Eqn (2-1). The duty cycle is the result of dividing the pulse width (1.5 ms) by the pulse repetition time (1/8000 Hz) or multiplying the pulse width by the pulse repetition frequency. (1.5 x 10-6 seconds) x (8 x 103 second-1) = 0.012 The duty cycle is the ratio of average power to peak power. Therefore, if the duty cycle is 0.012, and the peak power is 500 kilowatts, the average power can be obtained by multiplying the duty cycle by the peak power. (0.012)(500 x 103 watts) = 6 kilowatts. From Figure 2-1, the resting time is the difference be- tween the pulse repetition time (the reciprocal of the pulse repetition frequency) and the pulse width. Rest Time = 1/8000 Hz - 1.5 x 10-6 s = 1.24 x 10-4 s = 124 ms
Eqn (2-1) For pulse repetition time (PRT) substitute rest time plus pulse width (RT + PW). Solve for PW. PW = 6.18 ms
Eqn (2-3) Solve equation (2-3) for R to find the range indicated by the full width of the screen. R = 1/2(3.0 x 108 m/s)(900 x 10-6 s ) = 135 Km A deflection halfway across the indicator represents half of 135 km, or 67.5 km.
From appendix B: 1 knot = .508 m/s, so 200 kts = 10160 cm/s Use Eqn (2-4) Frequency shift = (2)(10160 cm/s)/(5cm) = 4.06 kHz for a course change of 45o, the velocity component in the closure direction becomes 10160cos(45o)=7184cm/s Frequency shift = 2(7184 cm/s)/(5 cm) = 2.87 kHz
Decrease, from (Eqn 2-4), if S is "negative", that is, going away, then the frequency shift is a negative value.
Range is measured by noting the round trip time for a pulse of radar energy. Since a CW radar transmits continuously, there is no means of measuring the round trip time and therefore no direct means of measuring range. The principle advantage of a CW radar is its inherent ability to recognize moving targets. It is safe to say that any target which represents some threat will also be in motion.
Since signal strength changes are more pronounced as the target moves toward the edge of the beam, the closer to the edge of the target is while being tracked the easier it is to detect a change in angle. Those large changes of signal strength occur closer to the axis in a narrow beam than in a wide beam resulting in better tracking accuracy. Wide beam. Because a narrow beam would cover too small an area in space to be useful in initially locating potential threats. A narrow beam is much better suited for target tracking and fire control solutions where accuracy is the main consideration. Search radars do not require any particular degree of resolution since their main function is to determine target presence only. Once this is accomplished, a narrow beam would be used for tracking.
In a broadside array, all elements are transmitting in phase. If each element were stimulated out of phase with its immediate neighbor, by an amount equal to the portion of a wavelength that separates adjacent elements, an end-fire array would result.
The parasitic reflector is employed in some radar antenna systems as a means of concentrating the radiated energy in a desired direction. Without a device to serve this purpose, an array type radar antenna would broadcast two equal main beams in opposite directions. When a conducting element is placed in an electric field, a voltage will be induced in the element. When this field varies, as around a radiating dipole, the variation in the induced voltage in the conductor (driven element) will cause it to radiate as well, but with a phase shift. Proper positioning of the driven element in the vicinity of the array will therefore set up interference which will tend to cancel the field in one direction and rein- force it in another, thereby allowing for a single, stronger main beam.
Single lobe systems are suitable for determining target bearing as long as no requirement exists for a high degree of accuracy. Once a strong signal is received, indicating a roughly centered beam, the single lobe system is at the limit of its value. This is because the teardrop shape of a radar beam makes it extremely difficult to fine tune; the antenna azimuth to the point of exact maximum return. A double lobe system does not require the determination of maximum return, but rather the azimuth at which the return signal is equal for each lobe. From an operator standpoint, it is considerably less difficult to match two signals where there is a constant reference than to determine the maximum of one signal where there is a variable reference. The double lobe system has an additional advantage in that, when the azimuth setting is close to correct, the return from each lobe is coming from nearer the side of the lobes where variations in signal strength are more pronounced. This makes it considerably less difficult to fine tune the antenna azimuth.
The greatest percentage of noise is generated with the electronic circuitry of the radar, particularly in the input to the receiver. A common source of this noise results from the random motion of electrons within electrical elements. Since there will be motion at any temperature above absolute zero, it is difficult to eliminate such noise (Johnson noise). The wider the bandwidth, the greater the degree of noise that will be input to the receiver. Since noise exists at all frequencies, the broader the frequency range to which the receiver is tuned, then the higher the intensity level if the noise and the lower the signal-to-noise ratio.
Use Eqn (2-3) For t, substitute the reciprocal of PRF, and solve for R. R = 1/2(3.0 x 108m/sec)(600 s-1)-1 = 250 km To determine the required PRF to achieved a maximum unambiguous range of 350 Km, substitute 350 km for R in Eqn (2-3) and solve for PRT. The required PRF is the reciprocal that results. PRF = [(2)(350x103cm)/(3.0x108m/sec)]-1 = 428.6 Hz
Use Eqn (2-3) Substitute PW for t. R = 1/2 (3.0 x 108m/s)(5.0 x 10-6 s) = 750 meters Assume a pulse compression ratio of 1.0. RRES = Rmin = 750 meters
Duty cycle is the ratio of average power to peak power. A low duty cycle would indicate low average power and high peak power. The low average power is desirable from the standpoint of equipment size and the high peak power is important for maximum return signal strength.
The probability of noise being generated in the same time space (i.e. Range) 10 or more times in a row is extremely low, therefore assuring returns are a target.
Use Eqn (2-5) Substitute -83 dBm for receiver sensitivity and solve for Smin. Smin = log-110[1/10(-83)]=5.012 x 10-9 mW = 5.012x10-12 W
Use Eqn (2-6) Substitute 3o for qB, 48o/sec for qS and 200 Hz for PRF. Solve for NB. NB = (3o)(200 sec-1)/(48o/sec) = 12.5 pulses
Use Eqn (2-8) Convert the beam widths to radius by dividing each value by 57.3o. Substitute the beam widths into Eqn (2-8) and solve for GD. GD = [(4)( p)]/[(2/57.3)(4/57.3)] = 5157.4
The radar cross-section of any target will vary with aspect and reflecting qualities. A plan form view of an A-4 will appear much larger than a nose-on view of an A-3, for example.
Use Eqn (2-17) Use Eqn (2-17) to find a ratio of values for Rmax under the two conditions. Under the new conditions, Pt(new) = 5Pt(old). All other factors remain the same and cancel. Ro(new)/Ro(old)=[5Pt(old)/Pt(old)]1/4 = 1.495 Therefore increasing Pt by a factor of 5 results in a 49.5 % increase in range.
Use Eqn (2-17) A 3dB loss in sensitivity equates to a revised sensitivity that's 1/2 the original sensitivity, or the revised Smin being twice the original Smin. As in problem 2-22, find a ratio of Rmax under the two conditions of Smin. Under the new conditions Smin(new) = 2 Smin(old). All other factors cancel. Rnew/Rold=[1/(2Smin(old)/1/Smin]1/4=(1/2)1/4=0.841. Therefore, the revised range is 84.1 Km
A 6dB improvement in sensitivity is a 4-fold improvement. Therefore, Smin(new) = 1/4 Smin(old) Rnew/Rold=[(lm2/1/4Smin(old)/(3m2/Smin(old)]1/4=1.074 f Rold=40 miles, Rnew=(1.074)(40 miles)=42.98 miles
Use Eqn (2-3) The maximum unambiguous range is 1/2 the distance that can be travelled by the radar energy during the time of one sweep from 400 to 800 MHz. Substitute 10 ms for t in Eqn (2-3) and solve for R. R = 1/2(3.0x108 m/sec)(10x10-6 sec) = 1.5KM.
The pulse-echo radar is capable only of measuring the elapsed travel time of individual pulses, however, where the Pulse-Doppler/MTI radars are capable of measuring a shift in the frequency of the RF energy in the return pulse and are therefore sensitive to target velocity.
Pulse-Doppler radars use filters to selectively pass Doppler frequencies. MTI radars use delay line and canceller to eliminate low/no Doppler frequency shifts.
f D = n PRF = n (4000 HZ) = 4000 HZ, 8000 HZ, 12000 HZ, etc.
Use Eqn (2-18) Find l using Eqn (1-1). l = (3.0x108 m/sec)/(9x109 sec-1)=3.33 cm Solve Eqn (2-18) for Vt = n (4000 Hz) (3.33x10-2 m)/2 = n 66.67 m/sec. Any integer multiple of 66.67 m/sec is undetectable by this system.
2000 KTS = 1016 m/s Let n = 1 Use Eqn (2-18) Substitute 1016 m/sec for Vt and 1 for n. Solve equations (2-18) for PRF. PRF=[(2)(1016 msec)]/[(1)(3.33x10-2 m)]=61.02 kHz
Use Eqn (2-3) Substitute the reciprocal of the PRF for t and solve for R. R=1/2(3.0x108 m/sec)(6.102 x 104sec-1)-1=2.46 km
Reduce frequency so that higher PRT's are possible.
Determine the angle between the fighter's course and its line of sight (LOS) to the target. (0o). Determine the angle between the target's course and the LOS from fighter to target (120o). The relative radial is determined by summing the velocity components along the LOS. S = Sret+Stgt = (600 m/s) cos(0o) + ( 300 m/s) cos(120o) = 450 m/s
Use Eqn (2-4) Substitute the relative radial velocity just obtained into Eqn (2-4) and solve for the change in frequency. (Use Eqn (1-1) to determine lambda). Frequency shift = (2)(450 m/sec)/0.375 m) = 2.4 kHz
Using Eqn (2-3, substitute PW for t, and solve for Rmin. Rmin = 1/2(3x108 m/sec)(3x10-6 sec) = 450m RRes = Rmax/PCR = 300 km/100 = 4.5 m
Millimeter systems approach "light" wavelengths and there fore can use "lens" to create beams. Microwave radars require larger antennas and therefore are harder to focus or create beams.
|frequency, f||5600 MHz|
|wave length, l||_____|
|pulse width, PW||1.3m sec|
|pulse repetition frequency, PRF||_____|
|pulse repetition time, PRT||_____|
|duty cycle||8.3 x 10-4|
|Antenna rotation rate, q||16 RPM|
|Horizontal beamwidth, qB||_____|
|vertical beam width, fB||4o|
|effective aperture, Ae||0.9 m2|
|power gain, G||3940|
|directive gain, GD||_____|
|number of returns per sweep, NB||9.9|
|minimum discernible signal, MDS||-83dBm|
|receiver sensitivity , Smin||_____|
|maximum unambiguous range, Runamb||_____|
|maximum theoretical range, Rmax||50 km|
|minimum range, Rmin||_____|
|range resolution, Rres||_____|
|radar cross-section, s||5 m2|
Use Eqn (1-1)
l= (3.0 x 108 m/sec)/(5600 x 106 Hz) = 5.36 cm
Use Eqn (2-1)
PRT = PW/DC = 1.3 msec/8.3x10-4 = 1.57x103 msec
PRF = 1/PRT = (1.57x10-3sec)-1 = 638.5 Hz
Hint, skip this and continue on to the end then solve.)
Use Eqn (2-17)
Rearrange Eqn (2-17) and solve for PT. (Eqn (2-17) requires Smin; that's why you had to skip this)
PT = (Smin)(4p R2)2/GAes = (5.012x10-12)[(4 p)(50x103m)2]2/(3940)(0.9)(5.0)
PT = 279KW
Solve Eqn (2-1) for the average power ( Pave ). Pave = Pt
DC = (279KW)(8.3 x 10-4) = 231.6W
Use Eqn (2-6) and (2-8)
Rearrange Eqn (2-6) to solve for qB, the horizontal beam width. Change the scan rate (16 RPM) to o/sec as follows:
scan rate =(16 revolutions/min)(360o/revolution) (60 sec/min)-1 = 96o/sec.
Solve Eqn (2-6) for q B:
q B=(NB)(scan rate)/PRF=(9.9)(96o/sec)/ (638.5 sec-1) = 1.49o
Solve Eqn (2-8) for GD, changing the beam widths to radians.
GD= (4p )/(4.0/57.3)(1.49/57.3) = 6929
Use Eqn (2-5) Substitute -83 dB for receiver sensitivity and solve for Smin = 10[1/10(-83)] = 5.012 x 10-12 W
Use Eqn (2-3) Substitute PRT for t, solve for R. R = 1/2(3.0x108m/sec)(1.57x10-3sec) = 235.5 km
Use Eqn (2-3) Substitute PW for t, solve for R. R = 1/2(3.0x108m/sec)(1.3x10-6sec) = 195 m
Assume no pulse compression. Rres = Rmin = 195m