Download Mine Detection - EECS @ Michigan

Overview: two parts
to this presentation
• Physics-Derived Basis Pursuit
for Buried Object Identification
• Fast 3D Blind Deconvolution of
Even Point Spread Functions
• Andrew E. Yagle, January 2005
Part I of this
Presentation
Springfield, VA, January 2005
Metal
Detector
Coil
Voltage
Response
Buried
Object
Physics Derived
Basis Pursuit
in Buried Object
Identification
Jay A. Marble and Andrew E. Yagle
Metal Detector
Phenomenology
• An Electromagnetic Induction (EMI) Metal Detector utilizes
a coil of wire to generate a magnetic field.
• This magnetic field interacts with a buried object inducing “swirling”
electrical (eddy) currents. These induced currents form secondary
sources, which can be modeled as vertical and horizontal dipoles.
• The metal detector coil is then switched from a transmitter to a receiver.
Source
Air
Primary
Magnetic
Field
Air
Coil
+
-
Secondary
Magnetic
Field
Horizontal Dipole
Ground
Vertical Dipole
Ground
Transmitted Fields
Induced
Sources
Fields at
Buried Object
Buried Object
Response
Induced Sources
(approximation)
EMI Phenomenology
EMI Wire Coil
+V
Current
Source
Electronics
& Sampler
_
Simplified EMI
System Concept
Data
Storage
I
Air
Source
Air
Primary
Magnetic
Field
Secondary
Magnetic
Field
Ground
Ground
Source H-field
Buried
Sphere
Incident Field at Object
Metal Object Reaction
EMI Phenomenology
m0
H 0 z ( x, y, z ) 
2

u 3 ( 1d  2h )
J 0 (ru )du
u 0  1   2 e
 m0
H 0 r ( x, y, z ) 
2

 1u 2 ( d  h )
J1 (ru )du
u 0  1   2 e
 1  u 2  ( 2 11  j1 1 )
1
2
Air
(x,y,h)
 2  u  (  2 2  j 2 2 )
2
Source
2
Ground
(x,y,-d)
Source H-field
EMI Phenomenology
mz  2a P( ,  s ,  s , a) H 0 z ( xs , ys , z s )
3
mr  2a 3 P( ,  s ,  s , a) H 0 r ( xs , ys , z s )
2 s (sinh(  )   cosh( ))   0 (sinh(  )   cosh( )   2 sinh(  ))
P( ,  s ,  s , a)  
 s (sinh(  )   cosh( ))   0 (sinh(  )   cosh( )   2 sinh(  ))
  a (i s s )
mz
H zz ( x, y, z ) 
2
m
H rz ( x, y, z )  r
2
* Model assumes a solid spherical target.
Secondary
Magnetic
Field

u 3 (  1z  2h )
J 0 (ru )du
u0  1   2 e

 1u 2 (  z  h )
J1 (ru )du
u 0  1   2 e
1
pz
pr
2
Metal Object Reaction
EMI Phenomenology
v  H 0 x H xz px  H 0 z H zz pz
* Model no longer assumes a solid spherical target.
H0x – Horizontal magnetic field at the center of the
target produced by the source magnetic dipole.
Hxz – Vertical magnetic field at the receive coil produced
by the horizontal induced magnetic dipole.
H0z – Vertical magnetic field at the center of the target
produced by the source magnetic dipole.
 px 
p 
 z
Target
Magnetic
Polarizability
Vector
Induced
Magnetic
Sources
pz
px
Hzz – Vertical magnetic field at the receive coil produced
by the vertical induced magnetic dipole.
Physics Derived
Basis Functions
• The horizontal dipole produces
the W basis.
0.6
0.6
0.4
0.4
0.2
0
-0.2
The L Basis Function
The W Basis Function
0.2
0
-0.2
-0.4
-0.4
-1.5
Voltage Response
Voltage Response
• The vertical dipole produces
the L basis.
-1
-0.5
0
0.5
Sample Location [m]
Vertical Dipole
1
1.5
-1.5
-1
-0.5
0
0.5
Sample Location [m]
1
Horizontal Dipole
1.5
Physics Derived
Basis Functions
s( x)  aL d x   bWd ( x)
• The spatial signal is
composed of the L(x)
and W(x) basis functions.
• The basis functions are
parameterized by depth.
• Any object at the same depth
will have the same basis.
• The object’s shape affects the
weighting coeffs “a” and “b”.
d – depth of buried object
a – Polarizability of object
in Z-direction.
b – Polarizability of object
in X-direction.
Canonical Depths
• These 3 signals come from identical spheres at different depths.
Sphere at 0.0m
Sphere at 0.25m
6
0.2
4
2
0
1.5
x 10
1
0.15
Voltage Response
Voltage Response
Voltage Response
0.25
Sphere at 1.0m
-3
Sphere at 0.25m
Sphere at 0.0m
8
0.1
0.05
0
0.5
0
-0.5
-2
-4
-1.5
Sphere at 1.0m
-0.05
-1
-0.5
0
0.5
Sample Location [m]
1
1.5
-0.1
-1.5
-1
-0.5
0
0.5
Sample Location [m]
1
1.5
-1
-1.5
All objects simulated at 0.25m depth.
-1
-0.5
0
0.5
Sample Location [m]
1
1.5
Canonical Depths
• These 3 signals come from identical spheres at different depths.
Sphere at 0.0m
Sphere at 0.25m
5
aL Component
2
1
0
-2
-1.5
5
0.08
0.06
0.04
0.02
0
-1
-1
-0.5
0
0.5
Sample Location [m]
1
1.5
-0.04
-1.5
4
3
2
1
0
-1
-0.02
bW Component
aL Component
6
Voltage Response
3
x 10
aL Component
0.1
Voltage Response
Voltage Response
0.12
Sphere at 1.0m
-4
Sphere at 0.25m
Sphere at 0.0m
4
Sphere at 1.0m
-2
bW Component
-1
-0.5
0
0.5
Sample Location [m]
1
1.5
-1.5
bW Component
-1
-0.5
0
0.5
Sample Location [m]
1
1.5
2D Signal
Subspace
Wd
Higher
Dimensional
Space
62°
Ld
Wd
Ld
2D Plane
Spanned by
Ld and Wd
• The natural Ld and Wd
bases are non-orthogonal.
• An interesting fact is that
the Ld and Wd bases form
an angle of 62° regardless
of object depth d.
• All metal objects at this depth
will exist in this signal
subspace.
Subspaces for
Objects at
Different Depths
Wd2
Second
Depth
Subspace
Wd1
Ld2
Ld1
First
Depth
Subspace
• The bases of objects at a
second depth, Ld2 and Wd2,
span a second plane that
is non-orthogonal to
the plane spanned by the
first depth bases, Ld1 and Wd1.
Canonical Shapes
• All 3 of these signals are represented by a vector in the same 2D subspace.
Sphere
Cylinder
Cylinder
Flat Plate
0.6
0.6
0.4
0.4
0.4
0.2
0
-0.2
-0.4
-1.5
Voltage Response
0.6
Voltage Response
Voltage Response
Sphere
Flat Plate
0.2
0
-0.2
-0.4
-1
-0.5
0
0.5
Sample Location [m]
1
1.5
-1.5
0.2
0
-0.2
-0.4
-1
-0.5
0
0.5
Sample Location [m]
1
1.5
-1.5
All objects simulated at 0.25m depth.
-1
-0.5
0
0.5
Sample Location [m]
1
1.5
Canonical Shapes
• All 3 of these signals are represented by a vector in the same 2D subspace.
Sphere
Cylinder
Sphere
Cylinder
0.15
0.1
0.05
0
-0.05
-1
-0.5
0
0.5
Sample Location [m]
1
1.5
aL Component
0.1
0.05
0
-0.05
bW Component
0.15
Voltage Response
aL Component
Voltage Response
Voltage Response
Flat Plate
0.15
0.1
-0.1
-1.5
Flat Plate
-0.1
-1.5
-1
-0.5
0
0.5
Sample Location [m]
0.05
0
-0.05
bW Component
1
1.5
aL Component
-0.1
-1.5
All objects simulated at 0.25m depth.
bW Component
-1
-0.5
0
0.5
Sample Location [m]
1
1.5
Effect of Object
Shape, Size,
and Content
2D Subspace for
Objects at Depth “d”
a
b
angle
sphere
1
1
45°
cylinder
1
0.5
30°
flat plate
1
0.176
10°
Wd
Larger or More
Conductive Sphere
45°-Sphere
62°
30°-Cylinder
Larger or More
Conductive Cylinder
Larger or More
Conductive Flat Plate
10°-Flat Plate
Ld
• The object’s polarizability
(the a and b coeffs)
determines the angle
of the signal in the 2D
subspace.
• Increasing the object’s
size increases the
weightings, a and b.
• More conductive metal
also increases the
weightings, a and b.
Subspace
Identification
Using Projections
H1  [L1 W1 ]
1
Pshallow  H1 ( H H1 ) H
T
1
T
1
1
T
1
T
Pmid  H 2 ( H 2 H 2 ) H 2
T
Pdeep  H 3 ( H 3 H 3 ) H 3
T
Subspace
Identification
Performance
0.3
0.5
Deep
Mid
Sphere
Depth
Sphere
0.25
0.4
Table 3a: Norm After Projection into Subspace (No Noise)
0.2
0.3
Spheres
Shallow
Shallow
Mid
1.00
0.80
Flat Plates
Deep
0.79
Shallow
Mid
1.00
0.99
Cylinders
Deep
0.93
Shallow
Mid
1.00
0.95
0.15
Deep
0.94
Mid
0.69
1.00
0.79
0.92
1.00
0.99
0.81
1.00
0.96
Deep
0.31
0.51
1.00
0.63
0.85
1.00
0.35
0.71
1.00
0.2
0.1
0.1
0.05
0
0
-0.1
-0.05
-0.1
-0.2
-0.15
-0.3
-1.5
-1.5
-1
-1
-0.5
-0.5
0
0
0.5
0.5
1
1
1.5
1.5
0
0
0.5
0.5
1
1
1.5
1.5
0
0
0.5
0.5
1
1
1.5
1.5
0.5
0.4
0.4
0.3
Table 3b: Norm After Projection into Subspace (Noise Var: 0.01)
Spheres
0.3
Flat Plates
0.2
0.2
Cylinders
0.1
0.1
Shallow
Mid
Deep
Shallow
Mid
Deep
Shallow
Mid
Deep
Mid
Deep
Depth
Sphere
Sphere
00
Shallow
0.98
0.75
0.73
0.98
0.97
0.92
0.99
0.93
0.92
-0.1
-0.1
Mid
0.66
0.97
0.71
0.92
0.98
0.97
0.84
0.98
0.94
-0.2
-0.2
Deep
0.32
0.54
0.93
0.64
0.83
0.98
0.38
0.72
0.96
-0.3
-1.5
-0.3
-1.5
-1
-1
-0.5
-0.5
0.5
0.5
0.4
0.4
Table 3c: Norm After Projection into Subspace (Noise Var: 0.05)
Spheres
0.3
0.3
Flat Plates
0.2
0.2
Cylinders
0.1
0.1
Shallow
Mid
Deep
Shallow
Mid
Deep
Shallow
Mid
Deep
Shallow
0.70
0.49
0.39
0.84
0.81
0.58
0.73
0.68
0.57
Mid
0.41
0.59
0.39
0.79
0.80
0.62
0.64
0.71
0.55
Deep
0.04
0.39
0.44
0.58
0.67
0.65
0.28
0.30
0.55
Mid
Deep
Depth
Sphere
Sphere
0
0
-0.1
-0.1
-0.2
-0.2
-0.3
-0.3
-0.4
-1.5
-1.5
-1
-1
-0.5
-0.5
New EMI
Modality
0.2
0.2
0.15
0.1
0.05
0
-0.05
-0.2
-1.5
0
-0.05
-0.1
-0.15
-1
-0.5
0
0.5
Sample Location [m]
1
60kHz
0.1
0.05
0
-0.05
-0.1
-1
-0.5
0
0.5
Sample Location [m]
1
-0.2
-1.5
1.5
-1
-0.5
0
0.5
Sample Location [m]
1
1.5
0.2
real
imag*10
10kHz
0.15
Voltage Response
Voltage Response
real
imag*10
-0.15
-0.2
-1.5
1.5
0.1
0.05
0
-0.05
-0.1
real
imag*10
40kHz
0.1
0.05
0
-0.05
-0.1
-0.15
-0.15
-0.2
-1.5
0.15
0.05
0.2
0.15
30kHz
0.1
-0.1
-0.15
Georgia Tech EMI
0.2
real
imag*10
Voltage Response
1kHz
Voltage Response
Voltage Response
0.15
real
imag*10
-1
-0.5
0
0.5
Sample Location [m]
1
1.5
-0.2
-1.5
-1
-0.5
0
0.5
Sample Location [m]
1
1.5
0.2
0.2
20kHz
Voltage Response
Voltage Response
0.15
0.15
real
imag*10
0.1
0.05
0
-0.05
-0.1
0.1
0.05
0
-0.05
-0.1
-0.15
-0.15
-0.2
-1.5
real
imag*10
50kHz
-0.2
-1.5
-1
-0.5
0
0.5
Sample Location [m]
1
1.5
-1
-0.5
0
0.5
Sample Location [m]
1
1.5
600Hz to 60kHz
Part II of this
Presentation
Fast 3D Blind Deconvolution of
Even Point Spread Functions
Andrew Yagle and Siddharth Shah
The University of Michigan, Ann Arbor
Motivation
Many blind
deconvolution
algorithms
need an initial
PSF estimate
BUT
Can be tedious
and problematic
to measure PSF
accurately
Blind Deconvolution
(Don’t need PSF)
But Blind Deconvolution algorithms tend to be slow !
Also many still need initial PSF estimate
What we need
A fast algorithm that performs blind deconvolution
We will show:
An algorithm that performs blind deconvolution that is
• Fast
• Parallelizable
• A linear algebraic formulation
• Non iterative (at least for the Least Squares solution)
Assumptions
1. Point Spread Function or
PSF h(x,y,z) is even in 3-D.
2. We
Reasonable in optics. PSFs
are symmetric in x, y, and z.
Hence h(x,y,z) = h(-x,-y,-z)
know the PSF or image support size.
3. Image
has compact support.
Potential Problem: Asymmetric PSFs due to optical aberrations
Can these be solved too ?
YES (later)
Formulation
DATA
PSF
OBJECT
NOISE
y(i1,i2,i3) = h(i1,i2,i3) *** u(i1,i2,i3) + n(i1,i2,i3)
where u(i1,i2,i3) ≠ 0 for 0 ≤ i1,i2,i3 ≤ M-1
h(i1,i2,i3) ≠ 0 for 0 ≤ i1,i2,i3 ≤ L-1
y(i1,i2,i3) ≠ 0 for 0 ≤ i1,i2,i3 ≤ N-1 N=L+M-1
h(i1,i2,i3) = h(L-i1,L-i2,L-i3) (even PSF)
n(i1,i2,i3) is zero mean white Gaussian noise.
PROBLEM
Given only data y(i1,i2,i3) reconstruct the object u(i1,i2,i3)
and the PSF h(i1,i2,i3)
1-D Solution
y ( n)  h( n)u ( n), h(n) is even
Taking z transform
1
Y ( z )  H ( z )U ( z )  z L H ( )U ( z )
z
1
1
M
N 1
Y ( z ) z U ( )  z Y ( )U ( z )
z
z
Equating coefficients, we get the following matrix
 y (0)
 y (1)

 

 
 

 
0
0
y * ( N  1)


y * ( N  2) 








 y ( N  2)


y ( N  1)
0
0
0
0
Toeplitz Structure
0   u ( M  1)  0
  
  

  
   u (0)    

 
    u (0)    
  
y * (1)  


  
y * (0)  u ( M  1) 0
2-D Problem
Y ( z1 , z 2 )  H ( z1 , z2 )U ( z1 , z 2 )  ( z1 z 2 ) L H (
1 1
, )U ( z1 , z 2 )
z1 z 2
or
1 1
1 1
N
Y ( z1 , z2 )( z1 z2 ) U ( , )  ( z1 z2 ) Y ( , )U ( z1 , z2 )
z1 z2
z1 z2
M
Example
Solve:
 3 10 8 
11 28 18  h0

 h
10 19 7   1
h1  u0
* *

h0  u2
u1 

u3 
2-D Example
10
11

3

0
19

28
10

0
7

18

8
0

0
0

0
0

0
0
0
8
0
0
10
0
0 18
8
0
11
0
0
7 18
0
3
0
0
0
7
0
0 10
0
8
0 10
19 11 10 28 10 18
28
3 11 19 28 7
10
0
3
0 19
0 19
0
3
7
28 19 11
0
0 10
3
28
18 10 28 10 11 19
8
0 10
0 10
0
0
7
0
0
0
3
0 18
7
0
0
11
0
8 18
0
0 10
0
0
0
0
8
0
0
0 
0 
0 
 

0 
0

 
0
0 
0   u1  0

 
8   u3  0
18   u0  0
   
7   u 2  0 

0    u 2  0 
   
10   u0  0

 
28   u3  0
 
19    u1  0

 
0
0 
0 
3

 
11
0 
0 
10 
 
Solution
u0 u1  3 4
u u   5 7

 2 3 
Toeplitz Block Toeplitz structure
Size of matrix
(2M + L- 2)2 X (2M2)
3-D Solution
Y ( z1 , z2 , z3 )( z1 z2 z3 ) M U (
1 1 1
1 1 1
, , )  ( z1 z2 z3 ) N Y ( , , )U ( z1 , z 2 , z3 )
z1 z 2 z3
z1 z2 z3
Equating coefficients we would get a doubly nested Toeplitz matrix
Matrix size: (2M + L-2)3 X (2M)3
Q: So we have solved the 3D problem ?
A: Not quite !! If M=5 and L=3 then the matrix size is 4913 X1024
It will be intractable to use this method “as is” in 3D !
Fourier
Decomposition
Let xk  e j 2k / M , yk  e j 2k / M , zk  e j 2k / M
0 ≤ k ≤ M-1
Then
Y ( z1 , yk , z3 )( z1 yk z3 ) M U (
1 1 1
1 1 1
, , )  U ( z1 , yk , z3 )( z1 yk z3 ) N Y ( , , )
z1 yk z3
z1 yk z3
Using conjugate symmetry
1 1 1
1 1 1
1
1
U ( , , )  U *( ,
, )  U * ( , yk , )
z1 yk z3
z1 * yk * z3 *
z1 *
z3 *
1
1
1
1
N
Y ( z1 , yk , z3 )( z1 yk z3 ) U * ( , yk , )  U ( z1 , yk , z3 )( z1 yk z3 ) Y ( , yk , )
z1 *
z3 *
z1 *
z3 *
M
The point ?
The last equation is decoupled into a set of M 2D problems !
2-D to 1-D
So we broke down a huge 3D problem to M simpler 2D problems
What next ?
Substitute for xk in each 2D problem and you would get M 1D problems in z
Y ( xk , yk , z3 )( xk yk z3 ) M U * ( xk , yk ,
1
1
)  U ( xk , yk , z3 )( xk yk z3 ) N Y ( xk , yk , )
z3 *
z3 *
To summarize
We broke up a large 3D problem into M2 1D problems
2-D scale factors
Note that each 1D problem will be correct upto a scale factor.
Decoupling 2D to 1D
Each row is solved to a scaled factor.
How do we get the whole 2D solution correctly ?
Solve along columns and compare coefficients
1D FT
along columns
d4U(:,4)
d3U(:,3)
d2U(:,2)
d1U(:,1)
solve
1D FT
along rows
c1U(1,:)
solve
c2U(2,:)
U(x,y)
c3U(3,:)
c4U(4,:)
Compare and scale
3-D scale factors
We just learned how a 2D problem could be correctly scaled
Solve 3D
Decouple 3D to 2D
Scale 2D Solns
Decouple 2D to 1D
2D solutions
Solve 1D
Scale 1D Solns
Solve 2D problems
d1
Decouple along x
d2
U(x,y,z)
Decouple along z
2D problems
c1
Solve 2D problems
c2
Compare
and
scale
Stochastic Case
In presence of noise the nullspace of the toeplitz structure no longer exists.
We can find “nearest” nullspace using Least Squares (LS) (fast)
Can use structure of matrix to solve by structured least squares (STLS)
(slow but more accurate)
We can show that such norm minimization will give us the
Maximum Likelihood Estimate of the object u(x,y,z)
Simulations
Synthetic bead image (30X30X30), (3X3X3) PSF, no noise
Stochastic case:
STLN vs. LS
Least Squares v/s STLN comparison
7X7X7 image. 3X3X3 PSF.
50 iterations per SNR
Least squares does well at high
SNRs but at low and medium
SNRS STLN is better.
Comparison with
Lucy Richardson
SNR
MSE
Our algorithm gave a lower MSE. In LR
Final accuracy even in absence of noise
depends on initial PSF estimate.
SNR
Time
Our algorithm need only a fixed amount
of time to solve independent of the SNR.
LR needs more time and time to solve
depends on the SNR.