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CSCI 510/EENG 510 Image and Multidimensional Signal Processing Homework Assignment 0 - SOLUTIONS This assignment is a review of some pre-requisite topics. It is not necessary to turn it in, although you should do these problems to make sure you understand the concepts. The first four problems are a review of some mathematical concepts. You should be able to do these problems by hand, but it is ok to check with a computer. 4 − 2 . 1. Consider the matrix A = 1 1 a) Find the transpose of A b) Find the determinant of A c) Find the trace of A d) Find the inverse of A. e) Which of the following vectors is an eigenvector of A? x = (-1 2)T x = (2 1)T x = (0 1)T x = (1 0)T Solution: 4 1 a) A T = − 2 1 b) |A| = det(A) = (4)(1) – (-2)(1) = 6. c) tr(A) = 4+1 = 5. 1/ 6 1/ 3 d) You can use the “cofactor” method. A −1 = − 1 / 6 2 / 3 e) The definition of an eigenvector is Ax = λx, where λ is the eigenvalue for that eigenvector. Trying each of the possibilities, we see that A (2 1)T = (6 3) T = 3(2 1)T . So (2 1)T is an eigenvector, and λ = 3 is the corresponding eigenvalue. 2. Consider the vectors x = (1 2 3)T and y = (-1 2 -3)T. a) Compute the inner (dot) product x∙y. b) Compute the vector (cross) product x × y . Solution: a) -6 b) (-12 0 4)T 3. Consider the 8-bit binary number B = 11000001. a) Find the decimal value of B if it represents an unsigned binary number b) Find the decimal value of B if it represents a two’s complement binary number. Solution: a) 1*2^7 + 1*2^6 + 0 + 0 + 0 + 0 + 0 + 1 = 193. b) -1*2^7 + 1*2^6 + 0 + 0 + 0 + 0 + 0 + 1 = -63. 1 CSCI 510/EENG 510 Image and Multidimensional Signal Processing 4. The faces of a 10-sided die are numbered 0 through 9. a) What is the probability that the value of the roll is a prime number? b) What is the expected value of the roll? c) If the die is rolled twice, what is the probability that the same number is obtained both times? Solution: a) The possible prime numbers in the range 0..9 are 2, 3, 5, and 7. There are four outcomes (out of 10), so the probability is 0.4. 10 b) The expected value is xavg = E ( x ) = ∑ xi p( xi ) = (0)(.1)+(1)(.1)+(2)(.1)+...+(9)(.1) = i =1 4.5. c) The probability that any particular number (such as a “1”) is rolled twice is (0.1)(0.1)=0.01. There are 10 possible outcomes like this, so the total probability is (10)(0.01) = 0.1. 5. Matlab exercises. a) Create a matrix A, any size and shape you want, and fill it with arbitrary values. Compute AAT and ATA. The results should be square, symmetric matrices. b) Write a Matlab program to produce the first 30 Fibonacci numbers. Solution: a) >> A = rand(3,5) A= 0.0357 0.6787 0.8491 0.7577 0.9340 0.7431 >> A*A' ans = 1.1164 0.8287 0.8287 1.7353 0.8384 1.5572 >> A'*A ans = 1.5946 1.3617 1.3617 1.5871 0.7305 0.8901 0.3109 0.7091 0.8532 0.7169 0.3922 0.7060 0.0462 0.6555 0.0318 0.0971 0.1712 0.2769 0.8235 0.8384 1.5572 2.2087 0.7305 0.8901 0.6128 0.3452 0.2227 0.3109 0.7091 0.3452 0.5762 0.2637 0.8532 0.7169 0.2227 0.2637 0.6896 b) % Generate Fibonacci numbers. % Fibonacci numbers are defined as the sequence F(n) = F(n-1) + F(n-2), % starting with the numbers F(0)=0 and F(1)=1. % Good idea to have these two instructions at the beginning of any program. clear all % Clear workspace of any old variables 2 CSCI 510/EENG 510 close all Image and Multidimensional Signal Processing % Close all open figures n1 = 0; fprintf('1 %d\n', n1); % To print to console, use "fprintf" or "disp" n2 = 1; fprintf('2 %d\n', n2); for i=3:30 temp = n1 + n2; fprintf('%d %d\n', i, temp); n1 = n2; n2 = temp; end Output is: 10 21 31 42 53 65 78 8 13 9 21 10 34 11 55 12 89 13 144 14 233 15 377 16 610 17 987 18 1597 19 2584 20 4181 21 6765 22 10946 23 17711 24 28657 25 46368 26 75025 27 121393 28 196418 29 317811 30 514229 3