George W. Stimson introduction to Airborne Radar (Se)

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Why Ghosts? Although we can tell from the continuity

in the plots of frequency versus time which frequency dif-

ferences belong to the same target, the continuity is not vis-

ible to the radar. As will be explained in detail in Chap. 18,

for a radar to discern small frequency differences such as

are normally encountered in FM ranging, it must receive

echoes from a target for an appreciable period of time. In

essence, all the radar observes are two frequency differences

at the end of the first segment and two (probably different)

frequency differences at the end of the second segment.

This is illustrated in Fig. 8. There, the first two differ-

ences are referred to as A and B; the second two, as x and y.

Without some further information it is impossible to tell for

sure how these differences should be paired—whether A

and x pertain to the same target or A and y.

Identifying Ghosts. In applications where ghosts are apt

to be encountered, they may be eliminated by adding

another segment to the modulation cycle—much as they

are by adding another PRF when PRF switching is used.

A representative three-slope cycle consists of equal

increasing and decreasing frequency segments—such as we

just considered—plus a constant-frequency segment

(Fig. 9). The latter, of course, provides a direct measure of

the targets’ doppler frequencies.

There is, however, no direct way of pairing the measured

doppler frequencies with A and B or x and y, either. But

knowing the doppler frequencies, the correct pairing of A

and B with x and y can quickly be found. Just as doppler

frequency cancels out when we add the frequency differ-

ences for positively and negatively sloping segments, so

transit time cancels out when we subtract the differences.

The result then is twice the doppler frequency.

∆f

= kt

+ f

– (∆f

= kt

– f

)

∆f

– ∆f

= 0 + 2f

Therefore, by subtracting A (or B) from x (or y) and

comparing the results with the measured doppler frequen-

cies, we can tell which of the two possible pairings is cor-

rect (Fig. 10). If, we say, (x – A) is twice one of the mea-

sured doppler frequencies, then the pairing should be as

follows:

x with A

y with B

Otherwise, y should be paired with A and x with B.

CHAPTER 14 FM Ranging

181

9. The problem is solved by adding a third segment in which

doppler frequencies are separately measured.

10. Knowing the two doppler frequencies, the radar can readily tell

whether x and A or y and A should be paired.

8. All the radar sees are two frequency differences at the end of

each segment. Radar has no way of telling whether A should

be paired with x or y.

X – A =2 f

20 – 10 =10

y – A =2 f

36 – 10 =26

GHOST

PART III Radar Fundamentals

182

Doppler Frequency Greater Than kt

. In the illustra-

tions shown so far, the doppler frequency has been less

than the frequency difference due to the ranging time,

. While this is true of applications such as altimetry, it

is not true of air-to-air applications. For these, the rate of

change of the transmitter frequency is generally made

low enough so that the maximum value of kt

will be

only a small fraction of the highest doppler frequency

normally encountered. In that case, a plot of the fre-

quency of the echoes from a closing target during the ris-

ing-frequency portion of the modulation cycle appears as

in Fig. 11.

The relationships between the measured frequency dif-

ferences for two or more simultaneously detected targets

can then be seen more clearly if the differences for each

segment of the cycle are plotted on separate horizontal

scales—one above the other—as in Fig. 12 (below). The

differences for the rising-frequency segment (A and B in

the figure) appear on the negative half of the frequency

scale; and the differences for the falling-frequency seg-

ment (x and y), on the positive half. The differences can

be paired by drawing horizontal arrows, between them,

of lengths corresponding to the doppler frequencies mea-

sured in the third segment of the cycle.

11. When doppler frequency (f

) is greater than frequency differ-

ence due to ranging time (kt

), echo frequencies are

higher

than transmitter frequency during the rising frequency

segment.

12. Relationships between frequency differences measured during rising and falling slopes can be seen more clearly if plotted on separate line

for each slope.

Click for high-quality image

In this case, we find that y is separated from A by two

lengths of the arrow, f

. These frequency differences, there-

fore, belong to the same target. The point where they abut

corresponds to the frequency difference due to the transit

time for the target, kt

A + fd2 = ktrA

Similarly, x is separated from B by two lengths of the

arrow, f

, and the point where they abut corresponds to

the frequency difference due to the transit time for the tar-

get, kt

Three Targets Detected Simultaneously. Figure 13

(below) plots the measured frequency differences for three

targets. They, too, can be paired easily. Comparing C with

x, y, and z, we find that it is separated from z, by 2f

There are still two possible combinations of A and B with x

and y: A with x and B with y, or A with y and B with x.

But, with C out of the way, we can readily tell which of

these are ghosts—just as we did when only two targets

were detected to begin with.

Certain combinations of ranges and doppler frequencies

may occur, however, for which more than one pairing of

A, B, and C with x, y, and z are possible. One of these is

CHAPTER 14 FM Ranging

183

13. When three targets are detected simultaneously, once one combination of frequency differences has been paired, the others may be paired

in the same way as when only two targets are detected.

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PART III Radar Fundamentals

184

illustrated in Fig. 14 (above). The frequency differences

shown there can readily be paired as follows.

A + 2f

= y

B + 2f

= x

C + 2f

= z

But a second pairing is also possible.

A + 2f

= x

B + 2f

= z

C + 2f

= y

The ranges indicated by one or the other of these pairings

are ghosts, and with only three PRFs, we cannot tell which.

As the number of simultaneously detected targets increases,

the number of these potential ghost-producing combina-

tions, though small, goes up.

They can be eliminated by adding more slopes to the

cycle. As with PRFs in pulse delay ranging, if N is the num-

ber of slopes, all possible combinations of N – 1 simultane-

ously detected targets can be deghosted. But the problem of

ghosts is generally much less severe with FM ranging. For,

in situations where it is normally used, neither range nor

doppler frequency is ambiguous.

14. With only three slopes, certain combinations of three targets will leave unresolved ghosts, but these combinations are rare.

Click for high-quality image

Recognizing that there will always be some possibility of

encountering unresolved ghosts, a three-slope modulation

cycle usually suffices.

Performance

The accuracy of FM ranging depends upon two basic

factors: the rate, k, at which the transmitter frequency is

changed, and the accuracy with which the frequency differ-

ences are measured.

The greater k is, the greater the frequency difference that

a given transit time will produce. The greater this difference

and the greater the accuracy with which frequency can be

measured, the more accurately the range will be deter-

mined.

Frequency Measurement Accuracy. This increases with

the length of time t

int

over which the measurement is

made—the length of the segments of the modulation cycle

(Fig. 15).

In search operation, the length of the segments is limited

by the length of time the antenna beam takes to scan across

a target: time on target, t

CHAPTER 14 FM Ranging

185

15.

Frequency measurement accuracy is limited by time-on-target,

16. In altimeters, slope of modulation curve can be made steep

enough to provide highly accurate range measurement.

Since the time-on-target is generally fixed by other

considerations, the steepness of the sloping segments of the

cycle—the rate, k—becomes the controlling factor for range

measurement accuracy.

Steepness of Slope, k. In applications such as low-alti-

tude altimeters, k can be made sufficiently high to provide

extremely precise range measurements (Fig. 16).

However, as will be explained in detail in Chap. 27, in

air-to-air applications, the value of k is severely limited. As

k is increased, ground return—which may be received

from ranges out to hundreds of miles—is smeared over an

increasingly broad band of frequencies. A point is quickly

reached, where the clutter blankets the targets, even though

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PART III Radar Fundamentals

186

the doppler frequencies of targets and clutter may be quite

different (Fig. 17).

Because of the limitations on k, in these applications FM

ranging is fairly imprecise. Whereas pulse-delay ranging

yields accuracies on the order of feet, FM ranging yields

accuracies on the order of miles.

Reduction in Detection Sensitivity. As with PRF switch-

ing, FM ranging reduces the detection sensitivity from what

it would be with no ranging. The reduction is primarily due

to the fact that if a target is to be detected at all, it must be

detected independently on each segment of the modulation

cycle. The time-on-target for each segment is the total time-

on-target divided by the number of segments. Also, each

time the slope of the transmitter frequency curve is

changed, the return from a considerable number of trans-

mitted pulses must be discarded.

In many situations, though, the reduction in detection

range is an acceptable price to pay for being able to range.

Summary

With FM ranging, the time lag between transmission and

reception is converted to a frequency shift. By measuring

this shift, the range is determined. Typically, the transmitter

frequency is changed at a constant rate. The change is con-

tinued over a considerable period of time so the frequency

difference can be accurately measured.

To cancel the contribution of the target’s doppler fre-

quency to the measured frequency difference, a second

measurement is made. This is done either while transmit-

ting at a constant frequency or while changing the transmit-

ter frequency in the opposite direction. The second mea-

surement is then subtracted from the first.

To resolve ambiguities occurring when two targets are

detected simultaneously, a third measurement may be

made.

For long range applications, FM ranging is more compli-

cated and generally less accurate than pulse-delay ranging

and reduces the radar’s detection range. But it enables the

high PRF waveform to be mechanized while still ranging.

17. For air-to-air applications, slopes must be made shallow to

avoid smearing the spectrum of the ground return. The result

is low accuracy.

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Doppler

Effect

189

1. A wave radiated from a point source when stationary

(a) and when moving (b). Wave is compressed in direction

of motion, spread out in opposite direction, and

unaffected in direction normal to motion.

y sensing doppler frequencies, a radar not only

can measure range rates, but also can separate tar-

get echoes from clutter, or produce high resolu-

tion ground maps. Since these are important

functions of many of today’s airborne radars, one of the

keys to understanding their operation is a good under-

standing of the doppler effect.

Accordingly, in this chapter, we will look at the doppler

shift more closely—first, in terms of the compression or

expansion of wavelength and, second, in terms of the con-

tinuous shift of phase. We will then pinpoint the factors

which determine the doppler frequencies of the return

from both aircraft and the ground. Finally, we will consider

the special case of the doppler shift of a target’s echoes as

observed by a semiactive missile.

Doppler Effect and Its Causes

The doppler effect is a shift in the frequency of a wave

radiated, reflected, or received by an object in motion. As

illustrated in Fig. 1, a wave radiated from a point source is

compressed in the direction of motion and is spread out in

the opposite direction. In both cases, the greater the

object’s speed, the greater the effect will be. Only at right

angles to the motion is the wave unaffected. Since frequen-

cy is inversely proportional to wavelength, the more com-

pressed the wave is, the higher its frequency is, and vice

versa. Therefore, the frequency of the wave is shifted in

direct proportion to the object’s velocity.

In the case of a radar, doppler shifts are produced by the

relative motion of the radar and the objects from which the

radar’s radio waves are reflected. If the distance between

PART IV Pulse Doppler Radar

190

the radar and a reflecting object is decreasing, the waves are

compressed. Their wavelength is shortened and their fre-

quency is increased. If the distance is increasing, the effect

is just the opposite.

With ground-based radars, any relative motion is due

entirely to movement of the radar’s targets. Return from

the ground has no doppler shift (Fig. 2). Differentiating

between ground clutter and the echoes of moving targets,

therefore, is comparatively easy.

With airborne radars, on the other hand, the relative

motion may be due to the motion of either the radar or the

targets, or both. Except in such aircraft as hovering heli-

copters, the radar is always in motion. Consequently, both

target echoes and ground return have doppler shifts. This

greatly complicates the task of separating target echoes from

ground clutter. A radar can differentiate between the two

only on the basis of differences in the magnitudes of their

doppler shifts.

Before discussing that, however, let’s take a close look at

how the shift actually occurs.

Where and How the Doppler Shift Takes Place

If both radar and target are moving, the radio waves may

be compressed (or stretched) at three points in their travel:

transmission, reflection, and reception.

The compression in wavelength occurring in the simple

case of a radar closing on a target, head-on, is illustrated in

Fig. 3 at the top of the facing page.

In these simplified diagrams, the slightly curved vertical

lines represent planes (viewed edge-on) at every point on

which the phase of the wave’s fields is the same. These

planes are called wavefronts. Those shown here, we’ll say,

are planes on which the fields have their maximum intensi-

ty in a positive direction—they represent wave “crests.”

Two successive wavefronts are shown at each of the points

in question. So that you can keep track of the wavefronts

easily, they are color coded—wavefront No. 1, red; wave-

front No. 2, blue.

For the sake of readability, the diagrams have not been

drawn to scale. In reading them, you must keep two

things in mind. First, the wavelength—spacing between

successive wavefronts of the same phase—is generally

only a small fraction of the length of the aircraft. Second,

since the speed of light is 162,000 nautical miles per sec-

ond, in a given period of time the aircraft would travel

only a minuscule fraction of the distance traveled by the

waves.

The diagram at top left in Fig. 3 illustrates the compres-

sion in wavelength occurring when a wave is transmitted.

2. With a ground-based radar, relative motion is due entirely to

the target’s motion. With airborne radar, it is due to motion of

both radar and target.

Click for high-quality image

The radar is a point A when it transmits wavefront No. 1.

By the time it transmits wavefront No. 2, it has advanced to

point B, decreasing the wavelength by a distance equal to

the velocity of the radar (V

) times the time between trans-

mission of the two wavefronts. That time, of course, is the

period of the wave (T). The space between wavefronts as

the wave travels out to the target, therefore, is (λ – V

T).

The diagram on the right in Fig. 3 illustrates the com-

pression occurring when the wave is reflected by the tar-

get. When wavefront No. 1 is reflected, the target is at

point D and wavefront No. 2 is at point C. By the time

wavefront No. 2 is reflected, the target has advanced to

point E, shortening the distance the wavefront has had to

travel from point C to reach the target by an amount equal

to the velocity of the target, V

, times the period, T.

Meanwhile, the reflection of wavefront No. 1 has traveled

an equal distance (D to F). But the target’s advance has

reduced the separation between this reflected wavefront

and the reflection of wavefront No. 2, which is just now

leaving the target, by V

The space between wavefronts of the reflected wave as it

travels back to the radar, therefore is λ –(V

+ 2V

)T.

The third diagram in Fig. 3 illustrates the reception of

the two wavefronts by the radar. The radar is at point G

when it receives wavefront No. 1. Wavefront No. 2 is one

compressed wavelength away. But by the time this wave-

front is received, the radar has advanced to point H. Thus,

CHAPTER 15 The Doppler Effect

191

3. Compression in wavelength occurs during transmission, reflection, and reception.

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PART IV Pulse Doppler Radar

192

during reception, the wavelength is still further compressed

by a distance V

T—the same as during transmission.

In all, the wavelength is compressed by twice the sum of

the two velocities times the period of the transmitted wave, T.

Total compression = 2(V

+ V

Since T is very short, the compression is extremely slight.

For an X-band radio wave and values of V

and V

of 600

knots, the compression is only about 5 millionths of an

inch. Nevertheless, since the radio frequency of an X-band

wave is very high (10 gigahertz), the resulting frequency

shift is more than 40 kilohertz.

Magnitude of the Doppler Frequency

Although we can get a physical feel for the doppler effect

by observing the compression in wavelength due to the rel-

ative motion of a radar and a target, we can calculate the

doppler frequency much more simply on the basis of the

shift in phase of the received wave.

Frequency, a Continuous Phase Shift. While not general-

ly though of in this way, a change in the frequency of a wave

is tantamount to a continuous shift in phase. This is illus-

trated in Fig. 4. It shows a one-second sample of two waves,

A and B. Their frequencies are 10 hertz and 11 hertz,

respectively. At 11 hertz, B completes one more cycle per

second than A. In other words, every second, the phase of B

relative to A advances 360˚. Since A completes 10 cycles every

second, the gain in phase per cycle of A is 360˚ ÷ 10 = 36 ˚.

If we wish to shift the frequency of B down to 10 hertz,

therefore, we can do so simply by inserting a time delay

equivalent to 36˚ of phase between successive wavefronts

(Fig. 5).

4. Wavefronts of two waves of slightly different frequency.

Frequency difference is tantamount to a continuous shift in

phase—here, 36˚ per cycle.

5. By inserting 36˚ phase shift between wavefronts, frequency is

decreased by 1 hertz from 11 hertz to 10 hertz. When

insertion is

discontinued, wave reverts to its original frequency.

As long as we continue shifting the wave’s phase, the fre-

quency shift will persist. But if we stop, B will revert to its

original frequency. By shifting phase in the opposite direc-

tion—i.e., decreasing the time between wavefronts, we can

similarly increase a wave’s frequency.