CN100517298C

Movatterモバイル変換

Info

Publication number: CN100517298C
Application number: CNB2004800350550A
Authority: CN
Inventors: 黄海滨; 林晓; S·拉哈尔贾; 俞容山
Original assignee: Agency for Science Technology and Research Singapore
Current assignee: Agency for Science Technology and Research Singapore
Priority date: 2003-09-29
Filing date: 2004-05-06
Publication date: 2009-07-22
Anticipated expiration: 2024-05-06
Also published as: CN1918562A; CN100570597C; CN1886737A; CN1882938A; CN100520765C

Abstract

A method for transforming a digital signal from the time domain to the frequency domain and vice versa using a transform function, said transform function comprising a transform matrix, said digital signal comprising a plurality of digital symbols grouped into a plurality of blocks, wherein each block comprises a predetermined number of said digital symbols, said method comprising: transforming two blocks of the digital signal with one transform element, wherein the transform element corresponds to a block diagonal matrix comprising two sub-matrices, wherein each sub-matrix comprises the transform matrix, the transform element comprises a plurality of lifting stages, and wherein each lifting stage comprises processing of a block of the digital signal by an auxiliary transform and rounding unit.

Description

Translated fromChinese

将数字信号从时域变换到频域及其反向变换的方法Method for transforming digital signal from time domain to frequency domain and its reverse transformation

相关申请的交叉引用Cross References to Related Applications

本申请要求2003年9月29日提交的美国临时申请No.60/507,210以及2003年9月29日提交的美国临时申请No.60/507,440的优先权，在此将每个的内容全文引入作为参考，以用于所有目的。This application claims priority to U.S. Provisional Application No. 60/507,210, filed September 29, 2003, and U.S. Provisional Application No. 60/507,440, filed September 29, 2003, the contents of each of which are incorporated herein in their entirety as reference for all purposes.

此外，下述共同拥有的申请一起同时提交，在此全文引入：In addition, the following commonly owned applications are filed concurrently and are hereby incorporated in their entirety:

“Method for Performing a Domain Transformation of a DigitalSignal from the Time Domain into the Frequency Domain and vice Versa”，代理案卷号No.P100442，以及"Method for Performing a Domain Transformation of a Digital Signal from the Time Domain into the Frequency Domain and vice Versa", Agent Docket No.P100442, and

“Process and Device for Determining a Transforming Element for aGiven Transformation Function，Method and Device for Transforming aDigital signal from the Time Domain into the Frequency Domain and viceVersa and Computer Readable Medium”，代理案卷号No.P100452。"Process and Device for Determining a Transforming Element for a Given Transformation Function, Method and Device for Transforming a Digital signal from the Time Domain into the Frequency Domain and viceVersa and Computer Readable Medium", agency case number No.P100452.

背景技术Background technique

本发明涉及用于将数字信号从时域变换到频域以及从频域变换到时域的方法。The invention relates to a method for transforming a digital signal from the time domain to the frequency domain and from the frequency domain to the time domain.

域变换，例如离散余弦变换(DCT)，被广泛地应用于当今信号处理工业。近年来，因为其在无损编码应用中的重要角色，称为整数DCT的DCT的变形已经吸引了许多研究兴趣。术语“无损”意味着解码器可以根据已编码的比特流产生源信号的确切复制。Domain transforms, such as the discrete cosine transform (DCT), are widely used in today's signal processing industry. In recent years, a variant of the DCT known as the integer DCT has attracted much research interest because of its important role in lossless coding applications. The term "lossless" means that a decoder can produce an exact replica of the source signal from the encoded bitstream.

所述DCT是实值块变换。即使所述输入块仅仅包括整数，所述DCT的输出块可以包括非整数分量。为了简便，所述输入块被称为输入矢量，而输出块被称为输出矢量。如果矢量仅仅包括整数分量，则该矢量被称为整数矢量。对照于DCT，所述整数DCT根据整数输入矢量产生整数输出矢量。对于同一整数输入矢量，整数DCT的整数输出矢量非常接近于DCT的实输出矢量。因此，整数DCT在频谱分析时保持所述DCT的所有良好的特性。The DCT is a real-valued block transform. Even if the input block only includes integers, the output block of the DCT may include non-integer components. For simplicity, the input blocks are called input vectors and the output blocks are called output vectors. A vector is called an integer vector if it only includes integer components. In contrast to DCT, the integer DCT produces integer output vectors from integer input vectors. For the same integer input vector, the integer output vector of the integer DCT is very close to the real output vector of the DCT. Thus, the integer DCT retains all the good properties of said DCT during spectral analysis.

所述整数DCT的重要特性是可逆性。可逆性意味着存在整数离散余弦反变换(IDCT)，使得如果所述整数DCT根据输入矢量x产生输出矢量y，则所述整数IDCT可以根据矢量y恢复出矢量x。有时整数DCT也被称为正向变换，整数IDCT被称为反向变换或反变换。An important property of the integer DCT is reversibility. Reversibility means that there is an Inverse Integer Discrete Cosine Transform (IDCT) such that if the integer DCT produces an output vectory from an input vectorx , the integer IDCT can recover a vectorx from vectory . Sometimes the integer DCT is also called the forward transform, and the integer IDCT is called the inverse transform or inverse transform.

称为整数改进离散余弦变换(IntMDCT)的变换近年被提出且被用于ISO/IEC MPEG-4音频压缩中。所述IntMDCT源于其原型---改进离散余弦变换(MDCT)。H.S.Malvar在1992年的“Signal Processing With lappedTransforms”中的公开通过利用DCT-IV块来级联一系列的Givens旋转来有效地实现MDCT。已经熟知的是，Givens旋转可以被分解为三个提升步骤，用于将整数映射到整数。例如，参见R.Geiger、T.Sporer、J.Koller、K.Brandenburg在2001年9月在美国纽约AES第111次会议上的“AudioCoding based on Integer Transforms”。A transform called the Integer Modified Discrete Cosine Transform (IntMDCT) was recently proposed and used in ISO/IEC MPEG-4 audio compression. The IntMDCT is derived from its prototype - Modified Discrete Cosine Transform (MDCT). H.S. Malvar's disclosure in "Signal Processing With lapped Transforms" in 1992 efficiently implements MDCT by utilizing a DCT-IV block to cascade a series of Givens rotations. It is already well known that the Givens rotation can be decomposed into three lifting steps for mapping integers to integers. For example, see "AudioCoding based on Integer Transforms" by R.Geiger, T.Sporer, J.Koller, K.Brandenburg at the 111th AES Conference in New York, USA in September 2001.

因此，IntMDCT的实现依赖于整数DCT-IV的有效实现。Therefore, the implementation of IntMDCT depends on the efficient implementation of integer DCT-IV.

通过利用三个提升步骤替换每个Givens旋转，可以从其原型直接转换整数变换。由于在每个提升步骤中存在一个四舍五入操作，整数变换的总四舍五入次数是原型变换的Givens旋转次数的3倍。对于离散三角变换(例如离散傅立叶变换(DFT)或离散余弦变换(DCT))，所涉及的Givens旋转的次数通常为Nlog₂N级，其中N是所述块的大小，即每个块中包括的所述数字信号被划分成的数据符号的量。因此，对于直接转换的整数变换的家族，所述总四舍五入次数也为Nlog₂N级。由于所述四舍五入，整数变换仅仅近似于其浮点原型。所述近似误差随着四舍五入的次数的增加而增加。Integer transformations can be directly converted from their prototypes by replacing each Givens rotation with three lifting steps. Since there is one rounding operation in each lifting step, the total number of roundings for the integer transformation is 3 times the number of Givens rotations for the prototype transformation. For discrete triangular transforms (such as discrete Fourier transform (DFT) or discrete cosine transform (DCT)), the number of Givens rotations involved is usually Nlog₂ N levels, where N is the size of the block, that is, each block includes The number of data symbols that the digital signal is divided into. Thus, for the family of directly converted integer transforms, the total number of roundings is also Nlog₂ N levels. Because of the rounding, integer transformations only approximate their floating-point prototypes. The approximation error increases with the number of roundings.

因此，所需要的是用于以更为有效的方式来对数字信号进行域变换的系统和方法。What is needed, therefore, are systems and methods for domain transforming digital signals in a more efficient manner.

发明内容Contents of the invention

本发明提供用于对数字信号进行域变换，由此在同一操作中同时对两个数据输入块进行域变换的系统和方法。这种配置减少了有效四舍五入操作的次数，并且因此减少近似误差。The present invention provides systems and methods for domain transforming digital signals, whereby two data input blocks are simultaneously domain transformed in the same operation. This configuration reduces the number of effective rounding operations, and thus reduces approximation errors.

在本发明的一个实施例中，呈现本发明的一种方法，该方法使用变换函数来将数字信号从时域变换到频域以及从频域变换到时域。所述变换函数包括变换矩阵，所述数字信号包括被分组为多个块的多个数据符号，每个块包括预定数目的数据符号。所述方法包括利用一个变换元素来变换数字信号的两个块，其中所述变换元素对应于包括两个子矩阵的块对角矩阵，其中每个子矩阵包括变换矩阵，而变换元素包括多个提升级(liftingstage)，其中每个提升级包括利用辅助变换和四舍五入单元来对数字信号的块进行处理。In one embodiment of the invention, a method of the invention is presented which uses a transform function to transform a digital signal from the time domain to the frequency domain and from the frequency domain to the time domain. The transform function includes a transform matrix, the digital signal includes a plurality of data symbols grouped into a plurality of blocks, each block including a predetermined number of data symbols. The method includes transforming two blocks of a digital signal with a transform element, wherein the transform element corresponds to a block diagonal matrix comprising two sub-matrices, wherein each sub-matrix comprises a transform matrix, and the transform element comprises a plurality of lifting stages (lifting stage), where each lifting stage includes processing blocks of digital signals with auxiliary transform and rounding units.

当按照附图和具体实施例的详细描述来观看时，本发明的这些和其他特征将更好理解。These and other features of the invention will be better understood when viewed in light of the accompanying drawings and detailed description of specific embodiments.

附图说明Description of drawings

图1示出了根据本发明的实施例的音频编码器的体系结构；Figure 1 shows the architecture of an audio encoder according to an embodiment of the invention;

图2示出了根据本发明的实施例的音频解码器的体系结构，其对应于图1中示出的音频编码器；Figure 2 shows the architecture of an audio decoder according to an embodiment of the invention, which corresponds to the audio encoder shown in Figure 1;

图3示出了根据本发明的方法的实施例的流程图；Figure 3 shows a flow chart of an embodiment of the method according to the invention;

图4说明了根据本发明的方法的实施例，其使用DCT-IV作为变换函数；Figure 4 illustrates an embodiment of the method according to the invention using DCT-IV as the transformation function;

图5说明了用于根据图4中说明的本发明的方法的实施例的反变换的算法；FIG. 5 illustrates an algorithm for inverse transformation according to an embodiment of the method of the invention illustrated in FIG. 4;

图6示出了根据本发明的实施例的图像归档系统的体系结构；Figure 6 shows the architecture of an image archiving system according to an embodiment of the present invention;

图7示出了用于估计所述提出的系统和方法的性能的正变换编码器和反变换编码器。Fig. 7 shows the forward transform coder and the inverse transform coder used to estimate the performance of the proposed system and method.

发明详述Detailed description of the invention

图1示出了根据本发明的实施例的音频编码器100的体系结构。所述音频编码器100包括基于改进离散余弦变换(MDCT)的常规感知基本层编码器(perceptual base layer coder)和基于整数改进离散余弦变换(IntMDCT)的无损增强编码器(enhancement coder)。Fig. 1 shows the architecture of anaudio encoder 100 according to an embodiment of the present invention. Theaudio encoder 100 includes a conventional perceptual base layer coder based on Modified Discrete Cosine Transform (MDCT) and a lossless enhancement coder based on Integer Modified Discrete Cosine Transform (IntMDCT).

例如，将由麦克风110提供且由模/数转换器111进行数字化的音频信号109提供给音频编码器100。所述音频信号109包括多个数据符号。所述音频信号109被分为多个块，其中每个块包括数字信号的多个数据符号，并且由改进离散余弦变换(MDCT)设备101对每个块进行变换。所述MDCT系数由量化器103借助于感知模型102来进行量化。所述感知模型按照这样一种方式控制所述量化器103，使得由量化误差产生的声音失真低。随后由比特流编码器104对已量化的MDCT系数进行编码，该比特流编码器104产生有损的感知编码的(perceptually coded)输出比特流112。For example, anaudio signal 109 provided by amicrophone 110 and digitized by an analog-to-digital converter 111 is provided to theaudio encoder 100 . Theaudio signal 109 includes a plurality of data symbols. Theaudio signal 109 is divided into a plurality of blocks, wherein each block comprises a plurality of data symbols of a digital signal, and each block is transformed by a Modified Discrete Cosine Transform (MDCT)device 101 . The MDCT coefficients are quantized by aquantizer 103 with the aid of aperceptual model 102 . The perceptual model controls thequantizer 103 in such a way that the sound distortion caused by quantization errors is low. The quantized MDCT coefficients are then encoded by abitstream encoder 104 which produces a lossy perceptually codedoutput bitstream 112 .

所述比特流编码器104利用诸如Huffman编码或游程(Run-Length)编码的标准方法无损地压缩其输入以产生一输出，该输出的平均比特率要低于其输入的平均比特率。所述输入音频信号109也被输送到产生IntMDCT系数的IntMDCT设备105中。由量化器103的输出的已量化MDCT系数被用于预测所述IntMDCT系数。所述已量化MDCT系数被输送到逆-量化器106中，并且所述输出(已恢复或非量化的MDCT系数)被输送到四舍五入单元107。Thebitstream encoder 104 losslessly compresses its input using standard methods such as Huffman coding or Run-Length coding to produce an output with an average bit rate lower than the average bit rate of its input. Theinput audio signal 109 is also fed to anIntMDCT device 105 which generates IntMDCT coefficients. The quantized MDCT coefficients output by thequantizer 103 are used to predict the IntMDCT coefficients. The quantized MDCT coefficients are fed into an inverse-quantizer 106 and the output (recovered or non-quantized MDCT coefficients) is fed to arounding unit 107 .

所述四舍五入单元将所述提供的MDCT系数四舍五入到一个整数值，并且由熵编码器108对残余的IntMDCT系数进行熵编码，所述残余的IntMDCT系数是整数值MDCT和IntMDCT系数之差。所述熵编码器，类似于比特流编码器104，无损地减少它的输入的平均比特率，并且产生无损增强比特流113。所述无损增强比特流113和感知编码比特流112一起承载必需的信息，以重构具有最小误差的输入音频信号109。The rounding unit rounds the supplied MDCT coefficients to an integer value and entropy codes residual IntMDCT coefficients, which are the difference between integer-valued MDCT and IntMDCT coefficients, by anentropy encoder 108 . The entropy encoder, similar to thebitstream encoder 104 , losslessly reduces the average bitrate of its input and produces a lossless enhancedbitstream 113 . The losslessenhanced bitstream 113 and the perceptually encodedbitstream 112 together carry the necessary information to reconstruct theinput audio signal 109 with minimal errors.

图2示出了包括本发明的实施例的音频解码器200的体系结构，其对应于图1中示出的音频编码器100。所述感知编码比特流207被提供给比特流解码器201，该比特流解码器201执行图1的比特流编码器104的操作的逆操作，产生已解码的比特流。所述已解码的比特流被提供给逆-量化器202，该逆-量化器202的输出(已恢复的MDCT系数)被提供给改进离散余弦反变换(反MDCT)设备203。因此，获得重构的感知编码音频信号209。FIG. 2 shows the architecture of anaudio decoder 200 comprising an embodiment of the invention, which corresponds to theaudio encoder 100 shown in FIG. 1 . The perceptually encodedbitstream 207 is provided to abitstream decoder 201 which performs the inverse of the operation of thebitstream encoder 104 of FIG. 1 to produce a decoded bitstream. The decoded bitstream is supplied to an inverse-quantizer 202 whose output (recovered MDCT coefficients) is supplied to an inverse modified discrete cosine transform (inverse MDCT)device 203 . Thus, a reconstructed perceptually encodedaudio signal 209 is obtained.

所述无损增强比特流208被提供给熵解码器204，该熵解码器204执行图1中的熵编码器108的操作的逆操作，产生相应的残余IntMDCT系数。由四舍五入设备205对逆-量化器202的输出进行四舍五入，以产生整数值MDCT系数。所述整数值MDCT系数被加到所述残余IntMDCT系数，由此产生所述IntMDCT系数。最后，由所述整数改进离散余弦反变换(反IntMDCT)设备206对所述IntMDCT系数进行所述整数改进离散余弦反变换，以产生所述重构的无损的已编码音频信号210。The losslessenhanced bitstream 208 is provided to anentropy decoder 204, which performs the inverse of the operation of theentropy encoder 108 in FIG. 1, producing corresponding residual IntMDCT coefficients. The output of the inverse-quantizer 202 is rounded by a roundingdevice 205 to produce integer-valued MDCT coefficients. The integer-valued MDCT coefficients are added to the residual IntMDCT coefficients, thereby generating the IntMDCT coefficients. Finally, the Inverse Integer Modified Discrete Cosine Transform (Inverse IntMDCT)device 206 performs the Inverse Integer Modified Discrete Cosine Transform on the IntMDCT coefficients to generate the reconstructed lossless encodedaudio signal 210 .

图3示出了根据本发明的方法的实施例的流程图300，该方法使用DCT-IV作为变换以及使用三个提升级，第一提升级301、第二提升级302以及第三提升级303。这个方法优选在图1的IntMDCT设备105和图2的反IntMDCT设备206中使用，以分别完成IntMDCT和反IntMDCT。在图3中，x₁和x₂分别是数字信号的第一块和第二块。z是中间信号，而y₁和y₂分别是与数字信号的第一块和第二块对应的输出信号。Figure 3 shows aflowchart 300 of an embodiment of a method according to the invention using DCT-IV as a transform and using three lifting stages, afirst lifting stage 301, asecond lifting stage 302 and athird lifting stage 303 . This method is preferably used in theIntMDCT device 105 of FIG. 1 and theinverse IntMDCT device 206 of FIG. 2 to perform IntMDCT and inverse IntMDCT, respectively. In Figure 3,x1 andx2 are the first and_second blocks_of the digital signal, respectively.z is the intermediate signal, andy1 andy2 are_the output signals corresponding_to the first and second blocks of the digital signal, respectively.

如上所述，DCT-IV算法在无损音频编码中扮演重要角色。As mentioned above, the DCT-IV algorithm plays an important role in lossless audio coding.

所述DCT-IV的变换函数包括变换矩阵C_N^IV。根据本发明的这个实施例，所述变换元素对应于包括两个块的块对角矩阵，其中每个块包括变换矩阵C_N^IV。The transformation function of the DCT-IV includes a transformation matrixC_N^IV . According to this embodiment of the invention, said transformation elements correspond to a block-diagonal matrix comprising two blocks, wherein each block comprises a transformation matrixC_N^IV .

因此，在这个实施例中，与根据本发明的变换元素对应的变换矩阵是：Thus, in this embodiment, the transformation matrix corresponding to the transformation elements according to the invention is:

$[\begin{matrix} \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} \\ \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} \end{matrix}]$

在这个实施例的上下文中，C_N^IV自此以后应该被称作变换矩阵。In the context of this embodiment_,CN^IV shall henceforth be referred to as a transformation matrix.

在本发明的这个实施例中，提升矩阵的数目，以及变换元素中的提升级的数目为3，其中DCT-IV是变换函数。In this embodiment of the invention, the number of lifting matrices, and the number of lifting levels in the transform elements is 3, where DCT-IV is the transform function.

N点实输入序列x(n)的DCT-IV被如下定义：The DCT-IV of an N-point real input sequence x(n) is defined as follows:

$y (m) = \sqrt{\frac{2}{N}} Σ_{n = 0}^{N - 1} x (n) \cos (\frac{(m + 1 / 2) (n + 1 / 2) π}{N})$ m，n＝0，1，...，N-1 (1) $the y (m) = \sqrt{\frac{2}{N}} Σ_{no = 0}^{N - 1} x (no) \cos (\frac{(m + 1 / 2) (no + 1 / 2) π}{N})$ m,n=0,1,...,N-1 (1)

假设C_N^IV是DCT-IV的变换矩阵，即，SupposeC_N^IV is the transformation matrix of DCT-IV, i.e.,

$\underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} = = \sqrt{\frac{22}{N N}} {[[cos cos ((\frac{((m m + + 11 / / 22)) ((n no + + 11 / / 22)) π π}{N N}))]]}_{m m,, n no = = 0,1 0,1,, . . . .,, N N - - 11} - - - - - - ((22))$

对于反DCT-IV矩阵，下述关系成立，For the inverse DCT-IV matrix, the following relation holds,

${((\underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}}))}^{- - 11} = = \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} - - - - - - ((33))$

特别地，矩阵C_N^IV是自逆矩阵(involutory)。In particular, the matrixC_N^IV is an involutory.

当x＝[x(n)]_{n＝0，1，...，N-1}和y＝[y(m)]_{m＝0，1，...，N-1}时，等式(1)可以表述为Whenx =[x(n)]_{n=0,1,...,N-1} andy =[y(m)]_{m=0,1,...,N-1} , equation (1 ) can be expressed as

$\underset{&OverBar; &OverBar;}{y the y} = = \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} \underset{&OverBar; &OverBar;}{x x} - - - - - - ((44))$

现在，假设x₁和x₂是两个整数N×1列矢量。所述列矢量x₁和x₂对应于数字信号的两个块，根据本发明，利用一个变换元素对该两个块进行变换。x₁和x₂的DCT-IV变换分别为y₁和y₂。Now, supposex1 andx2 are two N-_by -1 column vectors_of integers. Said column vectorsx1 andx2 correspond to two blocks of a digital signal which_are transformed according to the invention with one_transform element. The DCT-IV transforms ofx₁ andx₂ arey₁ andy₂ , respectively.

$\underset{&OverBar; &OverBar;}{{y the y}_{11}} = = \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV} {x x}_{11}} - - - - - - ((55))$

$\underset{&OverBar; &OverBar;}{{y the y}_{22}} = = \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV} {x x}_{22}} - - - - - - ((66))$

合并(5)和(6)：Combine (5) and (6):

$[\begin{matrix} \underset{&OverBar; &OverBar;}{{y the y}_{11}} \\ \underset{&OverBar; &OverBar;}{{y the y}_{22}} \end{matrix}] = = [\begin{matrix} \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} \\ \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} \end{matrix}] [\begin{matrix} \underset{&OverBar; &OverBar;}{{x x}_{11}} \\ \underset{&OverBar; &OverBar;}{{x x}_{22}} \end{matrix}] - - - - - - ((77))$

上述对角矩阵是根据本发明的变换元素对应的块对角矩阵。The above diagonal matrix is a block diagonal matrix corresponding to the transformed elements according to the present invention.

如果利用简单的代数修正来改变上述等式，例如导致If the above equation is changed using a simple algebraic correction, e.g. resulting in

$[\begin{matrix} \underset{&OverBar; &OverBar;}{{y the y}_{11}} \\ \underset{&OverBar; &OverBar;}{{y the y}_{22}} \end{matrix}] = = [\begin{matrix} \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} \\ \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} \end{matrix}] [\begin{matrix} \underset{&OverBar; &OverBar;}{{x x}_{22}} \\ \underset{&OverBar; &OverBar;}{{x x}_{11}} \end{matrix}] - - - - - - ((88))$

则仍在本发明的范围内。It is still within the scope of the present invention.

假设T_2N是(8)中的反(counter)对角矩阵，则Assuming thatT_2N is the inverse (counter) diagonal matrix in (8), then

$\underset{&OverBar; &OverBar;}{{T T}_{22 N N}} = = [\begin{matrix} \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} \\ \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} \end{matrix}] - - - - - - ((99))$

矩阵T_2N可被如下分解The matrixT_2N can be decomposed as follows

$\underset{&OverBar; &OverBar;}{{T T}_{22 N N}} = = [\begin{matrix} \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} \\ \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} \end{matrix}] = = [\begin{matrix} \underset{&OverBar; &OverBar;}{{I I}_{N N}} \\ - - \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} & \underset{&OverBar; &OverBar;}{{I I}_{N N}} \end{matrix}] [\begin{matrix} - - \underset{&OverBar; &OverBar;}{{I I}_{N N}} & \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} \\ \underset{&OverBar; &OverBar;}{{I I}_{N N}} \end{matrix}] [\begin{matrix} \underset{&OverBar; &OverBar;}{{I I}_{N N}} \\ \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} & \underset{&OverBar; &OverBar;}{{I I}_{N N}} \end{matrix}] - - - - - - ((1010))$

其中I_N是N×N的单位矩阵。whereIN is_an N×N identity matrix.

使用等式(3)中的DCT-IV的特性可以容易地验证等式(10)。使用等式(10)，等式(8)可以被表述为Equation (10) can be easily verified using the properties of DCT-IV in Equation (3). Using equation (10), equation (8) can be expressed as

$[\begin{matrix} \underset{&OverBar; &OverBar;}{{y the y}_{11}} \\ \underset{&OverBar; &OverBar;}{{y the y}_{22}} \end{matrix}] = = [\begin{matrix} \underset{&OverBar; &OverBar;}{{I I}_{N N}} \\ - - \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} & \underset{&OverBar; &OverBar;}{{I I}_{N N}} \end{matrix}] [\begin{matrix} - - \underset{&OverBar; &OverBar;}{{I I}_{N N}} & \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} \\ \underset{&OverBar; &OverBar;}{{I I}_{N N}} \end{matrix}] [\begin{matrix} \underset{&OverBar; &OverBar;}{{I I}_{N N}} \\ \underset{&OverBar; &OverBar;}{{C C}_{N N}^{IV IV}} & \underset{&OverBar; &OverBar;}{{I I}_{N N}} \end{matrix}] [\begin{matrix} \underset{&OverBar; &OverBar;}{{x x}_{22}} \\ \underset{&OverBar; &OverBar;}{{x x}_{11}} \end{matrix}] - - - - - - ((1111))$

等式(11)中的三个提升矩阵对应于图3中的三个提升级。The three boosting matrices in equation (11) correspond to the three boosting stages in FIG. 3 .

根据等式(11)，可以得到下述整数DCT-IV算法，该算法使用一个变换元素来计算两个整数DCT-IV。According to equation (11), the following integer DCT-IV algorithm can be obtained, which uses one transform element to calculate two integer DCT-IVs.

图4例示了根据本发明的方法的实施例，该方法使用DCT-IV作为变换函数。这个实施例被用于图1中示出的音频编码器100中，以实现IntMDCT。类似于图3中，x₁和x₂分别是所述输入数字信号的两个块。z是中间信号，而y₁和y₂分别是输出信号的相应块。Fig. 4 illustrates an embodiment of a method according to the invention using DCT-IV as a transformation function. This embodiment is used in theaudio encoder 100 shown in Fig. 1 to implement IntMDCT. Similar to Figure 3,x1_andx2 are the two blocks_of the input digital signal respectively.z is the intermediate signal, whiley1 andy2 are the_respective blocks_of the output signal.

图4中例示的三个提升级对应于等式(11)中的三个提升矩阵。The three boosting stages illustrated in FIG. 4 correspond to the three boosting matrices in equation (11).

如图4所示，利用下述方案来确定时域到频域整数变换：As shown in Figure 4, the following scheme is used to determine the time-domain to frequency-domain integer transformation:

在第一级401中，利用DCT-IV变换来对x₂进行变换402，对DCT-IV系数进行四舍五入403。随后将经过四舍五入后的DCT-IV系数加到x₁404。由此，产生中间信号z。因此，中间信号z满足等式：In the first stage 401,x2 is transformed 402 using a DCT-IV transform and the DCT-IV coefficients are_rounded 403. The rounded DCT-IV coefficients are then added tox₁ 404 . From this, an intermediate signalz is generated. Therefore, the intermediate signalz satisfies the equation:

在第二级405中，利用DCT-IV变换来对z进行变换406，对DCT-IV系数进行四舍五入407。随后从经过四舍五入后的DCT-IV系数中减去x₂。由此，产生输出信号y₁。因此，输出信号y₁满足等式：In the second stage 405z is transformed 406 using a DCT-IV transform and the DCT-IV coefficients are rounded 407 .x2_is then subtracted from the rounded DCT-IV coefficients. From this, an output signaly₁ is generated. Therefore,_{the output signal y1}satisfies the equation:

在第三级409中，利用DCT-IV变换来对y₁进行变换410，对DCT-IV系数进行四舍五入411。随后从z中减去经过四舍五入后的DCT-IV系数。由此，产生输出信号y₂。因此，输出信号y₂满足等式：In the third stage 409,y1 is transformed 410 using a DCT-IV transform and the DCT-IV coefficients are rounded 411_. The rounded DCT-IV coefficients are then subtracted fromz . From this, an output signaly₂ is generated. Therefore,_{the output signal y2}satisfies the equation:

其中表示四舍五入操作。in Indicates a rounding operation.

图5说明了根据本发明的方法的实施例的反变换的算法，该方法使用DCT-IV作为变换函数。这个实施例被用于图2中示出的音频解码器200中，以实现反IntMDCT。图5中例示的算法是图4中例示的算法的逆运算。不同信号的表示y₁，y₂，x₁，x₂以及z被选择为对应于图4中的表示。Fig. 5 illustrates the algorithm of the inverse transformation according to an embodiment of the method of the present invention using DCT-IV as the transformation function. This embodiment is used in theaudio decoder 200 shown in Fig. 2 to implement the inverse IntMDCT. The algorithm illustrated in FIG. 5 is the inverse of the algorithm illustrated in FIG. 4 . The representationsy₁ ,y₂ ,x₁ ,x₂ andz of the different signals are chosen to correspond to the representations in FIG. 4 .

如图5所示，利用下述方法来确定频域到时域的整数变换：As shown in Figure 5, the following method is used to determine the integer transformation from the frequency domain to the time domain:

在第一级501中，利用DCT-IV变换来对y₁进行变换502，对DCT-IV系数进行四舍五入503。随后将经过四舍五入后的DCT-IV系数加到y₂504。由此，产生中间信号z。因此，中间信号z满足等式：In the first stage 501,y1 is transformed 502 with a DCT-IV transform and the DCT-IV coefficients are rounded 503_. The rounded DCT-IV coefficients are then added toy₂ 504 . From this, an intermediate signalz is generated. Therefore, the intermediate signalz satisfies the equation:

在第二级505中，利用DCT-IV变换来对z进行变换506，对DCT-IV系数进行四舍五入507。随后从经过四舍五入后的DCT-IV系数中减去y₁。由此，产生信号x₂。因此，信号x₂满足等式：In a second stage 505z is transformed 506 using a DCT-IV transform and the DCT-IV coefficients are rounded 507 .y₁ is then subtracted from the rounded DCT-IV coefficients. From this, a signalx₂ is generated. Therefore,_the signalx2 satisfies the equation:

在第三级509中，利用DCT-IV变换来对x₂进行变换510，对DCT-IV系数进行四舍五入511。随后从z中减去经过四舍五入后的DCT-IV系数。由此，产生信号x₁。因此，信号x₁满足等式：In the third stage 509,x2 is transformed 510 using a DCT-IV transform and the DCT-IV coefficients are_rounded 511 . The rounded DCT-IV coefficients are then subtracted fromz . From this, a signalx₁ is generated. Therefore,_the signalx1 satisfies the equation:

可以看出，根据等式(13a)到(13c)的算法是根据等式(12a)到(12c)的算法的逆。因此，如果在图1和图2中例示的编码器和解码器中使用，则所述算法提供用于无损音频编码的方法和装置。It can be seen that the algorithm according to equations (13a) to (13c) is the inverse of the algorithm according to equations (12a) to (12c). Thus, if used in the encoder and decoder illustrated in Figures 1 and 2, the algorithm provides a method and apparatus for lossless audio coding.

在下述解释的本发明的实施例中，将上述方法用于图像归档系统。In an embodiment of the present invention explained below, the above method is applied to an image filing system.

等式(12a)到(12c)和(13a)到(13c)进一步示出，为了计算两个N×N的整数DCT-IV，需要三次N×N的DCT-IV、三次N×1的四舍五入以及三次N×1的加法。因此，对于一个N×N的整数DCT-IV，平均值为：Equations (12a) to (12c) and (13a) to (13c) further show that to compute two N×N integer DCT-IVs, three N×N DCT-IVs, three N×1 roundings are required and three N×1 additions. Thus, for an N×N integer DCT-IV, the mean is:

RC(N)＝1.5N (14)RC(N)＝1.5N (14)

$AC AC ((N N)) = = 1.5 1.5 AC AC (({C C}_{N N}^{IV IV})) + + 1.5 1.5 N N - - - - - - ((1515))$

其中RC(.)是总的四舍五入次数，而AC(.)是算法操作的总次数。与直接转换的整数DCT-IV算法相比，所述提出的整数DCT-IV算法将RC从Nlog₂ N数量级减少到N。where RC(.) is the total number of roundings and AC(.) is the total number of arithmetic operations. The proposed integer DCT-IV algorithm reduces RC from the order of Nlog₂ N to N compared to the direct conversion integer DCT-IV algorithm.

如等式(15)所示，所述提出的整数DCT-IV算法的复杂度多于DCT-IV算法的复杂度约50％。然而，如果还考虑RC，则所述提出的算法的组合复杂度(AC+RC)并未大大超过直接转换的整数算法的复杂度。所述算法的复杂度的精确分析取决于所使用的DCT-IV算法。As shown in equation (15), the complexity of the proposed integer DCT-IV algorithm is about 50% more than that of the DCT-IV algorithm. However, if RC is also considered, the combinatorial complexity (AC+RC) of the proposed algorithm does not greatly exceed that of the directly converted integer algorithm. The exact analysis of the complexity of the algorithm depends on the DCT-IV algorithm used.

如图4和5中所示，所述提出的整数DCT-IV算法简单且在结构上模块化。在其DCT-IV计算块中，其可以使用任何现有DCT-IV算法。所述提出的算法适合于要求IntMDCT的应用，例如在MPEG-4音频扩展3参考模型0中。As shown in Figures 4 and 5, the proposed integer DCT-IV algorithm is simple and modular in structure. In its DCT-IV calculation block, it can use any existing DCT-IV algorithm. The proposed algorithm is suitable for applications requiring IntMDCT, eg in MPEG-4 Audio Extension 3 Reference Model 0.

图6示出了根据本发明的实施例的图像归档系统的体系结构。FIG. 6 shows the architecture of an image filing system according to an embodiment of the present invention.

在图6中，图像源601，例如照相机，提供模拟图像信号。由模/数转换器602来对该图像信号进行处理，以提供相应的数字图像信号。由无损图像编码器603对该数字图像信号进行无损编码，其包括从时域到频域的变换。在这个实施例中，时域对应于所述图像的坐标空间。所述无损编码后的图像信号被存储在存储设备604中，例如硬盘或DVD。当需要所述图像时，从所述存储设备604中取出所述无损编码后的图像信号，并且将其提供给与无损图像编码器603对应的无损图像解码器605，该无损图像解码器605对无损编码后的图像信号进行解码，并且重构所述原始图像信号而不会出现数据丢失。In Fig. 6, animage source 601, such as a camera, provides an analog image signal. The image signal is processed by an analog-to-digital converter 602 to provide a corresponding digital image signal. The digital image signal is losslessly encoded by thelossless image encoder 603, which includes transformation from the time domain to the frequency domain. In this embodiment, the temporal domain corresponds to the coordinate space of said image. The lossless encoded image signal is stored in thestorage device 604, such as hard disk or DVD. When the image is needed, the image signal after the lossless encoding is taken out from thestorage device 604, and provided to thelossless image decoder 605 corresponding to thelossless image encoder 603, thelossless image decoder 605 pairs The lossless encoded image signal is decoded, and the original image signal is reconstructed without data loss.

例如，在所述图像是半导体晶片的误差图且必须被存储来以用于以后分析的情况下，图像信号的此种无损归档是重要的。Such lossless archiving of image signals is important, for example, where the image is an error map of a semiconductor wafer and must be stored for later analysis.

在本发明的这个实施例中，图3到图5中例示的方法的实施例用于无损图像编码器603和无损图像解码器605中。如上所述，图3到图5中例示的方法的实施例提供一种可逆的变换，因此特别提供了一种用于无损图像编码的方法。In this embodiment of the invention, the embodiments of the methods illustrated in FIGS. 3 to 5 are used in thelossless image encoder 603 and thelossless image decoder 605 . As mentioned above, the embodiment of the method illustrated in Fig. 3 to Fig. 5 provides a reversible transformation, thus in particular a method for lossless image coding.

根据本发明的方法不限于音频图像信号。还可以利用根据本发明的方法来对例如视频信号的其他数字信号进行变换。The method according to the invention is not limited to audio image signals. Other digital signals such as video signals can also be transformed with the method according to the invention.

在下面，对根据本发明的用于将数字信号从时域变换到频率域和从频率域变换到时域的方法的又一实施例进行解释。In the following, a further embodiment of a method for transforming a digital signal from the time domain to the frequency domain and from the frequency domain to the time domain according to the present invention is explained.

在本发明的这个实施例中，所述域变换是DCT变换，由此块大小N是某一整数。在一个实施例中，N是2的幂。In this embodiment of the invention, the domain transform is a DCT transform, whereby the block size N is some integer. In one embodiment, N is a power of 2.

假设C_N^II是DCT的N×N变换矩阵(也被称为II型DCT)：Suppose C_N^II is the N×N transformation matrix of DCT (also known as type II DCT):

$\begin{matrix} {C C}_{N N}^{II II} = = \sqrt{22 / / N N} [[{k k}_{m m} cos cos ((m m ((n no + + 11 / / 22)) π π / / N N))]] \\ m m,, n no = = 0,1 0,1,, . . . . . .,, N N - - 11 \end{matrix} - - - - - - ((1616))$

其中in

${k k}_{m m} = = \{\begin{matrix} 11 / / \sqrt{22} & if if & m m = = 00 \\ 11 & if if & m m &NotEqual; &NotEqual; 00 \end{matrix} - - - - - - ((1717))$

并且N是变换大小。m和n是矩阵索引(index)。And N is the transform size. m and n are matrix indices (index).

假设C_N^IV是IV型DCT的DCT的N×N变换矩阵，已经如上定义：Suppose C_N^IV is the N×N transformation matrix of DCT of type IV DCT, which has been defined as above:

$\begin{matrix} {C C}_{N N}^{IV IV} = = \sqrt{22 / / N N} [[cos cos ((((m m + + 11 / / 22)) ((n no + + 11 / / 22)) π π / / N N))]] \\ m m,, n no = = 0,1 0,1,, . . . . . .,, N N - - 11 \end{matrix} - - - - - - ((1818))$

如上，使用多个提升矩阵，在这个实施例中，所述提升矩阵是具有下述形式的2N×2N矩阵：As above, multiple lifting matrices are used, which in this embodiment are 2Nx2N matrices of the form:

${L L}_{22 N N} = = [\begin{matrix} &PlusMinus; &PlusMinus; {I I}_{N N} & {A A}_{N N} \\ {O o}_{N N} & &PlusMinus; &PlusMinus; {I I}_{N N} \end{matrix}] - - - - - - ((1919))$

其中I_N是N×N的单位矩阵，O_N是N×N的零矩阵，而A_N是任意的N×N矩阵。where_IN is an N×N identity matrix, O_N is an N×N zero matrix, and A_N is an arbitrary N×N matrix.

对于每个提升矩阵L_2N，按照与引入的下述参考文献中描述的2×2提升步骤相同的方式来实现提升级可逆的整数到整数的映射，所述参考文献是朗讯科技贝尔实验室的I.Daubechies和W.Sweldens在1996年的Tech.Report(技术报告)“Factoring Wavlet Transforms into Lifting Steps”。仅有的区别在于四舍五入被应用于矢量，而不是应用于单个变量。For each lifting matrix L_2N , the lifting-level reversible integer-to-integer mapping is implemented in the same manner as the 2×2 lifting step described in the incorporated reference, Lucent Technologies Bell Laboratories I.Daubechies and W.Sweldens in 1996 Tech.Report (technical report) "Factoring Wavlet Transforms into Lifting Steps". The only difference is that the rounding is applied to vectors, not to individual variables.

在其他实施例的上述描述中，已经详细地描述了如何为一个提升矩阵实现一个提升级，因此，在下面将省略与提升矩阵对应的提升级的解释。In the above descriptions of other embodiments, how to implement a boosting level for a boosting matrix has been described in detail, so the explanation of the boosting level corresponding to the boosting matrix will be omitted below.

可以看出，L_2N的转置L_2N^T也是提升矩阵。It can be seen that the transpose L_2N^T of L_2N is also a lifting matrix.

在这个实施例中，所述变换元素对应于矩阵T_2N，其被按照下述方式定义为2N×2N矩阵：In this embodiment, the transformation elements correspond to the matrix T_2N , which is defined as a 2N×2N matrix in the following way:

${T T}_{22 N N} = = [\begin{matrix} {C C}_{N N}^{IV IV} & {O o}_{N N} \\ {O o}_{N N} & {C C}_{N N}^{IV IV} \end{matrix}] - - - - - - ((2020))$

将矩阵T_2N分解为提升矩阵具有下述形式：Decomposing the matrix T_2N into a lifting matrix has the following form:

T_2N＝P3·L8·L7·L6·P2·L5·L4·L3·L2·L1·P1 (21)T_2N ＝P3·L8·L7·L6·P2·L5·L4·L3·L2·L1·P1 (21)

在下面解释组成上述等式的右手侧的矩阵。The matrices making up the right-hand side of the above equation are explained below.

P1是由下述等式给出的第一置换矩阵P1 is the first permutation matrix given by

$P P 11 = = [\begin{matrix} {O o}_{N N} & {D D.}_{N N} \\ {J J}_{N N} & {O o}_{N N} \end{matrix}] - - - - - - ((22 twenty two))$

其中J_N是由下面给出的反索引矩阵(counter index matrix)where J_N is the counter index matrix given below

而D_N是其中对角元素交替为1和-1的N×N的对角矩阵：And D_N is an N×N diagonal matrix in which the diagonal elements are alternately 1 and -1:

P2是第二置换矩阵，其例子由下述MATLAB脚本语言产生：P2 is the second permutation matrix, an example of which is generated by the following MATLAB script language:

作为例子，当N为4时，P2是8×8矩阵，如下给出As an example, when N is 4, P2 is an 8×8 matrix given by

$P 2 = [\begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \end{matrix}]$ 其中N＝4 (25) $P 2 = [\begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \end{matrix}]$ where N=4 (25)

P3是第三置换矩阵，其例子由下述MATLAB脚本语言产生：P3 is the third permutation matrix, an example of which is generated by the following MATLAB script language:

作为例子，当N为4时，P3是8×8矩阵，如下给出As an example, when N is 4, P3 is an 8×8 matrix given by

$P 3 = [\begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{matrix}]$ 其中N＝4 (26) $P 3 = [\begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{matrix}]$ where N=4 (26)

L1是第一提升矩阵L1 is the first lifting matrix

$L L 11 = = [\begin{matrix} {I I}_{N N} & {O o}_{N N} \\ {Z Z 11}_{N N} & {I I}_{N N} \end{matrix}] - - - - - - ((2727))$

其中Z1_N是如下给出的N×N反对角矩阵：where Z1_N is an N×N anti-diagonal matrix given by:

L2是第二提升矩阵L2 is the second lifting matrix

$L L 22 = = [\begin{matrix} {I I}_{N N} & {Z Z 22}_{N N} \\ {O o}_{N N} & {I I}_{N N} \end{matrix}] - - - - - - ((2929))$

其中Z2_N是如下给出的N×N反对角矩阵：where Z2_N is an N×N anti-diagonal matrix given by:

L3是第三提升矩阵L3 is the third lifting matrix

$L L 33 = = [\begin{matrix} {I I}_{N N} & {O o}_{N N} \\ {Z Z 33}_{N N} & {I I}_{N N} \end{matrix}] - - - - - - ((3131))$

其中in

${Z Z 33}_{N N} = = \sqrt{22} {C C}_{N N}^{IV IV} + + {I I}_{N N} + + {Z Z 11}_{N N} - - - - - - ((3232))$

L4是第四提升矩阵L4 is the fourth lifting matrix

$L L 44 = = [\begin{matrix} - - {I I}_{N N} & {Z Z 44}_{N N} \\ {O o}_{N N} & {I I}_{N N} \end{matrix}] - - - - - - ((3333))$

其中in

${Z Z 44}_{N N} = = {C C}_{N N}^{IV IV} / / \sqrt{22} - - - - - - ((3434))$

L5是第五提升矩阵L5 is the fifth lifting matrix

$L L 55 = = [\begin{matrix} {I I}_{N N} & {O o}_{N N} \\ {Z Z 55}_{N N} & {I I}_{N N} \end{matrix}] - - - - - - ((3535))$

其中in

${Z Z 55}_{N N} = = - - ((\sqrt{22} {C C}_{N N}^{IV IV} + + {I I}_{N N})) - - - - - - ((3636))$

L6是第六提升矩阵L6 is the sixth lifting matrix

$L L 66 = = [\begin{matrix} {I I}_{N N} & {O o}_{N N} \\ {Z Z 66}_{N N} & {I I}_{N N} \end{matrix}] - - - - - - ((3737))$

其中Z6_N是如下给出的N×N反对角矩阵：where Z6_N is an N×N anti-diagonal matrix given by:

L7是第七提升矩阵L7 is the seventh lifting matrix

$L L 77 = = [\begin{matrix} {I I}_{N N} & {Z Z 77}_{N N} \\ {O o}_{N N} & {I I}_{N N} \end{matrix}] - - - - - - ((3939))$

其中Z7_N是如下给出的N×N反对角矩阵：where Z7_N is an N×N anti-diagonal matrix given by:

L8是第八提升矩阵：L8 is the eighth lifting matrix:

L8＝L6 (41)L8＝L6 (41)

由此，导致如(42)中所示的因式分解：This leads to a factorization as shown in (42):

T_2N＝P3·L8·L7·L6·P2·L5·L4·L3·L2·L1·P1 (42)T_2N ＝P3·L8·L7·L6·P2·L5·L4·L3·L2·L1·P1 (42)

其中P1、P2和P3是三个置换矩阵，L_j是八个提升矩阵，其中j从1到8。where P1, P2, and P3 are three permutation matrices, and L_j are eight lifting matrices, where j ranges from 1 to 8.

提升矩阵L3、L4和L5包括辅助变换矩阵，在这种情况下，其为变换矩阵C_N^IV自身。The lifting matrices L3, L4 and L5 comprise auxiliary transformation matrices, in this case the transformation matrix C_N^IV itself.

根据等式(42)，可以为大小为N×1的两个输入信号计算整数DCT。According to equation (42), the integer DCT can be calculated for two input signals of size Nxl.

由于等式(42)提供描述DCT-IV变换域的提升矩阵因式分解，所以其提升矩阵可被按照这里示出的方式用来计算所施加的输入信号的域变换。Since equation (42) provides a lifting matrix factorization describing the DCT-IV transform domain, its lifting matrix can be used in the manner shown here to compute the domain transform of the applied input signal.

可以按照下述方式得到等式(42)。Equation (42) can be obtained as follows.

可以使用下述公开来得到下述分解，该公开是Wang，Zhongde在1985十月的IEEE Transactions on Acoustic，Speech and Signal Processing(声学、语音和信号处理学报)，Vol.ASSP-33，No.4上发表的“On Computingthe Discrete Fourier and Cosine Transforms”。The following decomposition can be obtained using the publication of Wang, Zhongde, IEEE Transactions on Acoustic, Speech and Signal Processing, October 1985, Vol.ASSP-33, No.4 "On Computing the Discrete Fourier and Cosine Transforms" published in .

${C C}_{N N}^{IV IV} = = {(({B B}_{N N}))}^{T T} \cdot &Center Dot; {(({P P}_{N N}))}^{T T} \cdot &Center Dot; [\begin{matrix} {C C}_{N N / / 22}^{II II} \\ \overset{\overset{&OverBar; &OverBar;}{&OverBar; &OverBar;}}{{S S}_{N N / / 22}^{II II}} \end{matrix}] \cdot &Center Dot; {T T}_{N N} - - - - - - ((4343))$

$= = {(({B B}_{N N}))}^{T T} \cdot &Center Dot; {(({P P}_{N N}))}^{T T} \cdot &Center Dot; [\begin{matrix} {C C}_{N N / / 22}^{II II} \\ {C C}_{N N / / 22}^{II II} \end{matrix}] \cdot &Center Dot; {P P}_{DJ DJ} \cdot &Center Dot; {T T}_{N N}$

是已知的，其中S_N/2^II表示II型离散正弦变换的变换矩阵。is known, where S_N/2^II represents the transformation matrix of the type II discrete sine transform.

${P P}_{DJ DJ} = = [\begin{matrix} I I \\ D D. \cdot \cdot J J \end{matrix}]$

P_N是如下给出的N×N置换矩阵P_N is the N×N permutation matrix given by

$\underset{&OverBar; &OverBar;}{{P P}_{N N}} = = [\begin{matrix} \underset{&OverBar; &OverBar;}{{I I}_{N N / / 22}} \\ \underset{&OverBar; &OverBar;}{{J J}_{N N / / 22}} \end{matrix}] - - - - - - ((4444))$

和and

等式(85)可以与下述等式合并Equation (85) can be combined with the following equation

${C C}_{N N}^{IV IV} = = {R R}_{PO PO} \cdot &Center Dot; [\begin{matrix} {C C}_{N N / / 22}^{IV IV} \\ {C C}_{N N / / 22}^{IV IV} \end{matrix}] \cdot \cdot {R R}_{PR PR} \cdot &Center Dot; {P P}_{D D.} \cdot \cdot {P P}_{EO EO} - - - - - - ((4545))$

其中P_EO是偶奇置换矩阵，where P_EO is the even-odd permutation matrix,

$\underset{&OverBar; &OverBar;}{{R R}_{pr pr}} = = \frac{11}{\sqrt{22}} [\begin{matrix} \underset{&OverBar; &OverBar;}{{I I}_{N N / / 22}} & \underset{&OverBar; &OverBar;}{{I I}_{N N / / 22}} \\ \underset{&OverBar; &OverBar;}{{I I}_{N N / / 22}} & \underset{&OverBar; &OverBar;}{{- - I I}_{N N / / 22}} \end{matrix}]$

R_PO 等于T_N，R_PO is equal to T_N ,

${\underset{&OverBar; &OverBar;}{P P}}_{D D.} = = [\begin{matrix} \underset{&OverBar; &OverBar;}{{I I}_{\frac{N N}{22}}} \\ \underset{&OverBar; &OverBar;}{{D D.}_{\frac{N N}{22}}} \end{matrix}]$

在转置等式(45)转换为(46)后：After transposing equation (45) into (46):

${C C}_{N N}^{IV IV} = = {(({P P}_{EO EO}))}^{T T} \cdot \cdot {(({P P}_{D D.}))}^{T T} \cdot \cdot {R R}_{PR PR} \cdot \cdot [\begin{matrix} {C C}_{N N / / 22}^{IV IV} \\ {C C}_{N N / / 22}^{IV IV} \end{matrix}] \cdot \cdot {(({R R}_{PO PO}))}^{T T} - - - - - - ((4646))$

$= = {(({P P}_{EO EO}))}^{T T} \cdot &Center Dot; {(({P P}_{D D.}))}^{T T} \cdot \cdot \frac{11}{\sqrt{22}} [\begin{matrix} {C C}_{N N / / 22}^{IV IV} & {C C}_{N N / / 22}^{IV IV} \\ {C C}_{N N / / 22}^{IV IV} & - - {C C}_{N N / / 22}^{IV IV} \end{matrix}] \cdot \cdot {(({R R}_{PO PO}))}^{T T}$

组合等式(43)和(46)得到：Combining equations (43) and (46) yields:

$[\begin{matrix} {C C}_{N N / / 22}^{II II} \\ {C C}_{N N / / 22}^{II II} \end{matrix}] = = {P P}_{N N} \cdot \cdot {B B}_{N N} \cdot &Center Dot; {(({P P}_{EO EO}))}^{T T} {(({P P}_{D D.}))}^{T T} \cdot \cdot \frac{11}{\sqrt{22}} [\begin{matrix} {C C}_{N N / / 22}^{IV IV} & {C C}_{N N / / 22}^{IV IV} \\ {C C}_{N N / / 22}^{IV IV} & - - {C C}_{N N / / 22}^{IV IV} \end{matrix}] \cdot &Center Dot; {(({R R}_{PO PO}))}^{T T} \cdot &Center Dot; {T T}_{N N} \cdot &Center Dot; {(({P P}_{DJ DJ}))}^{T T} - - - - - - ((4747))$

$= = {P P}_{33} \cdot &Center Dot; {R R}_{22} \cdot &Center Dot; {P P}_{22} \cdot &Center Dot; \frac{11}{\sqrt{22}} [\begin{matrix} {C C}_{N N / / 22}^{IV IV} & {C C}_{N N / / 22}^{IV IV} \\ {C C}_{N N / / 22}^{IV IV} & - - {C C}_{N N / / 22}^{IV IV} \end{matrix}] \cdot \cdot {R R}_{11} \cdot \cdot {P P}_{11}$

其中：in:

P₁＝(P_DJ)^TP₁ =(P_DJ )^T

P₂＝(P_EO)^T·(P_D)^T＝(P_D·P_EO)^TP₂ ＝(P_EO )^T ·(P_D )^T ＝(P_D ·P_EO )^T

P₃＝P_NP₃ =P_N

R₁＝(R_PO)^T·T_NR₁ ＝(R_PO )^T · T_N

R₂＝B_NR₂ =B_N

根据等式(47)，可以容易地得到等式(42)。From equation (47), equation (42) can be easily obtained.

在这个实施例中，域变换的计算仅仅需要4N次四舍五入操作，如同现在将要描述。In this embodiment, the calculation of the domain transformation requires only 4N rounding operations, as will now be described.

假设α(*)是实加法的次数，μ(*)是实乘法的次数，而γ(*)是实四舍五入的次数。对于所述提出的IntDCT算法，可以得到：Suppose α(*) is the number of real additions, μ(*) is the number of real multiplications, and γ(*) is the number of real roundings. For the proposed IntDCT algorithm, it can be obtained:

α(IntDCT)＝11N+3α(DCT-IV)α(IntDCT)=11N+3α(DCT-IV)

μ(IntDCT)＝9N+3μ(DCT-IV)μ(IntDCT)=9N+3μ(DCT-IV)

γ(IntDCT)＝8Nγ(IntDCT)=8N

因为所述提出的IntDCT算法对它们一起进行处理，所以上述结果是针对数据采样的两个块的。由此，对于数据采样的一个块，所述计算的次数被减半，其为The above results are for two blocks of data samples because the proposed IntDCT algorithm processes them together. Thus, for one block of data samples, the number of calculations is halved, which is

α₁(IntDCT)＝5.5N+1.5α(DCT-IV)α₁ (IntDCT)=5.5N+1.5α(DCT-IV)

μ₁(IntDCT)＝4.5N+1.5μ(DCT-IV)μ₁ (IntDCT)=4.5N+1.5μ(DCT-IV)

γ₁(IntDCT)＝4Nγ₁ (IntDCT) = 4N

其中α₁、μ₁和γ₁分别是针对采样的一个块的实加法的次数、实乘法的次数以及实四舍五入次数。where α₁ , μ₁ and γ₁ are the number of real additions, real multiplications and real roundings for one block of samples, respectively.

对于DCT-IV计算，可以使用在并入的参考文献H.S.Malvar，1992年由Norwood，MA.Artech House出版“Signal Processing With lappedTransforms”的第199-201页上描述的基于FFT的算法，根据该算法For DCT-IV calculations, the FFT-based algorithm described on pages 199-201 of the incorporated reference H.S. Malvar, "Signal Processing With lapped Transforms", Norwood, MA. Artech House, 1992, can be used, according to which

α(DCT-IV)＝1.5Nlog₂ Nα(DCT-IV)=1.5Nlog₂ N

μ(DCT-IV)＝0.5Nlog₂ N+Nμ(DCT-IV)=0.5Nlog₂ N+N

因此得到：and thus get:

α₁(IntDCT)＝2.25Nlog₂ N+5.5Nα₁ (IntDCT)=2.25Nlog₂ N+5.5N

μ₁(IntDCT)＝0.75Nlog₂ N+6Nμ₁ (IntDCT)=0.75Nlog₂ N+6N

在下面，对根据本发明的用于将数字信号从时域变换到频移和从频率域变换到时域的方法的又一实施例进行解释。In the following, a further embodiment of a method according to the invention for transforming a digital signal from the time domain to the frequency shift and from the frequency domain to the time domain is explained.

在这个实施例中，将离散快速傅立叶变换(FFT)用作域变换。In this embodiment, a discrete Fast Fourier Transform (FFT) is used as the domain transform.

假设F是具有归一化的FFT的N×N变换矩阵，Suppose F is an N×N transform matrix with normalized FFT,

$\begin{matrix} F f = = \sqrt{\frac{11}{N N}} [[exp exp ((\frac{- - j j 22 πmn πmn}{N N}))]] \\ m m,, n no = = 0,1 0,1,, . . . . . .,, N N - - 11 \end{matrix} - - - - - - ((4848))$

其中N是变换大小。m和n是矩阵索引。where N is the transform size. m and n are matrix indices.

在这个实施例中，维数为N×N的置换矩阵P是包括索引0或1的矩阵。在将其与N×1维矢量(输入信号的矩阵表示)相乘后，所述矢量中的元素的顺序被改变。In this embodiment, the permutation matrix P of dimension N×N is a matrix including index 0 or 1. After multiplying it with an Nxl dimensional vector (matrix representation of the input signal), the order of the elements in the vector is changed.

在这个实施例中，提升矩阵被定义为具有下述形式的2N×2N矩阵。In this embodiment, the lifting matrix is defined as a 2N×2N matrix having the following form.

$L L = = [\begin{matrix} {P P}_{11} & A A \\ O o & {P P}_{22} \end{matrix}] - - - - - - ((4949))$

其中P₁和P₂是两个置换矩阵，O是N×N零矩阵，A是任意N×N矩阵。对于提升矩阵L，按照与上述并入的I.Daubechies的参考文献中的2×2提升步骤相同的方式来实现可逆整数到整数映射。然而，如上所述，将四舍五入应用于矢量而不是应用于单个变量。显而易见的是，所述L的转置LT也是提升矩阵。where_P1 and_P2 are two permutation matrices, O is an N×N zero matrix, and A is an arbitrary N×N matrix. For the lifting matrix L, the reversible integer-to-integer mapping is implemented in the same way as the 2x2 lifting step in the above incorporated reference by I. Daubechies. However, as mentioned above, rounding is applied to vectors and not to individual variables. It is obvious that the transpose LT of L is also a lifting matrix.

此外，假设T是2N×2N变换矩阵：Also, assuming T is a 2Nx2N transformation matrix:

$T T = = [\begin{matrix} O o & F f \\ F f & O o \end{matrix}] - - - - - - ((5050))$

因此，改进的变换矩阵T(并且相应地所述域变换本身)可以被表示为提升矩阵因子分解形式：Thus, the improved transformation matrix T (and accordingly the domain transformation itself) can be expressed in the lifting matrix factorization form:

$T T = = [\begin{matrix} I I & O o \\ - - Q Q \cdot \cdot F f & I I \end{matrix}] \cdot &Center Dot; [\begin{matrix} - - Q Q & F f \\ O o & I I \end{matrix}] \cdot &Center Dot; [\begin{matrix} I I & O o \\ F f & I I \end{matrix}] - - - - - - ((5151))$

其中I是N×N的单位矩阵，而Q是如下给出的N×N的置换矩阵where I is the N×N identity matrix and Q is the N×N permutation matrix given by

$Q Q = = [\begin{matrix} 11 & {O o}_{11 xN xN - - 11} \\ {O o}_{N N - - 11 x x 11} & J J \end{matrix}] - - - - - - ((5252))$

并且O_1xN-1和O_N-1x1分别是具有N-1个零的行矢量和列矢量。And O_1xN-1 and O_N-1x1 are respectively row and column vectors with N-1 zeros.

J是如下给出的(N-1)×(N-1)反索引矩阵J is the (N-1)×(N-1) inverse index matrix given by

在等式(53)中，方括号中的空白处表示所有零矩阵元素。In equation (53), the blank spaces in the square brackets represent all zero matrix elements.

从等式(51)中可以看出，提升矩阵因子分解形式可以被用来使用这里描述的方法来为两个N×1复矢量计算整数FFT。As can be seen from equation (51), the lifted matrix factorization form can be used to compute integer FFTs for two Nxl complex vectors using the method described here.

在这个实施例中，域变换的计算仅仅需要3N次四舍五入操作，如同现在将要描述的。In this embodiment, the calculation of the domain transformation requires only 3N rounding operations, as will now be described.

分别假设α(*)是实加法的次数，Assuming respectively that α(*) is the number of real additions,

μ(*)是实乘法的次数，以及μ(*) is the number of real multiplications, and

γ(*)是实四舍五入的次数。γ(*) is the number of times of real rounding.

对于所述提出的IntFFT算法，可以得到：For the proposed IntFFT algorithm, it can be obtained:

α(IntFFT)＝6N+3α(FFT)α(IntFFT)=6N+3α(FFT)

μ(IntFFT)＝3μ(FFT)μ(IntFFT)=3μ(FFT)

γ(IntFFT)＝6Nγ(IntFFT)=6N

因为所述提出的IntFFT算法对它们一起进行处理，所以上述结果是针对数据采样的两个块的。由此，对于数据采样的一个块，所述计算的次数被减半，其为The above results are for two blocks of data samples because the proposed IntFFT algorithm processes them together. Thus, for one block of data samples, the number of calculations is halved, which is

α₁(IntFFT)＝3N+1.5α(FFT)α₁ (IntFFT)=3N+1.5α(FFT)

μ₁(IntFFT)＝1.5μ(FFT)μ₁ (IntFFT) = 1.5μ(FFT)

γ₁(IntFFT)＝3Nγ₁ (IntFFT)=3N

其中α₁、μ₁和γ₁分别是针对采样的一个块的实加法的次数、实乘法的次数以及实四舍五入操作次数。where α₁ , μ₁ and γ₁ are the number of real additions, real multiplications and real rounding operations for one block of samples, respectively.

对于FFT计算，可以使用分裂基FFT(SRFFT)的算法，根据该算法For FFT calculations, the algorithm of Split-Basic FFT (SRFFT) can be used, according to which

α(SRFFT)＝3Nlog₂ N-3N+4α(SRFFT)=3Nlog₂ N-3N+4

μ(SRFFT)＝Nlog₂ N-3N+4μ(SRFFT)=Nlog₂ N-3N+4

结果，我们得到：As a result, we get:

α₁(IntFFT)＝4.5Nlog₂ N-1.5N+6α₁ (IntFFT)=4.5Nlog₂ N-1.5N+6

μ₁(IntFFT)＝1.5Nlog₂ N-4.5N+6μ₁ (IntFFT)=1.5Nlog₂ N-4.5N+6

图7示出了用于评定上述DCT变换技术和上述FFT域变换的变换精确度的正变换编码器和反变换编码器。所述测试涉及根据在这里引入的2003年三月泰国的ISO/IEC JTC 1/SC 29/WG 11 N5778 Pattaya，“Codingof Moving Pictures and Audio：Work plan for Evaluation of Integer MDCTfor FGS to Lossless Experimentation Framework”中描述的由MPEG-4无损音频编码组提出的评估标准来测量变换的平均方差(MSE)。Figure 7 shows the forward transform coder and the inverse transform coder used to evaluate the transform accuracy of the above-mentioned DCT transform technique and the above-mentioned FFT domain transform. The tests involved are based on ISO/IEC JTC 1/SC 29/WG 11 N5778 Pattaya, Thailand, March 2003, "Coding of Moving Pictures and Audio: Work plan for Evaluation of Integer MDCT for FGS to Lossless Experimentation Framework" introduced here Describes the evaluation criteria proposed by the MPEG-4 Lossless Audio Coding Group to measure the mean variance (MSE) of the transformation.

具体地，IntDCT和整数反DCT(IntIDCT)的MSE如下给出Specifically, the MSEs of IntDCT and integer inverse DCT (IntIDCT) are given as follows

$MSE MSE = = \frac{11}{K K} {Σ Σ}_{j j = = 00}^{K K - - 11} \frac{11}{N N} {Σ Σ}_{i i = = 00}^{N N - - 11} {e e}_{i i}^{22} - - - - - - ((5454))$

其中，对于IntDCT，误差信号e是e_j；对于IntIDCT，误差信号e是e_i，如图1中所示。K是所述评估中使用的采样块的总数。Wherein, for IntDCT, the error signal e is e_j ; for IntIDCT, the error signal e is e_i , as shown in FIG. 1 . K is the total number of sample blocks used in the evaluation.

IntFFT和整数反FFT(IntIFFT)的MSE如下给出The MSE of IntFFT and Integer Inverse FFT (IntIFFT) is given by

$MSE MSE = = \frac{11}{K K} {Σ Σ}_{j j = = 00}^{K K - - 11} \frac{11}{N N} {Σ Σ}_{i i = = 00}^{N N - - 11} {| | | | {e e}_{i i} | | | |}^{22} - - - - - - ((5555))$

其中，对于IntFFT，误差信号e是e_j；对于IntIFFT，误差信号e是e_i，如图1中所示。||*||表示复数值的模。K是所述评估中使用的采样块的总数。Wherein, for IntFFT, the error signal e is e_j ; for IntIFFT, the error signal e is e_i , as shown in FIG. 1 . ||*|| denotes the modulus of a complex value. K is the total number of sample blocks used in the evaluation.

对于两种域变换，在48kHz/16比特测试组中使用具有15个不同类型的音乐文件的总共450秒。表I示出了所述测试结果。For both domain transformations, a total of 450 seconds with 15 different types of music files were used in the 48kHz/16bit test set. Table I shows the test results.

从表1中可以看出，使用本发明的系统和方法产生的MSE非常小，并且不像常规系统，基本上与处理块的大小无关。参照DCT-IV域变换，在将块大小N增大到多达4096个比特时，所述MSE仅仅稍微增加。所述FFT的MSE甚至更好，对于块大小增大到4096个比特，显示出稳定的MSE 0.4。当根据所呈现的能力和对更长块大小的需求的增长来看本发明所展示的性能时，本发明的优势更加明显。As can be seen from Table 1, the MSE produced using the system and method of the present invention is very small and, unlike conventional systems, is substantially independent of the processing block size. Referring to the DCT-IV domain transform, the MSE increases only slightly when increasing the block size N up to 4096 bits. The MSE of the FFT is even better, showing a stable MSE of 0.4 for block sizes up to 4096 bits. The advantages of the invention are even more apparent when looking at the performance demonstrated by the invention in terms of the capabilities presented and the growing demand for longer block sizes.

NN IntDCT-IVIntDCT-IV IntIDCT-IVIntIDCT-IV IntFFTIntFFT IntIFFTIntIFFT

8 8 0.5370.537 0.5370.537 0.4560.456 0.3710.371 1616 0.5460.546 0.5460.546 0.4800.480 0.4120.412 3232 0.5490.549 0.5480.548 0.4610.461 0.3910.391 6464 0.5500.550 0.5500.550 0.4620.462 0.3930.393 128128 0.5510.551 0.5510.551 0.4610.461 0.3910.391 256256 0.5520.552 0.5520.552 0.4610.461 0.3910.391 512512 0.5520.552 0.5520.552 0.4610.461 0.3910.391 10241024 0.5520.552 0.5520.552 0.4600.460 0.3910.391 20482048 0.5520.552 0.5520.552 0.4610.461 0.3910.391 40964096 0.5530.553 0.5520.552 0.4610.461 0.3910.391

表ITable I

引入文献Introduce literature

通过参考在这里引入下述文献：The following documents are hereby incorporated by reference:

H.S.Malver，“Signal Processing with Lapped Transforms”ArtechHouse，1992；H.S.Malver, "Signal Processing with Lapped Transforms" ArtechHouse, 1992;

R.Geiger，T.Sporer，J.Koller，K.Brandenburg，“Audio Coding basedon Integer Transforms”AES 111^th Convention，New York，USA，Sept.2001；R.Geiger, T.Sporer, J.Koller, K.Brandenburg, "Audio Coding based on Integer Transforms"AES 111^th Convention, New York, USA, Sept.2001;

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