Shannon Source Coding Theorem

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香农奈奎斯特采样定理

香农奈奎斯特采样定理

香农奈奎斯特采样定理
香农-奈奎斯特采样定理(Shannon-Nyquist Sampling Theorem)是一项基本的信号处理原理,它规定了一个连续时间信号的采样频率应该至少是该信号中最高频率成分的两倍,以便在离散时间中完整地重构原始信号。

这个定理是由克劳德·香农(Claude Shannon)和哈里·奈奎斯特(Harry Nyquist)在20世纪初提出的。

具体来说,香农-奈奎斯特采样定理表述如下:
如果一个连续时间信号的最高频率成分为f_max,那么为了在离散时间中准确地重建原始信号,采样频率f_s(采样率)必须满足:
f_s ≥ 2 * f_max
这意味着采样频率应至少是信号中最高频率的两倍。

如果采样频率不满足这个条件,就会出现所谓的"混叠"或"奈奎斯特折叠",导致信号在离散时间中无法准确还原。

香农-奈奎斯特采样定理在数字信号处理、通信系统、音频处理、图像处理和各种数据采集应用中具有重要作用。

它强调了适当选择采样频率的重要性,以避免信息丢失和混叠问题,确保准确的信号重建。

因此,合理的采样频率选择是数字信号处理的基本原则之一。

简述采样定理的基本内容

简述采样定理的基本内容

简述采样定理的基本内容采样定理,也被称为奈奎斯特定理(Nyquist theorem)或香农-奈奎斯特采样定理(Shannon-Nyquist sampling theorem),是在信号处理领域中至关重要的一条基本原理。

它对数字信号处理、通信系统以及采样率等方面具有重要的指导意义。

1. 采样定理的基本内容采样定理表明,如果要正确恢复连续时间信号的完整信息,就需要以至少两倍于信号最高频率的采样频率对信号进行采样。

采样频率应该大于等于信号最高频率的两倍,即Fs >= 2 * Fmax。

采样定理的原理基于奈奎斯特频率,奈奎斯特频率是指信号频谱中的最高频率成分。

如果采样频率小于奈奎斯特频率的两倍,那么采样信号中将出现混叠现象,即频谱中的不同频率成分相互干扰,导致原信号无法准确恢复。

2. 采样定理的应用采样定理在多个领域都有广泛的应用,以下是几个常见的应用领域:音频处理:在音频信号的数字化处理中,采样定理保证了通过合适的采样率可以准确还原原始音频信号,同时避免了音频信号的混叠现象。

这就是为什么音频 CD 的采样率是44.1kHz,超过人类可听到的最高频率20kHz的两倍。

通信系统:在数字通信系统中,为了正确传输模拟信号,信号需要经过模数转换(采样)和数模转换两个过程。

采样定理确保了在采样时不会丢失信号的信息,同时在接收端通过恢复出原始信号。

这对于保证通信质量和准确传输数据来说非常关键。

图像处理:在数字图像采集中,采样定理用于设置合适的采样率,以避免图片出现信息丢失和混叠现象。

在数字摄影中,也需要根据采样定理来选择适当的像素密度,以保证图像的质量和细节。

3. 采样定理的局限性和改进采样定理的一个重要前提是信号是带限的,即信号的频谱有一个上限,超过这个上限的频率成分可以被忽略。

然而,在实际应用中,许多信号并不是严格带限的,因此采样定理可能无法完全适用。

为了克服采样定理的局限性,一种常见的方法是使用过采样(oversampling)技术。

香农采样定理英文描述

香农采样定理英文描述

香农采样定理英文描述Shannon Sampling Theorem or Nyquist Sampling Theorem is an important concept in digital signal processing. Thetheorem states that the sample rate must be equal to or more than twice the highest frequency component of a signal to accurately reconstruct it. In simpler terms, it means that we need to sample a signal at a rate equal to or higher thantwice its maximum frequency.The theorem was proposed by Claude Shannon in 1949, andit has been widely used in many areas of science, engineering, and technology. It has been used for signal processing in music, telecommunications, image processing, and many other fields.The following are the steps involved in understandingand using the Shannon Sampling Theorem:1. Definition of the theorem: The Shannon Sampling Theorem is a mathematical principle that describes the minimum sampling rate required to accurately represent a continuous-time signal in the digital domain.2. Understanding the importance of sampling rate: Sampling rate is the number of times a signal is sampled per unit of time. It is essential to have a higher sampling rateto capture the details of a signal accurately. If thesampling rate is not high enough, the reconstructed signalmay not be accurate.3. Finding the Nyquist frequency: The Nyquist frequencyis half of the sampling rate. It represents the maximum frequency component that can be accurately represented. Ifthe maximum frequency of a signal is above the Nyquist frequency, aliasing occurs, and the signal cannot be accurately reconstructed.4. Applying the theorem: Once the Nyquist frequency is known, it can be used to determine the minimum sample rate required to accurately reproduce the signal. This is done using the formula Fs = 2B, where Fs is the sample rate and B is the bandwidth of the signal.5. Examples: Here is a simple example that demonstrates the importance of the Shannon Sampling Theorem. Suppose we have a signal that has a maximum frequency of 100 Hz. According to the theorem, the minimum sampling rate required to accurately reproduce the signal is 200 Hz (2*100). Any lower sampling rate would lead to aliasing, and the signal cannot be reconstructed accurately.In conclusion, the Shannon Sampling Theorem is a fundamental principle in digital signal processing. It is essential to understand its importance and apply it correctly to avoid errors and inaccuracies in signal processing.。

量子力学07

量子力学07

For the next several lectures we will be discussing the von Neumann entropy and various concepts relating to it.This lecture is intended to introduce the notion of entropy and its connection to compression.7.1Shannon entropyBefore we discuss the von Neumann entropy,we will take a few moments to discuss the Shannon entropy.This is a purely classical notion,but it is appropriate to start here.The Shannon entropy of a probability vector p∈RΣis defined as follows:p(a)log(p(a)).H(p)=−∑a∈Σp(a)>0Here,and always in this course,the base of the logarithm is2.(We will write ln(α)if we wish to refer to the natural logarithm of a real numberα.)It is typical to express the Shannon entropy slightly more concisely asp(a)log(p(a)),H(p)=−∑a∈Σwhich is meaningful if we make the interpretation0log(0)=0.This is sensible given thatαlog(α)=0.limα↓0There is no reason why we cannot extend the definition of the Shannon entropy to arbitrary vectors with nonnegative entries if it is useful to do this—but mostly we will focus on probability vectors.There are standard ways to interpret the Shannon entropy.For instance,the quantity H(p)can be viewed as a measure of the amount of uncertainty in a random experiment described by the probability vector p,or as a measure of the amount of information one gains by learning the value of such an experiment.Indeed,it is possible to start with simple axioms for what a measure of uncertainty or information should satisfy,and to derive from these axioms that such a measure must be equivalent to the Shannon entropy.Something to keep in mind,however,when using these interpretations as a guide,is that the Shannon entropy is usually only a meaningful measure of uncertainty in an asymptotic sense—as the number of experiments becomes large.When a small number of samples from some experi-ment is considered,the Shannon entropy may not conform to your intuition about uncertainty,as the following example is meant to demonstrate.Example7.1.LetΣ={0,1,...,2m2},and define a probability vector p∈RΣas follows:p(a)= 1−1m2−m21≤a≤2m2.56nn∑j=1Y j−E[Y j] ≥ε =0,which is true by the(weak)law of large numbers.uu∗, Ξ⊗1L(Z) (uu∗)represents one way of measuring the quality with whichΦacts trivially on the state uu∗.Now,any purification u∈W⊗Z ofσmust take the formu=vec √k∑j=1√σB 2=k ∑j=1 σ,A j 2.Another expression for the channelfidelity isF channel(Ξ,σ)=ρ⊗n,B j A i。

解密模数转换器(ADC)分辨率和采样率

解密模数转换器(ADC)分辨率和采样率

解密模数转换器(ADC)分辨率和采样率分辨率和采样率是选择(模数转换器)((ADC)) 时要考虑的两个重要因素。

为了充分理解这些,必须在一定程度上理解量化和奈奎斯特准则等概念。

在选择模数转换器((AD)C) 的过程中要考虑的两个最重要的特性可能是分辨率和采样率。

在进行任何选择之前,应仔细考虑这两个因素。

它们将影响选择过程中的一切,从价格到所需模数转换器的底层架构。

为了为特定应用正确确定正确的分辨率和正确的采样率,应该对这些特性有一个合理的了解。

下面是与模数转换相关的术语的一些数学描述。

数学很重要,但它所代表的概念更重要。

如果您能忍受数学并理解所介绍的概念,您将能够缩小适合您应用的ADC 的数量,并且选择将变得容易得多。

量化(Quan(ti)sation)模数转换器将连续(信号)(电压或(电流))转换为由离散逻辑电平表示的数字序列。

术语量化是指将大量值转换为较小值集或离散值集的过程。

在数学上,ADC 可以被描述为量化具有大域的函数以产生具有较小域的函数。

上面的等式在数学上描述了模数转换过程。

在这里,我们将输入电压V in描述为一系列位b N-1 ...b 0。

在这个公式中,2 N 代表量化级别的数量。

直观的是,更多的量化级别会导致原始(模拟)信号的更精确的数字表示。

例如,如果我们可以用1024 个量化级别而不是256 个级别来表示信号,我们就提高了ADC 的精度,因为每个量化级别代表一个更小的幅度范围。

(Vr)ef 表示可以成功转换为精确数字表示的最大输入电压。

因此,重要的是V ref 大于或等于V in的最大值。

但是请记住,比V in值大得多的值将导致表示原始信号的量化级别更少。

例如,如果我们知道我们的信号永远不会增加到 2.4 V 以上,那么使用5 V 的电压参考将是低效的,因为超过一半的量化电平将被使用。

量化误差(Quantisation Error)量化误差是一个术语,用于描述原始信号与信号的离散表示之间的差异。

逻辑综合中的大体概念

逻辑综合中的大体概念

1. 逻辑综合(Logic Synthesis)EDA工具把数字电路的功能描述(或结构描述)转化为电路的结构描述。

实现上述转换的同时要满足用户给定的约束条件,即速度、功耗、成本等方面的要求。

2. 逻辑电路(Logic Circuit)逻辑电路又称数字电路,在没有特别说明的情况下指的是二值逻辑电路。

其电平在某个阈值之上时看作高电平,在该阈值之下时看作低电平。

通常把高电平看作逻辑值1;把低电平看作逻辑值0。

3. 约束(restriction)设计者给EDA工具提出的附加条件,对逻辑综合而言,约束条件一般包括速度、功耗、成本等方面的要求。

4. 真值表(Truth Table)布尔函数的表格描述形式,描述输入变量每一种组合情况下函数的取值。

输入变量组合以最小项形式表示,函数的取值为真或假(1 或0)。

5. 卡诺图(Karnaugh Map)布尔函数的图形描述形式,图中最小方格和最小项对应,两个相邻的最小方格所对应的最小项只有一个变量的取值不同。

卡诺图适合于用观察法化简布尔函数,但是当变量的个数大于4时,卡诺图的绘制和观察都变得很困难。

6. 单输出函数(Single-output Function)一个布尔函数的单独描述。

7. 多输出函数(Multiple-output Function)输入变量相同的多个布尔函数的统一描述。

8. 最小项(Minterm)设a1,a2,…a i,…a n是n个布尔变量,p为n个因子的乘积。

若是在p中每一变量都以原变量a i或反变量的形式作为因子出现一次且仅出现一次,则称p 为n个变量的一个最小项。

最小项在卡诺图中对应于最小的方格;在立方体表示中对应于极点。

9. 蕴涵项(Implicant)布尔函数f的"与-或"表达式中的每一乘积项都叫作f的蕴涵项。

例如:f=+中的乘积项和都是函数f的蕴涵项。

蕴涵项对应于立方体表示法中的立方体。

10.质蕴涵项(Prime Implicant,PI)设函数f有多个蕴涵项,若某个蕴涵项i所包含的最小项集合不是任何别的蕴涵项所包含的最小项集合的子集的话,则称i为函数f的质蕴涵项。

信息论英文课后部分习题答案

信息论英文课后部分习题答案

本答案是英文原版的配套答案,与翻译的中文版课本题序不太一样但内容一样。

翻译的中文版增加了题量。

2.2、Entropy of functions. Let X be a random variable taking on a finite number of values. What is the (general) inequality relationship of ()H X and ()H Y if(a) 2X Y =?(b) cos Y X =?Solution: Let ()y g x =. Then():()()x y g x p y p x ==∑.Consider any set of x ’s that map onto a single y . For this set()():():()()log ()log ()log ()x y g x x y g x p x p x p x p y p y p y ==≤=∑∑,Since log is a monotone increasing function and ():()()()x y g x p x p x p y =≤=∑.Extending this argument to the entire range of X (and Y ), we obtain():()()log ()()log ()x y x g x H X p x p x p x p x =-=-∑∑∑()log ()()yp y p y H Y ≥-=∑,with equality iff g if one-to-one with probability one.(a) 2X Y = is one-to-one and hence the entropy, which is just a function of the probabilities does not change, i.e., ()()H X H Y =.(b) cos Y X =is not necessarily one-to-one. Hence all that we can say is that ()()H X H Y ≥, which equality if cosine is one-to-one on the range of X .2.16. Example of joint entropy. Let (,)p x y be given byFind(a) ()H X ,()H Y .(b) (|)H X Y ,(|)H Y X . (c) (,)H X Y(d) ()(|)H Y H Y X -.(e) (;)I X Y(f) Draw a Venn diagram for the quantities in (a) through (e).Solution:Fig. 1 Venn diagram(a) 231()log log30.918 bits=()323H X H Y =+=.(b) 12(|)(|0)(|1)0.667 bits (/)33H X Y H X Y H X Y H Y X ==+===((,)(|)()p x y p x y p y =)((|)(,)()H X Y H X Y H Y =-)(c) 1(,)3log3 1.585 bits 3H X Y =⨯=(d) ()(|)0.251 bits H Y H Y X -=(e)(;)()(|)0.251 bits=-=I X Y H Y H Y X(f)See Figure 1.2.29 Inequalities. Let X,Y and Z be joint random variables. Prove the following inequalities and find conditions for equality.(a) )ZHYXH≥X(Z()|,|(b) )ZIYXI≥X((Z);,;(c) )XYXHZ≤Z-H-XYH),(,)(((X,,)H(d) )XYIZIZII+-XZY≥Y(););(|;;(Z|)(XSolution:(a)Using the chain rule for conditional entropy,HZYXHZXH+XH≥XYZ=),(|(Z)(||,()|)With equality iff 0YH,that is, when Y is a function of X andXZ,|(=)Z.(b)Using the chain rule for mutual information,ZIXXIZYX+=,I≥IYZ(|;)X);)(,;;(Z)(With equality iff 0ZYI, that is, when Y and Z areX)|;(=conditionally independent given X.(c)Using first the chain rule for entropy and then definition of conditionalmutual information,XZHYHIXZYX==-H-XHYYZ)()(;Z)|,|),|(X(,,)(XHXZH-Z≤,=,)()()(X|HWith equality iff 0ZYI, that is, when Y and Z areX(=|;)conditionally independent given X .(d) Using the chain rule for mutual information,);()|;();,();()|;(Z X I X Y Z I Z Y X I Y Z I Y Z X I +==+And therefore this inequality is actually an equality in all cases.4.5 Entropy rates of Markov chains.(a) Find the entropy rate of the two-state Markov chain with transition matrix⎥⎦⎤⎢⎣⎡--=1010010111p p p p P (b) What values of 01p ,10p maximize the rate of part (a)?(c) Find the entropy rate of the two-state Markov chain with transition matrix⎥⎦⎤⎢⎣⎡-=0 1 1p p P(d) Find the maximum value of the entropy rate of the Markov chain of part (c). We expect that the maximizing value of p should be less than 2/1, since the 0 state permits more information to be generated than the 1 state.Solution:(a) T he stationary distribution is easily calculated.10010*********,p p p p p p +=+=ππ Therefore the entropy rate is10011001011010101012)()()()()|(p p p H p p H p p H p H X X H ++=+=ππ(b) T he entropy rate is at most 1 bit because the process has only two states. This rate can be achieved if( and only if) 2/11001==p p , in which case the process is actually i.i.d. with2/1)1Pr()0Pr(====i i X X .(c) A s a special case of the general two-state Markov chain, the entropy rate is1)()1()()|(1012+=+=p p H H p H X X H ππ.(d) B y straightforward calculus, we find that the maximum value of)(χH of part (c) occurs for 382.02/)53(=-=p . The maximum value isbits 694.0)215()1()(=-=-=H p H p H (wrong!)5.4 Huffman coding. Consider the random variable⎪⎪⎭⎫ ⎝⎛=0.02 0.03 0.04 0.04 0.12 0.26 49.0 7654321x x x x x x x X (a) Find a binary Huffman code for X .(b) Find the expected codelength for this encoding.(c) Find a ternary Huffman code for X .Solution:(a) The Huffman tree for this distribution is(b)The expected length of the codewords for the binary Huffman code is 2.02 bits.( ∑⨯=)()(i p l X E )(c) The ternary Huffman tree is5.9 Optimal code lengths that require one bit above entropy. The source coding theorem shows that the optimal code for a random variable X has an expected length less than 1)(+X H . Given an example of a random variable for which the expected length of the optimal code is close to 1)(+X H , i.e., for any 0>ε, construct a distribution for which the optimal code has ε-+>1)(X H L .Solution: there is a trivial example that requires almost 1 bit above its entropy. Let X be a binary random variable with probability of 1=X close to 1. Then entropy of X is close to 0, but the length of its optimal code is 1 bit, which is almost 1 bit above its entropy.5.25 Shannon code. Consider the following method for generating a code for a random variable X which takes on m values {}m ,,2,1 with probabilities m p p p ,,21. Assume that the probabilities are ordered so thatm p p p ≥≥≥ 21. Define ∑-==11i k i i p F , the sum of the probabilities of allsymbols less than i . Then the codeword for i is the number ]1,0[∈i Frounded off to i l bits, where ⎥⎥⎤⎢⎢⎡=i i p l 1log . (a) Show that the code constructed by this process is prefix-free and the average length satisfies 1)()(+<≤X H L X H .(b) Construct the code for the probability distribution (0.5, 0.25, 0.125, 0.125).Solution:(a) Since ⎥⎥⎤⎢⎢⎡=i i p l 1log , we have 11log 1log +<≤i i i p l pWhich implies that 1)()(+<=≤∑X H l p L X H i i .By the choice of i l , we have )1(22---<≤ii l i l p . Thus j F , i j > differs from j F by at least il -2, and will therefore differ from i F is at least one place in the first i l bits of the binary expansion of i F . Thus thecodeword for j F , i j >, which has length i j l l ≥, differs from thecodeword for i F at least once in the first i l places. Thus no codewordis a prefix of any other codeword.(b) We build the following table3.5 AEP. Let ,,21X X be independent identically distributed random variables drawn according to theprobability mass function {}m x x p ,2,1),(∈. Thus ∏==n i i n x p x x x p 121)(),,,( . We know that)(),,,(log 121X H X X X p n n →- in probability. Let ∏==n i i n x q x x x q 121)(),,,( , where q is another probability mass function on {}m ,2,1.(a) Evaluate ),,,(log 1lim 21n X X X q n-, where ,,21X X are i.i.d. ~ )(x p . Solution: Since the n X X X ,,,21 are i.i.d., so are )(1X q ,)(2X q ,…,)(n X q ,and hence we can apply the strong law of large numbers to obtain∑-=-)(log 1lim ),,,(log 1lim 21i n X q n X X X q n 1..))((log p w X q E -=∑-=)(log )(x q x p∑∑-=)(log )()()(log )(x p x p x q x p x p )()||(p H q p D +=8.1 Preprocessing the output. One is given a communication channel withtransition probabilities )|(x y p and channel capacity );(max )(Y X I C x p =.A helpful statistician preprocesses the output by forming )(_Y g Y =. He claims that this will strictly improve the capacity.(a) Show that he is wrong.(b) Under what condition does he not strictly decrease the capacity? Solution:(a) The statistician calculates )(_Y g Y =. Since _Y Y X →→ forms a Markov chain, we can apply the data processing inequality. Hence for every distribution on x ,);();(_Y X I Y X I ≥. Let )(_x p be the distribution on x that maximizes );(_Y X I . Then__)()()(_)()()();(max );();();(max __C Y X I Y X I Y X I Y X I C x p x p x p x p x p x p ==≥≥===.Thus, the statistician is wrong and processing the output does not increase capacity.(b) We have equality in the above sequence of inequalities only if we have equality in data processing inequality, i.e., for the distribution that maximizes );(_Y X I , we have Y Y X →→_forming a Markov chain.8.3 An addition noise channel. Find the channel capacity of the following discrete memoryless channel:Where {}{}21Pr 0Pr ====a Z Z . The alphabet for x is {}1,0=X . Assume that Z is independent of X . Observe that the channel capacity depends on the value of a . Solution: A sum channel.Z X Y += {}1,0∈X , {}a Z ,0∈We have to distinguish various cases depending on the values of a .0=a In this case, X Y =,and 1);(max =Y X I . Hence the capacity is 1 bitper transmission.1,0≠≠a In this case, Y has four possible values a a +1,,1,0. KnowingY ,we know the X which was sent, and hence 0)|(=Y X H . Hence thecapacity is also 1 bit per transmission.1=a In this case Y has three possible output values, 0,1,2, the channel isidentical to the binary erasure channel, with 21=f . The capacity of this channel is 211=-f bit per transmission.1-=a This is similar to the case when 1=a and the capacity is also 1/2 bit per transmission.8.5 Channel capacity. Consider the discrete memoryless channel)11 (mod Z X Y +=, where ⎪⎪⎭⎫ ⎝⎛=1/3 1/3, 1/3,3 2,,1Z and {}10,,1,0 ∈X . Assume thatZ is independent of X .(a) Find the capacity.(b) What is the maximizing )(*x p ?Solution: The capacity of the channel is );(max )(Y X I C x p =)()()|()()|()();(Z H Y H X Z H Y H X Y H Y H Y X I -=-=-=bits 311log)(log );(=-≤Z H y Y X I , which is obtained when Y has an uniform distribution, which occurs when X has an uniform distribution.(a)The capacity of the channel is bits 311log /transmission.(b) The capacity is achieved by an uniform distribution on the inputs.10,,1,0for 111)( ===i i X p 8.12 Time-varying channels. Consider a time-varying discrete memoryless channel. Let n Y Y Y ,,21 be conditionally independent givenn X X X ,,21 , with conditional distribution given by ∏==ni i i i x y p x y p 1)|()|(.Let ),,(21n X X X X =, ),,(21n Y Y Y Y =. Find );(max )(Y X I x p . Solution:∑∑∑∑∑=====--≤-≤-=-=-=-=ni i n i i i n i i ni i i ni i i n p h X Y H Y H X Y H Y H X Y Y Y H Y H X Y Y Y H Y H X Y H Y H Y X I 111111121))(1()|()()|()(),,|()()|,,()()|()();(With equlity ifnX X X ,,21 is chosen i.i.d. Hence∑=-=ni i x p p h Y X I 1)())(1();(max .10.2 A channel with two independent looks at Y . Let 1Y and 2Y be conditionally independent and conditionally identically distributed givenX .(a) Show );();(2),;(21121Y Y I Y X I Y Y X I -=. (b) Conclude that the capacity of the channelX(Y1,Y2)is less than twice the capacity of the channelXY1Solution:(a) )|,(),(),;(212121X Y Y H Y Y H Y Y X I -=)|()|();()()(212121X Y H X Y H Y Y I Y H Y H ---+=);();(2);();();(2112121Y Y I Y X I Y Y I Y X I Y X I -=-+=(b) The capacity of the single look channel 1Y X → is );(max 1)(1Y X I C x p =.Thecapacityof the channel ),(21Y Y X →is11)(211)(21)(22);(2max );();(2max ),;(max C Y X I Y Y I Y X I Y Y X I C x p x p x p =≤-==10.3 The two-look Gaussian channel. Consider the ordinary Shannon Gaussian channel with two correlated looks at X , i.e., ),(21Y Y Y =, where2211Z X Y Z X Y +=+= with a power constraint P on X , and ),0(~),(221K N Z Z ,where⎥⎦⎤⎢⎣⎡=N N N N K ρρ. Find the capacity C for (a) 1=ρ (b) 0=ρ (c) 1-=ρSolution:It is clear that the two input distribution that maximizes the capacity is),0(~P N X . Evaluating the mutual information for this distribution,),(),()|,(),()|,(),(),;(max 212121212121212Z Z h Y Y h X Z Z h Y Y h X Y Y h Y Y h Y Y X I C -=-=-==Nowsince⎪⎪⎭⎫⎝⎛⎥⎦⎤⎢⎣⎡N N N N N Z Z ,0~),(21ρρ,wehave)1()2log(21)2log(21),(222221ρππ-==N e Kz e Z Z h.Since11Z X Y +=, and22Z X Y +=, wehave ⎪⎪⎭⎫⎝⎛⎥⎦⎤⎢⎣⎡++++N N P N N P N Y Y P P ,0~),(21ρρ, And ))1(2)1(()2log(21)2log(21),(222221ρρππ-+-==PN N e K e Y Y h Y .Hence⎪⎪⎭⎫⎝⎛++=-=)1(21log 21),(),(21212ρN P Z Z h Y Y h C(a) 1=ρ.In this case, ⎪⎭⎫⎝⎛+=N P C 1log 21, which is the capacity of a single look channel.(b) 0=ρ. In this case, ⎪⎭⎫⎝⎛+=N P C 21log 21, which corresponds to using twice the power in a single look. The capacity is the same as the capacity of the channel )(21Y Y X +→.(c) 1-=ρ. In this case, ∞=C , which is not surprising since if we add1Y and 2Y , we can recover X exactly.10.4 Parallel channels and waterfilling. Consider a pair of parallel Gaussianchannels,i.e.,⎪⎪⎭⎫⎝⎛+⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛212121Z Z X X Y Y , where⎪⎪⎭⎫ ⎝⎛⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛222121 00 ,0~σσN Z Z , And there is a power constraint P X X E 2)(2221≤+. Assume that 2221σσ>. At what power does the channel stop behaving like a single channel with noise variance 22σ, and begin behaving like a pair of channels? Solution: We will put all the signal power into the channel with less noise until the total power of noise+signal in that channel equals the noise power in the other channel. After that, we will split anyadditional power evenly between the two channels. Thus the combined channel begins to behave like a pair of parallel channels when the signal power is equal to the difference of the two noise powers, i.e., when 22212σσ-=P .。

香农公式

香农公式
X L x1 ,x 2 ,...,x M L = L p( ) p( X ) px1, px 2 ,..., px M L
给定有D个元素的码符号集,对扩展信源编码,总可以找 到一种唯一可译码,使码长 n L 满足:
X
Y
联 合 熵
交 互 熵
X
Y
X
Y
将定理3.3推广到L次扩展信源---
香农第一定理:变长编码定理
X x1 ,x2 ,...,xM 定理3.4 给定熵为H(X)的离散无记忆信源 p x , p x ,..., p x p( X ) M 1 2 其L次扩展信源的熵记为H(X)
nL n L
信源符号对应 的平均码字数
HX H U L ,limn RD logD L logD n
信息传输速率
这是信息传输速率 RD 能达到的极限值,对应于等概分布。
Shannon第一定理的物理意义:
信源编码时,应使编码后的码集中各码字尽可能等概 分布,若将该码集看成一个新的信源,此时新信源所含信 息量最大。
限定理都有其共性,也有个性。所给出的指导作用也各
不相同,但其证明方式都采用随机编码方式证明。 所谓存在性,是指定理仅给出是否存在着一种(至少
一种)编码方式可以满足要求;但如何编码则无可奉告。
它们的逆定理则给出了不存在性,这是它们的共性。 所谓构造性,是指定理不仅指出了存在性,而且还 给出了最佳码字的结构特性,如码长、代码形式等。
有噪信道编码逆定理
离散、无记忆、平稳信道,信道容 量为C,如果信息率R>C,则肯定找不 到一种信道编码方法,使得码长N足够 大时,平均差错率任意接近于零。
信道编码的指导意义

《信息技术英语》-unit3

《信息技术英语》-unit3

Unit 3Transition to Modern Information ScienceChapter One&Part4 Extensive Reading @Part 1 Notes to Text@Part5Notes to Passage & Part 2 Word Study@Part3 Practice on Text @Part6 Practice on Passage@Part 1 Notes to TextTransition to Modern Information Science1)With the 1950‘s came increasing awareness of the potentialof automatic devices for literature searching and informationstorage and retrieval.随着二十世纪五十年代的来临,人们对用于文献资料搜索、信息储存与检索的自动装置的潜力认识日益增长。

注释:该句是一个完全倒装句。

主语是awareness;介词短语With the 1950‘s是状语,修饰谓语动词came。

2)As these concepts grew in magnitude and potential, so did thevariety of information science interests. 由于这些概念的大量增长,潜移默化,对信息科学研究的各种兴趣也亦如此。

注释:介词短语in magnitude and potential作方式状语,意思是“大量地,潜移默化地”;后面的主句因为so放在句首而倒装。

So指代前文的grew in magnitude and potential。

3) Grateful Med at the National Library of Medicine美国国家医学图书馆数据库注释:Grateful Med是对另一个NLM(国家医学图书馆)基于网络的查询系统的链接。

简述采样定理的基本内容

简述采样定理的基本内容

采样定理的基本内容1. 什么是采样定理采样定理(Sampling Theorem)是数字信号处理中的一个基本理论,也被称为奈奎斯特定理(Nyquist Theorem)或香农定理(Shannon Theorem)。

它描述了如何在连续时间域中对信号进行采样,以便在离散时间域中能够完全还原原始信号。

2. 采样定理的基本原理采样定理的基本原理是:当一个信号的带宽不超过采样频率的一半时,我们可以通过对信号进行采样并以一定的频率进行重建,从而完整地恢复原始信号。

3. 采样定理的数学表达采样定理可以用数学方式表达如下: - 一个信号的最高频率为B,则采样频率Fs 应满足Fs > 2B,即采样频率必须是信号最高频率的2倍以上。

- 采样频率过低会导致混叠现象,也称为折叠现象(Aliasing),即原始信号的高频部分在采样后被混叠到低频部分。

- 采样频率过高不会引起混叠现象,但会浪费存储和计算资源。

4. 采样定理的应用采样定理在数字信号处理中有着广泛的应用,包括但不限于以下几个方面:4.1 通信系统在通信系统中,采样定理保证了信号的完整传输。

发送端将模拟信号进行采样,并通过数字信号处理技术将其转换为数字信号,然后通过传输介质传输到接收端。

接收端将数字信号还原为模拟信号,以便接收者能够恢复原始信息。

4.2 数字音频在数字音频领域,采样定理被广泛应用于音频录制和播放。

音频信号在录制过程中通过模拟转换器(ADC)进行采样,并以数字形式存储。

在播放过程中,数字音频通过数字转换器(DAC)转换为模拟信号,以便音箱或耳机能够播放出声音。

4.3 数字图像在数字图像处理中,采样定理被用于图像的采集和显示。

采样定理保证了图像的细节在数字化过程中不会丢失。

图像传感器将连续的光信号转换为数字图像,然后在显示器上以像素的形式显示出来。

4.4 数据压缩采样定理对数据压缩也有重要意义。

在信号的采样过程中,我们可以通过降低采样频率来减少数据量,从而实现信号的压缩。

数字通信中的信源编码和信道编码【精选文档】

数字通信中的信源编码和信道编码【精选文档】

数字通信中的信源编码和信道编码摘要:如今社会已经步入信息时代,在各种信息技术中,信息的传输及通信起着支撑作用.而对于信息的传输,数字通信已经成为重要的手段。

本论文根据当今现代通信技术的发展,对信源编码和信道编码进行了概述性的介绍。

关键词:数字通信;通信系统;信源编码;信道编码Abstract:Now it is an information society。

In the all of information technologies,transmission and communication of information take an important effect。

For the transmission of information,Digital communication has been an important means。

In this thesis we will present an overview of source coding and channel coding depending on the development of today’s communica tion technologies.Key Words:digital communication; communication system; source coding; channel coding1.前言通常所谓的“编码”包括信源编码和信道编码。

编码是数字通信的必要手段。

使用数字信号进行传输有许多优点, 如不易受噪声干扰,容易进行各种复杂处理,便于存贮,易集成化等。

编码的目的就是为了优化通信系统.一般通信系统的性能指标主要是有效性和可靠性.所谓优化,就是使这些指标达到最佳。

除了经济性外,这些指标正是信息论研究的对象.按照不同的编码目的,编码可主要分为信源编码和信道编码。

在本文中对此做一个简单的介绍.2.数字通信系统通信的任务是由一整套技术设备和传输媒介所构成的总体—-通信系统来完成的.电子通信根据信道上传输信号的种类可分为模拟通信和数字通信.最简单的数字通信系统模型由信源、信道和信宿三个基本部分组成.实际的数字通信系统模型要比简单的数字通信系统模型复杂得多。

有噪信道编码定理

有噪信道编码定理

有噪信道编码定理
噪声信道编码定理(Noise channel coding theorem)是通信理论中的一个重要定理,也被称为香农编码定理(Shannon's coding theorem)。

它说明了在有噪声的信道中,通过适当的编码和解码技术,可以实现任意小的误码率。

具体来说,噪声信道编码定理提供了用于传输信息的信道容量的上限,称为香农容量(Shannon capacity)。

香农容量表示了在给定的信道条件下,所能传输的最大有效数据速率。

根据该定理,如果某个编码方案的数据速率小于香农容量,则可以通过适当的编码和解码技术实现任意小的误码率。

噪声信道编码定理的核心思想是通过错误检测和纠正编码,将原始的输入符号转化为冗余的编码符号,这些编码符号可以对信道中的噪声进行纠正或者检测错误。

通过正确的编码和解码过程,接收端可以恢复出原始的输入符号,并降低误码率。

噪声信道编码定理的应用非常广泛,包括在无线通信、有线通信、光纤通信等各种通信系统中。

它为信道编码提供了理论指导,对于提高通信系统的可靠性和容量具有重要的意义。

信道容量的一般计算方法

信道容量的一般计算方法

信道容量的一般计算方法
信道容量是指在给定带宽条件下,信道可以传输的最大数据速率。

信道容量的计算是通过信道的带宽和信噪比之间的关系来确定的。

Step 1: 确定信道带宽(B)
信道带宽是指信道能够传输信号的频率范围,通常以赫兹(Hz)为单位。

确定信道带宽是计算信道容量的第一步。

Step 2: 确定信噪比(SNR)
信噪比是指信号和噪声的比例,以分贝(dB)为单位。

信噪比越高,信道传输的可靠性越高。

信噪比的计算需要根据具体信道的特性和环境条件进行。

Step 3: 计算信道的最大传输速率(C)
根据香农定理(Shannon's theorem),信道的最大传输速率(C)可以通过以下公式计算:
C = B * log2(1 + SNR)
其中,B为信道的带宽,SNR为信噪比。

这个公式表明,信道容量与信道带宽和信噪比的对数成正比。

Step 4: 优化信噪比以提高信道容量
为了提高信道容量,可以采取一些措施来优化信噪比,例如增加发射功率、减少噪声源、改善接收设备等。

Step 5: 考虑误码率和纠错编码
实际的信道容量还需要考虑误码率和纠错编码。

误码率是指在信道传
输过程中出现错误比特的概率,而纠错编码是一种冗余编码技术,可以在
接收端纠正部分错误。

综上所述,信道容量的计算方法主要包括确定信道带宽、信噪比和使
用香农定理计算最大传输速率。

通过优化信噪比和考虑误码率和纠错编码,可以进一步提高信道容量。

这些方法可以用于计算各种无线通信系统、光
纤通信系统等的信道容量,并对系统性能进行评估和优化。

奈奎斯特定理和香农定理解析

奈奎斯特定理和香农定理解析

奈奎斯特定理(Nyquist's Theorem)和香农定理(Shannon's Theorem)是网络传输中的两个基本定理。

学习之前先了解一下下面几个定义:波特率(baud rate)、比特率(bit rate)、带宽(bandwidth)、容量(capacity)。

波特率:波特率指的是信号每秒钟电平变化的次数,单位是Hz:比如一个信号在一秒钟内电平发生了365次变化,那么这个信号的波特率就是365Hz;比特率:比特率是信号每秒钟传输的数据的位数,我们知道在计算机中,数据都是用0,1表示的,所以比特率也就是每秒钟传输0和1的个数,单位是bps(bit per second)。

波特率和比特率之间有什么关系呢?我们可以假设一个信号只有两个电平,那么这个时候可以把低电平理解为“0”,高电平理解为“1”,这样每秒钟电平变化的次数也就是传输的0,1个数了,即比特率 = 波特率。

但是有些信号可能不止两个电平,比如一个四电平的信号,那么每个电平就可以被理解成“00”,“01”,“10”,“11”,这样每次电平变化就能传输两位的数据了,即比特率= 2 × 波特率。

一般的,bit rate = buad rate × log2L,这里L就是信号电平的个数。

下面再来看看带宽和容量的概念。

一般信道都有一个最高的信号频率(注意不是波特率哦,频率是指每秒钟的周期数,而每个周期都会有几次电平变化。

)和最低的信号频率,只有在这两个频率之间的信号才能通过这个信道,这两个频率的差值就叫做这个信道的带宽,单位是Hz。

信道的容量又是怎么回事呢?我们知道数据在信道中传输会有他们的速度——比特率,这里面最高的比特率就叫做这个信道的容量,单位是bps。

就好象每条公路都有他们的最高限速,那么所有在里面开的车都不会超过这个速度。

口语中也会把信道容量叫做“带宽”的,比如“带宽10M的网络”,“网络带宽是10M”等等。

Nyquist-Shannon sampling theorem

Nyquist-Shannon sampling theorem
按照 Fourier 变换原理,任何连续的周期函数,都可以通过Fourier 变换,变成由许多不同频率的正弦波组成的级ne2),但是如果正弦波的频率是采样频率的两倍(如sine1),采样的结果就会误认为sine1是与sine2同频率的正弦波,这就是发生的“混叠失真”现象。所以,采样的频率必须是最大频率的两倍。而更高的采样频率是多余的。
If a function s(x) has a Fourier transfrom F[s(x)] = S(f) = 0 for |f| >
W, then it is completely determined by giving the value of the function at a series of points spaced 1/(2W) apart. The values s = s(n/(2W)) are n
注意:这里有一个我们课上容易误解的地方,正是为了避免发生上面的“混叠失真”,才需要一个模拟的“低通滤波器”在采样前来滤掉那些频率高于采样频率1/2的正弦波,而不是因为Nyquist-Shannon Sampling Theorem,使过滤。这就是所谓的“抗混淆滤波器”。
也就是因为Nyquist-Shannon Sampling Theorem的上述原理,在达到人们应用要求的情况下,它被用来减少我们的信号采样频率,如电话上的语音采样频率。还被用来减少对现有的数字信号的采样频率(Fourier Fast Transform在对巨型离散数据的处理中
the Nyquist-Shannon Interpolation Formula.
Undersampling
It has to be noted that even if the concept of "twice the highest frequency" is the more commonly used idea, it is not absolute. In fact the theorem stand for "twice the bandwidth", which is totally different. Bandwidth

Homework

Homework

Discussion TopicsLiang Hao 142107201811. What the basic functional blocks are included in a digital communication system?Source encoder, channel encoder, digital modulator, digital demodulator, channel demodulator, source decoder.2. What are the average mutual information, entropy, and conditional entropy? What is the relationship of these three concepts?For two discrete random variables X and Y, The mutual information is defined as()()()()()()()j i j i n i mj j i n i mj j i j i y P x P y x P y x P y x I y x P Y X I ,log,;,;1111∑∑∑∑======Where p(x, y) is the joint probability distribution function of X and Y, and p(x) and p(y) are the marginal probability distribution functions of X and Y respectively. An important characteristic of the average mutual information is that()0;≥Y X I .When X and Y are statistically independent, we define average self-information as:()()()()()i nii i n i i x P x P x I x P X H log 11∑∑==-==When X represents the alphabet of possible output letters from a source, represents the average self-information per source letter, and it is called the entropy of the source.An average conditional self-information is called conditional entropy. It’s def ined as:()()()j i n i mjj i y x P y x P Y X H |1log ,|11∑∑===The relationship between the mutual information, entropy and conditional entropy is:()()()Y X H X H Y X I \;-=3. What is the Shannon source coding theorem?In information theory, Shannon's source coding theorem (or noiseless coding theorem ) establishes the limits to possible data compression, and the operationalmeaning of the Shannon entropy.The source coding theorem shows that (in the limit, as the length of a stream of independent and identically-distributed random variable (i.i.d.) data tends to infinity) it is impossible to compress the data such that the code rate (average number of bits per symbol) is less than the Shannon entropy of the source, without it being virtually certain that information will be lost. However it is possible to get the code rate arbitrarily close to the Shannon entropy, with negligible probability of loss.4. How to represent the band-pass signal using low-pass signal?Firstly develop a mathematical representation of such signals construct a signal that contains only the positive frequencies in s (t). ()()()f S f u f S 2=+ The signal S t (t) is called the analytic or the pre-envelope of s(t) :()()()()()t s t j t s t s t j t t s *1*ππδ+=⎥⎦⎤⎢⎣⎡+=+ Define ()t s ˆ as ()()()τττππd t s t s tt s ⎰∞∞--==1*1ˆ the signal may be viewed as the output of the filter with impulse response h (t).The analytic ()t s + is a band-pass signal ,we may obtain an equivalent low-pass representation by performing a frequency translation :Define S l (f ) as ()()c l f f S f S +=+. The equivalent time-domain relation is()()()()[]t f j t f j l c c e t s j t s e t s t s ππ22ˆ--++==. In general ,the signal ()t s l is complex-valued ,and may be expressed as:()()()t jy t x t s l +=. The expression is the desired form for the representationof a band-pass signals.5. How to express the energy in the band-pass signal in terms of the equivalent low-pass signal?The energy in signal is defined as ()()[]{}⎰⎰∞∞-∞∞-==dt et s dtt s E t f j lc 222Re π. Theenergy in band-pass signal is expressed in terms of equivalent low-pass signal is()⎰∞∞-=dt t s E l 221.6. What is the AWGN noise? Make discussion in details.Additive white Gaussian noise (AWGN) is a basic noise model used in Information theory to mimic the effect of many random processes that occur in nature. The modifiers denote specific characteristics:(1) Additive because it is added to any noise that might be intrinsic to the information system.(2) White refers to the idea that it has uniform power across the frequency band for the information system. It is an analogy to the color white which has uniform emissions at all frequencies in the visible spectrum.(3) Gaussian because it has a normal distribution in the time domain with an average time domain value of zero.7. What is the modulated carrier signal of the minimum-shift keying modulation?Minimum Shift Keying (MSK) is a special from of binary CPFSK where the modulation index h = 1/2. The carrier-modulated signal corresponding may beexpressed as ()()[]0;2cos 2φφπ++=I t f TEt s c8. What is the expression of the power density spectrum if the sequence is real and mutual uncorrelated?When the information symbols in the sequence are real and mutually uncorrelated. The autocorrelation function can be expressed as()()()⎩⎨⎧≠=+=00222m m m ii i ii μμσφ. We represent the()m ii φ in()()∑∞-∞=-=ΦmfmTj ii ii e m f πφ2 we can get the result()∑∞-∞=-+=ΦmfmTj ii ii e f πμσ222.9. Discussion on the optimal receiver configuration over AWGN channels?The optimal receiver means we can minimize the probability of making error. So we can conveniently subdivided the receiver into two parts: signal demodulator and detector. There are two ways to realize the signal demodulator, one is based on the use of signal correlators, and the other is based on the use of matched filters. Normally we have two demodulator: one is correlation type demodulator and the other is matched-filter type demodulator. Both the two demodulator can produce the vector which contains all the necessary information. The optimum detector that follows the signal demodulator is designed to minimize the probability of error. To achieve this, we can apply the MAP rule or ML rule on our detector.10. What are the optimum decision criterions of the optimal receiver over AWGN channels?There are two optimum decision criterions of the optimal receiver over AWGN channels, one is the MAP criterion-minimum distance detection. The decision criterion is based on selecting the signal corresponding to the maximum of the set of posterior probabilities such a criterion is called as maximum a posteriori probability (MAP) criterion. The other is maximum-likelihood (ML) criterion.11. What is the Shannon capacity of the band-limited AWGN waveform channel with a band-limited and average power-limited input? And make the discussion in detail about this Shannon capacity .The Shannon capacity is ⎪⎪⎭⎫⎝⎛+=01log WN P W C av . Where the average received power is av P and the noise power spectral density is 0N .is known as received signal-to-noise ratio (SNR). When the SNR is large (SNR >>0dB), the capacity C ≈⎪⎪⎭⎫⎝⎛=0log WN P W C av is logarithmic in power and approximatelylinear in bandwidth. This is called the bandwidth-limited regime. When the SNR is small (SNR << 0dB), the capacity C ≈ e N P C av log 2=is linear in power butinsensitive to bandwidth. This is called the power-limited regime.。

香农定理 调制格式

香农定理 调制格式

香农定理调制格式Shannon's theorem, also known as Shannon's capacity theorem, is a fundamental result in information theory. It states that there is a maximum rate at which information can be transmitted over a communication channel with a specified bandwidth and error rate. 香农定理,也称为香农容量定理,在信息论中是一个基本的结果。

它指出在指定的带宽和误码率下,信息可以被传输的最大速率。

The theorem was formulated by Claude Shannon in 1948 and has had a profound impact on the development of modern communication systems. 香农定理是由克劳德·香农在1948年提出的,对现代通信系统的发展产生了深远的影响。

One key concept in Shannon's theorem is the notion of channel capacity, which represents the maximum amount of information that can be reliably transmitted over a communication channel. 香农定理中一个关键的概念是信道容量,它表示可以可靠传输的信息的最大量。

The formula for channel capacity is given by C = Blog₂(1 + S/N), where C is the channel capacity in bits per second, B is the channelbandwidth in hertz, and S/N is the signal-to-noise ratio of the channel. 信道容量的公式为C = Blog₂(1 + S/N),其中C是信道容量,B 是信道的带宽,S/N是信道的信噪比。

奈奎斯特采样定理和香农采样定理

奈奎斯特采样定理和香农采样定理

奈奎斯特采样定理和香农采样定理
一、奈奎斯特采样定理
1、奈奎斯特采样定理(Nyquist Sampling Theorem)指出,对
任何一个连续的时间函数,如果它在时间轴上有频率不超过一个上限,则只要把它采样频率设计在该上限的两倍以上即可完全重建出这个
函数。

奈奎斯特采样定理是数字信号处理的基本原理之一,该定理指出如果采样频率大于两倍最高信号频率,则可以完全重建出信号的完整信息。

该定理的意义在于,在信号数字化时,我们只需要采样频率大于信号最高频率两倍即可精确无损地重建信号,因此也可称其为“无损采样定理”。

2、基于奈奎斯特采样定理,在模拟信号转换为数字信号时,需
要将模拟信号先做低通滤波,使阻带范围不超过采样频率的一半,被称为“奈奎斯特限制频率”,与此同时,将采样频率设置在奈奎斯特
限制频率的两倍以上,这样可以保证数字信号重建时无损传输。

二、香农采样定理
1、香农采样定理(Shannon Sampling Theorem)又称“总变换
定理”,由Shannon于1949年提出,表明任何一个带宽有限的连续信号都可以通过取样的方式近似表示,而且取样频率满足一定条件时,信号可以完整的重建。

2、香农采样定理的条件是采样频率为该信号的频率范围的两倍
以上,并且频率范围的宽度要大于频谱中峰值频率的两倍,此时采样
时的取样频率叫做重建阈值,即信号可以完整重建所需要的最低采样频率。

香农采样定理是分析数字信号的基础原理,它解决了模拟信号数字化的问题,指出任何一个带宽有限的连续信号都可以通过取样的方式近似表达,并且只要实现正确的采样取样频率,就可以完整重建数字信号。

b为什等于2w_奈奎斯特定理_概述及解释说明

b为什等于2w_奈奎斯特定理_概述及解释说明

b为什等于2w 奈奎斯特定理概述及解释说明1. 引言1.1 概述奈奎斯特定理,也被称为奈奎斯特-香农采样定理,是信号处理和通信领域中的一项重要定理。

该定理阐述了在进行连续时间信号采样和离散时间信号重构时的基本原则与条件。

根据奈奎斯特定理,为了避免采样和重构过程中出现混叠现象(aliasing),采样频率必须大于信号的最高频率成分的两倍。

1.2 文章结构本文将首先介绍奈奎斯特定理的原理及应用,并解释其在通信领域中的实际应用示例。

随后,我们将回顾相关的理论背景和发展历程,包括早期关于信号采样的限制条件分析以及奈奎斯特提出的采样定理及其后续研究进展。

接下来,我们将详细解释和讨论奈奎斯特定理并提供数学推导与证明过程概述、定义和解释采样率与信号带宽之间关系以及为什么b等于2w是满足奈奎斯特定理条件的解释。

最后,我们将给出本文的结论。

1.3 目的本文旨在提供关于奈奎斯特定理的概述和解释,并阐明其在信号处理和通信领域中的重要性和应用价值。

通过对奈奎斯特定理及其背后的原理进行详细讲解,读者将能够全面了解信号采样与重构过程中需要考虑的关键因素,以及如何避免混叠现象并确保信号准确地恢复。

希望本文能够为读者提供有关奈奎斯特定理基本概念和相关原理的清晰认识,并促进对该定理进一步研究和实际应用的探索。

2. 奈奎斯特定理的原理及应用2.1 信号采样与重构的基本概念在通信领域中,我们经常需要对连续时间的信号进行采样和重构。

采样是指将连续时间信号转换为离散时间信号,而重构则是将离散时间信号还原为连续时间信号。

在进行信号采样时,我们需要选择一个适当的采样率。

采样率是指每秒钟对信号进行采集的样本数。

根据奈奎斯特定理,我们知道采样率必须至少是信号中最高频率的两倍,也就是说要满足采样率大于等于2倍的最高频率。

然后,在对被采样的离散时间信号进行重构时,我们使用插值方法来还原连续时间信号。

插值方法可以通过填充缺失数据点来恢复原始连续时间信号。

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PT ≡ p(x) ≥ 1 − .
(18)
x∈T
The total probability PT represents the probability for a randomly chosen sequence x to lie in the typical set T . Now consider the special random variable
p(ak) = pk ∈ (0, 1], k = 1, . . . , K
(1)
where k pk = 1. Each given string x of a random message is an instance or realization of the message ensemble
X ≡ X1 · · · XN , where each random letter Xn is identical to a fixed letter ensemble X,
PT ≡ p(x) ≥ 1 − .
(23)
x∈T
If we encode only typical sequences, the probability of error
Perr := 1 − PT ≤
(24)
can be made arbitrarily small by choosing N large enough. Now let us determine how many typical sequences there are. The lefthand side of (22) gives
possible messages
typical messages
code words
FIG. 1: Lossy coding.
x ≡ x1 · · · xN of letters, which are independently drawn from an alphabet A = {a1, . . . , aK } with a priori probabilities
R variable and the law of large numbers. Given the letter
ensemble X, the function f : A → defines a discrete, real random variable. The realizations of f (X) are the real numbers f (x), x ∈ A. The average of f (X) is defined as
∗Electronic address: bostroem@
Now consider a very long message x. Typically, the letter ak will appear with the frequency Nk ≈ N pk. Hence, the probability of such typical message is roughly
(15)
N
The relative standard deviation of A yields
∆A 1 ∆f (X)
=√
.
(16)
A
Байду номын сангаас
N f (X)
Concluding, in the limit of large N the arithmetic average of the sequence f (X) and the ensemble average of f (X) coincide. This is the law of large numbers. It is responsible for the validity of statistical experiments. Without this law, we could never verify statistical properties of a system by performing many experiments. In particular, quantum mechanics would be free of any physical meaning.
1
I ≡ N IN = H(X).
(7)
This is Shannon’s source coding theorem in a nutshell. Now let us get a bit more into detail. In order to rigorously prove the theorem we need the concept of a random
1N
A= N
f (Xn) = f (X) ,
(11)
i=1
2
and the variance of A reads
∆2A = A2 − A 2
(12)
1
= N2
f (Xn)f (Xm)
n,m
1
−N2
f (Xn) f (Xm)
(13)
n,m
1 = N2
f 2(Xn) − f (Xn) 2
(14)
n
= 1 ∆2f (X).
Let us reformulate the law of large numbers in the , δlanguage. For δ > 0 we define the typical set T of a random sequence X as the set of realizations x ≡ x1 · · · xN such that
|T | 2−N(H−δ) ≥ 1 −
(29)
⇔ |T | ≥ (1 − ) 2N(H−δ).
(30)
Relations (28) and (30) can be combined into the crucial relation
(1 − ) 2N(H−δ) ≤ |T | ≤ 2N(H+δ).
(31)
1N
H −δ ≤ − N
log p(xn) ≤ H + δ,
(21)
n=1
or equivalently
2−N(H+δ) ≤ p(x) ≤ 2−N(H−δ),
(22)
where H ≡ H(X). By the law of large numbers, the probability for a randomly drawn message x to be a member of T reads
1N
A := N
f (Xn),
(10)
n=1
which is also a random variable. Since the Xn are identical copies of the letter ensemble X, the average of A is equal to the average of f (X),
K
f (X) := p(x) f (x) = pk f (ak), (8)
x∈A
k=1
and the variance is given by
∆2f (X) := f 2(X) − f (X) 2.
(9)
For the sequence f (X) ≡ f (X1), . . . , f (XN ) we define its arithmetic average as
Xn = X, n = 1, . . . , N.
(2)
A particular message x = x1 · · · xN appears with the probability
p(x1 · · · xn) = p(x1) · · · p(xn),
(3)
which expresses the fact that the letters are statistically independent from each other.
Shannon’s Source Coding Theorem
Kim Bostr¨om Institut fu¨r Physik, Universit¨at Potsdam, 14469 Potsdam, Germany ∗
The idea of Shannon’s famous source coding theorem [1] is to encode only typical messages. Since the typical messages form a tiny subset of all possible messages, we need less resources to encode them. We will show that the probability for the occurence of non-typical strings tends to zero in the limit of large message lengths. Thus we have the paradoxical situation that although we “forget” to encode most messages, we loose no information in the limit of very long strings. In fact, we make use of redundancy, i.e. we do not encode “unnecessary” information represented by strings which almost never occur. Recall that a random message of length N is a string
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