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Fundamental Principles of Radar 1st Edition By Habibur Rahman (Solution Manual)

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Fundamental Principles of Radar, 1e Habibur Rahman (Solution Manual) Fundamental Principles of Radar, 1e Habibur Rahman (Solution Manual)

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  • June 9, 2023
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  • 2022/2023
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  • Fundamental Principles of Radar, 1e Habibur Rahma
  • Fundamental Principles of Radar, 1e Habibur Rahma
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Solutions Manual FUNDAMENTAL PR INCIPLES OF RADAR
Habibur Rahman, 1 Chapter 2: Radar Fundamentals Solutions 2.1 The distance of the moon from the radar transmitter located on the surface of the earth is
83.84×10m. Calculate the elapsed round trip time of a radar signal transmitted from radar the antenna. Solution: 8
82 2(3.84 10 )2.56 s3 10dRtc= = =
2.2 Consider a low PRF pulsed radar with a PRF of 1500 pps and a bandwidth of 0.5 MHz. Calculate the maximum unambiguous range, pulse width, range resolution, and the duty factor. Solution: 8
6
8 6)
633 10100 km2 2 1500
115μs.2 10
(3 10 )(2 10300 m22
(2 10 )(1500) 3 10u
r
trcRf
B
cR
df

−
−−= = =
= = =
 = = =
= =  = 
2.3 A C-band radar transmits a peak power of 1 MW at a frequency of 5.5 GHz with the pulse length of 1
μsand the PRF of 200 Hz. (a)Find the average transmitted power.
(b)Find the bandwidth and the range resolution of the radar.
Solution: (a) 6 6)(10 )(10 )(200) 200 Wav t t rP P P fT−= = = =
(b) 6
86111 MHz10
(3 10 )(10 )150 m22B
cR
−
−= = =
 = = = 2 2.4 A pulsed radar has a PRF of 1 500 pps and transmit rectangular pulse train of duration 15
s . (a) What maximum range can a target have if no part of its first time around returned echo is to overlap any part of the transmitted pulse? (b)What is the minimum distan ce of separation so that targets can be identified? Solution: (a) 83 10100 km2 2 (2)(1500)u
rcT cRf= = = =
(b) 86(3 10 )(15 10 )2250 m22cR− = = =
2.5 The speed of a missile toward a radar is 300 m/s. Assume an X -band radar operating at a frequency of 12 GHz. (a)Calculate the exact Doppler frequency at the receiver.
(b)Calculate the receiver Doppler frequency assuming
rvc . Solution: (a) 0
0 0 0 0
9)
82
(2)(300)(12 1024000.024 Hz3 10 300d
dc v vff f f f fc v c v
f+=  = − =−−
==−
(b) 9
0
82 (2)(300)(12 10 )24000 Hz3 10dvffc = =
2.6 Assuming that the target is receding (opening), derive the expres sion for the Doppler shift Follow the article 2.4 by assuming that the target is receding and thereby changing the polarity of v . 2.7 For an approaching (closing) target whose radial velocity is 300 m/s, find the Doppler shift and the unambiguous range when the PRF is 8000 pps and the transmitting frequency is 15 GHz. Solution: 3 9
0
82 (2)(300)(15 10 )30 kHz3 10dvffc= = =
83 1018.75 km2 (2)(800)u
rcRf= = =2.8 Consider an S -band pulsed radar transmitting 250 kW of peak power with a pulse width of 1.5
s and a PRF of 500 pps. The radar is transmitting at a frequency of 3000 MHz (a)Calculate the maximum unambiguous range of this radar, range resolution, and
duty factor
(b)Calculate the average transmitted power and the energy radiated in first 10 ms.
(c)Calculate the Doppler shift for a target approaching the radar with a radial velocity
of 30 m/s.
Solutions: (a) The maximum unambiguous range, range resolution, and duty factor are calculated
as follows:
83 10300 km2 2 (2)(500)u
rcT cRf= = = =
86(3 10 )(1.5 10 )2 225 m22cR− = = =
64(1.5 10 )(500) 7.5 10trdf−−= =  = 
(b) The average transmitted power and the energy radiated in first 10 ms.
34
46(250 10 )(7.5 10 ) 187.5 W
(total time)=(187.5)(10 ) 1.875 10 Jav t t
avP Pd
EP−= =   =
=  = 
(c) The Doppler shift is calculated as
9
0
82 (2)(30)(3 10 )600 Hz3 10r
dvffc= = =
2.9 Find an expression for the Doppler shift when the radial velocity of the approaching target is in km. Plot the Doppler frequency as a function of radial velocity of the target for various transmitted frequencies.

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