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Chapter 2: Radar Fundamentals
Solutions
2.1 The distance of the moon from the radar transmitter located on the surface of the earth is m. Calculate the elapsed round trip time of a radar signal transmitted from radar the antenna. Solution: 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: 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 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) (b)
2.4 A pulsed radar has a PRF of 1500 pps and transmit rectangular pulse train of duration 15. (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) (b) 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. Solution: (a) (b) 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. 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:
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) (b) 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 target velocity vector makes angle of , and with the radar line of sight. In each case, calculate the time dilation factor. Solution: For and for ______________________________________________________________________
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 with in each case. Solution: We have 3.2 Calculate the maximum gain of 2 m radius antenna operating in the L-, S-, and C-bands. Assume with in each case. Solution: The maximum gain of an antenna is directly proportional to the maximum value in corresponding band.
Therefore 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 The minimum detectable signal is The peak power and the pulse width of the radar 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 m^2 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 where
Thus 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 Pt = 10 kW, the antenna aperture area is 2.0 m^2 , aperture efficiency is = 0.80, radar cross section is = 2 m^2 , and overall loss is L = 5 dB. Calculate the maximum range. Solution: For this radar The (3.10) can be modified by introducing the overall loss L as 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 produce an available received signal power of Pr = W. Solution: Here Again by introducing the total system loss as in Problem 3.8, we can modify (3.10) as
with aperture efficiency. Find the diameter of the antenna in order to produce an available receiver power W. Solution: 3.12 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 , pulse repetition frequency PRF = 250 kHz, antenna gain G = 35 dB, radar cross section of the target =10 m^2 , receiver noise figure F = 3 dB, and the overall system loss including the propagation path loss L = 5 dB. (a) 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. (b) Repeat part (a) for 0 dB SNR. Solution: (a) From (3.18): where Then, for
(b) and for 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 m^2 (a) Find the signal-to-noise ratio at target ranges of 32 and 160 kms. (b) Find the target range at unity signal-to-noise ratio. Solution: (a) Using (3.16) gives where Thus for R = 32 km, and for R = 160 km, (b) Using (3.19):
(b) We have 3.16 A millimeter wave (MMW) search radar has the following specifications: Pt = 5 W, PRF = 12 KHz, pulse width = s, overall system loss L = 10 dB, circular aperture antenna with diameter D = 0.3048 m, target RCS = 25 m^2 , noise figure F = 6.17 dB, azimuth scan , elevation scan , and (a) Find the power aperture product. (b) Find the signal-to-noise ratio (SNR) to detect a target at a range of 10 km. Solution: We have (a) The power aperture product is (b) The SNR can be calculated using (3.35) as where 3.17 A typical MMW search radar operating at a frequency of 94 GHz is used in a sector defined by azimuth and 3.18 elevation scan, and has the following specifications:
Antenna Gain 40 dB Antenna diameter 0.25 m Radar cross section 25 m^2 System losses 10 dB Noise figure 3 dB Transmit peak power 5 W 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, (a) The single pulse radar signal-to-noise ratio equation for this case can written using (3.16) as
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(b) The angular coverage. Then the antenna coverage rate is degrees/s The time on-target or the dwell time is (c) The number of integrated pulses is
3.20 The radar in problem 3.17 is now subject to stand-off jammer (SOJ) with the following parameters: Pj = 200 W, Gj = 20 dB, Lj = 3 dB, =10 dB, and Rj =20 km. (a) Find the cross over range for a target of RCS σ = 5 m^2. (b) Find the detection range if the required SNR for detection is 10 dB. Solution: We have additional parameters, (a) From (3.58): (b) From (3.60): 3.21 Work Example 3.7 to find the signal-to-noise ratio at the missile of the bistatic system when the atmospheric attenuation of 0.08 dB/km during the propagation. Solution: From Example 3.7 we have SNR = 16.33 dB. Therefore, the SNR is decreased by the atmospheric attenuation as
3.22 In a bistatic radar the two stations use identical antennas at 40 GHz with
gain of 30 dB. In this system Pt = 55 kW , F = 1.66 dB , B = 10 MHz , total
loss , and (SNR)r = 13.01 dB, and target RCS σ = 5 m^2.
What are the target ranges Rt and Rr if it is found that Rt = 1.65 Rr?
Solution: We have
From (3.42): , We get
