Relationship Between Weight of Antenna and Transmitter in Minimum Total Weight for Fixed Range

1. If the weight of a transmitt4er is proportional to the transmitter power (i.e WT = KT Pt) and if the weight of an antenna is proportional to its volume (so that we can say its weight is proportional to the 3/2 power of the antenna aperture area) A or WA = KA A3/2), what is the relationship between the weight of the antenna and weight of the transmitter that makes the total weight W = WT + WA a minimum assuming fixed range?

2. A radar mounted on an automobile is to be used to determine the distance to a vehicle traveling directly in front of it. The radar operates at a frequency of 9375 MHz with a pulse width of 10 ns. The max. range is to be 500ft. • a) What is the pulse repetition frequency that corresponds to a range 500ft? • b) What is the range resolution (in meters )? • c) If the antenna beam width were 60, what would be the cross-range resolution (meters) at a range of 500fts.? • d) If the antenna dimensions were 1 ft by 1 ft and the antenna efficiency were 0.6, what would be the antenna gain (dB)? • e) Find the average power required to detect a 10 m2 radar cross-section vehicle at a range of 500ft, if the minimum detectable signal is 510-13W.

3. What is the highest freq. that a radar can be operated if it is required to have a maximum unambiguous range of 200 nmi and no blind speeds less than 600 Kt. • How can the transmission of N constant prf radar waveforms, each at different RF freq. be used to avoid blind speeds? Derive an expression for V1 /Vcf where (v1 is the first blind speed when N different RF frequencies are transmitted, all dt the same prf and Vcf is the blind speed when only a single RF frequency is transmitted equal to the average of N RF – frequencies.

4. v w r • If a rotating reflector antenna with dimension D, what is the relationship between the spread in doppler freq. due to finite time on target and the doppler freq. shift from the tip of the rotating antenna.? Antenna scanning vote in red/sec.

5. The cost of a radar is the cost of power plus the cost of aperture plus a constant. The cost of power is the cost /kilowatts multiplied by the no. of kilowatts; Cp = CKW NKW . And the cost of aperture is the cost/square meter multiplied by the no. of square meters : CA = Cm2 Nm2. Show that for minimum cost. • CP = CA • The anti-Ballistic Missile Defense Treaty of the strategic. Arms limitations accords with Russia limits radars at ABM sites in the two countries to 3106 watts-meter-squared of power –aperture product. Assuming such a radar surveyed are eighth of a hemisphere in 10 seconds, required an S/N of 10, had a system noise temperature of 1000 degrees k and system losses of 10, what would be its range against a 1 m2 target? ( Ans. 1200 Km). • F = ? • Scan time, ts=to /0, to =n/fp is the dwells time on the target • n-no. of pulses received as the antenna scans past the target • fp =PRF.

6.  – angular region to be scorched 0 – solid angular beamwidth  G e G = 4/0

7. Consider a transmitter with peak power of 1.5 KW with a gain of 10 dB when propagating at 3.0 GHz. Calculate (a) the magnitude of the signal received at room temperature by the receiver of 8 m2 aperture of the radar is upward looking, 20.9 dB for all extraneous search losses and the target of 1.5 m2 cross-section is viewed at about 100 Km away from the receiver. Consider a noise factor of 5 dB and noise bandwidth to match the receiver’s bandwidth. (b) a mismatch between the noise and receiver bandwidth was noticed during observation. If the noise bandwidth was given as 1 KHz, Calculate the magnitude of the signal received. (c ) If the radar scans at 100 rpm, calculate the time frame required for the scan. (d) calculate the antenna elevation beamwidth for an equal azimathal beamwidth. (e) How long can the radar dwell on the target?

10. 1. a. What should be the pulse repetition frequency of a radar in order to achieve a maximum unambiguous range of 60 nmi? b. How long does it take for the radar signal to travel out and back when the target is at maximum unambiguous range? c. If the radar has a pulse width of 1.5 s, what is the extent (in meters) of the pulse energy in space in the range coordinate? d. How for apart in range (meter) must two equal-size targets be separated in order to be certain they are completely resolved by a pulse width of 1.5 s? e. If the radar has a peak power of 800 KW. What is its average power? f. What is the duty cycle of this radar?

11. An airbone radar with two-way antenna beamidth of 2.90 in both axes and range resolution of 75 m (2.0 MHz BW) looks down at target of radar cross-section 1.0 m2. The grazing angle is 20 The minimum signal-to-clutter ratio for reliable detection is 20 dB. Find (a) the maximum detector range for a single hit on this target in sea clutter with 0 of-27dB. And (b) the max, detection range in the same clutter if signal processing applies an MTI-I or 35 dB.

12. A radar operates with a STALO at 5.535 GHz and a COHO at 320 MHz. The radar tracks a target which is moving radically outbound at 14.733 Smph (6586.2 m/s). Assume that all the frequencies given are exact, the single sideband modulator produces the upper sideband (the sum frequency) that the receive mixer produces a frequency which is the difference between the two inputs and that the velocity of propagation is 3108 m/s. Find the following frequencies • The transmit frequency • The received echo’s freq. • The freq. out of the RF signal processor. • The freq. out of the mixer. • The freq. out of the IF amplifier. • The rate of change of the signal out of the I/Q demodulator. • A second receiver input freq. which will produce exactly the same output from I/Q demodulator.

13. Ans. i) The trams. Freq. = freq. of STALO + COHO. = 5.855,000,000 GHz ii) The radial velo. Is 6586.2 m/s and the doppler shift is 257,080 Hz. Since the target is outbound, the doppler shift is negative ad the freq. received is less than that transmitted. The received freq. is 5.854,742, 920 GHz. iii) The RF signal processor dues not change the freq. of the signal upon which it acts. The output freq. is 5.854,742,920 GHz. iv) The freq. of the mixer output is the difference between its inputs from the RF signal processor one from the STALO, in that order. The mixer output freq. is 319.742,920 MHz. v) The IF amplifier does not change the signals freq. and its output is at 319.742, 920 MHz. vi) The output of the I/Q demodulator changes at the difference freq. between the IF amplifier output and the “c OHO. This freq. is -257,080 Hz or the Doppler shift. vii) Visualize and interfering signal whose freq. is the transmit freq. less twice the COHO freq. plus Doppler shift 9as though from an inbound target). The freq. would be 5.215,257,080 GHz. It will produce exactly the same I/Q output as the target in our example. It is called the image freq.

14. A gated CW waveform has a center freq. of exactly 1.250 GHz, a pulse width of 5.0 s and a PRF of 400 pps, find the matched bandwidth, the spacing of the spectral lines and describe the spectral envelope. Solution The BW of this wave is related to pulse width. The main spectral lobe extends from the centre freq. minus the pulse width reciprocal (1,250,000 GHz – 200 KHz = 1,249, 800 KHz) to the center freq. plus the reciprocal of the pulse width (1,250,200 KHz). The matched bandwidth is approx. Half the width of the main spectral lobe  or 200 KHz. The spectral lines are separated by PRF or 400 Hz. The five spectral lines closest to the center of the spectrum have. Frequencies of respectively, 1,249,999.2KHz. (center freq-2PRF0, 1,249,999.6 KHz (center –PRF), 1,250,000 KHz (center freq.), 1,250,000.4 KHz (center plus PRF) and 1,250,000.8 KHz (center plus 2 PRF). The envelope of the spectrum is a sine function. Sin[(f-1,250,000,000)0.000,005]/[(f-1,250,000,000)0.000,005].

15. Determine (a) the peak power (watts) and (b) the antenna physical area (m2 ) which make the cost of the following radar a minimum. • Freq. : 1230 MHz. • Antenna aperture efficiency = 0.6 . • Receiver minimum detectable sighnal : 3 10-13 W. • Units cost of transmitter : $2.20 per watt of peak power. • Unit cost of antenna : $1400 per square meter of physical area. • Cost of receiver and other items : $ 1,000,000 • The radar must detect a target of 2 m2 cross-section at a range of 200 nmi. • What is the cost of the antenna and the cost of the transmitter? • Pt = 1.89 MW, A = 54.58 m2 ).

16. T=1/fp T=2/fp t = 0 Rumamb • A radar at a frequency of 1.35 GHz has an antenna of width D = 32ft, a maximum unambiguous range of 220 nmi and an antenna scan time of 10s. • What is the no. of echo pulses per scan received by the radar from a point target? • A radar measures an apparent range of 7 nmi when the prf is 4000 Hz but it measures an apparent range of about 18.6 nmi when the prf is 3500 Hz. What is the true range (nmi)?

17. Rtrue = R1 or (R1 +Run ) or (R1 + 2 Run ) or R1 = apparent range , Run1 –unambiguous range for prf 1. R2 = apparent range, Run2 ………….for prf2. Rtrue = R2 +Run2 ) or (R2 +2Run2 ) or …..

18. 10 s pulse requires range gate of 1.5 Km. • At 30 Km, the physical area of sea illuminated is • roughly = 3104 1.5103 0.3 m2 = 135,000,00 m2 • Clutter RCS = area  reflectivity = 135105 10-3 m2 = 13500 m2 =41 dB m2 • The radar receiver must be able to detect noise signal 13 dB below the target (for 13 dB S/N ratio detections, so it must be capable of detecting -38 dB m2 signals (-13 dB -25 dB m2 ) • Dynamic range required by the receiver must be at least 938 dB m2 +41 dB m2 ) = 79 dB. • At 60 m, dynamic range reqd. is (79+3) =82 dB. • (Since clutter is 3 dB more and target size remains same).