MAP 6385 Convolution and Image Processing
k = 1/2
Denition 1. Let z, w l2 (R), the set of sequences indexed on Z such that z (m)2
The convolution z w of z and w is the sequence dened by the equation
(z w)(m) =
z (k )w(m k )
September 9, 2009
WARNING: There are some big Mathematica problems on here. One of them is an openended question that requires experimentation. You probably shouldnt wait until the last
minute to do these.
0) In class, we found the gen
MAP 6385 Homework 2
September 2, 2009
1) Graph the rst ve Legendre polynomials (P0 (x) through P4 (x) on the same graph. Then plot the rst ve Chebyshev polynomials (T0 (x) through T4 (x) on the same graph. Compare the two results. What similarities do you
August 27, 2009
1) The following table gives a list of data points. Find the least squares polynomials of
degrees 1, 2, and 3 for this data set. For each degree, calculate the error E , and plot the
polynomial versus the data set.
1) You are allowed to work together. HOWEVER, if you do work with someone else,
then when you submit your homework, I want to know with whom you worked. If
someone else solved a problem and then helped you, tell me. Give credit whe
Steps for the Huffman Compression 1) Write down all of the letters appearing in the text in a list, most frequent to least frequent. For each letter, put the frequency as a subscript (it will look like a scrabble tile). For our example
MAP 6385 Exam 2
November 5, 2009
Same rules apply as from Exam 1. The exam is due on November 12 at 4:30 PM. Note that Problems 7 and 8 do not have explicit answers: they are more experimentation problems. For credit, you need to describe what you tried t
MAP 6385 DWT, Part 3
Denition 18. Let M be an even number. Dene the down operator D : CM CM/2 by D(z )(k ) = z (2k 1) Furthermore, if M is any number, then dene the up operator U : CM C2M by k+1 z ( 2 ) if k is odd U (z )(k ) = 0 if k is even Theorem 19.
MAP 6385 DWT, Part 2
Example 14. Let M be a number divisible by 4. The First Stage Shannon Wavelet Basis is the wavelet basis generated by the father and mother wavelets u and v satisfying: 2 if 1 n M or 3M + 1 n M 4 4 u(n) = 0 if M + 1 n 3M 4 4 0 if 1 n
MAP 6385 DWT
Definition 1. A vector z CM is time localized if its entires with large magnitudes are clustered around a particular index. A vector z is frequency localized if the entries of z with ^ large magnitudes are clustered around a particular index.
MAP 6385 DFT, Part 2
Denition: Let z and w be vectors in CM . The convolution of z and w, written as z w, is the vector dened by
z w(m) =
z (m n + 1)w(n)
Theorem: Let z , w, and y be vectors in CM , and let c C. Then: i) z w = w z ii) (z w) y = z (w
MAP 6385 DFT
Definition: Let CM be the vector m-dimensional complex vector space. We represent a typical element by z, written out as: z = [z(1), z(2), . . . , z(M )] The inner product on CM is defined as
z, w =
We make our vector space cyc
September 18, 2009
1) By hand (yes, by hand), calculate z , where z = [1, 0, 3, 2].
2) I have said that z z is a linear transformation. This means we can represent it as a
matrix multiplication. Find the matrix for when M = 4.
3) a) As