10 range and doppler measurements in radar systems

1

Range & DopplerMeasurements

in RADAR Systems

SOLO HERMELIN

Updated: 09.10.08http://www.solohermelin.com

2

Range & Doppler Measurements in RADAR SystemsSOLO

Table of Contents

RADAR RF Signal

Radar Signals

Waveform Hierarchy

Doppler Effect due to Target Motion .( )0≠R

Continuous Wave Radar (CW Radar)

Basic CW Radar

Frequency Modulated Continuous Wave (FMCW)

Linear Sawtooth Frequency Modulated Continuous Wave

Sinusoidal Frequency Modulated Continuous Wave

Multiple Frequency CW Radar (MFCW)

Phase Modulated Continuous Wave (PMCW)

3


Table of Contents (continue – 1) Pulse Waves

Pulse Compression Techniques

Stepped Frequency Waveform (SFWF)

Phase Coding

Resolution

Range Measurement Unambiguity

Unambiguous Range and Velocity

Coherent Pulse Doppler Radar

4

RADAR RF SignalsSOLO

The transmitted RADAR RF Signal is:

( ) ( ) ( )[ ]ttftEtEt 0000 2cos ϕπ +=E0 – amplitude of the signal

f0 – RF frequency of the signal φ0 –phase of the signal (possible modulated)

The returned signal is delayed by the time that takes to signal to reach the target and toreturn back to the receiver. Since the electromagnetic waves travel with the speed of lightc (much greater then RADAR andTarget velocities), the received signal is delayed by

c

RRtd

21 +≅

The received signal is: ( ) ( ) ( ) ( )[ ] ( )tnoisettttfttEtE dddr +−+−−= ϕπα 00 2cos

To retrieve the range (and range-rate) information from the received signal thetransmitted signal must be modulated in Amplitude or/and Frequency or/and Phase.

ά < 1 represents the attenuation of the signal

5

SOLO

The received signal is:

( ) ( ) ( ) ( )[ ] ( )tnoisettttfttEtE dddr +−+−−= ϕπα 00 2cos

( ) ( ) tRRtRtRRtR ⋅+=⋅+= 222111 &

We want to compute the delay time td due to the time td1 it takes the EM-wave to reachthe target at a distance R1 (at t=0), from the transmitter, and to the time td2 it takes the EM-wave to return to the receiver, at a distance R2 (at t=0) from the target. 21 ddd ttt +=

According to the Special Relativity Theorythe EM wave will travel with a constant velocity c (independent of the relative velocities ).21 & RR

The EM wave that reached the target at time t was send at td1 ,therefore

( ) ( ) 111111 ddd tcttRRttR ⋅=−⋅+=− ( )1

111 Rc

tRRttd

+⋅+=

In the same way the EM wave received from the target at time t was reflected at td2 , therefore

( ) ( ) 222222 ddd tcttRRttR ⋅=−⋅+=− ( )2

222 Rc

tRRttd

+⋅+=

RADAR RF Signals

6

SOLO


( ) ( ) ( ) ( )[ ] ( )tnoisettttfttEtE dddr +−+−−= ϕπα 00 2cos

21 ddd ttt += ( )1

111 Rc

tRRttd

+⋅+= ( )

2

222 Rc

tRRttd

+⋅+=

( ) ( )2

22

1

1121 Rc

tRR

Rc

tRRtttttttt ddd

+⋅+−

+⋅+−=−−=−

+

−+−+

+

−+−=−

2

2

2

2

1

1

1

1

2

1

2

1

Rc

Rt

Rc

Rc

Rc

Rt

Rc

Rctt d

From which:

or:

Since in most applications we canapproximate where they appear in the arguments of E0 (t-td), φ (t-td),however, because f0 is of order of 109 Hz=1 GHz, in radar applications, we must use:

cRR <<21,

1,2

2

1

1 ≈+−

+−

Rc

Rc

Rc

Rc

( )

−⋅

++

−⋅

+=

−⋅

−⋅+

−⋅

−⋅≈− 2

.

201

.

1022

011

00 2

1

2

1

2

121

2

121

21

D

RalongFreqDoppler

DD

RalongFreqDoppler

Dd ttffttffc

Rt

c

Rf

c

Rt

c

Rfttf

( ) ( ) ( ) ( ) ( )[ ] ( )tnoisettttffttEtE ddDdr +−+−⋅+−≈ ϕπα 00 2cos

where 212

21

1212

021

01 ,,,,2

,2

dddddDDDDD tttc

Rt

c

Rtfff

c

Rff

c

Rff +=≈≈+=−≈−≈

Finally

Doppler Effect

RADAR RF Signals

7

SOLO

The received signal model:

( ) ( ) ( ) ( ) ( )[ ] ( )tnoisettttffttEtE ddDdr +−+−⋅+−≈ ϕπα 00 2cos

Delayed by two-way trip time

Scaled downAmplitude Possible phase

modulated

CorruptedBy noise

Dopplereffect

We want to estimate:

• delay td range c td/2

• amplitude reduction α

• Doppler frequency fD

• noise power n (relative to signal power)

• phase modulation φ

Table of Content

RADAR RF Signals

8

RADAR SignalsSOLO

Waveforms

( ) ( ) ( )[ ]tttats θω += 0cos

a (t) – nonnegative function that represents any amplitude modulation (AM)

θ (t) – phase angle associated with any frequency modulation (FM)

ω0 – nominal carrier angular frequency ω0 = 2 π f0

f0 – nominal carrier frequency

Transmitted Signal

( ) ( ) ( )[ ]{ }ttjtats θω += 0exp

Phasor (complex, analytic) Transmitted Signal

9

RADAR SignalsSOLO

Quadrature Form( ) ( ) ( )[ ]

( ) ( )[ ] ( ) ( ) ( )[ ] ( )tttattta

tttats

00

0

sinsincoscos

cos

ωθωθθω

−=+=

where: ( ) ( ) ( )[ ]( ) ( ) ( )[ ]ttats

ttats

Q

I

θθ

sin

cos

==

( ) ( ) ( ) ( ) ( )ttsttsts QI 00 sincos ωω −=

One other form: ( ) ( ) ( )[ ] ( ) ( ) ( )[ ]tjtjtjtj eeta

tttats θωθωθω −−+ +=+= 00

2cos 0

( ) ( ) ( )[ ]tjtj etgetgts 00 *

2

1 ωω −+= ( ) ( ) ( ) ( ) ( )tjQI etatsjtstg θ=+=:

Complex envelope

10

RADAR SignalsSOLO

Spectrum

Define the Fourier Transfer F

( ) ( ){ } ( ) ( )∫+∞

∞−

−== dttjtstsS ωω exp:F ( ) ( ){ } ( ) ( )∫+∞

∞−

==πωωωω

2exp:

dtjSSts -1F

( ) ( ) ( )[ ]tjtj etgetgts 00 *

2

1 ωω −+= ( ) ( ) ( )[ ]0*

02

1 ωωωωω −−+−= GGS-1FF

-1FF

( ) ( ) ( ) ( ) ( )tjQI etatsjtstg θ=+=:

( ) ( ) ( )[ ]tttats θω += 0cosInverse Fourier Transfer F -1

Complex envelope

11

RADAR SignalsSOLO

Energy ( ) ( ) ( )[ ]tttats θω += 0cos

( ) ( ) ( )[ ]{ } ( )∫∫∫+∞

∞−

+∞

∞−

+∞

∞−

≈++== dttadttttadttsEs2

022

2

122cos1

2

1: θω

Parseval’s Formula

Proof:

( ) ( ) ( ) ( )∫∫+∞

∞−

+∞

∞−

= ωωωπ

dFFdttftf 2*

12*

1 2

1

( ) ( ) ( )∫+∞

∞−

−= dttjtfF ωω exp11

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∫∫ ∫∫ ∫∫+∞

∞−

+∞

∞−

+∞

∞−

+∞

∞−

+∞

∞−

+∞

∞−

=−=−=πωωω

πωωω

πωωω

22exp

2exp 2

*

112*

2*

12*

1

dFF

ddttjtfFdt

dtjFtfdttftf

( ) ( ) ( )∫+∞

∞−

−=πωωω

2exp*

2

*

2

dtjFtf

If s (t) is real, than s (t) = s*(t) and

( ) ( ) ( )∫∫∫+∞

∞−

+∞

∞−

+∞

∞−

=== ωωπ

dSdttsdttsEs

222

2

1:

12

RADAR SignalsSOLO

Energy (continue – 1) ( ) ( ) ( )[ ]tttats θω += 0cos

( ) ( ) ( )∫∫∫+∞

∞−

+∞

∞−

+∞

∞−

=== ωωπ

dSdttsdttsEs

222

2

1:

( ) ( ) ( ) ( )[ ] ( ) ( )[ ]( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

−−−+−−−+

−−−−+−−=

−−+−−−+−=

−

−−

00

0000

0

*

0

*2

00

0

*

00

*

0

00

*

0

*

0

*

4

1

4

1

ϕϕ

ϕϕϕϕ

ωωωωωωωωωωωωωωωω

ωωωωωωωωωω

jj

jjjj

eGGeGG

GGGG

eGeGeGeGSS

For finite band signals (see Figure)

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )∫∫∫

∫∫∞+

∞−

∞+

∞−

∞+

∞−

+∞

∞−

−+∞

∞−

=−−−−=−−

≈−−−=−−−

ωωωωωωωωωωωωω

ωωωωωωωωωω ϕϕ

dGGdGGdGG

deGGdeGG jj

*

0

*

00

*

0

2

0

*

0

*2

00 000

( ) ( )∫∫+∞

∞−

+∞

∞−

≈= ωωπ

ωωπ

dGdSEs

22

2

1

2

1

2

1:

Table of Content

13

SOLO Waveform Hierarchy

Radar Waveforms

CW Radars Pulsed Radars

FrequencyModulated CW

PhaseModulated CW

bi – phase & poly-phase

Linear FMCWSawtooth, or

Triangle

Nonlinear FMCWSinusoidal,

Multiple Frequency,Noise, Pseudorandom

Intra-pulse Modulation

Pulse-to-pulse Modulation,

Frequency AgilityStepped Frequency

FrequencyModulate Linear FM

Nonlinear FM

PhaseModulatedbi – phase poly-phase

Unmodulated CW

Multiple FrequencyFrequency

Shift Keying

Fixed Frequency

14


( )tf

2

τ2

τ−

A

∞→t

2τ+T

2

τ−T

A

2

τ+−T2

τ−−T

A

t←∞−

T TA

t

A

t

A

LINEAR FM PULSECODED PULSE

T T

PULSED (INTRAPULSE CODING)

t

( )tf

A

2

τ2

τ−T

AA

T T

A

22

τ+T2

2τ−T

A

T T

A

2

τ− 2

τ+T

TN

t

( )tf

A

2

τ2

τ−T

AA

T T

A

22

τ+T2

2τ−T

A

T T

A

2

τ− 2

τ+T

TN

PHASE CODED PULSES HOPPED FREQUENCY PULSES

PULSED (INTERPULSE CODING)

t

( )tf

A

T

2/τ−

LOW PRFMEDIUM PRF

PULSED( )tf

T T T T

2/τ+

τ

HIGH PRF

TT T T

A Partial List of the Family of RADAR Waveforms

15

Range & Doppler Measurements in RADAR SystemsSOLORadar Waveforms and their Fourier Transforms

16

Range & Doppler Measurements in RADAR SystemsSOLORadar Waveforms and their Fourier Transforms

17


Table of Content

18

Doppler Effect due to Target Motion .

SOLO

ΔT – time from point A to travel from radar to target at range R0 (at transmission time t0) is

c

TRRT

∆+=∆

0

Rc

RT

−=∆ 0

Total round-trip is 2ΔT . Therefore point A returns to radar at

Rc

RTT

−+= 0

01

2

Point B returns to radar atRc

RTTT RF −

++= 102

2

where andRFRF

RF fT

ωπ21 ==

RFTRRR += 01

( )

−+=

−+=

−−+=−=

Rc

RcT

Rc

TRT

Rc

RRTTTT RF

RFRFRF

22

:' 0112

λ

λ Rf

c

Rf

cRcR

fT

f RF

fc

RF

c

R

RF

RF

221

1

1

'

1'

/1

−=

−≈

+

−==

=<<

λR

fDoppler2−=

Two WayDoppler Frequency Shift

( )0≠R Range & Doppler Measurements in RADAR Systems

19



( ) ( ) ( ) ( )[ ] ( )

( ) ( ) ( )tnoisec

RRtRRtftE

tnoisettttfttEtE

fc

dddr

+

+−++−=

+−+−−==

2121

0000

/

00

22cos

2cos

00

ϕλππα

ϕπαλ

If we consider only (c = speed of light) then the frequency of the electromagneticwave that reaches the receiver is given by:

ctd

Rd <<

+

−≈

+

+−=

+−+

+−=

c

td

Rd

td

Rd

f

c

tdRd

tdRd

ud

dff

c

RRt

c

RRtf

td

df

21

0

21

0~

00

21210

1

2

1

22

1

ϕπ

ϕππ

λ

+

−=tdRd

tdRd

fd

21

is the doppler frequency shift at the receiver

Christian Johann Doppler first observed the effect in acoustics.

20

2-Way Doppler Shift Versus Velocity and Radio Frequency SOLO

Table of Content

λλ logloglog −=⇒−=td

Rdf

td

Rdf dd

21

SOLO

• Transmitter always on

• Range information can be obtained by modulating EM wave [e.g., frequency modulation (FM), phase modulation (PM)]

• Simple radars used for speed timing, semi-active missile illuminators, altimeters, proximity fuzes.

• Continuous Wave Radar (CW Radar)

Table of Content

22

SOLO • Continuous Wave Radar (CW Radar)

The basic CW Radar will transmit an unmodulated (fixed carrier frequency) signal.

( ) [ ]00cos ϕω += tAtsThe received signal (in steady – state) will be.

( ) ( ) ( )[ ]00cos ϕωωα +−+= dDr ttAtsα – attenuation factor

ωD – two way Doppler shiftc

RfRff

fc

DDD

0

/ 22&2

0

−=−===λ

λπω

The Received Power is related to the Transmitted Power by (Radar Equation):

4

1~RP

P

tr

rcv

One solution is to have separate antennas for transmitting and receiving.

For R = 103 m this ratio is 10-12 or 120 db. This means that we must have a good isolation between continuously transmitting energy and receiving energy.

Basic CW Radar

23

SOLO • Continuous Wave Radar (CW Radar)

The received signal (in steady – state) ( ) ( ) ( )[ ]002cos ϕπα +−⋅+= dDr ttffAts

We can see that the sign of the Doppler is ambiguous (we get the same result for positiveand negative ωD).

To solve the problem of doppler sign ambiguity we can split the Local Oscillator into two channels and phase shifting theSignal in one by 90◦ (quadrature - Q) with respect to other channel (in-phase – I). Both channels are downconverted to baseband.If we look at those channels as the real and imaginary parts of a complex signal, we get:

has the Fourier Transform: ( ){ } ( ) ( )[ ]DDv ts ωωδωωδπ ++−=F

After being heterodyned to baseband (video band), the signal becomes (after ignoring amplitude factors and fixed-phase terms): ( ) [ ]tts Dv ωcos=

( ) ( ) ( )[ ] tjDDv

Detjtts ωωω2

1sincos

2

1 =+= ( ){ } ( )Dv ts ωωδπ −=2

F

Table of Content

24

SOLO • Frequency Modulated Continuous Wave (FMCW)The transmitted signal is: ( ) ( )[ ]00cos ϕθω ++= ttAts

The frequency of this signal is: ( ) ( )

+= t

dt

dtf θω

π 02

1

For FMCW the θ (t) has a linear slope as seen in the figures bellow

Table of Content

25

SOLO • Frequency Modulated Continuous Wave (FMCW)


( ) ( ) ( ) ( )[ ]00cos ϕθωωα +−+−+= ddDr ttttAts

α – attenuation factor

( ) ( )

−++= dDr ttdt

dfftf θ

π2

10

ωD – two way Doppler shiftλ

πω Rff DDD

2&2 −==

td – two way time delay

c

Rtd

2=

The frequency of received signal is:

λ – mean value of wavelength


26


To extract the information we must subtract the received signal frequency fromthe transmitted signal frequency. This is done by mixing (multiplying) those signalsand use a Lower Side-Band Filter to retain the difference of frequencies

( ) ( ) ( ) ( ) ( ) Ddrb fttdt

dt

dt

dtftftf −

−−

=−= θ

πθ

π 2

1

2

1The frequency of mixed signal is:


( ) ( )[ ]00cos ϕθω ++= ttAts

( ) ( ) ( ) ( )[ ]

( ) ( ) ( ) ( )[ ]ddD

ddDdr

ttttttA

ttttttAts

−++−+++

−−+−−=

θθωωωα

θθωωα

002

02

cos2

1

cos2

1

Lower Side-BandFilter

Lower SB Filter


27


The returned signal has a frequency change due to:

• two way time delayc

Rtd

2=

• two way doppler additionλR

fD2−=

From Figure above, the beat frequencies fb (difference between transmitted to received frequencies) for a Linear Sawtooth Frequency Modulation are:

Dm

Ddm

b fRTc

fft

T

ff −∆=−∆=+ 4

2/

Dm

Ddm

b fRTc

fft

T

ff −∆−=−∆−=− 4

2/

( )28

−+ −∆

= bbm ff

f

TcR ( )

2

−+ +−= bbD

fff

We have 2 equations with 2 unknowns R and fD

with the solution:


28

SOLO• Frequency Modulated Continuous Wave (FMCW)


For R = 103 m this ratio is 10-12 or 120 db. This means that we must have a good isolation between continuously transmitting energy and receiving energy.

4

1~RP

P

tr

rcv

One solution is to have separate antennas for transmitting and receiving.


29

SOLO • Frequency Modulated Continuous Wave (FMCW)Linear Sawtooth Frequency Modulated Continuous Wave

Performing Fast Fourier Transform (FFT) we obtain fb+ and fb.

( )28

−+ −∆

= bbm ff

f

TcR

( )2

−+ +−= bbD

fff

From the Doppler Window we get fb+ and fb

-, from which:

30




α – attenuation factor

( ) ( )

−++= dDr ttdt

dfftf θ

π21

0

ωD – two way Doppler shiftλ

πω Rff DDD

2&2 −==

td – two way time delay

c

Rtd

2=

The frequency of received signal is:

λ – mean value of wavelength

Linear Triangular Frequency Modulated Continuous Wave

31


To extract the information we must subtract the received signal frequency fromthe transmitted signal frequency. This is done by mixing (multiplying) those signalsand use a Lower Side-Band Filter to retain the difference of frequencies

( ) ( ) ( ) ( ) ( ) Ddrb fttdt

dt

dt

dtftftf −

−−

=−= θ

πθ

π 2

1

2

1The frequency of mixed signal is:


( ) ( )[ ]00cos ϕθω ++= ttAts

( ) ( ) ( ) ( )[ ]

( ) ( ) ( ) ( )[ ]ddD

ddDdr

ttttttA

ttttttAts

−++−+++

−−+−−=

θθωωωα

θθωωα

002

02

cos2

1

cos2

1


Lower SB Filter


32


The returned signal has a frequency change due to:

• two way time delayc

Rtd

2=

• two way doppler additionλR

fD2−=

From Figure above, the beat frequencies fb (difference between transmitted to received frequencies) for a Linear Triangular Frequency Modulation are:

Dm

Ddm

b fRTc

fft

T

ff −∆=−∆=+ 8

4/

positiveslope

Dm

Ddm

b fRTc

fft

T

ff −∆−=−∆−=− 8

4/

negativeslope

( )28

−+ −∆

= bbm ff

f

TcR ( )

2

−+ +−= bbD

fff

We have 2 equations with 2 unknowns R and fD

with the solution:


33


The Range Unambiguity is given bythe FMCW time period Tm:

Range Resolution is a function of FMCW bandwidth and the linearity of FM:

msunambiguou Tc

R2

=

To preserve this Range Resolution the non-linearity must be:

For Linear Triangular FMCW the bandwidth is: fB ∆= 2

For a perfect Linear Triangular modulation the Range Resolution is given by:

f

c

B

cR

∆== 2δ

mmm

sunambiguou TfTBTcBc

R

Rtynonlineari

∆===<<2

11

2

2δ


34

SOLO• Frequency Modulated Continuous Wave (FMCW)


For R = 103 m this ratio is10-12 or 120 db. This means that we must have a good isolationbetween continuously transmittingenergy and receiving energy.

4

1~RP

P

tr

rcv

But solutions with a commonantenna for transmitting andreceiving, and with a goodisolation between them, do exist.

One solution is to have separateantennas for transmitting andreceiving.

35


One Target Detected

Performing FFT for the positive slope we obtain fb

+.

Performing FFT for the negative slope we obtain fb

-.

( )28

−+ −∆

= bbm ff

f

TcR

( )2

−+ +−= bbD

fff

Two Targets Detected

Performing FFT for each of the positive and negative slopes we obtain two Beats in each Doppler window and we cannot say what is the pair in the other window. A solution to solve this is to add an unmodulated segment (see next slide)

36

SOLO • Frequency Modulated Continuous Wave (FMCW)Two Targets Detected

Performing FFT for each of the positive, negative and zero slopes we obtain two Beats in each Doppler window.

To solve two targets we can use the Segmented Linear Frequency Modulation.

In the zero slope Doppler window, we obtain the Doppler frequency of the two targets fD1 and fD2.Since , it is easy to find the pair from Positive and Negative Slope Windows that fulfill this condition, and then to compute the respective ranges using:

( )2

−+ +−= bbD

fff

( )28

−+ −∆

= bbm ff

f

TcR

This is a solution for more than two targets.

One other solution that can solve also range and doppler ambiguities is to use manymodulation slopes (Δ f and Tm).

Table of Content

37


Sinusoidal Frequency Modulated Continuous Wave One of the practical frequency modulations is the Sinusoidal Frequency Modulation.

Assume that the transmitted signal is:

( ) ( )

∆+= tff

ftfAts m

m

ππ 2sin2sin 0

The spectrum of this signal is:

( ) ( )

( )[ ] ( )[ ]{ }

( )[ ] ( )[ ]{ }

( )[ ] ( )[ ]{ }

+

−++

∆+

−++

∆+

−++

∆+

∆=

tfftfff

fJA

tfftfff

fJA

tfftfff

fJA

tff

fJAts

mmm

mmm

mmm

m

32sin32sin

22sin22sin

2sin2sin

2sin

003

002

001

00

ππ

ππ

ππ

πwhere Jn (u) is the Bessel Functionof the first kind, n order and argument u.

Bessel Functions of the first kind

38


Sinusoidal Frequency Modulated Continuous Wave One of the practical frequency modulations is the Sinusoidal Frequency Modulation.

Assume that the transmitted signal is:

( ) ( )

∆+= tff

ftfAts m

m

ππ 2sin2sin 0

The transmitted and received signal are heterodyned in a mixer to give the differencefrequency


( ) ( ) ( ) ( )[ ]

−∆+−⋅+= dmm

dD ttff

fttffAtr ππα 2sin2sin 0


( )ts

( )tr( )[ ] ( )[ ] ( )

∆−−∆+−+ tf

f

fttf

f

fttftfA m

mdm

mdDd πππα 2sin2sin2cos 0

2

( )[ ] ( )

−∆−−+=

22cossin

22cos 0

2 dmdm

mdDd

ttftf

f

fttftfA πππα

39



Since td << Tm=1/fm we have

( ) ( )[ ] ( )

−∆−−+=

22cossin

22cos 0

2 dmdm

mdDd

ttftf

f

fttftfAtm πππα


( )ts

( )tr

( )tm

( ) ( )[ ]

−∆−−+≈

22cos22cos 0

2 dmddDd

ttftfttftfAtm πππα

The frequency is obtained by differentiating the argument of this equation with respect to time

( )[ ]

( )

−∆+=

−∆−−+=

22sin2

22cos22

2

10

dmmdD

dmddDdb

ttfftff

ttftfttftf

td

df

ππ

ππππ

( ) mm

dmm f

dmdmD

f tfd

mmdD

m

f

b

m

b

ttftfffdt

ttfftff

f

dtf

f

f2

1

0

2

1

0

12

1

0 22cos2

22sin2

211

211

−∆−≈

−∆+== ∫∫

< <

ππππ

The average of the beat frequency over one-half a modulating cycle is:

40




( )ts

( )tr

( )tm

Rc

ffftffff mDdmD

tf

b

dm ∆+=∆+≈=< <

+

84

1π


( ) ( )

∆+= tff

ftfAts m

m

ππ 2sin2sin 0

By changing the phase of the sinusoidal modulationby 180 degree each modulation cycle, we will get:

( ) ( )

∆−=− tff

ftfAts m

m

ππ 2sin2sin 0

Rc

ffftffff mDdmD

tf

b

dm ∆−=∆−≈=< <

−

84

1π


41




( )ts

( )tr

( )tmR

c

ffftffff mDdmD

tf

b

dm ∆+=∆+≈=< <

+

84

1π

A possible modulating is describe bellow, in which we introduce a unmodulated segmentto measure the doppler and two sinusoidal modulation segments in anti-phase.

From which we obtain:

Rc

ffftffff mDdmD

tf

b

dm ∆−=∆−≈=< <

−

84

1π

The averages of the beat frequency over one-half a modulating cycle are:

28−+

−∆

= bbmff

f

TcR

2−+

+= bb

D

fff

(must be the same as in unmodulated segment)

Note: We obtaind the same form as for Triangular Frequency Modulated CW

Table of Content

42

SOLO • Multiple Frequency CW Radar (MFCW)

Assume that the transmitted signal is: ( ) [ ]tfAts 02sin π=The received signal is: ( ) ( ) ( )[ ]dD ttffAtr −⋅+= 02sin πα

c

Rt

c

Rff dD

2,

210 ≈−≈

( ) ( )

⋅−⋅−⋅+=

c

Rf

c

RftffAtr DD

22

222sin 00 πππα

where:

Therefore:

We can see that the change in received phase Δφ is related to range R by:

2/2

22

22

22

/

00

00

λππππϕ

λ R

c

Rf

c

Rf

c

Rf

cfff

D

D =>>

=⋅≈⋅+⋅=∆

The maximum unambiguous range is given when Δφ=2π : 2/λ=sunambiguouR

( ) ( )GHzfmmBandLGHzfcm 956.1115 00 =÷→==λ

We can see that the maximum unambiguous range is too small, when we use a single transmitted frequency, for any practical applications.

43


Assume that the transmitted signal is: ( ) [ ]tfAts 02sin π=The received signal is: ( ) ( ) ( )[ ]dD ttffAtr −⋅+= 02sin πα

c

Rt

c

Rff dD

2,

210 ≈−≈

( ) ( )

⋅−⋅−⋅+=

c

Rf

c

RftffAtr DD

22

222sin 00 πππα

where:

Therefore:

We can see that the change in received phase Δφ is related to range R by:

2/2

22

22

22

/

00

00

λππππϕ

λ R

c

Rf

c

Rf

c

Rf

cfff

D

D =>>

=⋅≈⋅+⋅=∆

The maximum unambiguous range is given when Δφ=2π : 2/λ=sunambiguouR

( ) ( )GHzfmmBandLGHzfcm 956.1115 00 =÷→==λ

We can see that the maximum unambiguous range is too small, when we use a single transmitted frequency, for any practical applications.

The maximum unambiguous range can be increased by using multiple transmitted frequencies.

44

SOLO

Assume that the transmitter transmits n CW frequencies fi (i=0,1,…,n-1)

Transmitted signals are: ( ) [ ] 1,,1,02sin −== nitfAts iii πThe received signals are: ( ) ( ) ( )[ ]dDiiiii ttffAtr −⋅+= πα 2sin

c

Rt

c

Rf

c

Rfff d

i

jjDi

2,

2210

10 ≈−≈

∆+−≈ ∑

=

where:

1,,2,11 −=∆+= − nifff iii

Since we want to use no more than one antenna for transmitted signals and one antenna for received signals we must have

1,,2,101

−=<<∆∑=

niffi

jj

We can see that the change in received phase Δφi , of two adjacent signals, is related to range R by:

( )c

Rf

c

R

c

Rf

c

Rf

c

Rff

c

Rf i

cR

iiDDii ii

22

222

22

22

22

2

1⋅∆≈⋅⋅∆+⋅∆=⋅−+⋅∆=∆

<<

−πππππϕ

The maximum unambiguous range is given when Δφi=2π :

isunambiguou f

cR

∆=2

• Multiple Frequency CW Radar (MFCW)

45


Table of Content

46

SOLO • Phase Modulated Continuous Wave (PMCW)

Another way to obtain a time mark in a CW signal is by using Phase Modulation (PM).PMCW radar measures target range by applying a discrete phase shift every T secondsto the transmitted CW signal, producing a phase-code waveform. The returning waveformis correlated with a stored version of the transmitted waveform. The correlation processgives a maximum when we have a match. The time to achieve this match is the time-delaybetween transmitted and receiving signals and provides the required target range.

There are two types of phase coding techniques: binary phase codes and polyphase codes. In the figure bellow we can see a 7-length Barker binary phase code of the transmittedsignal

47

SOLO • Phase Modulated Continuous Wave (PMCW)

In the figure bellow we can see a 7-length Barker binary phase code of the receivedsignal that, at the receiver, passes a 7-cell delay line, and is correlated to a sampleof the 7-length Barker binary signal sample.

Digital CorrelationAt the Receiver the coded pulse enters a7 cells delay lane (from left to right),a bin at each clock.The signals in the cells are summed

-1 = -1

+1 -1 = 0

-1 +1 -1 = -1

-1 -1 +1-( -1) = 0

+1 -1 -1 –(+1)-( -1) = -1

+1 +1 -1-(-1) –(+1)-1= 0

+1+1 +1-( -1)-(-1) +1-(-1)= 8

+1+1 –(+1)-( -1) -1-( +1)= 0

+1-(+1) –(+1) -1-( -1)= -1

-(+1)-(+1) +1 -( -1)= 0

-(+1)+1-(+1) = -1

+1-(+1) = 0-(+1) = -1

0 = 0

-1-1 -1

clock

123456789

1011121314

+1+1+1+1

Table of Content

48


• Pulse Waves

• Range Resolution is determined by the system bandwidth

B

ccR

B

FilterMatched 22

/1 ττ ===∆

200 MHz 1 meter 325 MHz 2 feet 650 MHz 1 foot1300 MHz 6 inch

• Use short pulse (τ) for high resolution, or, bandwidth can be achieved by:

• Pulse Compression – intra-pulse coding

• Frequency Modulated Continuous Wave (FMCW)

• Stretch Processing

• Stepped Frequency Waveform (SFWF) – pulse-to-pulse coding

Table of Content

49


• Pulse Compression Techniques• Wave Coding

• Frequency Modulation (FM)

- Linear

• Phase Modulation (PM)]

- Non-linear

- Pseudo-Random Noise (PRN)

- Bi-phase (0º/180º)

- Quad-phase (0º/90º/180º/270º)

• Implementation

• Hardware

- Surface Acoustic Wave (SAW) expander/compressor

• Digital Control- Direct Digital Synthesizer (DDS)

- Software compression “filter”Table of Content

50


• Stepped Frequency Waveform (SFWF)

The Stepped Frequency Waveform is a Pulse Radar System technique for obtaining high resolution range profiles with relative narrow bandwidth pulses.

• SFWF is an ensemble of narrow band (monochromatic) pulses, each of which is stepped in frequency relative to the preceding pulse, until the required bandwidth is covered.

• We process the ensemble of received signals using FFT processing.

• The resulting FFT output represents a high resolution range profile of the Radar illuminated area.

• Sometimes SFWF is used in conjunction with pulse compression.

51


• Stepped Frequency Waveform (SFWF)

52

SOLO Waveform Hierarchy• Pulse Compression Techniques

53

SOLO Waveform Hierarchy• Steped Frequency Waveform (SFWF)

Table of Content

54


57


58


Phase CodingA transmitted radar pulse of duration T is divided in N sub-pulses of equal durationτ = T/N, and each sub-pulse is phase coded in terms of the phase of the carrier.

The complex envelope of the phase codedsignal is given by:

( ) ( ) ( )∑−

=

−=1

02/1

1 N

nn ntu

Ntg τ

τ where:

( ) ( ) ≤≤

=elsewhere

tjtu n

n 0

0exp τϕ

59

-1

Pulse bi-phase Barker coded of length 3

Digital Correlation At the Receiver the coded pulse enters a 3 cells delay lane (from left to right), a bin at each clock.The signals in the cells are multiplied according to ck* sign and summed.

clock

-1 = -11

+1 -1 = 02

-( +1) = -15

0 = 06

+1 +1-( -1) = 33

+1-( +1) = 04

SOLO Pulse Compression Techniques

1

2

3

4

5

6

0

+1+1

0 = 00

60

-1 = -1

+1 -1 = 0

-1 +1 -1 = -1

-1 -1 +1-( -1) = 0

+1 -1 -1 –(+1)-( -1) = -1

+1 +1 -1-(-1) –(+1)-1= 0

+1+1 +1-( -1)-(-1) +1-(-1)= 8

+1+1 –(+1)-( -1) -1-( +1)= 0

+1-(+1) –(+1) -1-( -1)= -1

-(+1)-(+1) +1 -( -1)= 0

-(+1)+1-(+1) = -1

+1-(+1) = 0

-(+1) = -1

0 = 0

-1-1 -1

Pulse bi-phase Barker coded of length 7

Digital CorrelationAt the Receiver the coded pulse enters a7 cells delay lane (from left to right),a bin at each clock.The signals in the cells are summed

clock

1

2

3

4

5

6

7

8

9

10

11

12

1314

SOLO Pulse Compression Techniques

+1+1+1+1

61

-1 = -1

-j +j = 0

+j -1-j = -1

+1 +1+1+1 = 4

-j-1+j = -1

+j - j = 0

Pulse poly-phase coded of length 4

At the Receiver the coded pulse enters a 3 cells delay lane (from left to right), a bin at each clock.The signals in the cells are multiplied by -1,+j,-j or +1 and summed.

clock

SOLOPoly-Phase Modulation

1

2

3

4

5

6

7

8

1+

1+j+

1+j+j−

1+j+j−1−

j+j−1−

j−1−

1−

1− 1+j+ j−

-1 = -1

0

Σ

62


Resolution

Resolution is the spacing (in range, Doppler, angle, etc.) we must have in order todistinguish between two different targets.

first targetresponse

second targetresponse

compositetarget

response

greather then 3 db

DistinguishableTargets

first targetresponse

second targetresponse

compositetarget

response

UndistinguishableTargets

less then 3 db

The two targets are distinguishable ifthe composite (sum) of the received signal has a deep (between the twopicks) of at least 3 db.

63


Pulse Range Resolution


Range Resolution

RADAR

τ

c

R

RR ∆+

Target # 1Target # 2

Assume two targets spaced by a range Δ R and a radar pulse of τ seconds.

The echoes start to be receivedat the radar antenna at times: 2 R/c – first target 2 (R+Δ R)/c – second target

The echo of the first target endsat 2 R/c + τ

τ τ

time from pulsetransmission

c

R2 ( )c

RR ∆+2τ+

c

R2

ReceivedSignals

Target # 1 Target # 2

The two targets echoes can beresolved if:

c

RR

c

R ∆+=+ 22 τ2

τcR =∆ Pulse Range Resolution

64


Pulse Range Resolution (continue)

time from pulse transmission

c

R2 ( )c

RR ∆+2τ+

c

R2

Received Signals

Target # 1 Target # 2Rcvτ Rcvτ

2

τcR =∆Pulse Range Resolution

To improve the Pulse Range Resolution we must decrease theReceived pulse duration τRcv.

This is done by Pulse Compressiontechnique:

• Linear or Nonlinear Frequency Modulation

• Phase Modulation (bi-phase, poly-phase)

The Pulse Range Resolution therefore is given by

1/

2 2

Rcv RcvBWRcv

Rcv

c cR

BW

ττ ≈

∆ = =

65


Angle Resolution


Angle Resolution

RADAR

Target # 1

Target # 2

R

R

3θ

2

cos 3θR

33

2sin2 θθ

RR ≈

Angle Resolution is Determined by Antenna Beamwidth.

33

2sin2 θθ

RRRC ≈

=∆

Angle Resolution is considered equivalent to the 3 db Antenna Beamwidth θ3.

The Cross Range Resolution is given by:

66


Doppler Resolution The Doppler resolution is defined bythe Bandwidth of the Doppler FiltersBWDoppler.

Doppler Dopplerf BW∆ =

67


Resolution Cell


Resolution Cell

RADAR

R∆ 3θR

3φR

The Volume Resolution Cell is the volume defined by the subtended solid angle and range resolution.

RRRRRRR

V

rrectangulaofarea

ellipseofarea

∆≈∆=∆

=∆

33

2

33

2

785.0

33

422φθφθπφθπ

Volume Resolution Cell increases with R2.

68

Range Measurement Unambiguity( )tf1

t

2

τ2

2τ−T

TA

T T T

2

τ−2

2τ+T

2

τ−T2τ+T

1 2 3c

Rt2

=

( )tf1

t

2

τ2

2τ−T

RA

T T T

2

τ−2

2τ+T

2

τ−T2

τ+T

1 2 3

Transmitted Pulses

Received Pulses

SOLO

The returned signal from the target located at a range R from the transmitter reaches the receiver (collocated with the transmitter) after

c

Rt2

=

To detect the target, a train of pulses must be transmitted.

PRT – Pulse Repetition Time PRF – Pulse Repetition Frequency = 1/PRT

To have an unanbigous target range the received pulse must arrive before the transmissionof the next pulse, therefore:

PRFPRT

c

Runabigous 1

2=<

PRF

cRunabigous

2<

Range & Doppler Measurements in RADAR Systems

69

Range measurementSOLO

70

SOLO

71

SOLO

72

Resolving Range Measurement Ambiguity

SOLO

To solve the ambiguity of targets return we must use multiple batches, each with different PRIs (Pulse Repetition Interval). Example: one target, use two batches

First batch: PRI 1 = T1

Target Return = t1-amb

R1_amb=2 c t1_amb


Second batch: PRI 2 = T2

Target Return = t2-amb

R2_amb=2 c t2_amb

To find the range, R, we must solve for the integers k1 and k2 in the equation:

( ) ( )ambamb tTkctTkcR _222_111 22 +=+=We have 2 equations with 3 unknowns: R, k1 and k2, that can be solved becausek1 and k2 are integers. One method is to use the Chinese Remainder Theorem .

For more targets, more batches must be used to solve the Range ambiguity.

See Tildocs # 763333 v1

73

SOLO

74

SOLO

75

SOLO

76

Doppler Frequency Shifts (Hz) for Various Radar Frequency Bands and Target Speeds

Band 1 m/s 1 knot 1 mph

L (1 GHz)S (3 GHz)C (5 GHz)X (10 GHz)

Ku (16 GHz)Ka (35 GHz)

mm (96 GHz)

6.6720.033.366.7107233633

3.4310.317.134.354.9120320

2.988.9414.929.847.7104283

RadarFrequency Radial Target Speed

SOLO

77

Coherent Pulse Doppler RadarSOLO

• STALO provides a continuous frequency fLO

• COHO provides the coherent Intermediate Frequency fIF

• Pulse Modulator defines the pulse width the Pulses Rate

Frequency (PRF) number of pulses in a batch • Transmitter/Receiver (T/R) (Circulator) - in the Transmission Phase directs the Transmitted Energy to the Antenna and isolates the Receiving Channel

• IF Amplifier is a Band Pass Filter in the Receiving Channel centered around IF frequency fIF.• Mixer multiplies two sinusoidal signals providing signals with sum or differences of the input frequencies

- in the Receiving Phase directs the Received Energy to the Receiving Channel

21 ff >>

2f

1f21 ff +

21 ff −

78

SOLO Coherent Pulse Doppler Radar

An idealized target doppler response will provide at IF Amplifier output the signal:

( ) ( )[ ] ( ) ( )[ ]tjtjdIFIF

dIFdIF eeA

tAts ωωωωωω +−+ +=+=2

cos

that has the spectrum:f

fIF+fd-fIF-fd

-fIF fIF

A2/4A2/4 |s|2

0

Because we used N coherent pulses ofwidth τ and with Pulse Repetition Time Tthe spectrum at the IF Amplifier output

f

-fd fd

A2/4A2/4|s|2

0

After the mixer and base-band filter:

( ) ( ) [ ]tjtjdd

dd eeA

tAts ωωω −+==2

cos

We can not distinguish between positive to negative doppler!!!

and after the mixer :

79


We can not distinguish between positive to negative doppler!!!

Split IF Signal:

( ) ( )[ ] ( ) ( )[ ]tjtjdIFIF

dIFdIF eeA

tAts ωωωωωω +−+ +=+=2

cos

( ) ( )[ ]

( ) ( )[ ]tAts

tA

ts

dIFQ

dIFI

ωω

ωω

+=

+=

sin2

cos2

Define a New Complex Signal:

( ) ( ) ( ) ( )[ ]tjQI

dIFeA

tsjtstg ωω +=+=2

ffIF+fd

fIF

A2/2|g|2

0

f

fd

A2/2|s|2

0

Combining the signals after the mixers

( ) tjd

deA

tg ω

2=

We now can distinguish between positive to negative doppler!!!

80


Split IF Signal:

( ) ( )[ ]

( ) ( )[ ]tAts

tA

ts

dIFQ

dIFI

ωω

ωω

+=

+=

sin2

cos2

Define a New Complex Signal:

( ) ( ) ( ) ( )[ ]tjQI

dIFeA

tsjtstg ωω +=+=2

ffd

A2/2|s|2

0

Combining the signals after the mixers

( ) tjd

deA

tg ω

2=

We now can distinguish between positive to negative doppler!!!

From the Figure we can see that in this case the doppler is unambiguous only if:

Tff PRd

1=<

Because we used N coherent pulses ofwidth τ and with Pulse Repetition Time Tthe spectrum after the mixer output is

81

Resolving Doppler Measurement Ambiguity

+=

+= ambDambD f

Tkf

TkV _2

22_1

11

1

2

1

2

λλ

SOLO

To solve the Doppler ambiguity of targets return we must use multiple batches, each with different PRIs (Pulse Repetition Interval). Example: one target, use two batches

First batch: PRI 1 = T1

Target Doppler Return in Range Gate i = fD1-amb

V1_amb=(λ/2) fD1_amb


To find the range-rate, V, we must solve for the integers k1 and k2 in the equation:

We have 2 equations with 3 unknowns: V, k1 and k2, that can be solved becausek1 and k2 are integers. One method is to use the Chinese Remainder Theorem .

Second batch: PRI 2 = T2

Target Doppler Return in Range Gate i = fD2-amb

V2_amb=(λ/2) fD2_amb

For more targets, more batches must be used to solve the Doppler ambiguity.

See Tildocs # 763333 v1

Return to Table of Content

82


83


Phase Comparison Monopulse

dα

Port A

Port B

S

Dαcos

d

AntennaBoresight

αλπψ cos2

d=

wavefrontfor a point

sourceat infinity

To illustrate the Monopulse Antennaassume thsat the RF is received throughonly two ports A and B.

When the rays are received froma direction ά relative to the Antennaboresight, we obtain a phase difference of between port A and port B:

αλπψ cos2

d=

AeB jψ= Let compute

( )2

cos2

sin2

cos2)sincos1(1:ψψψψψψ AjAjAeBAS j

+=++=+=+=

( ) ( )2

sin2

sin2

cos2)sincos1(1:ψψψψψψ AjAjjAejBAjDj j

+=−−=−=−=

=2

tanψ

SDj

84


Transmitted RF signal (in phasor form) is ( ) ( )tpetS tj

TrRFω=

p (t) - the pulse train function

At the front-end of the Antenna we receive a shifted and attenuated version of the transmitted pulse:

( ) ( ) ( )cRtpeVtS tj

cvTRF /2Re −= −ωω

ωRF - the RF angular velocity

ωT - the target’s Doppler shift

2 R/c time delay between transmission and reception

V – random complex voltage strengthc – velocity of light

We assume that from the Antenna emerge radar signal of the Sum S and Difference D

( ) ( )( ) ( ) ( )cRtpFeVD

cRtpeVStj

tj

TRF

TRF

/2

/2

−∆=

−=−

−

ψωω

ωω

85


Receiver

The Superheterodyne Receiver translates the high RF frequency ωRF to a lower frequency for a better processing. This is done my mixing (nonlinear multiplication) the input frequency ωRF- ωT with ωRF± ωIF to obtain ωIF - ωT

IFAmp

IFAmp

Band Passat IF

Band Passat IF

S

D'D

'S

( ) tjst IFRFeLO ωω ±1

Mixer

Mixer

First Intemediate Frequaency (1st IF)

( ) ( )( ) ( ) ( )cRtpFeVD

cRtpeVStj

tj

TRF

TRF

/2

/2

−∆=

−=−

−

ψωω

ωω

The Receiver translates the high RF frequency ωRF to a lower frequency to abetter processing. This is done my mixing (nonlinear multiplication) the input frequency ωRF- ωT with ωRF± ωIF to obtain ωIF - ωT .

The IF signal is amplified and bandpass filtered to produce an output at IF frequency( ) ( )

( ) ( ) ( )cRtpFeVD

cRtpeVStj

tj

TIF

TIF

/2''

/2''

−∆=

−=−

−

ψωω

ωω

If the mixing frequency is centered at ωRF± ωIF than the output is centered atωIF and at the image 2 ωRF± ωIF .

86


Receiver (continue – 1)

A second mixing frequency is sometimes added to avoid potential problems withimage frequency.

IFAmp

'S''S

( ) tjnd IFIFeLO ωω 22 ±

Mixer

Second Intemediate Frequaency (2nd IF)

IFAmp

'D

''D

Mixer

PhaseShifter

AGC

AGC Band Passat 2nd IF

Band Passat 2nd IF

( ) ( )( ) ( ) ( )cRtpFeVD

cRtpeVStj

tj

TIF

TIF

/2"

/2"2

2

−∆=

−=−

−

ψωω

ωω

The output of the Second Intermediate Frequency (2nd IF)

( ) ( )( ) ( ) ( )cRtpFeVD

cRtpeVStj

tj

TIF

TIF

/2''

/2''

−∆=

−=−

−

ψωω

ωω

87


Receiver (continue – 2)

A second mixing frequency stage the signal consists of sinusoidals that possessesan arbitrary phase relationship with respect to the radar’s phase reference.

"'IS

I/Q Detection

VideoAmplifier A/D

Mixer

VideoAmplifier A/D

Mixer

VideoAmplifier A/D

Mixer

VideoAmplifier A/D

Mixer

2/π

2/π

''S

''D

"'QS

"'QD

"'ID "'ijI

D

"'ijQ

D

"'ijQ

S

"'ijI

S

tj IFe 2ω−

[ ] ( )[ ] ( )cRtpeVS

cRtpeVStj

Q

tj

I

T

T

/2"Im'"

/2"Re'"

−=−=

ω

ω

For a coherent Doppler andmonopulse processing is necessary to digitize the signal.

I/Q Detection

To find the phase and reduce the signal frequency to Video

with two 2nd IF signals at 90◦

(cos => I = in phase, sin => Q = quadrature).

[ ] ( ) ( )[ ] ( ) ( )cRtpFeVD

cRtpFeVDtj

Q

tj

I

T

T

/2"Im'"

/2"Re'"

−∆=−∆=

ψψ

ω

ω

'"'"'" QI SjSS +=

'"'"'" QI DjDD +=

88

SOLO Coherent Pulse Doppler Conceptual Operation

89

SOLO Signal Processing

Range – Doppler Cells in Σ and ΔAz, ΔEl

After Fast Fourier Transform (FFT) of the signals of the Batch in each Range Gatewe obtain Σ, ΔAz, ΔEl Rang-Doppler Maps.

90

SOLO Signal Processing

Parameters of Σ , ΔAz, ΔEl Range – Doppler Maps

f

fM

R

RN sunambiguousunambiguou

∆=

∆= &

The Parameters defining the Range – Doppler Maps are:

Δ R – Map Range Resolution

Δ f – Map Doppler Resolution

RUnambiguous – Unambiguous Range

fUnambiguous – Unambiguous Doppler

Range – DopplerCell

Range – DopplerMap

Range Gates are therefore i = 1, 2, …, NNumber of Range-Doppler Cells = N x M

Doppler Gates are therefore j = 1, 2, …, M

Note: The Map Range & Doppler resolution (Δ R, Δ f) may change as function of Seeker task (Search, Detection, Acquisition, Track). This is done by choosingthe Pulse Repetition Interval (PRI) and the number of pulses in a batch.

resolutionresolution ffRR ≥∆≥∆ &

91

SOLO Signal Processing Generation of Σ , ΔAz, ΔEl Range – Doppler Maps (continue – 1)

( ) ( )[ ] ( ) ( )ttTktttTkttfCts ddkdkrk

rk ++≤≤++−= τθπ2cos

The received signal from the scatter k is:

Ckr – amplitude of received signal

td (t) – round trip delay time given by ( )2/c

tRRtt kk

d

+=

θk – relative phase The received signal is down-converted to base-band in order to extract the quadrature components. More precisely sk

r (t) is mixed with:( ) [ ] τθπ +≤≤+= TktTktfCty kkk 2cos

After Low-Pass filtering the quadrature components of Σk, ΔAz k or ΔEl k signals are:

( ) ( )( ) ( )

==

tAtx

tAtx

kkQk

kkIk

ψψ

sin

cos

( ) ( )

+−≅−=

c

tR

c

Rfttft kkkdkk

2222 ππψ

The quadrature samples are given by:( ) ( )

+−≅=

c

tR

c

RfjAjAtX kkkkkkk

222expexp πψ

Ak - amplitude of Σk, ΔAz k or ΔEl k signals ψk - phase of Σk, ΔAz k or ΔEl k signals

( )

+−

+≅+=

c

tR

c

RfAj

c

tR

c

RfAxjxtX kk

kkkk

kkQkIkk

222sin

222cos ππ

92

SOLO Signal Processing Generation of Σ , ΔAz, ΔEl Range – Doppler Maps (continue – 2)

The received signal from the scatter k is:

The energy of the received signal is given by: ( ) ( ) 2kkkk AtXtXP == ∗

( )

+−

+≅+=

c

tR

c

RfAj

c

tR

c

RfAxjxtX kk

kkkk

kkQkIkk

222sin

222cos ππ

where * is the complex conjugate.

Therefore:kk PA =

94


References on RADAR

Skolnik, M.I., “Introduction to Radar Systems”, McGraw Hill, 1962

Scheer, J.A., Kurtz, J.L., Ed., “Coherent Radar Performance Estimation”, Artech House, 1993

Schleher, D.C., “MTI and Pulsed Doppler Radar”, Artech House, 1991

Barton, D.K., Ward, H.R., “Handbook of Radar Measurements”, Artech House, 19

Morris, G.V., “Airborne Pulse Radar”, Artech House, 2nd Ed., 19

Maksimov, M.V., Gorgonov, G.I., “Electronic Homing Systems”, Artech House, 19

Wehner, D.R., “High Resolution Radar”, Artech House, 19

Hovanessian, S.A., “Introduction to Sensor Systems”, Artech House, 19

Barton, D.K., “Modern Radar System Analysis”, Artech House, 19

Berkowitz, R.S., “Modern Radar Analysis, Evaluation and System Design”, John Wiley & Sons, 1965

95

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 – 2013

Stanford University1983 – 1986 PhD AA

96


Chinese Remainder Theorem The original form of the theorem, contained in a third-century AD book by Chinese mathematician Sun Tzu and later republished in a 1247 book by Qin Jiushao.

Suppose n1, n2, …, nk are integers which are pairwise coprime. Then, for any given integers a1,a2, …, ak, there exists an integer x solving the system

1 1 1 1 1

2 2 2 2 2

1 2

0

0

0

, , , integersk k k k k

k

x n t a n a

x n t a n a

x n t a n a

t t t are

≡ + > >≡ + > >

≡ + > >L L L L L

L

or in modern notation

( )mod 1,2, ,i ix a n i k≡ = L ai is the reminder of x : ni

x

97


Chinese Remainder Theorem (continue – 1)

A Constructive Solution to Find x

( )mod 1,2, ,i ix a n i k≡ = L

x

Define 1 2: kN n n n= L

For each i, ni and N/ni are coprime.

Using the extended Eulerian algorithm we can therefore find integers ri and si such that

( )/ 1i i i irn s N n+ =Define

Therefore ei divided by ni has the remainder 1 and divided by nj (j≠i) has the remainder 0,because of the definition of N.

( ): / 1i i i i ie s N n rn= = −

( ) ( )1 mod 0 modi i i je n and e n i j= = ∀ ≠Because of this the solution is of the form

1

k

i ii

x a e=

= ∑ But also ( )1

modk

i ii

a e x N=

=∑

98


Chinese Remainder Theorem (continue – 2)

A Constructive Solution to Find x (Example)

( )mod 1,2, ,i ix a n i k≡ = L

1 2 3: 60N n n n= × =

( )( )( )

2 mod 3 ,

3 mod 4 ,

1 mod 5 .

x

x

x

≡

≡

≡1 2 33, 4, 5n n n= = =

1 2 3/ 20, / 15, / 12N n N n N n= = =

( ){ { { {

11 11

/

13 3 2 20 1sn N n

r

− + = ÷

( ){ { { {

2 2 22

/

11 4 3 15 1n s N n

r

− + = ÷

( ){ { ( )

{ {33

3 3/

5 5 2 12 1N nn

r s

+ − = ÷

( ): /i i ie s N n= ( )1 : 2 20 40e = = ( )2 : 3 15 45e = = ( )3 : 2 12 24e = − = −

1 2 32, 3, 1a a a= = =

( )1 1 2 2 3 3 2 40 3 45 1 24 191x a e a e a e= + + = × + × + × − =

Check:

191 63 3 2 47 4 3 38 5 1= × + = × + = × +

( )/ 1i i i irn s N n+ =Find ri and si such that:

Compute:

Therefore:

and ( )11 191 11 mod 60x N= ¬ = =

11 3 3 2 2 4 3 2 5 1= × + = × + = × +

102


• Continuous Wave (CW)

103



104



105



106



107



108



109



110



112

SOLO Waveform Hierarchy• Steped Frequency Waveform (SFWF)

3 – 4 GHz

6 – 7 GHz

3 GHz

4 GHz

3 – 4 GHz

10 range and doppler measurements in radar systems

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