George W. Stimson introduction to Airborne Radar (Se)

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So it is with the doppler shift in the signal a radar

receives from a target. Only in this case, phase is not shift-

ed through the arbitrary insertion or removal of increments

of time between wavefronts.

Rather, it is shifted as the result of the continuous change

in the time the radio waves take to travel from the radar to

the target and back

—

the change in the round-trip transit time.

Phasor Representation of Doppler Frequency. As with

so many radar concepts, the shift in phase may be most

easily visualized with phasors. The phase of the received

wave relative to the transmitted wave is portrayed by a sim-

ple phasor diagram in Fig. 6. Phasor T represents the trans-

mitted wave; phasor R, the received wave. (To make the

relationship between the two phasors easier to visualize,

we’ll assume that the radar transmits continuously, though

that is not necessary). At any one instant, the phase of the

received wave, R, lags that of the transmitted wave, T, by

the round-trip transit time, t

. If t

were zero, there would

be no lag, and the two phasors would coincide. If t

were

half the period of the transmitted wave, R would lag half a

revolution behind T.

Let us suppose, more realistically, that t

is 100,000

times the period of the transmitted wave plus some frac-

tion, φ (Fig. 7). The rotation of R, though 100,000 com-

plete revolutions behind the rotation of T, will be out of

phase with it by only the fraction of a complete revolution

(cycle), φ.

Now, if the transit time is constant (range rate = 0), the

phase lag, too, will be constant and the angle φ will remain

the same. The two phasors, therefore, will rotate at the

same rate. The frequencies of the transmitted and received

signals will be the same.

However, if the transit time decreases slightly, the total

phase lag will decrease, reducing the angle φ. If the

decrease continues (decreasing range), R will rotate coun-

terclockwise relative to T (Fig. 8). The frequency of the

received wave will be greater than that of the transmitted

wave.

Essentially, the same thing happens if the transit time

increases (positive range rate). The only difference is that

then the phase lag increases, and R (though still rotating

counterclockwise in absolute terms) rotates clockwise rela-

tive to T. The frequency of the received wave is less than

that of the transmitted wave (Fig. 8).

In either event, the difference in frequency between the

transmitted and received waves—the target’s doppler fre-

quency, f

—is proportional to the rate of change of φ.

CHAPTER 15 The Doppler Effect

193

6. Phase of received wave lags that of transmitted wave by

round-trip transit time.

7. If round-trip transit time is 100,000 times the period of the

transmitted wave plus a fraction φ, R will be out of phase with

T by only the fraction φ.

8. If range decreases,

φ will decrease, causing R to rotate counter-

clockwise relative to T, and so have a higher frequency.

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PART IV Pulse Doppler Radar

194

Equation for f

Derived. If the rate of change of the

phase angle, φ, is measured in whole revolutions per sec-

ond (1 revolution = 2π radians = 360˚ = 1 whole cycle per

second), the doppler frequency in hertz equals

φ. Since

phasor R makes one revolution relative to phasor T every

time the round-trip distance (d) to the target changes by

one wavelength (λ), the doppler frequency equals the rate

of change of d in wavelengths.

The minus sign accounts for the fact that, if d

˙ is negative

(closing target), the doppler frequency is positive.

Since d is twice the target’s range (d = 2R), the rate of

change of d (Fig. 9) is twice the range rate (d

˙ + 2R˙ ). The

target’s doppler frequency, therefore, is twice the range rate

divided by the wavelength.

= – 2

where

= doppler frequency, hertz

˙ = range rate, feet (or meters) per second

λ = transmitted wavelength, same units as R

Since wavelength equals the speed of light divided by the

frequency of the wave (Fig. 10), an alternative expression

for doppler frequency is

= – 2

˙ f

where f is the frequency of the transmitted wave and c is

the speed of light.

9. As illustrated by this simple mechanical analogy, the round-

trip distance from radar to target changes at twice the range

rate. If pulley A moves to right at rate R

., weight moves up at

rate d

, which is twice R

10. Reciprocal of wavelength (1/λ) is equal to f/c.

λ = c T

But T =

∴ λ = and =

DOPPLER SHIFT IN A NUTSHELL

For every half wavelength per second that a target’s range

decreases, the radio frequency phase of the received echo

advances by the equivalent of one whole cycle per second.

∴ f



–



– 2

λ/2 λ

where f

 doppler shift (positive for decreasing R)

R  radial component =of relative velocity

λ  wavelength

Doppler Frequency of an Aircraft

With either of these expressions, you can quickly and

accurately calculate the doppler frequency of any target for

any radar. Take an X-band radar (Fig. 11). Its wavelength is

0.1 foot. Suppose the radar is closing on a target at 1000

feet per second (R

= –1000 fps). The target’s doppler fre-

quency is (–2 x –1000) / 0.1 = 20,000 Hz, or 20 kHz.

If the wavelength were only half as long—0.05 foot,

instead of 0.1 foot—the same closing rate would produce

twice the doppler shift—40 kilohertz, instead of 20 kilo-

hertz.

The equations apply equally to targets whose range is

increasing. In this case, f

has a negative sign, signifying

that the radio frequency of the echoes is f

hertz less than

the transmitter frequency.

A simple rule of thumb for estimating doppler frequen-

cies for X-band radars is 1 knot of range rate produces 35

hertz of doppler shift (Fig. 12). By this rule, a target whose

closing rate is 600 knots would have a doppler frequency of

600 x 35 = 21 kHz. Turning the rule around, a target whose

doppler frequency is 7 kilohertz would have a range rate of

7000 ÷ 35 = 200 knots.

For other wavelengths, you simply scale the constants to

the wavelength: 10.5 hertz per knot for S-band (λ =10

cm); 21 hertz, for C-band (λ = 5 cm); etc.

Since for X-band (λ = 0.1 foot, another useful rule of

thumb is 1000 feet per second of range rate produces 20 kilo-

hertz of doppler shift.

A target’s range rate, of course, depends upon the veloci-

ties of both the radar and the target. For a radar approach-

ing a target head-on (Fig. 13a) the range rate is simply the

numerical sum of the magnitudes of the two velocities.

˙ = – (V

+ V

)

Consequently,

= – 2

⋅

= 2

+ V

λλ

For a target tail-on (Fig. 13b), the rate is the difference

between them. If the radar’s velocity is greater than the tar-

get’s, the range rate will be negative (decreasing range). If

the radar’s velocity is less than the target’s, the range rate

will be positive (increasing range). If the two velocities are

equal, the rate will be zero.

CHAPTER 15 The Doppler Effect

195

11. With this expression you can easily calculate the doppler fre-

quency of any target.

13a. For a target approaching nose-on, range rate is sum of the

magnitudes of aircraft velocities.

12. Rules of thumb for estimating doppler frequency.

DOPPLER FREQUENCIES FOR X-BAND

Closing Rate f

(Hz)

1 knot 35

1 mile/hour 30

1 kilometer/hour 19

1000 fps 20 X 10

= – 2 ( ) = 20 kHz

– 1000

0.1

13b. For tail-on approach, range rate is the difference between them.

PART IV Pulse Doppler Radar

196

For the more general case where the velocities are not

colinear, the range rate is the sum of the projections of the

radar velocity and the target velocity on the line of sight to

the target. As illustrated in Fig. 14, if the projection of the

target velocity is toward the radar, the range will be

decreasing. But if it is not, whether the range is decreasing

or increasing depends upon the relative magnitudes of the

two projections—as in the colinear tail-on case.

A target’s doppler frequency, therefore, can vary widely

depending on the operational situation. In nose-on

approaches, it is always high. In tail-on approaches, it is

generally low. In between, its value depends upon the look

angle and the direction the target is flying.

Doppler Frequency of Ground Return

The doppler frequency of the return from a patch of

ground is also proportional to the range rate divided by the

wavelength. The only difference: the range rate of a patch of

ground is due entirely to the radar’s own velocity (Fig. 15).

Therefore, the projection of the radar’s velocity on the

line of sight to the patch can be substituted for –R

˙ . For a

ground patch dead ahead, this projection equals the radar’s

full velocity, V

. For a ground patch directly to the side or

directly below, the projection is zero. In between, it equals

times the cosine of the angle, L, between V

and the line

of sight to the patch.

The doppler frequency of the return from a patch of

ground, therefore, is

= 2

cos L

where

= doppler frequency of ground patch, hertz

= velocity of radar, feet (meters) per second

L = angle between V

and line of sight to patch

λ = transmitted wavelength, same units as in V

Suppose, for example, the velocity and wavelength are such

that 2V

/λ = 10,000 and return is received from a patch at

15. Range rate of a ground patch is the magnitude of the projec-

tion of radar velocity on line of sight to patch.

14. In general, range rate of a target is sum of the magnitudes of

the projections of radar velocity and target velocity on line of

sight to target.

an angle of 60˚. The cosine of 60˚ being 0.5, the doppler

frequency is 10,000 x 0.5 = 5 kHz.

If the angle, L, is resolved into its azimuth and elevation

components, the term cos L in the above equation must be

replaced by the product of the cosines of the azimuth and

elevation angles of the patch (Fig. 16)

= 2

cos η cos ε

where

η = azimuth angle of patch

ε = lookdown angle of patch

As a rule, ground return is received, not from a single

small patch, but from a great many patches at a great many

different angles. The return therefore covers a broad spec-

trum of frequencies. This spectrum is discussed in Chap. 22.

Doppler Frequency Seen by a Semiactive Missile

A semiactive missile, you may recall, homes on the scat-

ter from a target which is illuminated by a radar carried in

the launch aircraft. Therefore, the doppler frequency of the

target as seen by the missile may be quite different from

that seen by the illuminating radar.

This is illustrated for a simple colinear case in Fig. 17.

The distance, d, from radar to target to missile changes at a

rate equal to the radar velocity plus two times the target

velocity plus the missile velocity (V

˙ = – (V

+ 2 V

+ V

)

The missile velocity, V

, equals the radar velocity plus

the incremental velocity of the missile relative to the radar

= V

+ ∆V

). With this substitution,

˙ = (2V

+ 2V

+ ∆V

)

The range rate of the target relative to the radar is

˙ = – (V

+ V

), and the doppler frequency is d˙/ λ.

Therefore, expressed in terms of relative velocities, the tar-

get’s doppler frequency as seen by the missile is

– 2R

˙ + ∆V

CHAPTER 15 The Doppler Effect

197

17. Doppler frequency of target as seen by semiactive missile is

proportional to rate of change of distance, d

, from radar to

target to missile.

16. Radar velocity V

, is projected onto line of sight to ground

patch in terms of azimuth angle, η, and depression angle, ε.

Projection of V

onto line of sight then equals V

cos η cos ε.

PART IV Pulse Doppler Radar

198

The foregoing equation, of course, applies only if the

velocities are all colinear and the missile is on the line of

sight from the radar to the target.

In the more general case, the rate at which the distance

from radar to target to missile changes equals the range rate

of the target relative to the radar plus the range rate of the

target relative to the missile (Fig. 18). The latter is the sum

of the projections of V

and – V

on the line of sight from

missile to target.

Initially, f

may be comparatively high. However, as the

attack progresses, f

may fall off considerably—particularly

if the missile is drawn into a tail chase, as shown in Fig. 19.

Summary

In the case of radar echoes, the doppler effect can be

visualized as the crowding (or spreading) of wavefronts due

to motion of the reflecting object relative to the radar. Since

frequency is tantamount to a continuous phase shift, the

resulting shift in frequency is equal to the rate (wavelengths

per second) at which the round-trip distance traveled by

the radio waves is changing—i.e., twice the range rate

divided by the wavelength.

Range rate of a moving target is determined by the veloc-

ities of the radar and target and by the angle of the line of

sight to the target relative to the direction of the radar’s

velocity. In nose-on approaches the range rate is usually

greater than the radar’s velocity; in tail-on approaches, less.

Range rate of a patch of ground is determined solely by

the radar’s velocity and the angle to the patch. Since

return may be received from ground patches in many

directions, the ground return generally covers a broad

band of frequencies.

• Doppler frequency of a target:

= – 2

Where R = range rate

λ = wavelength

• Doppler Frequency of a ground patch:

= – 2

Where V

= radar's velocity

L = look angle to the patch

• Doppler shifts at X band for common velocities:

1 knot = 35 Hz of doppler shift

1000 fps = 20 Hz

Some Relationships To Keep In Mind

cos L

19. Target’s doppler frequency as seen by missile may decrease

as attack progresses, particularly if missile is drawn into a tail

chase.

18. Rate of change of distance to missile is sum of range rate of

target relative to radar plus range rate of target relative to

missile.

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199

Spectrum of

Pulsed Signal

n Chap. 9, we considered the effect of pulsed transmis-

sion on a transmitter’s power output. But we did not

consider its crucial effect on the spectra of the transmit-

ted and received signals—i.e., on the distribution of their

energy over the range of possible radio frequencies.

It so happens that, if a radio wave of constant wavelength is

transmitted in short pulses, it can be detected by a receiver at

more than one radio frequency. As seen by the tuned circuit of

a receiver that can be tuned continuously over the complete

radio frequency spectrum, the wave’s power is spread over a

broad band of frequencies. This is true regardless of how

“sharply” the circuit is tuned.

On the surface, this behavior is perplexing. For if we plot

one or more complete cycles of the wave and measure the

spacing between zero crossings or even the rate of change of

the wave’s amplitude with time, we observe no change what-

soever in frequency as a result of having chopped the wave

into pulses. No matter how we define frequency—whether in

terms of wavelength, or period, or rate of change of phase

with time—as seen by us, the wave has only one frequency.

The reason for the difference between our perception and

the receiver’s will be explained in the next chapter. In this

chapter, we will accept that difference as a fact and concern

ourselves only with the nature of the spectrum observed by

the receiver. By performing a few simple experiments, we will

determine how the spectrum of a pulsed signal is influenced

by the pulse width, PRF, and duration of the signal. Along the

way, we will learn what coherence is and why it is so vital in a

doppler radar.

3. A train of independent pulses having a pulse width of 0.01

second and a constant PRF produces a receiver output that is

continuous over a band of frequencies 2000 hertz wide.

PART III Radar Fundamentals

200

Illustrative Experiments

To get a feel for the relationships in question, we will

perform a series of simple experiments—in our mind’s eye,

of course. All require just two pieces of equipment (Fig. 1).

One is a microwave transmitter, which for the initial

experiments consists simply of an oscillator. Its output sig-

nal, we’ll assume, has a constant amplitude and a constant,

highly stable wavelength. A key is provided with which we

can turn the transmitter on or off at any desired instant.

The other equipment is a microwave receiver with which

we can detect the transmitted signal, in much the same way

as one “tunes-in” a radio station on a broadcast receiver.

This receiver, however, is far more selective

and can be

tuned over a much broader band of frequencies. A meter

indicates the amplitude of the receiver’s output.

Bandwidth

To find what determines the bandwidth of a pulsed sig-

nal, we perform two experiments.

Experiment No. 1: CW Signal. In this, the control

experiment, we transmit a continuous wave at a frequency,

, and slowly tune—a hertz at a time—through the receiver’s

frequency range in search of the transmitted signal (Fig. 2).

As anyone might have predicted, the signal produces a

strong output from the receiver at a single point on our

hypothetical radio dial: the frequency f

. Though we search

the entire tuning range, we find no trace of the signal at any

other frequency. If we plot the amplitude of the receiver

output versus frequency, it appears as a vertical line.

Experiment No. 2: Stream of Independent Pulses. In our

second experiment, we periodically key the transmitter

“on” and “off” so that it transmits a continuous stream of

pulses having a constant PRF (Fig. 3). It should be noted,

however, that although the keying is as precise as we can

make it, the radio frequency phases of successive pulses are

not the same, but vary randomly from pulse to pulse.

Each pulse is exactly 1/1000th of a second long. While

1/1000th of a second (1000 microseconds) is a very short

time, bear in mind that it is on the order of a thousand

times longer than the pulses of a great many radars.

Because of the signal’s lower average power (the transmit-

ter is “on” only a fraction of the time), the receiver output is

not as strong as before. But it still occurs at the same point,

, on the dial. The plot of receiver output versus frequency,

however, is not quite as sharp as before. In fact, if we

expand it, we see that it is continuous over a band of fre-

quencies extending from 1000 hertz below f

to 1000 hertz

above it. The null-to-null bandwidth of 2 kilohertz.

1. To determine the effect of pulse modulation on radio frequen-

cy, we perform a series of simple experiments with a

microwave transmitter and receiver.

2. A continuous-wave signal produces an output from the receiv-

er only when it is tuned to a single frequency.

1. The receiver’s passband is

only one hertz wide; outside

this band, it’s sensitivity is

negligible.

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The signal also produces an output in a succession of

contiguous bands above and below this one. Within these

bands, which are half as wide as the central one, the output

is very much weaker, becoming more so, the farther the

bands are removed from f

. The plot of receiver output ver-

sus frequency (Fig. 4) has, in fact, the same sin x / x shape

as the radiation pattern of a uniformly illuminated linear-

array antenna. Although these spectral sidelobes are impor-

tant, for the time being will ignore them and concern our-

selves only with the central band.

Now, the width of this central band might be determined

by either the PRF or the pulse width, or by both.

To see if it is the PRF, we repeat the experiment at several

progressively lower PRFs. But, except for a reduction in

receiver output due to the lower duty factor, the receiver

output is unchanged. For a signal of the sort our simple

transmitter puts out, the PRF does not affect the spectrum.

Carrying this finding to a logical extreme in which the

interpulse period is stretched to days, we further conclude

that the spectrum of a single pulse is exactly the same as

that of a stream of independent pulses. Bandwidth, we con-

clude, is not determined by the PRF. What about pulse width?

To find the relationship between bandwidth and pulse

width, we repeat the experiment several times using pro-

gressively narrower pulses. The final pulse width is

1/1,000,000th of a second (1 microsecond).

The result of narrowing the pulses is striking. As the

pulse width decreases, the bandwidth increases tremen-

dously (Fig. 5). For the final pulse width—1 microsec-

ond—the band extends from 1,000,000 hertz below f

1,000,000 hertz above it. The total bandwidth, from null to

null, is 2 megahertz!

A frequency of 2 megahertz is 2 divided by 1 millionth

of a second. Similarly, a frequency of 2 kilohertz is 2 divid-

ed by 1/1000th of a second. Consequently, we conclude

that the null-to-null width of the spectral lobe of a stream

of independent pulses is

where

= null-to-null bandwidth

t = pulse width

The null-to-null bandwidth of a half-microsecond pulse,

for example, is 2 ÷ 0.5 µs = 4 MHz (Fig. 6).

But this raises a serious question. If at X-band the doppler

shift is only 35 kilohertz per thousand knots of closing rate,

the doppler frequencies encountered by most airborne

radars will be no more than a few hundred kilohertz.

CHAPTER 16 Spectrum of Pulsed Signal

201

4. Plot of receiver output versus frequency has sin x /x shape.

Sidelobes half the width of the central lobe and continuously

diminishing in amplitude extend above and below mainlobe.

5. For a 1 microsecond pulse width, the null-to-null bandwidth of

the central lobe is 2 megahertz.

6. The narrower the pulses, the wider the central spectral lobe.

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9. Pulses of master oscillator-power amplifier are in effect cut from

a continuous wave; hence are coherent.

PART III Radar Fundamentals

202

If a pulsed signal has a null-to-null bandwidth on the

order of a few megahertz—roughly 10 times the highest

doppler shift—then how can a pulsed radar ever detect

doppler frequencies? The answer is that it can’t, unless the

received pulses are, in some way, coherent.

Coherence

By coherence is meant a consistency, or continuity, in

the phase of a signal from one pulse to the next. There are

many forms of coherence. That used almost universally is

illustrated in Fig. 7. In it, the first wavefront in each pulse

is separated from the last wavefront of the same polarity in

the preceding pulse by some integral number of wave-

lengths. For example, if the wavelength is exactly 3 cen-

timeters: the separation may be 3,000,000 or 3,000,003 or

3,000,006 centimeters etc.; but not, say, 3,000,001 or

3,000,0033.15.

In the preceding experiment, you will recall, the pulses

were formed by keying the transmitter (oscillator) “on” for

each pulse. Although the keying was fairly precise, the

radio frequency phases of the individual pulses—their

“starting” phases—varied at random from pulse to pulse.

The transmitted signal was not coherent.

This is not surprising, when you consider that the period

of, say, an X-band signal is only 1/10,000th of a microsec-

ond and a degree of phase is only 1/3,600,000th of a

microsecond.

Achieving Coherence. With a somewhat more elaborate

transmitter, coherence can readily be achieved. The type of

transmitter most commonly used in doppler radars is called

a master oscillator–power amplifier (Fig. 8). It consists of an

oscillator, which produces a low-power signal of highly sta-

ble wavelength, and an amplifier, which amplifies the signal

to the power level needed for transmission.

The oscillator runs continuously; the power amplifier is

keyed on and off to produce the pulses. Although the key-

ing is no more precise than in the simple, noncoherent

(random starting phase) transmitter, the radio frequency

phases of successive pulses are exactly the same as if the

pulses had been cut from a continuous wave (Fig. 9). The

separation between the last wavefront in one pulse and the

first wavefront of the same polarity in the next pulse is thus

always exactly equal to a whole number of wavelengths.

The pulses are coherent.

Experiment No. 3: Effect of Coherence. To see what

effect coherence has on the bandwidth of a pulsed signal,

we perform Experiment No. 2 again. But this time, we use a

master oscillator–power amplifier transmitter.

7. Common form of coherence. First wavefront in second pulse

is separated from last wavefront of same phase in first pulse

by a whole number of wavelengths.

8. Coherent pulse train may be produced with master oscillator-

power amplifier. Oscillator runs continuously; amplifier is

keyed “on” to produce pulses.