0 V
0.5 V
O t
o a) to m a:
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.,. 1. N 0') tr Draw your waveforms to scaie on the two graphs provided above. the total current I (1) (sum of forwaxd and reverse Balms Question (5 points): What is
t at the point x = 20 cm (in the center of the line)? current waves) going from left to sigh
Draw the total current wavefonn in the graph beiow. 1(20 cm)
20 mA " — ..
10 mA
0 t
Q U)
C:
V 1ns
2ns
ans ...
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 Spring '11
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