- Calculate the duty cycle of a radar which transmits a 1.5
ms pulse at a PRF of 8 kHz. If the
peak power of this radar is 500 kilowatts, what is the average
power? What is the resting time?

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 10^{3}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 10^{3}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 - A pulsed radar has a duty cycle of .016. If the resting time
is 380 ms, what is the pulse width?
What is the PRF? What is the minimum range, in meters, of this
radar?

Eqn (2-1) For pulse repetition time (PRT) substitute rest time plus pulse width (RT + PW). Solve for PW. PW = 6.18 ms

- If a CRT is designed so that the spot traverses the indicator
face in 900 ms, what range could be
indicated by a spot deflection half way across the indicator?

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 10

^{8 }m/s)(900 x 10^{-6}s ) = 135 Km A deflection halfway across the indicator represents half of 135 km, or 67.5 km. - List the seven basic pulse radar components and briefly describe
the function of each.

- Synchronizer
- The timing unit of the system. As such it determines the PRF and coordinates the sweep on the sweep on the indicator with the firing of the transmitter.
- Transmitter
- Generates RF energy of the desired frequency and delivers it to the antenna system.
- Antenna System
- Broadcasts transmitter energy in desired pattern into space and receives return energy.
- Duplexer
- Consists of TR (Transmit-Receive) and a ATR (Anti-Transmit-Receive) device. Allows the use of a single antenna for Xmit and Receive.
- Receiver
- Amplifies weak return signal and transforms it into a form useful for dispaly on the indicator.
- Display
- Provides a method for presenting desired target information in a useful form for the operator.
- Power Supply
- Supplies power!

- A target is closing on a radial of a radar site with a relative
velocity of 200 knots. The radar transmits continuous wave energy
at a wavelength of 5 cm. What will the Doppler shift of the target
be? What will the Doppler shift be if the target alters its course
by 45
^{o}(closure speed reduced)?

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 45

^{o}, the velocity component in the closure direction becomes 10160cos(45^{o})=7184cm/s Frequency shift = 2(7184 cm/s)/(5 cm) = 2.87 kHz - Will the Doppler shift for an opening target indicate an increase
or decrease in frequency? Why?
Decrease, from (Eqn 2-4), if S is "negative", that is, going away, then the frequency shift is a negative value.

- Why can't a CW radar measure range? What is the main advantage
of a CW radar?

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.

- Why can a better angle measurement be achieved with a narrow
beam than a wide beam? Which would be more suitable for searching
for targets? Why?

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.

- What is the difference between a broadside and end-fire array?
Explain how the broadside array of Figure (2-17) could be modified
to become an end-fire array.

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.

- Explain the function of a parasitic reflector.

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.

- Discuss the advantages of a double lobe system over a single
lobe system in terms of azimuth determination.

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.

- List and briefly discuss three methods of determining target
elevation.

- Threshold-pickup A low-accuracy single lobe method which makes use of signals reflected from the earth's surface. The path length difference between the direct radar energy and the reflected radar energy causes cancellation at some points in space and reinforcement at others. The net result is a "layer of lobes" in space. This antenna would be fixed in elevation and an operator would have a fade chart to use which would give an elevation based upon range at which the first discernible signal appears.
- Signal-comparison Similar to threshold-pickup except that two lobes are utilized and the ratio of signal strengths as well as target range are used to determine altitude. This method is considerably more accurate than the threshold-pickup method.
- Tilted-antenna This method is similar to the single lobe method for determining azimuth. The antenna Œ is elevated such that no portion of the lobe is reflected from the surface. The elevation angle of the antenna at target acquisition and the range can be trigonometric- ally combined to determine target altitude.

- What is the primary source of radar noise? Discuss the
relationship between signal-to-noise ratio and bandwidth.

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.

- What is the maximum unambiguous range for a radar with
a PRF of 600 Hz? What PRF is required for maximum unambiguous
range of 350 km?

Use Eqn (2-3) For t, substitute the reciprocal of PRF, and solve for R. R = 1/2(3.0 x 10

^{8}m/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)(350x10^{3}cm)/(3.0x10^{8}m/sec)]-1 =*428.6 Hz* - What is the minimum range for radar with a PW of 5 ms?
What is the radar's range resolution?

Use Eqn (2-3) Substitute PW for t. R = 1/2 (3.0 x 10

^{8}m/s)(5.0 x 10^{-6 }s) = 750 meters Assume a pulse compression ratio of 1.0. R_{RES}= R_{min}=*750 meters* - Why is it advantageous to have a low duty cycle?

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.

- Why is there a requirement to have at least 10 return
pulses (or echoes) to rate the target as valid?

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.

- If the receiver has a receiver sensitivity of -83 dBm,
what is the value of the minimum discernible signal (S
_{min})?

Use Eqn (2-5) Substitute -83 dBm for receiver sensitivity and solve for S

_{min}. S_{min }= log-110[1/10(-83)]=5.012 x 10^{-9}mW = 5.012x10^{-12}W - Given an antenna beam width of 3", a scan rate of
48
^{o}/sec and a PRF of 200 Hz; how many pulses will be returned from a point target as the antenna scans through its beam width?

Use Eqn (2-6) Substitute 3

^{o}for q_{B}, 48^{o}/sec for q_{S}and 200 Hz for PRF. Solve for NB. NB = (3^{o})(200 sec-1)/(48^{o}/sec) = 12.5 pulses - What is the directive gain for an antenna with a horizontal
half power beam width of 2
^{o}and a vertical half power beam width of 4^{o}?

Use Eqn (2-8) Convert the beam widths to radius by dividing each value by 57.3

^{o}. Substitute the beam widths into Eqn (2-8) and solve for G_{D}. G_{D }= [(4)( p)]/[(2/57.3)(4/57.3)] = 5157.4 - Explain why the echo from an A-4 might be much stronger
than the echo from a larger aircraft at the same range.

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.

- Increasing the transmitter power of a radar by a factor
of 5 will increase the maximum range by what percent?

Use Eqn (2-17) Use Eqn (2-17) to find a ratio of values for R

_{max}under the two conditions. Under the new conditions, P_{t}(new) = 5P_{t}(old). All other factors remain the same and cancel. R_{o}(new)/R_{o}(old)=[5P_{t}(old)/P_{t}(old)]1/4 = 1.495 Therefore increasing P_{t}by a factor of 5 results in a 49.5 % increase in range. - For a given target a radar has a maximum range of 100
km. If the sensitivity of the radar receiver deteriorates by 3
dB, what is the degraded maximum range for that target?

Use Eqn (2-17) A 3dB loss in sensitivity equates to a revised sensitivity that's 1/2 the original sensitivity, or the revised S

_{min}being twice the original S_{min}. As in problem 2-22, find a ratio of R_{max}under the two conditions of S_{min}. Under the new conditions S_{min}(new) = 2 S_{min}(old). All other factors cancel. R_{new}/R_{old}=[1/(2S_{min}(old)/1/Smin]1/4=(1/2)1/4=0.841. Therefore, the revised range is 84.1 Km - A given radar with a S
_{min}= 10-9W can detect a target having a radar cross-section of 3 meter2 at a range of 40 miles. At what range can this same radar detect a 1 meter2 target if its Sensitivity is improved by 6 dB? Use Eqn (2-17)

A 6dB improvement in sensitivity is a 4-fold improvement. Therefore, S

_{min}(new) = 1/4 S_{min}(old) R_{new}/R_{old}=[(lm2/1/4Smin(old)/(3m2/Smin(old)]1/4=1.074 f Rold=40 miles, R_{new}=(1.074)(40 miles)=42.98 miles - A frequency modulated radar sweeps from 400 MHz to 800
MHz in 10 ms. What is the maximum unambiguous
range which can be measured by this radar?

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(3.0x10

^{8 }m/sec)(10x10^{-6}sec) = 1.5KM. - What is the main feature which differentiates pulsed
Doppler or MTI radars from pulse-echo radars?

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.

- Discuss the difference between pulse-Doppler and MTI
radars.

Pulse-Doppler radars use filters to selectively pass Doppler frequencies. MTI radars use delay line and canceller to eliminate low/no Doppler frequency shifts.

- A pulse-Doppler radar has a carrier frequency of 9 GHz
and a PRF of 4000 Hz.

- What are its "blind" Doppler frequencies?

f D = n PRF = n (4000 HZ) = 4000 HZ, 8000 HZ, 12000 HZ, etc.

- What radial target velocities would be undetected by
the radar?

Use Eqn (2-18) Find l using Eqn (1-1). l = (3.0x10

^{8}m/sec)/(9x10^{9}sec^{-1})=3.33 cm Solve Eqn (2-18) for V_{t}= 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. - What modification can be made to the radar that would
eliminate blind speeds below 2000 knots (use PRF)?

2000 KTS = 1016 m/s Let n = 1 Use Eqn (2-18) Substitute 1016 m/sec for V

_{t}and 1 for n. Solve equations (2-18) for PRF. PRF=[(2)(1016 msec)]/[(1)(3.33x10^{-2}m)]=61.02 kHz - What would the maximum unambiguous range of the modified
radar be?

Use Eqn (2-3) Substitute the reciprocal of the PRF for t and solve for R. R=1/2(3.0x10

^{8}m/sec)(6.102 x 10^{4}sec^{-1})-1=2.46 km - What additional modification(s) would allow for an increase
in maximum unambiguous range?

Reduce frequency so that higher PRT's are possible.

- What are its "blind" Doppler frequencies?
- A fighter aircraft in level flight on a heading of 090
^{o}T and at an airspeed of 600 m/sec is operating a pulse-Doppler radar at a frequency of 800 MHz. A target is detected at the same altitude, bearing 000^{o}R, heading 030^{o}T at a speed of 300 m/sec.

- What is the relative radial velocity between the fighter
and the target?

Determine the angle between the fighter's course and its line of sight (LOS) to the target. (0

^{o}). Determine the angle between the target's course and the LOS from fighter to target (120^{o}). The relative radial is determined by summing the velocity components along the LOS. S = S_{ret}+S_{tgt}= (600 m/s) cos(0^{o}) + ( 300 m/s) cos(120^{o}) = 450 m/s - What will the resulting Doppler shift be?

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

- What is the relative radial velocity between the fighter
and the target?
- A pulse radar propagates the pulse train shown below.
Each pulse has its frequency increased over the duration of the
pulse width resulting in a pulse compression ratio of 100:1. What
is the range resolution of this radar?

Using Eqn (2-3, substitute PW for t, and solve for R

_{min}. R_{min}= 1/2(3x10^{8}m/sec)(3x10^{-6}sec) = 450m R_{Res}= R_{max}/PCR = 300 km/100 = 4.5 m - Why is it possible to achieve a more narrow beamwidth
with a millimeter system than a microwave radar? (Microwave radar
frequencies cover 800-1000 MHz.)

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.

- The following table lists the characteristics of the
components of a pulse-echo type surface search radar. Using the
concepts presented in this chapter, complete this table.

frequency, f 5600 MHz wave length, l _____ pulse width, PW 1.3m sec pulse repetition frequency, PRF _____ pulse repetition time, PRT _____ peak power _____ average power _____ duty cycle 8.3 x 10 ^{-4}Antenna rotation rate, q 16 RPM Horizontal beamwidth, q _{B}_____ vertical beam width, f _{B}4 ^{o}effective aperture, A _{e}0.9 m ^{2}power gain, G 3940 directive gain, G _{D}_____ number of returns per sweep, N _{B}9.9 minimum discernible signal, MDS -83dBm receiver sensitivity , S _{min}_____ maximum unambiguous range, R _{unamb}_____ maximum theoretical range, R _{max}50 km minimum range, R _{min}_____ range resolution, R _{res}_____ radar cross-section, s 5 m ^{2}#### Wavelength

Use Eqn (1-1)

l= (3.0 x 10^{8}m/sec)/(5600 x 10^{6}Hz) = 5.36 cm#### PRT

Use Eqn (2-1)

PRT = PW/DC = 1.3 msec/8.3x10^{-4}= 1.57x10^{3}msec#### PRF

PRF = 1/PRT = (1.57x10

^{-3}sec)^{-1}= 638.5 Hz#### Peak Power

Hint, skip this and continue on to the end then solve.)

Use Eqn (2-17)

Rearrange Eqn (2-17) and solve for P_{T}. (Eqn (2-17) requires S_{min}; that's why you had to skip this)

P_{T}= (S_{min})(4p R^{2})2/GA_{e}s = (5.012x10^{-12})[(4 p)(50x10^{3}m)^{2}]2/(3940)(0.9)(5.0)

*P*_{T}= 279KW#### Average Power

Solve Eqn (2-1) for the average power ( P

_{ave}). P_{ave}= P_{t}

*DC = (279KW)(8.3 x 10-4) = 231.6W*#### Directive Gain

Use Eqn (2-6) and (2-8)

Rearrange Eqn (2-6) to solve for q_{B}, the horizontal beam width. Change the scan rate (16 RPM) to^{o}/sec as follows:

scan rate =(16 revolutions/min)(360^{o}/revolution) (60 sec/min)^{-1}= 96^{o}/sec.

Solve Eqn (2-6) for q_{B}:

q_{B}=(NB)(scan rate)/PRF=(9.9)(96^{o}/sec)/ (638.5 sec^{-1}) = 1.49^{o}

Solve Eqn (2-8) for G_{D}, changing the beam widths to radians.

*G*_{D}= (4p )/(4.0/57.3)(1.49/57.3) = 6929#### Minimum Discernible Signal

Use Eqn (2-5) Substitute -83 dB for receiver sensitivity and solve for S

_{min}= 10^{[1/10(-83)]}=*5.012 x 10*^{-12}W#### R

_{unamb}Use Eqn (2-3) Substitute PRT for t, solve for R. R = 1/2(3.0x10

^{8}m/sec)(1.57x10^{-3}sec) =*235.5 km*#### R

_{min}Use Eqn (2-3) Substitute PW for t, solve for R. R = 1/2(3.0x10

^{8}m/sec)(1.3x10^{-6}sec) =*195 m*#### R

_{res}Assume no pulse compression.

*Rres = Rmin = 195m*