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**Unformatted text preview: **Chapter lit Sections lllml to Jill-=4: Satellite Communication Systems Problem 10.1 A remote sensing satellite is in circular orbit around the earth at an
altitude of 1,000 km above the earth’s surface. What is its orbital period? Solution: The orbit’s radius is R9 = Re + h = 6,378 + 1000 : 7378 km. Rewriting
Eq. (10.6) for T: T _ 41933 “3 _ ' 47:2 x (7.378 x106)3 “2
_ GMe " [6.67 x 10‘“ X 5.98 x1024
: 63043439 5 = 105-08 minutes- i‘Drebleni W2 A transponder with a bandwidth of 500 MHZ uses polarization diversity. if the bandwidth allocated to transmit a single telephone channel is 4 ld-lz,
how many telephone channels can be carried by the tramponder? ZXSOOMI-lz 2X5>< 103
So ration: Number or" telephone channels 2 — = —— = 2.5 X 105 4 kHz 4 >< 10-3
channels. Problem 1&3 Repeat Problem 10.2 for TV channels, each requiring a bandwidth
of 6 MHz. 2x5x103 Sol ti :N rnbe oftele ho echan. ‘5:
non u r p n uei 6x106 = 166.67 x 166 channels. We need to round down becasue We cannot have a. partial channel. Problem 19.4 A geostationary satellite is at a distance of 40,000 km from a ground
receiving stations The satellite transznitting antenna is a circular aperture with a
l-rn diameter and the ground station uses a parabolic dish antenna with an effective
diameter of 30 em. if the satellite transmits 1 kW of power at 12 Gl-iz and the ground
receiver is character-ind by a syStem noise temperature of LOGO K, what would be
the signal-to-noise ratio of a received TV signal with a bandwidth of 6 Nil-l2? The
antennas and the atmosphere may be assumed lossless. Solo-tide; We are given azaxiolm, d;=1rn, a,=o.3m, a=1o3w,
f=12GHz, rsYFLOOOK, B=6ME-lz. W mass. ans-.1, z. __ CHAPTER 1 0 337 Atf=12GHz, A:c/j=3x103/12x109= 2.5x 10'21'11. Witt-1QI : g: 1, _ _ 411A: _ 41t(7td§/4)_ 41:an 1 _ a
G“ Di - 2.2 c 2.2 - - 15391-31
4am, * 4n(nd3/4) 411: x mos)?-
Gr~Dr* 713 — T - - 1421-22- Appiying Eq. (10.1 1) with 1(8) 2 1 gives: 511 _P,-_G:Gr( 2. )2 103x15,791.3?><1421.22 (2.5x10‘2 _ — — r; 7 .4 .
4-7:}? 4nx4x107) 6 0 9 ' 1.38 x10—33x103 X6X105 ‘ K2353 Sections 3&5 to 10-3: Rad-at” Sensors Psobﬁem 19.5 A coﬂisiou avoidance automotive rada: is designed to detect the
presence of vehicles 13p to a range of 1 km. What is the maximum usable PRF? Soiuﬁon: From Eq. (10.14), c _3><108 _ =1.5X105 H2.
2811 2X103 Frobtem 10.6 A 10—61-12 weather radar uses a 30—cm~diameter Eossless antenna. At
a distance of 1 km, what are the dimensions of the volume resolvable by the radar if
the pulse length is 1 ,us? Solution: Resolvable volume has dimensions ALA}; and AR. x 3x10-2
AI=A'= R=—R=——— 3: ‘
y B d 03 X10 100m,
(:1: 3x108 _6 Pmbiem 116.7 A radar system is characterized by the following parameters:
PI = 1 kW, ’5 = 0.1 ,us, G: 30 63, I3»: 3 cm, and Im = 1,500 K. The radar cross
section of a car is typicaily 30 m2. How far can the car be and remain detectable by
the radar with a minimum signal-to-noise ratio of 13 dB? 338 CHAPTER i 0 Seiuﬁon: 5min : 13 dB means 3ij = 20. G = 30 dB means G = 1000. Hence, by
Eq. (10-27), R _ RIGZFLZQ 11”
max— (411)3KT5y53mm 103 10-7x106x “xlﬂ’z'leo “'4
= :5753.12m=5.75 km. (4-7:)3 x133 X 10'33 x 1-5 x103 x 20 - rohiem 193.8 A 3-m—wave1ength radar is located at the origin of an x—y coordinate
system. A car located at x = 100 m and y = 200 m is heading east (At—direction) at a
speed of 120 kmfhr. What is the Doppler frequency measured by the radar? 3' Figure P103: Geometry of Froblem 10.8. Seluﬁon: 2 0 Gztan"1 {#0 = 63.43",
\EOU 1.2 x 105
= 2 : ——'—" = n . 3 1..
u 1 Okmfhr 3600 33 3 “an,
-2u —2 x 33.33 c0563.43° : -—993.88 Hz. fa: “A case: 3x10—2 ...

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