topic #32 pulse doppler processing and windowing in slow …• peak power in the doppler spectrum...
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Topic #32Pulse Doppler Processing and
Windowing in Slow Time
The Concept of Pulse Doppler
• Each frequency sample is subjected to a threshold detector– threshold based on
dominant interference at that frequency
fast
tim
e(r
ange
bin
#)
slow time (pulse #)
00 M-1
L-11 complex
(I&Q) samplein each cell
samplinginterval = PRI
sampling
interval ≅1/B
Digital Filteror
SpectrumAnalyzer
Spectrum frequency samples
FD
Alternative Approach
• Pulse Doppler processing is equivalent to passing the signal in each range bin through a bank of bandpass filters covering the Doppler range– separate detection
decision made for each filter output
filter #0
filter #1
filter #K-1
y[m]
y0[m]
y1[m]
yK-1[m]
passband of onefilter in the bank
FD
Range-Doppler Map
Ind
epe
nd
ent
FF
T o
n e
ach
ro
w
fast
tim
e (r
ange
bin
#)
slow time (pulse #)
00 M-1
L-1
fast
tim
e (r
ange
bin
#)
frequency (“Doppler”) bin
00 K-1
L-1
Pulse Doppler Pros and Cons
• Advantages are–less clutter, noise competing
with target–ability to register multiple
targets–ability to indicate speed
and direction• Disadvantage is greater
complexity, cost
PRF Regimes - 1• Pulse Doppler radars have many modes with
widely varying PRFs– typical of airborne multimode radars, e.g. APG-67,
-68, -70, etc.• “Low PRF” means no range ambiguities
– but low ambiguous velocity• “High PRF” means no velocity ambiguities
– but short unambiguous range• “Medium PRF” is ambiguous in both range
and velocity– but capable of good measurements of both
2 2uacT c
RPRF
= =2 2ua
PRFv
T
λ λ= =4ua ua
cR v
λ=
PRF Regimes - 2
0 50 100 1500
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
Range (km)
Vel
ocity
(km
/s) max velocity of interest
max range of interest
Medium PRFOperation
Low PRFOperation
High PRFOperation
F = 10 GHz
• The coherent receiver baseband output of an ideal, constant-radial-velocity moving target for an M-pulse dwell is a pure complex exponential:
[ ] 2 , 0, , 1Dj F mTy m Ae m Mπ= = −…
The Discrete-Time Fourier Transform of a Moving Target
( ) ( )( )
( )( )1sin
sinDD j M F F T
D
F F MTY F A e
F F Tππ
π− − −⎡ ⎤−⎣ ⎦=
⎡ ⎤−⎣ ⎦
DTFT of a Moving Target• The coherent receiver
baseband output of an ideal, constant-radial-velocity moving target for an M-pulse dwell is a pure complex exponential
• The discrete-time Fourier Transform (DTFT) is a “digital sinc” or “aliased sinc” function– peak at Doppler frequency– -13.2 dB sidelobes– Rayleigh
width = 1/MT Hz– -3 dB
mainlobewidth = 0.89/MT Hz
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.50
2
4
6
8
10
12
14
16
18
20
Doppler Frequency (multiples of PRF)
|Y(F
)|
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.50
2
4
6
8
10
12
14
16
18
20
Doppler Frequency (multiples of PRF)
|Y(F
)|
M = 20FD = PRF/4
fD = 0.25
[ ] 2 , 0, , 1Dj F mTy m Ae m Mπ= = −…
Windowing
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
time (microseconds)
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-1
-0.6
-0.20
0.2
0.6
1
time (microseconds)-2 -1.5-1 -0.5 0 0.5 1 1.5 20
0.2
0.4
0.6
0.8
1
time (microseconds)
Hamming window• We can window the
data in slow time to reduce Doppler sidelobes in the usual fashion
• Windows also widen mainlobe ⇒decreased Doppler resolution− but this also reduces
straddle loss
• Response shape changes from digital sinc function to the Fourier transform of the window:
( ) [ ] ( ) ( )1
2
0
D
Mj F F mT
Dm
Y F w m e W F Fπ−
− −
== = −∑
Effect of Windows on the Moving Target DTFT
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.50
2
4
6
8
10
12
Doppler Frequency (multiples of PRF)
|Y(F
)|
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.50
2
4
6
8
10
12
Doppler Frequency (multiples of PRF)
|Y(F
)|
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.50
2
4
6
8
10
12
14
16
18
20
Doppler Frequency (multiples of PRF)
|Y(F
)|
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.50
2
4
6
8
10
12
14
16
18
20
Doppler Frequency (multiples of PRF)
|Y(F
)| BEFORE AFTER
( ) ( )( )
sin
sin
D
D
F F MTY F A
F F T
ππ
⎡ ⎤−⎣ ⎦=⎡ ⎤−⎣ ⎦
Why Do We Care About Sidelobes?Sidelobe Masking
• Sidelobes from a strong target can mask the return from a nearby (same range, different Doppler) weaker target
• Solution: windowing for sidelobe suppression
without window with Hamming window-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5
-70
-60
-50
-40
-30
-20
-10
0
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-60
-50
-40
-30
-20
-10
0
Loss in Peak Gain• Windows reduce the peak of DTFT a loss in peak gain (LPG)
• Consider a slow time signal of form
• Peak power in the Doppler spectrum without a window is just
• With a window it is
• The ratio is the LPG:
• 0 dB for rectangular window, about -5.7 to -5.4 dB for Hamming (weak function of M)
( ) 2 2 2DY F A M=
( ) [ ] ( ) [ ]2 21 12 2 0 2
0 0
M Mj mT
Dm m
Y F A w m e A w mπ− −
−
= == =∑ ∑
[ ] 2 , 0, , 1Dj F mTy m Ae m Mπ= = −…
[ ]21
20
1 M
m
LPG w mM
−
== ∑
Processing Loss - 1• Windows also cause a modest signal to noise ratio
loss, called the processing loss, PL
• PL is the reduction in SNR due to use of a window on noisy data
• We can separate this into the window’s effect on the signal and its effect on the noise:
• … so we need to figure out the window’s effect on noise power …
( )( )
w ww w
w w
S NSNR S N NLPG
SNR S N S N N
⎛ ⎞ ⎛ ⎞⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
Processing Loss - 2• Suppose y[m] is zero mean stationary white noise
with variance σ2; integration of the windowed data gives a random variable with power
• If w[m] = 1 for all m (no window),
[ ] [ ] [ ] [ ]
[ ] [ ]
[ ]
1 12
0 0
1 2
0
1 22
0
cross terms
M M
wm l
M
m
M
wm
w m y m w l y l
w m y m
w m N
σ
σ
− −∗ ∗
= =
−
=
−
=
⎧ ⎫⎛ ⎞⎛ ⎞⎪ ⎪= ⎨ ⎬⎜ ⎟⎜ ⎟⎪ ⎪⎝ ⎠⎝ ⎠⎩ ⎭⎧ ⎫⎛ ⎞⎪ ⎪= +⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭
= =
∑ ∑
∑
∑
E
E
2 2w M Nσ σ= =
Processing Loss - 3• Putting it all together:
• Function of the window shapeand length only
• 0 dB for rectangular,-1.7 to -1.34 dB for Hamming(weak function of M)
[ ]
[ ]
212
01
22 2
0
M
mwM
w
m
w m MSNR N
LPGSNR N
M w m
σ
σ
−
=−
=
⎛ ⎞= =⎜ ⎟
⎝ ⎠
∑
∑
[ ]
[ ]
21
01
2
0
M
mM
m
w m
PL
M w m
−
=−
=
=∑
∑
Window
3 dB Mainlobe
Width (relative to rectangular
window)
LPG (dB) Peak
Sidelobe (dB)
PL (dB)
Rectangular 1.0 0.0 -13 0
Hann 1.62 -6.0 -32 -1.76
Hamming 1.46 -5.4 -43 -1.34
Kaiser, a = 2.0 1.61 -6.2 -46 -1.76
Kaiser, a = 2.5 1.76 -8.1 -57 -2.17
Dolph-Chebyshev ( 50 dB equiripple)
1.49 -5.5 -50 -1.43
Dolph-Chebyshev ( 70 dB equiripple)
1.74 -6.9 -70 -2.10
Some Characteristics of Some Common Windows
• Note: metrics are a weak function of window size; can vary significantly for very short windows. Numbers given are asymptotic values (large windows).
End of Topic #32
Topic #33Straddle Loss and Interpolation
Sampling the DTFT: The Discrete Fourier Transform
• The DFT is a sampled version of the DTFT– sampled at frequencies k(PRF/K) Hz, k = 0,...,K-1
• to get denser set of samples, increase K (zero padding)
• sample points are fixed on the frequency axis
[ ] [ ]
( )
12
0
, 0, , 1M
j mk K
m
F k PRF K
Y k y m e k K
Y F
π−
−
=
=
= = −
=
∑ …
• Recall vector formulation of matched filter:
• If interference is white, then
• For an ideal moving target, the target vector model is (counting forward in time)
• Therefore the ideal matched filter output is
• This is a DFT if FD = k(1/KT) for some k and K
( )2 12ˆ 1 DD j F M Tj F TA e e ππ − ′⎡ ⎤= ⎢ ⎥⎣ ⎦
t
1−= Ih S t *2nσ=IS I
The DFT as a Matched Filter - 1
[ ]1
2
0
D
Mj F m T
m
A y m e π−
−
=′ = ∑h y
• This is the kth output of a K-point DFT!– of an M point data sequence
• So DFT is a matched filter for the case where– target frequency is one of the DFT sample
frequencies, k/KT Hz– interference is white noise
• DFT effectively computes K matched filters at once: a filterbank!
[ ] [ ]1
2
0
Mj km K
m
AY k A y m e π−
−
=′ = = ∑h y
The DFT as a Matched Filter - 2
DFT Sampling of the DTFT - 1
• An ideal result for a pure complex exponential input!
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.50
5
10
15
20
25
|Y(f
)| an
d |Y
[k]|
fk10 12 14 16 18 0 2 4 6 8
y[m] = exp(j2πfDm), 0≤m≤19fD = 0.25K = 20
DFT Sampling of the DTFT - 2
• Much different result from very similar waveform!
0
5
10
15
20
25
|Y(f
)| an
d |Y
[k]|
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5k10 12 14 16 18 0 2 4 6 8f
y[m] = exp(j2πfDm), 0≤m≤19fD = 0.275
K = 20
Windows Reduce Straddle Loss
• With no data window, maximum loss is 3.9 dB and average loss is 1.5 dB.
• Hamming window reduces maximum loss to 1.7 dB, average to 0.65 dB
0
1
2
3
4
0 0.1 0.2 0.3 0.4 0.5Normalized deviation from bin center (bins)
Stra
ddle
loss
(dB
)
no window
Hamming window
• Recall straddle loss is apparent loss of gain when sinusoid frequency does not correspond to a DFT sample frequency– DFT samples miss the
peak of the DTFT sinc
• Maximum straddle loss occurs when sinusoid is 1/2 bin off center
Reducing Straddle Loss by Interpolation
• Interpolation can be used to reduce straddle loss• Method 1: zero pad and use a bigger FFT
– interpolates the whole spectrum even if only part needed– computationally expensive
• Method 2: use the DFT (not FFT) to compute single samples– only in region needed– restricted to frequencies of form 2π/K for some K
• Method 3: sinc interpolation– only in region needed
• Method 4: polynomial interpolation– low computational cost– reduced accuracy
Method 1: Bigger FFTs• Sampling density increased by zero padding
– underlying DTFT is not changed– merely sampled at more frequencies– resolution, SNR are not changed
K = 64FD = 0.25PRF
0
5
10
15
20
25
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4Doppler Frequency (multiples of PRF)
|Y[k]|
Method 2: DFT
• Pick a value of k and K that gives a frequency sample near that needed
• More efficient than K-point FFT if number of samples evaluated is less than logK
[ ] [ ]1
2
0
Kj k K
k
Y k y m e π−
−
== ∑
Method 3: Sinc Interpolation - 1• Given the K-point DFT of an M-point sequence y[m],
find the value of the DTFT Y(ω) at an arbitrary ω:
• This leads to the following ideal interpolation formula:
( ) [ ]
[ ]
[ ]
1
0
1 12
0 0
1 1
0 0
1
1 2exp
Mj m
m
M Kj mk K j m
m k
K M
k m
Y y m e
Y k e eK
kY k jm
K K
ω
π ω
ω
πω
−−
=− −
− −
= =
− −
= =
=
⎛ ⎞= ⎜ ⎟
⎝ ⎠⎧ ⎫⎛ ⎞⎛ ⎞= − −⎨ ⎬⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎩ ⎭
∑
∑ ∑
∑ ∑
( ) [ ] ( )1
,0
1,
K
M Kk
Y Y k Q kK
ω ω−
== ∑
• The kernel function is
• So interpolation uses all available DFT samples and asincinterpolating functions
• Can evaluate at any desired value of ω• This is a little complicated and expensive …
–have to evaluate at multiple values of ω to find the actual peak
Sinc Interpolation - 2
( )
( )
1
0
,
2exp
2sin 2
2exp 1 2
2sin 2
,
M
m
M K
kjm
K
kM
Kkj M
K kK
Q k
πω
πωπω
πω
ω
−
=
⎛ ⎞⎛ ⎞− − =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
⎛ ⎞⎛ ⎞−⎜ ⎟⎜ ⎟⎛ ⎞⎛ ⎞ ⎝ ⎠⎝ ⎠− − −⎜ ⎟⎜ ⎟ ⎛ ⎞⎛ ⎞⎝ ⎠⎝ ⎠ −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠≡
∑
Method 4: Polynomial Interpolation
• Fit a quadratic to three points centered on each local peak, then differentiate to find peak location and amplitude
- -
0 k0-1 k0 k0+1
k
X[k] Measured DFT samplesInterpolated peak ofquadratic fit
k’
A'
Δk
[ ]Y k
Solving for Δk and A' - 1• We want to fit the three points around a local
maximum with a parabola of the form
• We get three equations in three unknowns:
[ ] 20 1 2Y k a a k a k= + +
[ ][ ]
[ ]
( )
( )
20 00 0
20 0 0 1
220 0 0
1 1 11
1
1 1 1 1
k kY k a
Y k k k a
aY k k k
⎡ ⎤⎡ ⎤ − −− ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥+ ⎣ ⎦+ +⎢ ⎥⎣ ⎦ ⎣ ⎦
Solving for Δk and A' - 2• The coefficient matrix is a Vandermonde
matrix; it has a number of nice properties• Its determinant is
• … so we know a unique solution exists
( )
( )( )( ) ( )( ) ( ) ( )( )
20 0
20 0
20 0
0 0 0 0 0 0
1 1 1
1
1 1 1
1 1 1 1
2
k k
k k
k k
k k k k k k
− −
=
+ +
= − − + − + − −
=
Solving for Δk and A' - 3• Solve using a Lagrangian method• Assume the peak occurs at k' = k0+Δk ;the value
there is A' = |Y[k0+Δk]|• The solution is
[ ] [ ]{ }[ ] [ ] [ ]
0 0
0 0 0
11 1
21 2 1
Y k Y kk
Y k Y k Y k
− + − −Δ =
− − + +
[ ] ( ) [ ]{( )( ) [ ] ( ) [ ]}
0 0
0 0
11 1
2
2 1 1 1 1
Y k k k k Y k
k k Y k k k Y k
+ Δ = Δ − Δ −
− Δ − Δ + + Δ + Δ +
Problem with Quadratic Interpolation
• Both nearest neighbors of the local peak may not be on thesame lobe– In Nyquist
(minimally)sampled,unwindowedcase,this problemwill exist!
• Parabolic fit isthen a poorassumption! 0 2 4 6 8 10 12 14 16 18
0
2
4
6
8
10
12
14
16
18
20
DFT index k
Am
plitu
de (
|Y[k
]|)
Getting All 3 Samples on Same Lobe
• Use window to spread mainlobe, and/or• Oversample (DFT size K > data length M)
0 2 4 6 8 10 12 14 16 180
2
4
6
8
10
12
14
16
18
20
DFT index k
Am
plitu
de (
|Y[k
]|)
A Hybrid Algorithm• Use the quadratic interpolation to
estimate Δk, thus giving an estimate of the peak frequency ω0
• If only a frequency estimate is needed, can stop here at computation of ω0; otherwise …
• Then use the DFT interpolation method to compute Y(ω)– evaluate at only one value of ω– Intermediate computational cost
End of Topic #33
Topic #34Blind Zones, Staggered PRF, and
Ambiguity Resolution
Range Blind Zones: Eclipsing• Monostatic radar cannot receive while a
pulse is being transmitted• The periodic blind zones in range at all
Doppler shifts that result is called eclipsing
cT/2
Transmittedpulse
time, range
T 2TcT0
τ
receiver on receiver on
Near-In Clutter Eclipsing
• Strong near-in clutter may lengthen the effective blind zone in range
cT/2
Transmittedpulse
time, range
T 2TcT0
τ
receiver on receiver on
Doppler Blind Zones• Clutter creates blind zones around zero
Doppler• Periodicity of slow-time DTFT repeats the
blind zone at intervals of PRF Hz• Repeated at every range bin
– according to clutter width at that range bin
F
2
PRF−2
PRF+PRF− PRF+
… …
Blind Zone Map
• Eclipsing, near-end clutter, and blind speeds combine to produce a 2-D pattern of ranges and velocities where targets are undetectable– “Blind zone map”
cT/2
Tran
smitted
pu
lse
time,
rang
eT
2TcT0
τreceiver o
nreceiv
er o
n
cT/2
Tran
smitted
pu
lse
time,
rang
eT
2TcT0
τreceiver o
nreceiv
er o
n
F
2
PRF−2
PRF+PRF− PRF+
… …F
2
PRF−2
PRF+PRF− PRF+
… …
Parameters for Example Blind Zone Map
• Assume pulse compression pulse with β = 1 MHz,βτ = 11– range resolution = range bin spacing = c/2β = 150 m
– uncompressed pulse is 11 μs long = 11 range bins
• Assume PRF = 10.4 kHz– so PRI = 96 μs = 96 range bins
• Assume 104 Doppler “filters”– either explicit bandpass filters, or 104-point DFT
– so Doppler bins are 100 Hz wide
• Assume two-sided clutter spectrum width is 3.4 kHz– 34 Doppler bins
Blind Zone Map for 1 PRF
• No near-in clutter eclipsing considered
• Positive Doppler frequencies only shown
17cells
11cells
96cells
Slide courtesy of N. Levanon
34cells
PRF=10.4 kHz
Reminder: Staggered PRFs• Method to increase velocity coverage without significant
effect on unambiguous range– eliminate Doppler blind zones in pulse Doppler processing
• Concept: combine data from multiple PRFs– target is not simultaneously blind on all of them
• Pulse-to-pulse stagger– PRI varies with each pulse
– Generally used only for low-PRF modes
– Increased Doppler coverage on a single dwell
– Hard to analyze non-uniformly sampled data
– Large pulse-to-pulse amplitude changes due to ambiguous mainlobe clutter
• Pulse-to-pulse stagger– PRI varies with each pulse
– Generally used only for low-PRF modes
– Increased Doppler coverage on a single dwell
– Hard to analyze non-uniformly sampled data
– Large pulse-to-pulse amplitude changes due to ambiguous mainlobe clutter
• Block-to-block stagger– Common in airborne pulse-
Doppler systems– PRI constant within a dwell,
varies dwell-to-dwell, multiple dwells combined for detection
– Can use coherent MTI– System stability less critical– Overall velocity response may
be poor– Consumes large amount of
radar timeline
• Block-to-block stagger– Common in airborne pulse-
Doppler systems– PRI constant within a dwell,
varies dwell-to-dwell, multiple dwells combined for detection
– Can use coherent MTI– System stability less critical– Overall velocity response may
be poor– Consumes large amount of
radar timeline
Dwell-to-Dwell Stagger: Blind Zone Viewpoint
• Two or more coherent processing intervals (CPIs) transmitted and processed
• “M of N” logic applied to detections
• Problem is selecting M, N, and the set of PRIs
movingtarget
noiseclutter
F
1
2
PRF+ 1PRF+ 13
2
PRF+13
2
PRF− 1PRF− 1
2
PRF−
replication of clutter
Blind @ PRF #1
F
2
2
PRF+ 2PRF+ 23
2
PRF+23
2
PRF− 2PRF− 2
2
PRF−
Not blind @ PRF #2
Dwell-to-Dwell Stagger: Aliasing Viewpoint
• Velocity ambiguities complicate velocity estimate
Blind @ PRF #1
Not blind @ PRF #2
movingtarget
noise
clutter
F
1
2
PRF+ 1PRF+ 13
2
PRF+13
2
PRF− 1PRF− 1
2
PRF−
F
2
2
PRF+ 2PRF+ 23
2
PRF+23
2
PRF− 2PRF− 2
2
PRF−
aliases of moving target
aliases of moving target
“M of N” Blind Zone Map• Goal is to design a set of staggered PRFs
that collectively provide a large space in range and Doppler where detection is possible
• Usually based on multiple PRFs with an “Mout of N” detection logic– if the target is eclipsed or covered by clutter in 1 or
2 PRFs, it will hopefully be in the clear on other PRFs
– 8 PRFs with detection on at least 3 is common in multimode airborne radar
Example: “1 of 2” Detection - 1• 10 μs pulse length• 100 Hz Doppler resolution
• Blind zone map for PRI=100 μs
• 20 m/s clutter spread• Eclipsing due to pulse length only
• Blind zone map for PRI=120 μs
Example: “1 of 2” Detection - 2• Combined blind zone map • Regions that are clear on at
least 1 of 2 PRIs
• Note first pulse duration and DC clutter are always blind
Ambiguous Range of Targets• Once steady state reached, target appears at a
shorter, ambiguous apparent range after every pulse
2 tR c
2 uaR c
2 aR c
t
t
t
t
t
pulse #1
pulse #2
pulse #4
pulse #5
target from pulse #1
target from pulse #2
target from pulse #3
Ambiguous Range Formula• If true range Rt > Rua, then apparent range is
– ((·))x notation means “modulo x”
• Given the measurement Ra and known ambiguous range Rua,
• Normalizing to the range bin spacing ΔR,
( )( )ua
a t RR R=
for some t a uaR R kR k= +
( )( ) for some ( , , .)
equivalently,
ua tt a t
a t N
R Rn n kN k N n etc
R R
n n
= + = =Δ Δ
=
Range with Multiple PRFs
• Number of range bins in PRF #i is Ni (i=0,1,…):– so
• If we use multiple PRFs (thus multiple values of Rua):– assume range bins are the same size for each PRF
• Equivalently,
0 10 0 1 1=t a an n k N n k N= + = + …
iua iR N R= Δ
( )( )i i
a t Nn n=
Aside: Chinese Remainder Theorem• Given pairwise relatively prime integers
N0, N1, … ,Nr-1 and the system of equations (“congruences”)
• then there is a unique solution for nt (modulo N = N0N1···Nr-1) given by
( )( ) ( )( ) ( )( )0 1 10 1 1
, , ,r r
a t a t a tN N Nn n n n n n
− −= = =
( )( ) ( )( )
0 1 10 0 1 1 1 1
11
0,
, 1
r
ii
t a a r r a
r
i i j i i i i NNj j i
n k n k n k n
k N N N k k
β β β
β β
−− −
−−
= ≠
= + + +
= = = ⇒ =∏
…
Solution Using the Chinese Remainder Theorem
• For 3 PRFs, to make it more specific:– and because 3 PRFs is a common choice in
airborne radars for ambiguity resolution
( )( )0 1 20 1 2
0 1 2t a a aN N N
n n n nα α α= + +
( )2
1 1 0 20
. ., i i i i jjj i
k N e g N Nα β β α β=≠
= = =∏
( )( ) ( )( ) ( )( )0 1 2
0 1 2 1 0 2 2 0 1
are smallest integers such that
1, 1, 1
i
N N NN N N N N N
ββ β β= = =
CRT Example• True range cell # = nt = 19
• # range cells per PRF N0=11, N1=12, N2=13,
•
• β0 = 6, β1 = 11, β2 = 7– e.g., β0 satisfies
• α0 = 936, α1 = 1573, α2 = 924– e.g., α0 = 6·12·13 = 936
•
0 1 28, 7, 6a a an n n= = =
( )( )0 1112 13 1β ⋅ ⋅ =
( )( )0 1 21 2 3
0 1 2ˆ 19t a a aN N N
n n n nα α α= + + =
Graphical Equivalent of CRT Algorithm
• True cells= #6 (unambiguous), #11 (ambiguous) result in these measurements when N0 = 7, N1 = 8, N2 = 9
• Extend by concatenation:
#6 #11
CRT Problems
• Measurement errors in data can cause very large range errors
• Same CRT example, but suppose we measure
• Then it turns out – instead of 19
• A number of algorithms exist in the literature to solve the problem in the presence of errors
0 1 28, 7, 7a a an n n= = =
ˆ 943!tn =
erroneous!
Ghosts• In general, need N+1 PRFs to resolve N targets• Insufficient number of PRFs can lead to
ghostingtarget #1 target #2
Resolving Velocity Ambiguity
• It’s the same problem– Doppler shift is measured modulo the PRF
– DFT size and PRF establish the size of the Doppler (velocity) bins
• Some techniques published based on Nyquist sampling/reconstruction theory– usually put constraints on the PRFs that
can be used
End of Topic #34