Range & Doppler measurements using RADAR systems

1
Range & Doppler
Measurements
in RADAR Systems
SOLO HERMELIN
Updated: 09.10.08
http://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 to
return back to the receiver. Since the electromagnetic waves travel with the speed of light
c (much greater then RADAR and
Target velocities), the received signal
is delayed by
c
RR
td
21 +
≅
The received signal is: ( ) ( ) ( ) ( )[ ] ( )tnoisettttfttEtE dddr +−+−−= ϕπα 00 2cos
To retrieve the range (and range-rate) information from the received signal the
transmitted 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 reach
the 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 Theory
the 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
11
1
Rc
tRR
ttd 

+
⋅+
=
In the same way the EM wave received from the target at time t was reflected at td2 ,
therefore
( ) ( ) 222222 ddd tcttRRttR ⋅=−⋅+=−  ( )
2
22
2
Rc
tRR
ttd 

+
⋅+
=
RADAR RF Signals

6
SOLO
( ) ( ) ( ) ( )[ ] ( )tnoisettttfttEtE dddr +−+−−= ϕπα 00 2cos
21 ddd ttt += ( )
1
11
1
Rc
tRR
ttd 

+
⋅+
= ( )
2
22
2
Rc
tRR
ttd 

+
⋅+
=
( ) ( )
2
22
1
11
21
Rc
tRR
Rc
tRR
tttttttt ddd 



+
⋅+
−
+
⋅+
−=−−=−






+
−
+
−
+





+
−
+
−
=−
2
2
2
2
1
1
1
1
2
1
2
1
Rc
R
t
Rc
Rc
Rc
R
t
Rc
Rc
tt d 



From which:
or:
Since in most applications we can
approximate 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
.
10
22
0
11
00
2
1
2
1
2
12
1
2
12
1
21
D
Ralong
FreqDoppler
DD
Ralong
FreqDoppler
Dd ttffttff
c
R
t
c
R
f
c
R
t
c
R
fttf

( ) ( ) ( ) ( ) ( )[ ] ( )tnoisettttffttEtE ddDdr +−+−⋅+−≈ ϕπα 00 2cos
where 21
2
2
1
121
2
02
1
01 ,,,,
2
,
2
dddddDDDDD ttt
c
R
t
c
R
tfff
c
R
ff
c
R
ff +=≈≈+=−≈−≈

Finally
Doppler Effect
RADAR RF Signals

7
SOLO
The received signal model:
( ) ( ) ( ) ( ) ( )[ ] ( )tnoisettttffttEtE ddDdr +−+−⋅+−≈ ϕπα 00 2cos
Delayed by two-
way trip time
Scaled down
Amplitude Possible phase
modulated
Corrupted
By noise
Doppler
effect
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
ωθωθ
θω
−=
+=
where: ( ) ( ) ( )[ ]
( ) ( ) ( )[ ]ttats
ttats
Q
I
θ
θ
sin
cos
=
=
( ) ( ) ( ) ( ) ( )ttsttsts QI 00 sincos ωω −=
One other form: ( ) ( ) ( )[ ] ( ) ( ) ( )
[ ]tjtjtjtj
ee
ta
tttats θωθω
θω −−+
+=+= 00
2
cos 0
( ) ( ) ( )[ ]tjtj
etgetgts 00 *
2
1 ωω −
+= ( ) ( ) ( ) ( ) ( )tj
QI etatsjtstg θ
=+=:
Complex envelope

10
RADAR SignalsSOLO
Spectrum
Define the Fourier Transfer F
( ) ( ){ } ( ) ( )∫
+∞
∞−
−== dttjtstsS ωω exp:F ( ) ( ){ } ( ) ( )∫
+∞
∞−
==
π
ω
ωωω
2
exp:
d
tjSSts -1
F
( ) ( ) ( )[ ]tjtj
etgetgts 00 *
2
1 ωω −
+= ( ) ( ) ( )[ ]0
*
0
2
1
ωωωωω −−+−= GGS-1
F
F
-1
F
F
( ) ( ) ( ) ( ) ( )tj
QI etatsjtstg θ
=+=:
( ) ( ) ( )[ ]tttats θω += 0cos
Inverse Fourier Transfer F -1
Complex envelope

11
RADAR SignalsSOLO
Energy ( ) ( ) ( )[ ]tttats θω += 0cos
( ) ( ) ( )[ ]{ } ( )∫∫∫
+∞
∞−
+∞
∞−
+∞
∞−
≈++== dttadttttadttsEs
2
0
22
2
1
22cos1
2
1
: θω
Parseval’s Formula
Proof:
( ) ( ) ( ) ( )∫∫
+∞
∞−
+∞
∞−
= ωωω
π
dFFdttftf 2
*
12
*
1
2
1
( ) ( ) ( )∫
+∞
∞−
−= dttjtfF ωω exp11
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∫∫ ∫∫ ∫∫
+∞
∞−
+∞
∞−
+∞
∞−
+∞
∞−
+∞
∞−
+∞
∞−
=−=−=
π
ω
ωω
π
ω
ωω
π
ω
ωω
22
exp
2
exp 2
*
112
*
2
*
12
*
1
d
FF
d
dttjtfFdt
d
tjFtfdttftf
( ) ( ) ( )∫
+∞
∞−
−=
π
ω
ωω
2
exp
*
2
*
2
d
tjFtf
If s (t) is real, than s (t) = s*(t) and
( ) ( ) ( )∫∫∫
+∞
∞−
+∞
∞−
+∞
∞−
=== ωω
π
dSdttsdttsEs
222
2
1
:

12
RADAR SignalsSOLO
Energy (continue – 1) ( ) ( ) ( )[ ]tttats θω += 0cos
( ) ( ) ( )∫∫∫
+∞
∞−
+∞
∞−
+∞
∞−
=== ωω
π
dSdttsdttsEs
222
2
1
:
( ) ( ) ( ) ( )[ ] ( ) ( )[ ]
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) 





−−−+−−−+
−−−−+−−
=
−−+−−−+−=
−
−−
00
0000
0
*
0
*2
00
0
*
00
*
0
00
*
0
*
0
*
4
1
4
1
ϕϕ
ϕϕϕϕ
ωωωωωωωω
ωωωωωωωω
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
( ) ( )∫∫
+∞
∞−
+∞
∞−
≈= ωω
π
ωω
π
dGdSEs
22
2
1
2
1
2
1
:
Table of Content

13
SOLO
Waveform Hierarchy
Radar Waveforms
CW Radars Pulsed Radars
Frequency
Modulated CW
Phase
Modulated CW
bi – phase &
poly-phase
Linear FMCW
Sawtooth, or
Triangle
Nonlinear FMCW
Sinusoidal,
Multiple Frequency,
Noise, Pseudorandom
Intra-pulse
Modulation
Pulse-to-pulse
Modulation,
Frequency Agility
Stepped Frequency
Frequency
Modulate
Linear FM
Nonlinear FM
Phase
Modulated
bi – phase
poly-phase
Unmodulated
CW
Multiple Frequency
Frequency
Shift Keying
Fixed
Frequency

14
( )tf
2
τ
2
τ
−
A
∞→t
2
τ
+T
2
τ
−T
A
2
τ
+−T
2
τ
−−T
A
t←∞−
T T
A
t
A
t
A
LINEAR FM PULSECODED PULSE
T T
PULSED (INTRAPULSE CODING)
t
( )tf
A
2
τ
2
τ
−T
AA
T T
A
2
2
τ
+T
2
2
τ
−T
A
T T
A
2
τ
− 2
τ
+T
TN
t
( )tf
A
2
τ
2
τ
−T
AA
T T
A
2
2
τ
+T
2
2
τ
−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 PRF
MEDIUM PRF
PULSED
( )tf
T T T T
2/τ+
τ
HIGH PRF
T
T T T
A Partial List of the Family of RADAR Waveforms

15
Radar Waveforms and their Fourier Transforms

16
Radar 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
TRR
T
∆+
=∆

0
Rc
R
T
−
=∆ 0
Total round-trip is 2ΔT . Therefore point A
returns to radar at
Rc
R
TT
−
+= 0
01
2
Point B returns to radar at
Rc
R
TTT RF −
++= 1
02
2
where and
RFRF
RF
f
T
ω
π21
== RFTRRR += 01
( )






−
+
=
−
+=
−
−
+=−=
Rc
Rc
T
Rc
TR
T
Rc
RR
TTTT RF
RF
RFRF 




22
:' 01
12
λ
λ
R
f
c
R
f
c
R
c
R
f
T
f RF
fc
RF
c
R
RF
RF 

 
22
1
1
1
'
1
'
/
1
−=





−≈












+
−
==
=
<<
λ
R
fDoppler
2
−=
Two Way
Doppler Frequency Shift
( )0≠R
Range & Doppler Measurements in RADAR Systems

19
( ) ( ) ( ) ( )[ ] ( )
( ) ( ) ( )tnoise
c
RR
tRRtftE
tnoisettttfttEtE
fc
dddr
+










 +
−++−=
+−+−−=
=
21
21
0
000
/
00
2
2cos
2cos
00
ϕ
λ
π
πα
ϕπα
λ
If we consider only (c = speed of light) then the frequency of the electromagnetic
wave that reaches the receiver is given by:
c
td
Rd
<<




















+
−≈






+












+−=











 +
−+




 +
−=
c
td
Rd
td
Rd
f
c
td
Rd
td
Rd
ud
d
ff
c
RR
t
c
RR
tf
td
d
f
21
0
21
0~
00
2121
0
1
2
1
2
2
1

ϕ
π
ϕπ
π
λ






+
−=
td
Rd
td
Rd
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 FrequencySOLO
Table of Content
λλ logloglog −=⇒−=
td
Rd
f
td
Rd
f 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 += tAts
The received signal (in steady – state) will be.
( ) ( ) ( )[ ]00cos ϕωωα +−+= dDr ttAts
α – attenuation factor
ωD – two way Doppler shift
c
RfR
ff
fc
DDD

0
/
22
&2
0
−=−==
=λ
λ
πω
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 positive
and negative ωD).
To solve the problem of doppler sign ambiguity
we can split the Local Oscillator into two
channels and phase shifting the
Signal 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=
( ) ( ) ( )[ ] tj
DDv
D
etjtts ω
ωω
2
1
sincos
2
1
=+= ( ){ } ( )Dv ts ωωδ
π
−=
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
d
tf θω
π
0
2
1
For FMCW the θ (t) has a linear slope as seen in the figures bellow
Table of Content

25
( ) ( ) ( ) ( )[ ]00cos ϕθωωα +−+−+= ddDr ttttAts
( ) ( )



−++= dDr tt
dt
d
fftf θ
π2
1
0
λ
πω
R
ff DDD
2
&2 −==
td – two way time delay
c
R
td
2
=
The frequency of received signal is:
λ – mean value of wavelength

26
To extract the information we must subtract the received signal frequency from
the transmitted signal frequency. This is done by mixing (multiplying) those signals
and use a Lower Side-Band Filter to retain the difference of frequencies
( ) ( ) ( ) ( ) ( ) Ddrb ftt
dt
d
t
dt
d
tftftf −



−−



=−= θ
π
θ
π 2
1
2
1
The frequency of mixed signal is:
( ) ( )[ ]00cos ϕθω ++= ttAts
( ) ( ) ( ) ( )[ ]
( ) ( ) ( ) ( )[ ]ddD
ddDdr
ttttttA
ttttttAts
−++−+++
−−+−−=
θθωωωα
θθωωα
00
2
0
2
cos
2
1
cos
2
1
Lower Side-Band
Filter
Lower SB Filter

27
The returned signal has a frequency change due to:
• two way time delay c
R
td
2
=
• two way doppler addition λ
R
fD
2
−=
From Figure above, the beat frequencies fb (difference between transmitted to
received frequencies) for a Linear Sawtooth Frequency Modulation are:
D
m
Dd
m
b fR
Tc
f
ft
T
f
f −
∆
=−
∆
=
+ 4
2/
D
m
Dd
m
b fR
Tc
f
ft
T
f
f −
∆
−=−
∆
−=
− 4
2/
( )
28
−+
−
∆
= bbm ff
f
Tc
R ( )
2
−+
+
−= bb
D
ff
f
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
Performing Fast Fourier Transform (FFT) we obtain fb
+
and fb.
( )
28
−+
−
∆
= bbm ff
f
Tc
R
( )
2
−+
+
−= bb
D
ff
f
From the Doppler Window we get fb
+
and fb
-
, from which:

30
( ) ( )



−++= dDr tt
dt
d
fftf θ
π2
1
0
λ
πω
R
ff DDD
2
&2 −==
td – two way time delay
c
R
td
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 from
the transmitted signal frequency. This is done by mixing (multiplying) those signals
and use a Lower Side-Band Filter to retain the difference of frequencies
( ) ( ) ( ) ( ) ( ) Ddrb ftt
dt
d
t
dt
d
tftftf −



−−



=−= θ
π
θ
π 2
1
2
1
The frequency of mixed signal is:
( ) ( )[ ]00cos ϕθω ++= ttAts
( ) ( ) ( ) ( )[ ]
( ) ( ) ( ) ( )[ ]ddD
ddDdr
ttttttA
ttttttAts
−++−+++
−−+−−=
θθωωωα
θθωωα
00
2
0
2
cos
2
1
cos
2
1
Lower Side-Band
Filter
Lower SB Filter

32
The returned signal has a frequency change due to:
• two way time delay c
R
td
2
=
• two way doppler addition λ
R
fD
2
−=
From Figure above, the beat frequencies fb (difference between transmitted to
received frequencies) for a Linear Triangular Frequency Modulation are:
D
m
Dd
m
b fR
Tc
f
ft
T
f
f −
∆
=−
∆
=
+ 8
4/
positive
slope
D
m
Dd
m
b fR
Tc
f
ft
T
f
f −
∆
−=−
∆
−=
− 8
4/
negative
slope
( )
28
−+
−
∆
= bbm ff
f
Tc
R ( )
2
−+
+
−= bb
D
ff
f
We have 2 equations with 2 unknowns R and fD
with the solution:

33
The Range Unambiguity is given by
the FMCW time period Tm:
Range Resolution is a function of FMCW bandwidth and the linearity of FM:
msunambiguou T
c
R
2
=
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
c
R
∆
== 2δ
mm
m
sunambiguou TfTBT
c
B
c
R
R
tynonlineari
∆
===<<
2
11
2
2δ

34
SOLO
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
But solutions with a common
antenna for transmitting and
receiving, and with a good
isolation between them, do exist.
One solution is to have separate
antennas for transmitting and
receiving.

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
Tc
R
( )
2
−+
+
−= bb
D
ff
f
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
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
−+
+
−= bb
D
ff
f
( )
28
−+
−
∆
= bbm ff
f
Tc
R
This is a solution for more than two targets.
One other solution that can solve also range and doppler ambiguities is to use many
modulation slopes (Δ f and Tm).
Table of Content

37
One of the practical frequency
modulations is the Sinusoidal
Frequency Modulation.
Assume that the transmitted
signal is:
( ) ( )




 ∆
+= tf
f
f
tfAts m
m
ππ 2sin2sin 0
The spectrum of this signal is:
( ) ( )
( )[ ] ( )[ ]{ }
( )[ ] ( )[ ]{ }
( )[ ] ( )[ ]{ }
+
−++




 ∆
+
−++




 ∆
+
−++




 ∆
+





 ∆
=
tfftff
f
f
JA
tfftff
f
f
JA
tfftff
f
f
JA
tf
f
f
JAts
mm
m
mm
m
mm
m
m
32sin32sin
22sin22sin
2sin2sin
2sin
003
002
001
00
ππ
ππ
ππ
π
where Jn (u) is the Bessel Function
of the first kind, n order and argument
u.
Bessel Functions of the first kind

38
One of the practical frequency
modulations is the Sinusoidal
Frequency Modulation.
Assume that the transmitted
signal is:
( ) ( )




 ∆
+= tf
f
f
tfAts m
m
ππ 2sin2sin 0
The transmitted and received signal are heterodyned in a mixer to give the difference
frequency
( ) ( ) ( ) ( )[ ]






−
∆
+−⋅+= dm
m
dD ttf
f
f
ttffAtr ππα 2sin2sin 0
Lower Side-Band
Filter
( )ts
( )tr
( )[ ] ( )[ ] ( )





 ∆
−−
∆
+−+ tf
f
f
ttf
f
f
ttftfA m
m
dm
m
dDd πππα 2sin2sin2cos 0
2
( )[ ] ( )


















−
∆
−−+=
2
2cossin
2
2cos 0
2 d
mdm
m
dDd
t
tftf
f
f
ttftfA πππα

39
Since td << Tm=1/fm we have
( ) ( )[ ] ( )


















−
∆
−−+=
2
2cossin
2
2cos 0
2 d
mdm
m
dDd
t
tftf
f
f
ttftfAtm πππα
Lower Side-Band
Filter
( )ts
( )tr
( )tm
( ) ( )[ ]


















−∆−−+≈
2
2cos22cos 0
2 d
mddDd
t
tftfttftfAtm πππα
The frequency is obtained by differentiating the argument of this equation with respect
to time
( )[ ]
( ) 











−∆+=


















−∆−−+=
2
2sin2
2
2cos22
2
1
0
d
mmdD
d
mddDdb
t
tfftff
t
tftfttftf
td
d
f
ππ
πππ
π
( )
m
m
dm
m f
d
mdmD
f tf
d
mmdD
m
f
b
m
b
t
tftfffdt
t
tfftff
f
dtf
f
f
2
1
0
2
1
0
12
1
0
2
2cos2
2
2sin2
2
1
1
2
1
1


















−∆−≈


















−∆+== ∫∫
<<
πππ
π
The average of the beat frequency over one-half a modulating cycle is:

40
Lower Side-Band
Filter
( )ts
( )tr
( )tm
R
c
ff
ftffff m
DdmD
tf
b
dm ∆
+=∆+≈=
<<
+
8
4
1π
The average of the beat frequency over one-half a modulating cycle
is:
( ) ( )




 ∆
+= tf
f
f
tfAts m
m
ππ 2sin2sin 0
By changing the phase of the sinusoidal modulation
by 180 degree each modulation cycle, we will get:
( ) ( )




 ∆
−=− tf
f
f
tfAts m
m
ππ 2sin2sin 0
R
c
ff
ftffff m
DdmD
tf
b
dm ∆
−=∆−≈=
<<
−
8
4
1π
The average of the beat frequency over one-half a
modulating cycle is:

41
Lower Side-Band
Filter
( )ts
( )tr
( )tm
R
c
ff
ftffff m
DdmD
tf
b
dm ∆
+=∆+≈=
<<
+
8
4
1π
A possible modulating is describe bellow, in which we introduce a unmodulated segment
to measure the doppler and two sinusoidal modulation segments in anti-phase.
From which we obtain:
R
c
ff
ftffff m
DdmD
tf
b
dm ∆
−=∆−≈=
<<
−
8
4
1π
The averages of the beat frequency over one-half a modulating cycle are:
28
−+
−
∆
=
bbm
ff
f
Tc
R
2
−+
+
=
bb
D
ff
f
(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
R
t
c
R
ff dD
2
,
2
10 ≈−≈

( ) ( ) 



⋅−⋅−⋅+=
c
R
f
c
R
ftffAtr DD
2
2
2
22sin 00 πππα
where:
Therefore:
We can see that the change in received phase Δφ is related to range R by:
2/
2
2
2
2
2
2
2
/
00
00
λ
ππππϕ
λ
R
c
R
f
c
R
f
c
R
f
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
R
t
c
R
ff dD
2
,
2
10 ≈−≈

( ) ( ) 



⋅−⋅−⋅+=
c
R
f
c
R
ftffAtr DD
2
2
2
22sin 00 πππα
where:
Therefore:
We can see that the change in received phase Δφ is related to range R by:
2/
2
2
2
2
2
2
2
/
00
00
λ
ππππϕ
λ
R
c
R
f
c
R
f
c
R
f
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 π
The received signals are: ( ) ( ) ( )[ ]dDiiiii ttffAtr −⋅+= πα 2sin
c
R
t
c
R
f
c
R
fff d
i
j
jDi
2
,
22
10
1
0 ≈−≈







∆+−≈ ∑=
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,,2,10
1
−=<<∆∑=
niff
i
j
j 
We can see that the change in received phase Δφi , of two adjacent signals, is related
to range R by:
( ) c
R
f
c
R
c
R
f
c
R
f
c
R
ff
c
R
f i
cR
iiDDii ii
2
2
22
2
2
2
2
2
2
2
2
1
⋅∆≈⋅⋅∆+⋅∆=⋅−+⋅∆=∆
<<
−
πππππϕ

The maximum unambiguous range is given when Δφi=2π :
i
sunambiguou
f
c
R
∆
=
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 seconds
to the transmitted CW signal, producing a phase-code waveform. The returning waveform
is correlated with a stored version of the transmitted waveform. The correlation process
gives a maximum when we have a match. The time to achieve this match is the time-delay
between 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 transmitted
signal

47
SOLO • Phase Modulated Continuous Wave (PMCW)
In the figure bellow we can see a 7-length Barker binary phase code of the received
signal that, at the receiver, passes a 7-cell delay line, and is correlated to a sample
of the 7-length Barker binary signal sample.
Digital Correlation
At the Receiver the coded pulse enters a
7 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
1
2
3
4
5
6
7
8
9
10
11
12
13
14
+1+1+1+1
Table of Content

48
SOLO
Waveform Hierarchy
• Pulse Waves
• Range Resolution is determined by the system bandwidth
B
cc
R
B
Filter
Matched 22
/1 ττ =
==∆
200 MHz  1 meter
325 MHz  2 feet
650 MHz  1 foot
1300 MHz  6 inch
• Use short pulse (τ) for high resolution, or, bandwidth can be achieved by:
• Pulse Compression – intra-pulse coding
• Stretch Processing
• Stepped Frequency Waveform (SFWF) – pulse-to-pulse coding
Table of Content

49
SOLO
Waveform Hierarchy
• 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
SOLO
Waveform Hierarchy
• 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
SOLO
Waveform Hierarchy
• Stepped Frequency Waveform (SFWF)

52
SOLO
Waveform Hierarchy

53
SOLO
Waveform Hierarchy
• Steped Frequency Waveform (SFWF)
Table of Content

54
SOLO
Waveform Hierarchy

57
SOLO
Waveform Hierarchy

58
SOLO
Waveform Hierarchy
Phase Coding
A transmitted radar pulse of duration T is divided in N sub-pulses of equal duration
τ = T/N, and each sub-pulse is phase coded in terms of the phase of the carrier.
The complex envelope of the phase coded
signal is given by:
( )
( )
( )∑
−
=
−=
1
0
2/1
1 N
n
n ntu
N
tg τ
τ
where:
( )
( )


 ≤≤
=
elsewhere
tj
tu 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 Correlation
At the Receiver the coded pulse enters a
7 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
13
14
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
SOLO
Poly-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 to
distinguish between two different targets.
first target
response
second target
response
composite
target
response
greather then 3 db
Distinguishable
Targets
first target
response
second target
response
composite
target
response
Undistinguishable
Targets
less then 3 db
The two targets are distinguishable if
the composite (sum) of the received
signal has a deep (between the two
picks) of at least 3 db.

63
Pulse Range Resolution
Range Resolution
RADAR
τ
c
R
RR ∆+
Target # 1
Target # 2
Assume two targets spaced by a range
Δ R and a radar pulse of τ seconds.
The echoes start to be received
at the radar antenna at times:
2 R/c – first target
2 (R+Δ R)/c – second target
The echo of the first target ends
at 2 R/c + τ
τ τ
time from pulse
transmission
c
R2 ( )
c
RR ∆+2
τ+
c
R2
Received
Signals
Target # 1 Target # 2
The two targets echoes can be
resolved if:
c
RR
c
R ∆+
=+ 22 τ
2
τc
R =∆ Pulse Range Resolution

64
Pulse Range Resolution (continue)
time from pulse
transmission
c
R2 ( )
c
RR ∆+2
τ+
c
R2
Received
Signals
Target #1 Target #2
Rcvτ Rcvτ
2
τc
R =∆Pulse Range Resolution
To improve the Pulse Range
Resolution we must decrease the
Received pulse duration τRcv.
This is done by Pulse Compression
technique:
• Linear or Nonlinear Frequency Modulation
• Phase Modulation (bi-phase, poly-phase)
The Pulse Range Resolution therefore is given by
1/
2 2
Rcv RcvBW
Rcv
Rcv
c c
R
BW
τ
τ ≈
∆ = =

65
Angle Resolution
Angle Resolution
RADAR
Target # 1
Target # 2
R
R
3
θ 





2
cos 3θ
R
3
3
2
sin2 θ
θ
RR ≈





Angle Resolution is Determined by Antenna Beamwidth.
3
3
2
sin2 θ
θ
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 by
the Bandwidth of the Doppler Filters
BWDoppler.
Doppler Dopplerf BW∆ =

67
Resolution Cell
Resolution Cell
RADAR
R∆ 3
θR
3
φR
The Volume Resolution Cell is the volume defined by the subtended solid angle and
range resolution.

RRRRR
RR
V
rrectangulaofarea
ellipseofarea
∆≈∆=∆











=∆

  
33
2
33
2
785.0
33
422
φθφθ
πφθ
π
Volume Resolution Cell increases with R2
.

68
Range Measurement Unambiguity
( )tf1
t
2
τ
2
2
τ
−T
T
A
T T T
2
τ
−
2
2
τ
+T
2
τ
−T
2
τ
+T
1 2 3c
R
t
2
=
( )tf1
t
2
τ
2
2
τ
−T
RA
T T T
2
τ
− 2
2
τ
+T
2
τ
−T
2
τ
+T
1 2 3
TransmittedPulses
ReceivedPulses
SOLO
The returned signal from the target
located at a range R from the transmitter
reaches the receiver (collocated with the
transmitter) after
c
R
t
2
=
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 transmission
of the next pulse, therefore:
PRF
PRT
c
Runabigous 1
2
=<
PRF
c
Runabigous
2
<

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 because
k1 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

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)
6.67
20.0
33.3
66.7
107
233
633
3.43
10.3
17.1
34.3
54.9
120
320
2.98
8.94
14.9
29.8
47.7
104
283
Radar
Frequency 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
1f
21 ff +
21 ff −

78
SOLO Coherent Pulse Doppler Radar
An idealized target doppler response will
provide at IF Amplifier output the signal:
( ) ( )[ ] ( ) ( )
[ ]tjtj
dIFIF
dIFdIF
ee
A
tAts ωωωω
ωω +−+
+=+=
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 of
width τ and with Pulse Repetition Time T
the spectrum at the IF Amplifier output
f
-fd fd
A2
/4A2
/4
|s|2
0
After the mixer and base-band filter:
( ) ( ) [ ]tjtj
dd
dd
ee
A
tAts ωω
ω −
+==
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:
( ) ( )[ ] ( ) ( )
[ ]tjtj
dIFIF
dIFdIF
ee
A
tAts ωωωω
ωω +−+
+=+=
2
cos
( ) ( )[ ]
( ) ( )[ ]t
A
ts
t
A
ts
dIFQ
dIFI
ωω
ωω
+=
+=
sin
2
cos
2
Define a New Complex Signal:
( ) ( ) ( ) ( )[ ]tj
QI
dIF
e
A
tsjtstg ωω +
=+=
2
f
fIF+fd
fIF
A2
/2|g|2
0
f
fd
A2
/2
|s|2
0
Combining the signals after the mixers
( ) tj
d
d
e
A
tg ω
2
=
We now can distinguish
doppler!!!

80
Split IF Signal:
( ) ( )[ ]
( ) ( )[ ]t
A
ts
t
A
ts
dIFQ
dIFI
ωω
ωω
+=
+=
sin
2
cos
2
Define a New Complex Signal:
( ) ( ) ( ) ( )[ ]tj
QI
dIF
e
A
tsjtstg ωω +
=+=
2
f
fd
A2
/2
|s|2
0
Combining the signals after the mixers
( ) tj
d
d
e
A
tg ω
2
=
We now can distinguish
doppler!!!
From the Figure we can see that in this
case the doppler is unambiguous only if:
T
ff PRd
1
=<
Because we used N coherent pulses of
width τ and with Pulse Repetition Time T
the spectrum after the mixer output is

81
Resolving Doppler Measurement Ambiguity






+=





+= ambDambD f
T
kf
T
kV _2
2
2_1
1
1
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 because
k1 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
D
α
cos
d
Antenna
Boresight
α
λ
π
ψ cos
2
d=
wavefront
for a point
source
at infinity
To illustrate the Monopulse Antenna
assume thsat the RF is received through
only two ports A and B.
When the rays are received from
a direction ά relative to the Antenna
boresight, we obtain a phase
difference of
between port A and port B:
α
λ
π
ψ cos
2
d=
AeB jψ
=
Let compute
( )
2
cos
2
sin
2
cos2)sincos1(1:
ψψψ
ψψψ
AjAjAeBAS j






+=++=+=+=
( ) ( )
2
sin
2
sin
2
cos2)sincos1(1:
ψψψ
ψψψ
AjAjjAejBAjDj j






+=−−=−=−=






=
2
tan
ψ
SDj

84
Transmitted RF signal (in phasor form) is ( ) ( )tpetS tj
Tr
RFω
=
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
cv
TRF
/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 strength
c – velocity of light
We assume that from the Antenna emerge radar signal of the Sum S and
Difference D
( )
( )
( )
( ) ( )cRtpFeVD
cRtpeVS
tj
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
IF
Amp
IF
Amp
Band Pass
at IF
Band Pass
at IF
S
D
'D
'S
( ) tjst IFRF
eLO ωω ±
1
Mixer
Mixer
First Intemediate Frequaency (1st IF)
( )
( )
( )
( ) ( )cRtpFeVD
cRtpeVS
tj
tj
TRF
TRF
/2
/2
−∆=
−=
−
−
ψωω
ωω
The Receiver translates the high RF frequency ωRF to a lower frequency to a
better 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
cRtpeVS
tj
tj
TIF
TIF
/2''
/2''
−∆=
−=
−
−
ψωω
ωω
If the mixing frequency is centered at ωRF± ωIF than the output is centered at
ωIF and at the image 2 ωRF± ωIF .

86
Receiver (continue – 1)
A second mixing frequency is sometimes added to avoid potential problems with
image frequency.
IF
Amp
'S
''S
( ) tjnd IFIF
eLO ωω 2
2 ±
Mixer
Second Intemediate Frequaency (2nd IF)
IF
Amp
'D
''D
Mixer
Phase
Shifter
AGC
AGC
Band Pass
at 2nd IF
Band Pass
at 2nd IF
( )
( )
( )
( ) ( )cRtpFeVD
cRtpeVS
tj
tj
TIF
TIF
/2"
/2"
2
2
−∆=
−=
−
−
ψ
ωω
ωω
The output of the Second Intermediate Frequency (2nd
IF)
( )
( )
( )
( ) ( )cRtpFeVD
cRtpeVS
tj
tj
TIF
TIF
/2''
/2''
−∆=
−=
−
−
ψωω
ωω

87
Receiver (continue – 2)
A second mixing frequency stage the signal consists of sinusoidals that possesses
an arbitrary phase relationship with respect to the radar’s phase reference.
"'IS
I/Q Detection
Video
Amplifier A/D
Mixer
Video
Amplifier A/D
Mixer
Video
Amplifier A/D
Mixer
Video
Amplifier A/D
Mixer
2/π
2/π
''S
''D
"'QS
"'Q
D
"'I
D "'ijI
D
"'ijQ
D
"'ijQS
"'ijI
S
tj IF
e 2ω−
[ ] ( )
[ ] ( )cRtpeVS
cRtpeVS
tj
Q
tj
I
T
T
/2"Im'"
/2"Re'"
−=
−=
ω
ω
For a coherent Doppler and
monopulse 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
cRtpFeVD
tj
Q
tj
I
T
T
/2"Im'"
/2"Re'"
−∆=
−∆=
ψ
ψ
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 Gate
we obtain Σ, ΔAz, ΔEl Rang-Doppler Maps.

90
SOLO
Signal Processing
Parameters of Σ , ΔAz, ΔEl Range – Doppler Maps
f
f
M
R
R
N
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 – Doppler
Cell
Range – Doppler
Map
Range Gates are therefore i = 1, 2, …, N
Number 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 choosing
the 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 ddkdk
r
k
r
k ++≤≤++−= τθπ2cos
The received signal from the scatter k is:
Ck
r
– amplitude of received signal
td (t) – round trip delay time given by ( )
2/c
tRR
tt 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
ψ
ψ
sin
cos
( ) ( ) 





+−≅−=
c
tR
c
R
fttft kk
kdkk
22
22 ππψ
The quadrature samples are given by:
( ) ( ) 











+−≅=
c
tR
c
R
fjAjAtX kk
kkkkk
22
2expexp πψ
Ak - amplitude of Σk, ΔAz k or ΔEl k signals
ψk - phase of Σk, ΔAz k or ΔEl k signals
( ) 











+−











+≅+=
c
tR
c
R
fAj
c
tR
c
R
fAxjxtX kk
kk
kk
kkQkIkk
 22
2sin
22
2cos 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: ( ) ( ) 2
kkkk AtXtXP ==
∗
( ) 











+−











+≅+=
c
tR
c
R
fAj
c
tR
c
R
fAxjxtX kk
kk
kk
kkQkIkk
 22
2sin
22
2cos 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
Technion
Israeli Institute of Technology
1964 – 1968 BSc EE
1968 – 1971 MSc EE
Israeli Air Force
1970 – 1974
RAFAEL
Israeli Armament Development Authority
1974 – 2013
Stanford University
1983 – 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
, , , integers
k 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 i
i
x a e
=
= ∑ But also ( )
1
mod
k
i i
i
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 1
1
/
13 3 2 20 1
sn N n
r
 
− + = ÷
 
( ){ { { {
2 2 2
2
/
11 4 3 15 1
n s N n
r
 
− + = ÷
 
( ){ { ( ){ {
33
3 3
/
5 5 2 12 1
N 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
SOLO
Waveform Hierarchy
• Continuous Wave (CW)

103
SOLO
Waveform Hierarchy

104
SOLO
Waveform Hierarchy

105
SOLO
Waveform Hierarchy

106
SOLO
Waveform Hierarchy

107
SOLO
Waveform Hierarchy

108
SOLO
Waveform Hierarchy

109
SOLO
Waveform Hierarchy

110
SOLO
Waveform Hierarchy

112
SOLO
Waveform Hierarchy
• Steped Frequency Waveform (SFWF)
3–4GHz
6–7GHz
3GHz
4GHz
3–4GHz

Range & Doppler measurements using RADAR systems

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