MIT CSAIL - Computational Imagingpeople.csail.mit.edu/bkph/talkfiles/Computational...3×3 node grid example Here we have three input nodes (1, 2, 3), three output nodes (7, 8, 9),

COMPUTATIONAL IMAGING

Berthold K.P. Horn

What is Computational Imaging?

• Computation inherent in image formation



(1) Computing is getting faster and cheaper

—precision physical apparatus is not





(2) Can’t refract or reflect some radiation





(2) Can’t refract or reflect some radiation

(3) Detection is at times inherently coded

Computational Imaging System

Berthold K.P. HornLine

Examples of Computational Imaging:

(1) Synthetic Aperture Imaging

(2) Coded Aperture Imaging

(3) Exact Cone Beam Reconstruction

(4) Diaphanography—DiffuseTomography

(1) SYNTHETIC APERTURE IMAGING

Traditional approach:

• Coupling of resolution, DOF, FOV to NA• Precision imaging — “flat” illumination

with: Michael Mermelstein, Jekwan Ryu,Stanley Hong, and Dennis Freeman

Objective Lens Parameter Coupling

Synthetic Aperture Imaging

Traditional approach:

• Coupling of resolution, DOF, FOV to NA• Precision imaging — “flat” illuminationNew approach:

• Precision illumination — Simple imaging• Multiple images — Textured illumination

Synthetic Aperture Imaging

• Precision illumination — Simple imaging• Multiple images — Textured illumination

• Image detail in response to textures• Non-uniform samples in FT space

SAM M6

Creating Interference Pattern

Reflective Optics M6

Creating Interference Pattern

Fourier Transform of Texture Pattern

Interference Pattern Texture

Synthetic Aperture Microscopy

• Interference of many Coherent Beams• Amplitude and Phase Control of Beams

Amplitude and Phase Control

Synthetic Aperture Microscopy

• Interference of many Coherent Beams• Amplitude and Phase Control of Beams

• On the fly calibration• Non-uniform inverse FT Least Squares

Wavenumber Calibration using FT

Hough Transform Calibration

Least Squares Match in FT

Polystyrene Micro Beads (1µm)

(2) CODED APERTURE IMAGING

• Can’t refract or reflect gamma rays• Pinhole — tradeoff resolution and SNR

with: Richard Lanza, Roberto Accorsi,Klaus Ziock, and Lorenzo Fabris.

Coded Aperture Imaging

• Can’t refract or reflect gamma rays• Pinhole — tradeoff resolution and SNR

• Multiple pinholes• Complex masks can “cast shadows”

Coded Aperture Principle

Decoding Method Rationale

Berthold K.P. HornStamp

Coded Aperture Imaging

• Can’t refract or reflect gamma rays• Pinhole — tradeoff resolution and SNR• Complex masks can “cast shadows”

• Decoding by Correlation• Special Masks with Flat Power Spectrum

Mask Design — Inverse Systems

Masks — XRT Coarse

Mask Design — 1D

Definition: q is a quadratic residue (mod p)if ∃n s.t. n2 ≡ q(mod p)Legendre symbol(

ap

)={

1 if a is quadratic residue−1 otherwise

Correlation with zero shift (p − 1)/2Correlation with non-zero shift (p − 1)/4

Mask Design

• Auto Correlation

a(i) = (p − 1)4

(1+ δ(i))

• Power Spectrum

A(j) = (p − 1)4

(δ(j)+ 1)

Coded Aperture Extensions

• Imaging Nearby Objects• Mask / Countermask Combination* Dynamic Reconstruction

Coded Aperture Application

• Detection of Fissile Material• Large Area Detector Myth• Signal and Background Amplified

Large Area Alone Doesn’t Help

Imaging and Large Area Do!

Coded Aperture Example

• Imaging — 1/R instead of 1/R2

Coded Aperture Detector Array

Computational Imaging System

Berthold K.P. HornLine

Dynamic Reconstruction

Three weak, distant radioactive sources

Reconstruction Animation

http://csail.mit.edu/~bkph/images/Back-402.html

Coded Aperture Applications

• Detection of Fissile Material• Imaging — 1/R instead of 1/R2

• Increasing Gamma Camera Resolution• Replacing Rats with Mice

.

(4) EXACT CONE BEAM ALGORITHM

• Faster Scanning—Fewer Motion Artifacts• Lower Exposure—Uniform Resolution

with: Xiaochun Yang

Exact Cone Beam Reconstruction

• Faster Scanning—Fewer Motion Artifacts• Lower Exposure—Uniform Resolution

• Parallel Beam → Fan Beam• Planar Fan → Cone Beam

Parallel Beam to Fan Beam

Coordinate Transform in 2D Radon Space

Cone Beam Geometry — 3D

Radon’s Formula

• In 2D: ~ derivatives of line integrals• In 3D: derivatives of plane integrals• Can’t get plane integrals from projections∫ (∫

f(r , θ)dr)dθ

∫ ∫1rf(x,y)dx dy

Radon’s Formula in 3D

f(x) = − 18π2

∫S2∂2R f(l,β)

∂l2

∣∣∣∣∣l=x·β

dβ

where

R f(l,β) =∫f(x) δ(x · β− l)dV

Grangeat’s Trick

∂∂z

∫ ∫f(x,y, z)dx dy =

∂∂θ

∫ ∫f(r ,φ,θ)dr dφ


• Data Sufficiency Condition• Good “Orbit” for Radiation Source

Radon Space — 2D

Circular Orbit is Inadequate (3D)

Data Insufficiency

Good Source Orbit

http://csail.mit.edu/~bkph/movies/ball_seams_45.pdf


• Data Sufficiency Condition• Good “Orbit” for Radiation Source

• Practical Issue: Spiral CT Scanners• Practical Issue: “Long Body” Problem

.

(3) DIAPHANOGRAPHY

(Diffuse Optical Imaging)

• Highly Scattering — Low Absorption• Many Sources — Many Detectors

with: Xiaochun Yang, Richard Lanza, David Boas and Anna Custo.

Diaphanography

• Randomization of Direction

• Scalar Flux Density

mm

mm

Head coronal slice

60 80 100 120 140 160

30

40

50

60

70

80

90

100

110

120

scalp

white

gray

CSF

skull

(a)

(b)

10 15 20 25 30 35 40−8

−7.5

−7

−6.5

−6

−5.5

−5

−4.5

−4

−3.5

source−detector separation [mm]

Log 1

0 F

luen

ce

Total fluence [CW]0.010.11.0

15 20 25 30 35 40

−0.25

−0.2

−0.15

−0.1

−0.05

0

0.05


(MC

− M

Co)

/ M

Co

Deviation of fluence

0.0010.010.11.0

(a)

(b)

15 20 25 30 35 400

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1


Nor

mal

ized

PP

F

scalp−skull

brain

1.00.10.01

15 20 25 30 35 40−0.01

−0.005

0

0.005

0.01

0.015

0.02

0.025

0.03

detector separation [mm]

(PP

F −

PP

Fo)

/PP

Fo

Deviation of Sensitivity to Scalp−Skull

0.001 0.010.11.0

15 20 25 30 35 40−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

detector separation [mm]

(PP

F −

PP

Fo)

/PP

Fo

Deviation of Sensitivity to Brain

0.001

0.01

0.1

1.0

(c)

(a)

(b)

−20

−15

−10Total fluence [TD]

−20

−15

−17.5

Log 1

0TP

SF

0.5 1 1.5 2 2.5 3−20

−18

−16

time steps [ns]

0.01

0.1

1.020 mm

30 mm

40 mm

0

0.5

1Deviation of fluence

0

0.5

1

(MC

− M

Co)

/ M

Co

0.5 1 1.5 2 2.5 3−0.5

0

0.5

time steps [ns]

0.001 0.01 0.1 1.0

20 mm

30 mm

40 mm

Diaphanography

• Approximation: Diffusion Equation

∆v(x,y)+ ρ(x,y)c(x,y) = 0v(x,y) flux densityρ(x,y) scattering coefficientc(x,y) absorption coefficient

• Forward: given c(x,y) find v(x,y)

3 × 3 node grid example

Here we have three input nodes (1, 2, 3), three output nodes (7, 8, 9), and three interior nodes (4, 5, 6).In three experiments we apply currents to each of the input nodes in turn, each time reading out all of theoutput nodes, yielding a total of nine measurements. We try and recover the nine leakage conductances toground from each of the nine nodes.

It is natural to partition the conductance matrix as follows given that I4, I5, I6, I7, I8, and I9 are alwayszero, and that we do not measure V1, V2, V3, V4, V5, and V6.

⎛⎜⎜⎜⎜⎜⎜⎝

G11... G12

... G13· · · · · · · · · · ·G21

... G22... G23

· · · · · · · · · · ·G31

... G32... G33

⎞⎟⎟⎟⎟⎟⎟⎠

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

V1V2V3· · ·V4V5V6· · ·V7V8V9

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

I1I2I3· · ·I4I5I6· · ·I7I8I9

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

In detail

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

g′1 −g12 −g13... −g14 0 0

... 0 0 0

−g12 g′2 −g23... 0 −g25 0

... 0 0 0

−g13 −g23 g′3... 0 0 g36

... 0 0 0· · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

−g14 −0 0... g′4 −g45 −g46

... −g47 0 00 −g25 0

... −g45 g′5 −g56... 0 −g58 0

0 0 −g36... −g46 −g56 g′6

... 0 −0 −g69· · · · · · · · · · · · · · · · · · · · · · · · · · · · ·0 0 0

... −g47 0 0... g′7 −g78 −g79

0 0 0... 0 −g58 0

... −g78 g′8 −g890 0 0

... 0 0 −g69... −g79 −g89 g′9

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

V1V2V3· · ·V4V5V6· · ·V7V8V9

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

I1I2I3· · ·I4I5I6· · ·I7I8I9

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

where g′1 = (g1 + g12 + g13 + g14), g′2 = (g2 + g12 + g23 + g25), g

′3 = (g3 + g13 + g23 + g36), g

′4 = (g4 + g14 +

g45 + g46 + g47), g′5 = (g5 + g25 + g45 + g56 + g47), g′6 = (g6 + g36 + g46 + g56 + g69), g

′7 = (g7 + g47 + g78 + g79),

g′8 = (g8 + g58 + g78 + g89), and g′9 = (g9 + g69 + g79 + g89).

We note that G13 and G31 are all zeros, and G12 = G21, and G23 = G32 are diagonal. Also, thesub-matrices appearing on the diagonal are of Toeplitz form. Tpeolitz matrices can be inverted in order N2

(instead of order N3).

5

We can write the inverse as follows:

⎛⎜⎜⎜⎜⎜⎜⎝

C11... C12

... C13· · · · · · · · · · ·C21

... C22... C23

· · · · · · · · · · ·C31

... C32... C33

⎞⎟⎟⎟⎟⎟⎟⎠

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

I1I2I3· · ·I4I5I6· · ·I7I8I9

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

V1V2V3· · ·V4V5V6· · ·V7V8V9

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

We can use the formula for the inverse of matrix partitioned into four parts twice on this matrix partitionedinto nine parts. But it may be a bit much to expect to easily obtain explicit formulae the way we did forthe 2 × 2 case. . .

Note that we are only really interested in the bottom left corner (C31) of the inverse, given that I4, I5,I6, I7, I8, and I9 are always zero, and that we do not measure V1, V2, V3, V4, V5, and V6. Each experimentyields three measurements and thus three equations of the form

C31

⎛⎝ I1I2

I3

⎞⎠ =

⎛⎝ V7V8

V9

⎞⎠ .

By performing three experiments we can find all nine elements of the matrix C31. Each of these is apolynomial in the unknown leakage conductances g1, g2, g3, g4, g5, g6, g7, g8, and g9 (or rather, we cancross-multiply to obtain nine such polynomials).

The part of the inverse of this conductance matrix that we need is the lower left corner, C31. Using thedecomposition rule for partitioned 2 × 2 matrices twice, we get

C31 = G−133 G32(G22 − G23G−133 G21)−1G21(G11 − G12(G22 − G23G−133 G21)−1G21

)−1

Note that the term (G22 −G23G−133 G21)−1G21 appears twice. This can be exploited to save on computation.

6

FORWARDSOLUTION

UPDATE RULE -"ASSIGNING BLAME"

I’(r1,r2)I(r1,r2)

c(k)(r)

Diaphanography

• “Invert” Diffusion Equation

• Regions of Influence

.


(1) Synthetic Aperture Imaging

(2) Coded Aperture Imaging

(3) Exact Cone Beam Reconstruction

(4) Diaphanography—DiffuseTomography

Synthetic Aperture Lithography

• Create pattern — controlled interferenceExample: Two Dots

Example: Straight Line

• Destructive interference “safe zone”Example: Bessel Ring

.

http://www.csail.mit.edu/~bkph/images/Binary_Stars.gifhttp://csail.mit.edu/~bkph/images/Linear_Growth.gifhttp://csail.mit.edu/~bkph/images/Color_Ring.gifhttp://csail.mit.edu/~bkph/images/Binary_Stars.gifhttp://csail.mit.edu/~bkph/images/Linear_Growth.gifhttp://csail.mit.edu/~bkph/images/Color_Ring.gif

SAM M4

Amplitude and Phase Control

Interference Pattern Texture

Fourier Transform of Texture Pattern

Berthold K.P. HornRectangle

Uneven Fourier Sampling

Resolution Enhancement

• Reflective Optics IlluminationVaccum UV — Short Wavelength

Resolution Enhancement

• Reflective Optics IlluminationVaccum UV — Short Wavelength

• Fluorescence ModeResolution Determined by Illumination

Masks — Fresnel Camera

Masks — Legri URA

Masks — XRT Fine

Masks — Hexagonal

Maximizing SNR

minn∑i=1w2i subject to

n∑i=1wi = 1

yields wi = 1n

Spatially Varying Background

Coded Aperture Extensions

• Artifacts due to Finite Distance• Mask / Countermask Combination

• Multiple Detector Array Positions• “Synthetic Aperture” Radiography

Dynamic Reconstruction

http://csail.mit.edu/~bkph/images/Coded_Backprojection.html

Diaphanography

• Approximation: Diffusion Equation

• Leaky Resistive Sheet Analog (2D)

(a)

(b)

(c)

1 2 30

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

120mm

time steps [ns]

Nor

mal

ized

PP

F

1 2 3

30mm

1 2 3

40mm

scalp skull

brain

0.010.11.0

−0.05

0

0.05Deviation of Sensitivity to Scalp−Skull

−0.1

0

0.1

(PP

F −

PP

Fo)

/PP

Fo

0.5 1 1.5 2 2.5 3−0.2

0

0.2

time steps [ns]

0.001 0.01 0.1 1.0

20 mm

30 mm

40 mm

−1

0

10.5

−0.5

Deviation of Sensitivity to Brain

−1

0

1

−0.5

0.50.5

(PP

F −

PP

Fo)

/PP

Fo

0.5 1 1.5 2 2.5 3−1

0

1

0.5

−0.5−0.5

time steps [ns]

0.001 0.01 0.1 1.0

20 mm

30 mm

40 mm

Sample Resistive Grid Inversion

Berthold K.P. Horn — 1996 February 14th

Here we consider about the simplest possible case of the two-d resistive sheet to get some insight into themore general problem.

(draw your own picture here :)There are four nodes, arranged in N = 2 rows and M = 2 columns. On the left are the two nodes used

for input ((1) and (2)) and on the right are the two output nodes ((3) and (4)). Four ‘horizontal’ resistorswith conductance g12, g13, g24 and g34 connect these four nodes. These resistors represent scattering, andare assumed to be of known value. There is also a ‘vertical’ leakage path from each of the four nodes toground — with conductances g1, g2, g3 and g4. These represent absorption, and are the unknowns.

We are to recover the values of the unknown leakage resistors. We perform two experiments: First weinject current at node (1) and measure the potentials on nodes (3) and (4). Call the ‘trans-impedance’ (ratioof output potential to injected current) observed this way R3,1 and R4,1. Then we inject instead current atnode (2) and again measure the potential on nodes (3) and (4). Call the ‘trans-impedance’ observed thisway R3,2 and R4,2.

If the grid was N ×M instead of 2×2 then we would have performed N experiments, each time injectingcurrent on one of the N input nodes and reading out the potential on each of the N output nodes. We thentry to recover the N ×M unknown leakage conductances to ground. Clearly there is not enough constraint ifM > N since there are then more unknowns than measurements. Conversely if M < N , we have redundantinformation and may want to use a least squares approach to obtain the best possible answer.

Here we deal with the simple case where M = N = 2. The node equations in this case are:

I1 = g1V1 + g13(V1 − V3) + g12(V1 − V2)I2 = g2V2 + g12(V2 − V1) + g24(V2 − V4)I3 = g3V3 + g13(V3 − V1) + g34(V3 − V4)I4 = g4V4 + g24(V4 − V2) + g34(V4 − V3)

or ⎛⎜⎜⎜⎜⎜⎜⎜⎝

(g1 + g13 + g12) −g12... −g13 0

−g12 (g2 + g12 + g24)... 0 −g24

· · · · · · · · · · · · ·−g13 0

... (g3 + g13 + g34) −g340 −g24

... −g34 (g4 + g24 + g34)

⎞⎟⎟⎟⎟⎟⎟⎟⎠

⎛⎜⎜⎜⎝

V1V2· · ·V3V4

⎞⎟⎟⎟⎠ =

⎛⎜⎜⎜⎝

I1I2· · ·I3I4

⎞⎟⎟⎟⎠

Note that all off diagonal elements are negative, and that the unknown leakage conductances all appear onthe diagonal. Also, the matrix becomes singular if all leakages conductances are set to zero, since then eachrow adds up to zero.

Making use of the partitioning indicated above, we can write

⎛⎜⎝ G11

... G12· · · · · · ·G21

... G22

⎞⎟⎠

⎛⎜⎜⎜⎝

V1V2· · ·V3V4

⎞⎟⎟⎟⎠ =

⎛⎜⎜⎜⎝

I1I2· · ·I3I4

⎞⎟⎟⎟⎠

This partitioning is convenient, since in the experiments we always have I3 = 0 and I4 = 0, and since V1 andV2 are not known.

If we invert this set of equations we get:

⎛⎜⎝ C11

... C12· · · · · · ·C21

... C22

⎞⎟⎠

⎛⎜⎜⎜⎝

I1I2· · ·I3I4

⎞⎟⎟⎟⎠ =

⎛⎜⎜⎜⎝

V1V2· · ·V3V4

⎞⎟⎟⎟⎠

1

whereC11 = (G11 − G12G−122 G21)−1C21 = −G−122 G21C11C22 = (G22 − G21G−111 G12)−1C12 = −G−111 G12C22

Upon multiplying out we obtain

C11

(I1I2

)=

(V1V2

)

where the term involving C12 drops out because I3 = 0 and I4 = 0. This equation is of no interest since wedon’t know V1 and V2. But we also obtain

C21

(I1I2

)=

(V3V4

)

where the term involving C22 drops out because I3 = 0 and I4 = 0. The four unkowns (g1, g2, g3, g4) occur inthe matrix C21, but we obviously can’t solve for them using a single set of measurements. We can however,combine two sets of measurements and obtain:

C21

(I1,1 I1,2I2,1 I2,2

)=

(V3,1 V3,2V4,1 V4,2

)

where typically we would choose I1,1 = 1, I2,1 = 0 for the first experiment and I1,2 = 0, I2,2 = 1 for thesecond. We then obtain

C21 =(

R3,1 R3,2R4,1 R4,2

)

where, provided I2,1 = 0 and I1,2 = 0, R3,1 = V3,1/I1,1, R4,1 = V4,1/I1,1, and R3,2 = V3,2/I1,2, R4,2 =V4,2/I1,2. So from image measurements we can recover the matrix C21, and we know that

C21 = −G−122 G21C11

orC21C

−111 = −G−122 G21

then, sinceC11 = (G11 − G12G−122 G21)−1

we obtainC21(G11 − G12G−122 G21) = −G−122 G21

We need to manipulate this some more to try and isolate the two matrices G11 and G22, which contain theunknowns g1, g2, g3, and g4. We see that

C21G11 = C21G12G−122 G21 − G−122 G21

orC21G11G

−121 G22 = C21G12 − I

or finallyG11G

−121 G22 = G12 − C−121

Here C−121 is obtained from experimental measurements, while G11 and G22 contain the unknown leakageconductances. In our simple example, G12 and G21 are diagonal.

2

Numerical Example

Suppose that the ‘horizontal’ resistors have conductance as follows: g13 = 1, g12 = 2, g24 = 1, and g34 = 2.Next assum that the leakage or ‘vertical’ resistors have conductance g1 = 1, g2 = 2, g3 = 2, and g4 = 1.Then the conductance matrix is ⎛

⎜⎜⎜⎜⎜⎜⎜⎝

4 −2 ... −1 0−2 5 ... 0 −1· · · · · · · · · · · · ·−1 0 ... 5 −20 −1 ... −2 4

⎞⎟⎟⎟⎟⎟⎟⎟⎠

and so

G−111 =116

(5 22 4

)and G−122 =

116

(4 22 5

)

and so

G12G−122 G21 = G

−122 =

116

(4 22 5

)

while

G21G−111 G12 = G

−111 =

116

(5 22 4

)

so

G11 − G12G−122 G21 =116

(60 −34

−34 75)

and

G22 − G21G−111 G12 =116

(75 −34

−34 60)

So in the inverse we have

C11 =1

209

(75 3434 60

)and C22 =

1209

(60 3434 75

)

and

C21 =1

209

(23 1620 23

)and C12 =

1209

(23 2016 23

)

Finally

G−1 = C =1

209

⎛⎜⎜⎜⎜⎜⎜⎝

75 34... 23 20

34 60... 16 23

· · · · · · · · · · · · ·23 16

... 60 34

20 23... 34 75

⎞⎟⎟⎟⎟⎟⎟⎠

In the first experiment we have I1 = 1 and the other node currents are zero so⎛⎜⎝

V1V2V3V4

⎞⎟⎠ = 1209

⎛⎜⎝

75342320

⎞⎟⎠ .

In the second experiment we have I2 = 1 and the other node currents are zero so⎛⎜⎝

V1V2V3V4

⎞⎟⎠ = 1209

⎛⎜⎝

34601623

⎞⎟⎠ .

3

Note that we can only measure V3 and V4 in each case. This is the end of the ‘forward’ problem (findingtrans-impedance given leakage conductances).

The ‘inverse’ task is to recover the unknown leakage conductances. Extracting the relevant parts fromthe above ‘experimental’ data we see that

C21 =1

209

(23 1620 23

).

so

C−121 =(

23 −16−20 23

).

so

G12 − C−121 =( −24 16

20 24

).

and

G11 =(

g1 + 3 −2−2 g2 + 3

)and G22 =

(g3 + 3 −2

−2 g4 + 3)

.

So

G11G−121 G22 =

(g1 + 3 −2

−2 g2 + 3) (

g3 + 3 −2−2 g4 + 3

).

So that we get the following equations in the unknown leakage conductances:

(g1 + 3)(g3 + 3) + 4 = 242(g1 + 3) + 2(g4 + 3) = 162(g3 + 3) + 2(g2 + 3) = 20(g2 + 3)(g4 + 3) + 4 = 24

orḡ1ḡ3 = 20

ḡ1 + ḡ4 = 8ḡ2 + ḡ3 = 10

ḡ2ḡ4 = 20

where ḡ1 = g1 +3, ḡ2 = g2 +3, ḡ3 = g3 +3, and ḡ4 = g4 +3. These equations have only one solution: ḡ1 = 4,ḡ2 = 5, ḡ3 = 5, and ḡ4 = 4, that is g1 = 1, g2 = 2, g3 = 2, and g4 = 1.

Summary

While this shows a solution method for a 2 × 2 grid, some of the points noted here also apply to the moregeneral case, although an explicit solution cannot be expected then. In general, the matrix would have to bepartitioned into a 3 × 3 arrangement corresponding to the fact that in addition to input nodes and outputnodes there are then also interior nodes.

4

We can write the inverse as follows:

⎛⎜⎜⎜⎜⎜⎜⎝

C11... C12

... C13· · · · · · · · · · ·C21

... C22... C23

· · · · · · · · · · ·C31

... C32... C33

⎞⎟⎟⎟⎟⎟⎟⎠

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

I1I2I3· · ·I4I5I6· · ·I7I8I9

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

V1V2V3· · ·V4V5V6· · ·V7V8V9

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

We can use the formula for the inverse of matrix partitioned into four parts twice on this matrix partitionedinto nine parts. But it may be a bit much to expect to easily obtain explicit formulae the way we did forthe 2 × 2 case. . .

Note that we are only really interested in the bottom left corner (C31) of the inverse, given that I4, I5,I6, I7, I8, and I9 are always zero, and that we do not measure V1, V2, V3, V4, V5, and V6. Each experimentyields three measurements and thus three equations of the form

C31

⎛⎝ I1I2

I3

⎞⎠ =

⎛⎝ V7V8

V9

⎞⎠ .

By performing three experiments we can find all nine elements of the matrix C31. Each of these is apolynomial in the unknown leakage conductances g1, g2, g3, g4, g5, g6, g7, g8, and g9 (or rather, we cancross-multiply to obtain nine such polynomials).

The part of the inverse of this conductance matrix that we need is the lower left corner, C31. Using thedecomposition rule for partitioned 2 × 2 matrices twice, we get

C31 = G−133 G32(G22 − G23G−133 G21)−1G21(G11 − G12(G22 − G23G−133 G21)−1G21

)−1

Note that the term (G22 −G23G−133 G21)−1G21 appears twice. This can be exploited to save on computation.

6

Binder11Binder10Binder9Binder6Binder5Binder4Binder3Binder2Binder1scattering_CIcomputational_imaging_new.pdfBinder3.pdfcomputational_imaging.pdfBinder5.pdfBinder2.pdfBinder1.pdfcomputational_imaging_new.pdftemp.pdf

temp.pdf

temp.pdf

temp1.pdftemp2.pdftemp3.pdftemp5.pdftemp7.pdftemp8.pdftemp9.pdf

projection1.pdfprojection2.pdf

projection1.pdfprojection2.pdf

temp.pdf

recon_figures

figures-AO1

threed_grid

threed_bottom

leakage

threed_grid

leakage

diaphanography_update

APL_Cover

threed_perspective

Documents

MIT CSAIL - Computational Imagingpeople.csail.mit.edu/bkph/talkfiles/Computational...3×3 node grid example Here we have three input nodes (1, 2, 3), three output nodes (7, 8, 9),