Fundamental Principles of Radar Solution Manual PDF

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Fundamental Principles of Radar System Solution Manual...

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111Equation Chapter 1 Section 1

Solutions Manual FUNDAMENTAL PRINCIPLES OF

RADAR

Habibur Rahman, Ph.D. Saint Louis University

Chapter 2: Radar Fundamentals Solutions 2.1 The distance of the moon from the radar transmitter located on the surface of the 8 earth is 3.84×10 m. Calculate the elapsed round trip time of a radar signal transmitted from radar the antenna. Solution: td 

2 R 2(3.84 108 )  2.56 s c 3 108

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: c 3 10 Ru   100 km 2 f r 2 1500 8

1 1  5μs B .2 106 c (3108 )(2 106) R    300 m 2 2 d t  f r (2 10  6 )(1500) 3 10  3



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 μs and the PRF of 200 Hz. (a) Find the average transmitted power. (b) Find the bandwidth and the range resolution of the radar. Solution: (a)

Pav Pt

 Pt f r (106 )(10 6) )(200) 200 W T

(b) 1 1 B    6 1 MHz  10 c (3 108 )(10  6 ) R    150 m 2 2 1

2.4 A pulsed radar has a PRF of 1500 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 distance of separation so that targets can be identified? Solution: (a) Ru 

cT c 3 10 8   100 km 2 2 fr (2)(1500)

(b) R 

c (3 108 )(15 10 6 )  2250 m 2 2

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 vr  c . Solution: (a) 2vf 0  c v  f0  f0    f d  f0  f0  c v c v fd 

(2)(300)(12 109) 24000.024 Hz 3108  300

(b) fd 

2vf 0 (2)(300)(12 109 )   24000 Hz c 3 108

2.6 Assuming that the target is receding (opening), derive the expression 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: 2

fd 

2vf 0 (2)(300)(15 109 )  30 kHz 3 10 8 c

Ru 

c 3108  18.75 km 2 f r (2)(800)

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: Ru 

cT c 3 10 8   300 km 2 2 f r (2)(500)

c (3 10 8)(1.5 10  6) R  2   225 m 2 2

dt  f r (1.5 10  6 )(500) 7.5 10  4 (b)

The average transmitted power and the energy radiated in first 10 ms. 4 3 Pav Pd t t (250 10 )(7.5 10 ) 187.5 W

E Pav (total time)=(187.5)(104 ) 1.875 106 J

(c)

The Doppler shift is calculated as fd 

2v r f 0 (2)(30)(3109 )  600 Hz c 3 108

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.

3

Solution: Starting with Eq. (2.18), obtain the desired Doppler shift and plot the resulting expression. 2.10

An L-band radar capable of transmitting a peak power of 500 W at 1000 MHz is designed to provide an unambiguous range of 100 km and range resolution of at least 100 m. a) Find the maximum required pulse width and the PRF. b) Find the average transmitted power.

Solution: (a) R 

Ru 

c 2 R (2)(100)   .6667μs 2 3 108 c

c c 3 108  fr   1.5 kHz 2 fr 2 Ru (2)(100 103 )

(b) 6 3 Pav Pt f r (500)(.6667 10  )(1.5 10 ) .5 W

2.11

An L-band radar operates at a frequency of 1500 MHz. Find the Doppler shift associated with an outbound target moving at the velocity of 100 m/s when the o o target velocity vector makes angle of 45 , and 90 with the radar line of sight. In each case, calculate the time dilation factor.

o Solution: For   45

fd 

2v cos (2)(100cos 45o )(1.5109 ) f0  707.11 HZ 3 10 8 c

Dilation Factor, DF 

c  v cos 45o 3108  70.71 299999929.29   .99999974 c  v cos 45o 3108  70.71 300000070.71

o and for  90

v r 100cos 90 o 0  f d 0, and DF 1 _____________________________________________________________________ _

4

Chapter 3: Radar Equations Solutions 3.1 Calculate the maximum gain of an X-band antenna operating at 8 GHz and having a diameter of 1 m. Repeat this problem with the diameter changed to 1.5, 2.0 m. Assume Ae   a A with a  1in each case. Solution: We have G 

4 Ae 2

D 1 m :

 3 10 8 8 10 9 0.0375 m, Ae  ( D 2)2  (1 2)2 0.785 m2 Gmax 

4 A e (4 )(0.785)  7014.55 38.46 dB (0.0375)2 2min

D 1.5 m :

  3 10 8 8 10 9  0.0375 m, Ae  (D 2)2  (1.5 2)2 1.767 m2 Gmax 

4 Ae (4 )(1.767)  41.98 dB 2 min (0.0375)2

D 2 m :

  3 10 8 8 10 9  0.0375 m, Ae  (D 2) 2  (2 2) 2  3.1416 m2 Gmax 

4 Ae (4 )(3.1416)  44.48 dB 2 (0.0375)2 min

3.2 Calculate the maximum gain of 2 m radius antenna operating in the L-, S-, and C  bands. Assume Ae  a A with  a 1in each case. Solution: The maximum gain of an antenna is directly proportional to the maximum value in corresponding band.

5

Gmax 

4 Ae 4 Ae 4 A 2 2   2 e  fmax  Hfmax 2 2 (c fmax )  c  min

4 A H   2 e  c L-band: S-band: C-band:

2    (4 )( (2 1)    16    4.386510 8 2 (3 10 )    

2  (4.386510 16 )(2 10 9 ) 2 32.44 dB f max  2 GHz  G max Hf max 2  (4.3865 10  16 )(4 10 9 ) 2 38.46 dB f max  4 GHz  G max Hf max 2 16 9 2 f max 8 GHz  G max Hf max (4.3865 10  )(8 10 ) 44.48 dB

3.3 Find the size of a circular aperture antenna of X-band radar operating at f 0 10 GHz G 30, 40, 50 dBs . Assume Ae   a A with  a  0.7 in each to attain case. Solution: We have G

 D2 G 02  4 A     D 0 A 2 0 4 4  

G

0.03 G ( ) 0.7

D 1.14 10 2 G

G 30 dB 10 3 : D 1.14 10  2 10 3 .360 m G  40 dB 104 : D 1.14 10 2 104 1.14 m G 50 dB 10 5 : D 1.14 10  2 10 5 3.60 m 3.4 An L-band radar operates at highest gain of 30 dB. The radar duty factor is 0.2 and the average power transmitted is 30 kW. Find the size of the antenna and the  power density at a range of 55 km. Assume  1 . Solution: It can be shown that  D 0 G 9   (3108 2 109 )  0.15 m   where 0 at f 2 10 GHz for obtaining maximum gain. Thus .15 1000  D 0 G 1.51 m    Now

6

Pav Pd t t  Pt 

Pav 300 103  1.5 MW 0.2 dt

Therefore pˆ t 

PG (1.5 10 6 )(10 3) t t   39.45 mW m2 4 R 2 4 (55103 )2

3.5 An L-band radar operating at frequency 1.5 MHz with an antenna of gain 36 dB is designed to obtain a single pulse minimum signal-to-noise ratio of 20 dB. Assume the receiver bandwidth of 4 MHz, RCS of 10 2, noise figure of 10 dB, and the maximum range of 120 km. Find the minimum detectable signal, the peak power, and the pulse width for this radar. Solution: We have G 36 dB 4000, SNR 20 dB 100, =

3 108 0.2 m, F 10 dB 10 1.5 109

The minimum detectable signal is

S min FKT 0B (SNR) min (10)(1.38 10  23)(290)(4 10 6)(100) 16 10  12 W The peak power and the pulse width of the radar Pt 

4 (4  ) 3 Rmax S min (4  ) 3 (120 103 )4 (16 10 12 )  1.03 MW (4000)2 (0.2) 2 (10) G 2 2



1 1  .25μs B 4 106

3.6 A C-band radar operating at a frequency of 6 GHz with an antenna having a gain of 50 dB transmits a peak power of 1.5 MW. Assume the receiver bandwidth of 5 MHz, the minimum output signal-to-noise ratio (SNR) min of 20 dB, and the radar cross section of 0.2 m2 for this radar system. Find the maximum range for the receiver noise figure of 5 dB and overall radar loss of 0 dB. Solution: From (3.18) we have Rmax

2 2   PG  t   3  (4 ) FKT0 BL(SNR) min 

1/4

where G 50 dB 10 5, (SNR) min 20 dB 10 2 ,  =

3 10 8 0.05 m, F 5 dB 3.1623, L 0 dB 1 6 109

Thus 7

  (1.5 10 6 )(10 5) 2(0.05) 2(0.2) R max    23 3 6 2   (4 ) (3.1623)(1.38 10 )(290)(5 10 )(1)(10 ) 

1/4

156.33 km

3.7 Consider a C- band radar operating at a frequency of 4.6 GHz that must provide a minimum received signal power of 10 -12 W. Assume that P t = 10 kW, the antenna  aperture area is 2.0 m2, aperture efficiency is a = 0.80, radar cross section is  = 2 m2, and overall loss is L = 5 dB. Calculate the maximum range. Solution: For this radar



3 10 8 4 A (4 )(0.8)(2.0)  0.0652 m, G= 2   4729.71, L  5 dB  3.1623  4.6 10 9 (0.0652) 2

The (3.10) can be modified by introducing the overall loss L as 2 2  PG2 2  PG   Pr  t 3 4  Rmax  t 3  (4 ) R  (4 ) Smin L 

Rmax

 (10 103 )(4729.71) 2 (0.0652) 2 (2.0)    (4 ) 3(10  12 )(3.1623)  

14

14

23.463 km

3.8 A C-band radar operating at a frequency of 4 GHz with an antenna having a gain of 45 dB transmits a peak power of 50 kW. Assume a total system loss of 2 dB. For a target located at a range of 100 km, find the minimum radar cross section to  12 produce an available received signal power of P r = 2 10 W. G  45 dB  31622.78, L  2 dB 1.585,   Solution: Here

3 108  0.075 m 4 109

Again by introducing the total system loss as in Problem 3.8, we can modify (3.10) as 2 2   min PG Pr (4 )3 R 4L (2 10 12 )(4 )3 (100 103 )4 (1.585) t Pr    min   2 2 PG  (4 )3 R 4L (50 103 )(31622.78)2 (0.075)2 t

 min  2.24 m2 3.9 An X-band radar employs the same circular aperture antenna for both transmission and reception at 8 GHz with its diameter of 3 m, antenna efficiency of 0.8. The  14 radar is designed to produce an average received power of 3 10 W when the 8

radar cross section is 1 m 2 at a maximum range of 100 km. If the total system loss is 3 dB, what transmitter peak power is required? Solution: We have 3 108   10 0.03 m, Ae  a ( D 2) 2 0.8 (3 2) 2 5.655 m 2 L 3 dB  2 10 4 A 4 (5.655) G  2e  78958.7  (0.03) 2

From (3.10): Pr 

2 2  PG Pr (4 )3 R 4L (3 10  14 )(4 )3 (100 103 )4 (2) t    P t (4 ) 3 R 4L (78958.7) 2 (0.03) 2 (1) G 2 2

Pt  2.122 kW 3.10

A millimeter-wave (MMW) radar uses a single antenna at 35 GHz to transmit a peak power of 650 W. The diameter of the antenna is 1.2 m, and antenna efficiency is 0.6. If a target at a range of 50 km has radar cross section of 10 m 2, calculate the available received power. Assume that the overall system loss is 0 dB.

Solution: Here 3 108 4 a ( D 2 4)  a 2 0.6 2  0.0086 m, G=    80067.1, L 0 dB 1 35 109 0.00862 2 2 Therefore, Pr 

3.11

2 2 PG (650)(80067.1)2 (.0086)2 (10)  t   2.48510 13 W 3 4 3 3 4 (4 ) R L (4 ) (50 10 ) (1)

A C-band monostatic radar operating at a frequency of 5.4 GHz transmits a peak power of 1 MW has the following parameters: total system loss L = 3 dB, R = 120 km,  = 1.2 m2, and the antenna has a circular aperture with aperture efficiency a 0.6 . Find the diameter of the antenna in order to produce an available receiver power

Pr 2.0 10  14

W.

Solution: L 3 dB 2,  

3 108 0.052 m 5.8 109

9

 (4 )3 Pr R4 L  G   2  Pt   

12

 (4 )3 (2.0 10  14 )(120 10 3) 4 (2)    (10 6)(0.052) 2(1.2)  

 G 2  4 a ( D2 4)   G D  2  2   a 

3.12

(a) (b)

12

 (2252.23)(0.052) 2   2  0.6  

12

2252.23

12

1.014 m 1 m

A high PRF airborne radar operating at a frequency of 10.5 GHz transmit a peak  power of 10 kW has the following parameters: pulse width 1.2  s , pulse repetition frequency PRF = 250 kHz, antenna gain G = 35 dB, radar cross section of the target  =10 m2, receiver noise figure F = 3 dB, and the overall system loss including the propagation path loss L = 5 dB. Find the maximum range at which the radar can detect the target if the minimum signal-to-noise ratio (SNR) for detection is 15 dB. Repeat part (a) for 0 dB SNR.

Solution: (a)

From (3.18):

Rmax

2 2    PG t   3  (4 ) FKT0BL (SNR ) min 

1/4

where 3 10 0.0286 m, G 35 dB 3162.27, L 5 dB 3.162, 9 10.5 10 1 1 F 3 dB 2, B    6 10 6  10 8



Then, for ( SNR ) min 15 dB 31.62,

Rmax

(b)

and for

4 2 2   (10 )(3162.27) (0.0286) (10)  3 6)   23  (4 ) (2)(1.38 10 )(290)(10 (3.162)(31.62) 

1/4

26.79 km

(SNR) min 0 dB 1, 1/4

  (10 4 )(3162.27) 2(0.0286) 2(10) R max    63.53 km  23 3 6)  (4 ) (2)(1.38 10 )(290)(10 (3.162)(1) 

10

3.13

A Doppler radar with a 1.3 m diameter antenna transmits 1.2 kW of power at a frequency of 3 GHz. The equivalent noise bandwidth is 1 kHz and the noise figure is 4.4 dB, and the overall loss factor is 10 dB. Assume a radar cross section of 10 m2 Find the signal-to-noise ratio at target ranges of 32 and 160 kms. Find the target range at unity signal-to-noise ratio.

(a) (b)

Solution: (a) where

2 2  So  PG  t   3 4 (4 ) R FKT 0BL N Using (3.16) gives  o 



3 108 0.1 m, F 4.4 dB 2.754, L 10 dB 10 3 109 (4)(3.1416)(1.3 2)2 (3.1416) 1667.97 G (0.1) 2

Thus for R = 32 km,  So   No

 (1.2 10 3)(1667.97) 2 (0.1) 2 (10)  1455.75  31.63 dB  3 3 4  23  (4 ) (32 10 ) (2.754)(1.38 10 )(290)(1000)(10)

and for R = 160 km,  So  (1.2 103 )(1667.97)2 (0.1)2 (10)  2.33  3.67 dB   3 3 4 23  N o  (4 ) (160 10 ) (2.754)(1.38 10 )(290)(1000)(10)

Ro4  (b)

Using (3.19): 2 2  PG   t R0   3  (4 ) FKT0BL 

3.14

2 2 PG  t 3 (4 ) FKT0BL

14

14

3 2 2   (1.2 10 )(1667.97) (0.1) (10)   3  23  (4 ) (2.754)(1.38 10 )(290)(1000)(10) 

197.66 km

A high PRF radar operating at 5.4 GHz transmits a peak power of 10 kW, and has the following parameters: antenna gain G = 20 dB, overall loss L = 10 dB, noise figure F = 3 dB, time on target Ti = 2.5 s, duty factor d t = 0.25, radar cross section σ = 0.02 m2. For target range R = 45 km, find the single pulse SNR.

Solution: We have

11



3 108 3 0.06 m, and Pav  Pd t t (10 10 )(0.25) 2.5 kW, 5 109 L 10 dB 10, F 3 dB 2

From (3.30) for a high PRF radar, the output SNR is given by (SNR)o  3.15

(a) (b)

(2.5 103 )(2.5)(100) 2 (0.06) 2 (0.02) PavTiG 2  2   6.90 8.39 dB (4 ) 3 R 4 FKT0L (4 ) 3(45 10 3 ) 4 (2)(1.38 10  23 )(290)(10)

Consider an X-band radar operating at 10 GHz with the following parameters:   o antenna gain G = 50,  a  e 3 , scan time Ts 3.0 s, overall system loss L = 5 dB, noise figure F = 4.41 dB, radar cross section σ = 0.1 m2, SNR = 12 dB, and the range R = 275 km. Find the power aperture product. Find the transmitted power corresponding to dt 0.3 .

Solution: We have 3 10 8 3 3  0.03 m,  =  0.00274 steradian, G  50 dB 105  9 10 10 (57.3)2 F 4.41 dB 2.758, SNR 12 dB 15.85, L  5 dB  3.162 (a)

From (3.35), we can write (SNR)o 

Pav A ts 16R 4FKT0L (SNR) o P A   av 16R 4 FKFT0 L  t s

The power-aperture product is thus calculated as (16)(275 10 3) 4(2.758)(1.38 10 23)(290)(3.162)(0.00274)(15.85)  36.65 dB PavA  (0.1)(0.3)

(b)

We have G

Pav  3.16

4 A G 2 (105 )(0.03) 2 A    7.162 m2 2 4 4

(PavA ) 9248.48 P 1291.33  1291.33 W  Pt  av  4.30 kW A 7.162 dt 0.3

A millimeter wave (MMW) search radar has the following specifications: P t = 5 W, 8 PRF = 12 KHz, pulse width  = 6 10 s, overall system loss L = 10 dB, circular 12

aperture antenna with diameter D = 0.3048 m, target RCS  = 25 m2, noise 25o 3o figure F = 6.17 dB, azimuth scan a , elevation scan e , and t s 3.5 s. (a) (b)

Find the power aperture product. Find the signal-to-noise ratio (SNR) to detect a target at a range of 10 km.

Solution: We have  D2   0.30482  2   A      .073 m 4 4    

(a)

The power aperture product is 8) 2 PavA Pt ( PRF )A (5)(6 10  (12000)(0.073) .000263 W-m

(b)

The SNR can be calculated using (3.35) as

( SNR) 

 where (SNR ) 

PavA ts 4 16 R FKT0 L 

(25  25)(3) .0457 steradian, F 6.17 dB 4.138 (57.3) 2

(0.000263)(25)(3.5)  19.00 12.79 dB (16)(10 10 ) (4.138)(1.38 10 23 )(290)(10)(0.0457) 3 4

3.17

A typical MMW search radar operating at a frequency of 94 GHz is used in a o o sector defined by 30 azimuth and 4 3.18 elevation scan, and has the following specifications: Antenna Gain 40 dB Antenna diameter 0.25 m Radar cross section 25 m2 System losses 10 dB Noise figure 3 dB Transmit peak power 5W Pulse width 40 ns Pulse repetition frequency 10 kHz (a) Find the detection range for a signal-to-noise ratio of 10 dB. (b) Find the antenna coverage rate and the time on-target (dwell time) if the coverage is obtained in a radar frame time of 6 seconds. (c) Find the number of integrated pulses. (d) Find the detection range when an integration loss of 3 dB is included. (e) Justify that it is below the maximum unambiguous range. Solution: For this radar we have, 13

 3 10 8 (94 10 9) 0.0031 m, G 40 dB 10 4, LS 10 dB 10, LI 3 dB 2,  (0.0031)(57.3) 0.71o , B 1 (40 109 )  25 106 Hz SNR=10 dB 10,  BW   D .25 (a)

The single pulse radar signal-to-noise ratio equation for this case can written using (3.16) as SNR 

2 2 2 2   PG PG 4 t t R   1 3 4 3 (4  ) R 1 FKT 0BL S (4  ) FKT 0BL S (SNR)

2 2    PG t R1    3 4  (4 ) R FKT0BLS (SNR)  `

(b)

14

4 2 2   (5)(10 ) (0.0031) (25)   3 6 23  (4 ) (2)(1.38 10 )(290)(25 10 )(10) 

14

2.345 km

o The angular coverage  (30  30) 5 300 . Then the antenna coverage rate...