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Wavelets have widely been used in many signal and image processing applications. In this paper, a new
serial-parallel architecture for wavelet-based image compression is introduced. It is based on a 4-tap wavelet
transform, which is realised using some FIFO memory modules implementing a pixel-level pipeline
architecture to compress and decompress images. The real filter calculation over 4 · 4 window blocks is
done using a tree of carry save adders to ensure the high speed processing required for many applications.
The details of implementing both compressor and decompressor sub-systems are given. The primarily analysis
reveals that the proposed architecture, implemented using current VLSI technologies, can process a
video stream in real time.
標簽：
applications
processing
Wavelets
widely
上傳時間：
2014-01-22
上傳用戶：hongmo
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平均因子分解法，適用于正定矩陣First, let s recall the definition of the Cholesky decomposition: given a symmetric positive definite square matrix X, the Cholesky decomposition of X is the factorization X=U U, where U is the square root matrix of X, and satisfies:
(1) U U = X
(2) U is upper triangular (that is, it has all zeros below the diagonal).
It seems that the assumption of positive definiteness is necessary. Actually, it is "positive definite" which guarantees the existence of such kind of decomposition.
標簽：
分解
上傳時間：
2013-12-24
上傳用戶：啊颯颯大師的
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光學設計軟件zemax源碼：
This DLL models an nular aspheric
surface as described in:
"Annular surfaces in annular field systems"
By Jose M. Sasian
Opt. eng. 36 (12) P 3401-3401 December 1997
This surface is essentially an odd aspheric surface with an offset in the aspheric terms.
The sag is given by:
Z = (c*r*r) / (1+(1-((1+k)*c*c*r*r))^ 1/2 ) + a*(r-q)^2 + b*(r-q)^3 + c*(r-q)^4 + ...
Note the terms a, b, c, ... have units of length to the -1, -2, -3, ... power.
標簽：
described
aspheric
surfaces
Annular
上傳時間：
2014-01-08
上傳用戶：yyyyyyyyyy
-
zemax源碼：
This DLL models an anamorphic aspheric surface.
This surface is essentially an even aspheric surface with different terms for
the x and y directions.
The sag is given by:
Z = ((CX*x*x)+(CY*y*y)) / (1 + sqrt(1-((1+KX)*CX*CX*x*x)-((1+KY)*CY*CY*y*y)))
+ AR*( (1 - AP)*x*x + (1 + AP)*y*y )^2
+ BR*( (1 - BP)*x*x + (1 + BP)*y*y )^3
+ CR*( (1 - CP)*x*x + (1 + CP)*y*y )^4
+ DR*( (1 - DP)*x*x + (1 + DP)*y*y )^5
Note the terms AR, BR, CR, and DR ... have units of length to the -3, -5, -7, and -9 power.
The terms AP, BP, CP, and DP are dimensionless.
The surface is rotationally symmetric only if AP = BP = CP = DP == 0 and CX = CY and KX = KY.
標簽：
surface
This
essentially
anamorphic
上傳時間：
2015-07-25
上傳用戶：lht618
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LINEINTRGAL Line Integral in a 2D Vector Field.
LINEINTRGAL(X,Y,U,V,C) computes the line integral along the lines given in cell array C.
X and Y define the coordinates of a rectangular grid over which
U and V are defined. X and Y must be monotonic and 2D plaid as
％ produced by MESHGRID. X, Y, U, and V must all be the same size.
標簽：
LINEINTRGAL
Integral
computes
integral
上傳時間：
2014-01-13
上傳用戶：hwl453472107
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ACM試題An Easy Problem
Description
As we known, data stored in the computers is in binary form. The problem we discuss now is about the positive integers and its binary form. given a positive integer I, you task is to find out an integer J, which is the minimum integer greater than I, and the number of 1 s in whose binary form is the same as that in the binary form of I. For example, if "78" is given, we can write out its binary form, "1001110". This binary form has 4 1 s. The minimum integer, which is greater than "1001110" and also contains 4 1 s, is "1010011", i.e. "83", so you should output "83".
標簽：
Description
computers
Problem
binary
上傳時間：
2013-12-11
上傳用戶：libenshu01
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Listed below are the typographical conventions used in this guide.
– Example C++ code and commands to be typed by the user are in non-bold characters in typewriter
font.
– Items where the user has to supply a name or number are given in lower-case italic characters in
typewriter font.
– Sections marked with a ‡ describe features that are also available in ANSI C.
標簽：
typographical
conventions
commands
Example
上傳時間：
2013-12-20
上傳用戶：xiaoxiang
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This code was used for making the practical measurements in section 2.3 of my thesis. This Matlab code allows an OFDM signal to be generated based on an input data file. The data can be random data, a grey scale image, a wave file, or any type of file. The generated OFDM signal is stored as a windows wave file, allowing it to be viewed, listened to and manipulated in other programs. The modified wave file can then be decoded by the receiver software to extract the original data. This code was developed for the experiments that I performed in my honours thesis, and thus has not been fully debugged.
This is the original code developed for the thesis and so has several problems with it. The BER performance given by the simulations is infact Symbol Error Rate.
標簽：
This
measurements
practical
section
上傳時間：
2015-09-20
上傳用戶：tedo811
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acm HDOJ 1051WoodenSticks
Description:
There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to prepare processing a stick. The setup times are associated with cleaning operations and changing tools and shapes in the machine. The setup times of the woodworking machine are given as follows:
(a) The setup time for the first wooden stick is 1 minute.
(b) Right after processing a stick of length l and weight w , the machine will need no setup time for a stick of length l and weight w if l<=l and w<=w . Otherwise, it will need 1 minute for setup.
標簽：
WoodenSticks
Description
length
wooden
上傳時間：
2014-03-08
上傳用戶：netwolf
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Pursuant to the ETSI IPR Policy, no investigation, including IPR searches, has been carried out by ETSI. No guarantee
can be given as to the existence of other IPRs not referenced in ETSI SR 000 314 (or the updates on the ETSI Web
server) which are, or may be, or may become, essential to the present document.
標簽：
investigation
IPR
including
Pursuant
上傳時間：
2013-12-12
上傳用戶：英雄
