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448 Chapter 10: Satellite Communication Systems and Radar Sensors Lesson #70 and 71 Chapter — Section: 10-1 to 10-4 Topics: Communication systems Highlights: Geosynchronous orbit Transponders, frequency allocations Power budgets Antennas Lesson #72 and 73 Chapter — Section: 10-5 to 10-8 Topics: Radar systems Highlights: Acronym for RADAR Range and azimuth resolutions Detection of signal against noise Doppler Monopulse radar

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Page 1: 45795-UlabyISMCh10.pdf

448

Chapter 10: Satellite Communication Systems and Radar Sensors

Lesson #70 and 71 Chapter — Section: 10-1 to 10-4 Topics: Communication systems Highlights:

• Geosynchronous orbit • Transponders, frequency allocations • Power budgets • Antennas

Lesson #72 and 73 Chapter — Section: 10-5 to 10-8 Topics: Radar systems Highlights:

• Acronym for RADAR • Range and azimuth resolutions • Detection of signal against noise • Doppler • Monopulse radar

Page 2: 45795-UlabyISMCh10.pdf

CHAPTER 10 449

Chapter 10

Sections 10-1 to 10-4: Satellite Communication Systems

Problem 10.1 A remote sensing satellite is in circular orbit around the earth at analtitude of 1,100 km above the earth’s surface. What is its orbital period?

Solution: The orbit’s radius is R0 � Re h � 6 � 378 1 � 100 � 7478 km. RewritingEq. (10.6) for T :

T � �4π2R3

0

GMe 1 � 2 � � 4π2 � �7 478 � 106 � 3

6 67 � 10 � 11 � 5 98 � 1024 � 1 � 2� 4978 45 s � 82 97 minutes.

Problem 10.2 A transponder with a bandwidth of 400 MHz uses polarizationdiversity. If the bandwidth allocated to transmit a single telephone channel is 4 kHz,how many telephone channels can be carried by the transponder?

Solution: Number of telephone channels � 2 � 400 MHz4 kHz

� 2 � 4 � 108

4 � 103 � 2 � 105

channels.

Problem 10.3 Repeat Problem 10.2 for TV channels, each requiring a bandwidthof 6 MHz.

Solution: Number of telephone channels � 2 � 4 � 108

6 � 106 � 133 3 � 133 channels.

We need to round down becasue we cannot have a partial channel.

Problem 10.4 A geostationary satellite is at a distance of 40,000 km from a groundreceiving station. The satellite transmitting antenna is a circular aperture with a1-m diameter and the ground station uses a parabolic dish antenna with an effectivediameter of 20 cm. If the satellite transmits 1 kW of power at 12 GHz and the groundreceiver is characterized by a system noise temperature of 1,000 K, what would bethe signal-to-noise ratio of a received TV signal with a bandwidth of 6 MHz? Theantennas and the atmosphere may be assumed lossless.

Solution: We are given

R � 4 � 107 m � dt � 1 m � dr � 0 2 m � Pt � 103 W �f � 12 GHz � Tsys � 1 � 000 K � B � 6 MHz

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450 CHAPTER 10

At f � 12 GHz, λ � c � f � 3 � 108 � 12 � 109 � 2 5 � 10 � 2 m. With ξt � ξr � 1,

Gt � Dt � 4πAt

λ2 � 4π�πd2

t � 4 �λ2 � 4π � π � 1

4 � �2 5 � 10 � 2 � 2 � 15 � 791 37 �

Gr � Dr � 4πAr

λ2 � 4π�πd2

r � 4 �λ2 � 4π � π

�0 2 � 2

4 � �2 5 � 10 � 2 � 2 � 631 65

Applying Eq. (10.11) with ϒ�θ ��� 1 gives:

Sn � PtGtGr

KTsysB

�λ

4πR 2 � 103 � 15 � 791 37 � 631 651 38 � 10 � 23 � 103 � 6 � 106

�2 5 � 10 � 2

4π � 4 � 107 2 � 298 Sections 10-5 to 10-8: Radar Sensors

Problem 10.5 A collision avoidance automotive radar is designed to detect thepresence of vehicles up to a range of 0.5 km. What is the maximum usable PRF?

Solution: From Eq. (10.14),

fp � c2Ru

� 3 � 108

2 � 0 5 � 103 � 3 � 105 Hz Problem 10.6 A 10-GHz weather radar uses a 15-cm-diameter lossless antenna. Ata distance of 1 km, what are the dimensions of the volume resolvable by the radar ifthe pulse length is 1 µs?

Solution: Resolvable volume has dimensions ∆x � ∆y, and ∆R.

∆x � ∆y � βR � λd

R � 3 � 10 � 2

0 15� 103 � 200 m �

∆R � cτ2� 3 � 108

2� 10 � 6 � 150 m

Problem 10.7 A radar system is characterized by the following parameters:Pt � 1 kW, τ � 0 1 µs, G � 30 dB, λ � 3 cm, and Tsys � 1 � 500 K. The radarcross section of a car is typically 5 m2. How far can the car be and remain detectableby the radar with a minimum signal-to-noise ratio of 13 dB?

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CHAPTER 10 451

Solution: Smin � 13 dB means Smin � 20. G � 30 dB means G � 1000. Hence, byEq. (10.27),

Rmax � � PtτG2λ2σt�4π � 3KTsysSmin

� 1 � 4� � 103 � 10 � 7 � 106 � �

3 � 10 � 2 � 2 � 5�4π � 3 � 1 38 � 10 � 23 � 1 5 � 103 � 20 � 1 � 4 � 4837 8 m � 4 84 km

Problem 10.8 A 3-cm-wavelength radar is located at the origin of an x–y coordinatesystem. A car located at x � 100 m and y � 200 m is heading east (x-direction) at aspeed of 120 km/hr. What is the Doppler frequency measured by the radar?

θ

θ

u = 120 km/hr200 m

100 mx

y

Figure P10.8: Geometry of Problem 10.8.

Solution:

θ � tan � 1

�200100 � 63 43 � �

u � 120 km/hr � 1 2 � 105

3600� 33 33 m/s �

fd � � 2uλ

cosθ � � 2 � 33 333 � 10 � 2 cos 63 43 � � � 993 88 Hz