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Solution to ECE544 Homework 2: Image Processing and Statistical Modeling - Prof. Pierre Mo, Assignments of Electrical and Electronics Engineering

University of Illinois - Urbana-ChampaignElectrical and Electronics Engineering
Prof. Pierre Moulin

Solutions to problem 1 and 2 of the ece544 homework 2. It includes the estimation of mean and correlation matrices for given images, determination of appropriate models for each image based on correlation matrices, and the design of an energy function for favoring specific configurations in problem 2. The document also mentions the use of the gibbs sampler for simulating ising and 8-level potts models in problems 3 and 4.

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Pre 2010

Uploaded on 03/16/2009

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Solution to ECE544 HW2
Problem 1
(a) The mean and 4 ×4 correlation matrices can be estimated as follows:
ˆµ=1
NX
n
x(n)
ˆ
R(k) = 1
NX
n
x(n)x(n+k), k ∈ {0,1,2,3}2.
We obtain the following values for the five images considered. In each case, a least-
squares fit could be used to determine whether ρkkk(isotropic model) or ρ|k1|+|k2|(sepa-
rable model) is more appropriate.
Elaine: ˆµ= 136.3565, ˆ
R=
2121.1 2066.6 2014.5 1960.2
2058.4 2047.6 2000.1 1935.2
2025.1 2002.9 1958.2 1907.9
1969.4 1952.7 1931.6 1874.9
Man: ˆµ= 89.0075, ˆ
R=
3355.7 3278.3 3166.9 3070.8
3290.1 3241.0 3143.4 3053.6
3200.9 3168.7 3096.9 3020.3
3120.1 3097.8 3044.0 2980.2
Aerial: ˆµ= 180.5718, ˆ
R=
1555.9 1398.7 1151.6 955.8
1331.7 1240.8 1060.1 893.9
1041.1 1009.3 920.5 814.2
839.4 825.2 782.9 721.9
Clock: ˆµ= 185.9803, ˆ
R=
3277.5 3127.5 2973.0 2841.3
3190.7 3068.3 2932.4 2809.9
3055.2 2960.7 2855.1 2751.1
2935.7 2856.2 2773.2 2687.1
Couple: ˆµ= 123.1772, ˆ
R=
1604.9 1501.1 1401.5 1334.4
1411.9 1351.0 1277.8 1223.7
1261.7 1216.6 1159.6 1117.0
1144.7 1107.9 1059.5 1022.7
1
pf3
pf4
pf5

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Solution to ECE544 HW

Problem 1

(a) The mean and 4 × 4 correlation matrices can be estimated as follows:

μˆ =

N

n

x(n)

Rˆ(k) =

N

n

x(n)x(n + k), k ∈ { 0 , 1 , 2 , 3 }

2 .

We obtain the following values for the five images considered. In each case, a least-

squares fit could be used to determine whether ρ

‖k‖ (isotropic model) or ρ

|k 1 |+|k 2 | (sepa-

rable model) is more appropriate.

Elaine: ˆμ = 136.3565, Rˆ =

Man: ˆμ = 89.0075, ˆR =

Aerial: ˆμ = 180.5718, Rˆ =

Clock: ˆμ = 185.9803, Rˆ =

Couple: ˆμ = 123.1772, Rˆ =

(b) This is a separable AR(1) process:

x(n 1 , n 2 ) = 100 + y(n 1 , n 2 )

where

y(n 1 , n 2 ) = 1 − ρy(n 1 − 1 , n 2 ) − ρy(n 1 − 1 , n 2 ) + ρ

2 y(n 1 − 1 , n 2 − 1) + e(n 1 , n 2 )

and e(n 1 , n 2 ) is white noise with mean 0 and variance 30(1 − ρ

2 )

2

. Realizations of this

process may be obtained by initializing the boundary of the image x (e.g., with zeroes)

and recursively applying the formula above. A more efficient implementation would

exploit the separability nature of the process by implementing horizontal filtering in a

first step, and vertical filtering in a second step:

z(n 1 , n 2 ) = ρz(n 1 − 1 , n 2 ) + e(n 1 , n 2 )

y(n 1 , n 2 ) = ρy(n 1 , n 2 − 1) + z(n 1 , n 2 )

(c) For ρ = 0. 1 , 0. 5 , 0 .95 we obtain the following realizations:

ρ = 0. 1 ρ = 0. 5 ρ = 0. 95

Problem 3: Simulation of Ising model.

After 100 iterations of the Gibbs sampler, we obtain the following configurations:

T

− 1 = 0. 1 T

− 1 = 0. 5 T

− 1 = 0. 88

T

− 1 = 1. 5 T

− 1 = 10

Observe that clusters become larger as the temperature parameter T decreases.

Problem 4: Simulation of 8-level Potts model.

After 100 iterations of the Gibbs sampler, we obtain the following configurations:

T

− 1 = 0. 1 T

− 1 = 0. 5 T

− 1 = 0. 88

T

− 1 = 1. 5 T

− 1 = 10

Observe that clusters become larger as the temperature parameter T decreases.