Unit 5：The Sampling Theorem
电子与通信专业英语第二版李霞 课后翻译
y(6> Mathematical methods lie in the foundation of physics, mechanics, engineering and othernatural sciences. For all of them mathematics is a powerful theretical and pratical tool withoutwhich no scientific calculation and no enineering and technology are pssilie
一个信号被公式化成一一个包含或多个变量的函数，并用于传达物理现象本质中的信息
Y(4) Nyquist sampling theorem states that if the highest-frequency componentofa CT signalx(t) is fmax, then x(t) can be recovered from its sampled sequence x(nT) if the sampling frequencyfs is dosldsibeealons maavaenaaomqiamgasuvaon larger than 2fmax.
(4)When feedback is used around an operational amplifier, the closed loop gain of the circuitis
determined by a ratio involving the input and feedback impedances used.
外研社版学术英语综合Unit5译文
第五单元哲学Text A机器人有没有头脑？奎迈·安东尼·阿皮亚电脑在当今很多电影中都担当主演，它们有的能用合成的声音说话，有的则在屏幕上显示一连串文字；有的能够操作太空飞船；有的作为机器人的“首脑”，控制着它们自己的“身体”。
人与它们对话，它们不仅能够理解，还能够与人类互相问候，交换信息。
当然，以上这些到目前为止都还只属于科幻小说的情节。
但是，现实中的电脑协助律师办理案件，医生诊断病情，还能帮助工程师检测原子反应堆的状态。
无论是想象中的还是现实的电脑都足以让我们的祖父母惊奇不已，他们之前也许认为那些只能通过魔术实现。
然而，我们大多数人对此都已经习以为常，认为这在硅时代都是理所当然的事情。
当然，怀疑是仍旧存在的。
我们人类总自认为自己是很特殊的存在，并假设物质世界与精神世界存在某些对立。
如果电脑真的是一个“物质思维”，那么人类不仅必须重新认识这种特殊性，而且还会用自己创造的东西将它摧毁。
我们应当小心避免这样的想法，直到真正地对这个问题进行深入思考。
不管我们外表表现地多么自然，对于电脑会思考，能做事表现得理所当然，但是我们终归不该这样假设。
在哲学中我们常会发现，那些通常被人们认为是理所当然的事情——那些“常识”——往往会阻碍我们对事情的正确理解。
因此，让我们来看一下我在第一段关于电脑的描述是否准确。
我说过它们能够说话。
但是它们真的能像人类一样表达自己吗？甚至说它们能够做一些类似于讲话的事情都有些牵强。
因为磁带录音机也可以做到，但是它们并不能说话。
当人类说话时，他们用所说的话的内容来表达自己的意思。
为了表达意思，则需要具备理解句子的能力。
当然，我之前也提到电脑能够理解人类对它们说的话。
但它们是真的理解了吗？它们是将我们讲话的声音被转换成了电脉冲，又通过电脑的电路，利用音响合成器发出声音。
人类非常聪明，能够设计出完成这件事情的机器，但是我们有什么证据证明机器是真正理解了呢？一台机器可以具备理解能力吗？关于这个问题，有两个显而易见的答案。
抽样与参数统计
4
4,1
4,2
4,3
4,4
19 5-
统计学
STATISTICS
样本均值旳抽样分布
(例题分析)
计算出各样本旳均值，如下表。并给出样本均 值旳抽样分布
16个样本旳均值（x）
第一种 第二个观察值
观察值 1
2
3
4
1
1.0 1.5 2.0 2.5
2
1.5 2.0 2.5 3.0
3
2.0 2.5 3.0 3.5
=10
n= 4
x 5
n =16 x 2.5
= 50 X
总体分布
x 50
x
抽样分布
22 5-
统计学
STATISTICS
中心极限定理
(central limit theorem)
中心极限定理：设从均值为 ，方差为 2旳一种任意 总体中抽取容量为n旳样本，当n充分大时，样本均值 旳抽样分布近似服从均值为μ、方差为σ2/n旳正态分布
统计学
STATISTICS
第 5 章 抽样与参数估计
1 5-
统计学 第 5 章 抽样与参数估计
STATISTICS
5.1 抽样及其分布 5.2 抽样方法 5.3 参数估计
5.4 样本容量旳拟定 5.5 Excel旳应用
2 5-
统计学
STATISTICS
学习目的
1． 了解抽样和抽样分布旳基本概
念
2． 了解点估计旳概念和估计量旳
30 5-
统计学
STATISTICS
样本百分比旳抽样分布
(数学期望与方差)
• 样本百分比旳数学期望
E( p)
• 样本百分比旳方差
Richard组合数学第5版-第5章课后习题答案（英文版）
Richard组合数学第5版-第5章课后习题答案（英⽂版）Math475Text:Brualdi,Introductory Combinatorics5th Ed. Prof:Paul TerwilligerSelected solutions for Chapter51.For an integer k and a real number n,we shown k=n?1k?1+n?1k.First assume k≤?1.Then each side equals0.Next assume k=0.Then each side equals 1.Next assume k≥1.RecallP(n,k)=n(n?1)(n?2)···(n?k+1).We haven k=P(n,k)k!=nP(n?1,k?1)k!.n?1 k?1=P(n?1,k?1)(k?1)!=kP(n?1,k?1)k!.n?1k(n?k)P(n?1,k?1)k!.The result follows.2.Pascal’s triangle begins111121133114641151010511615201561172135352171182856705628811936841261268436911104512021025221012045101···13.Let Z denote the set of integers.For nonnegative n∈Z de?ne F(n)=k∈Zn?kk.The sum is well de?ned since?nitely many summands are nonzero.We have F(0)=1and F(1)=1.We show F(n)=F(n?1)+F(n?2)for n≥2.Let n be /doc/6215673729.htmling Pascal’s formula and a change of variables k=h+1,F(n)=k∈Zn?kk=k∈Zn?k?1k?1=k∈Zn?k?1k+h∈Zn?h?2h=F(n?1)+F(n?2).Thus F(n)is the n th Fibonacci number.4.We have(x+y)5=x5+5x4y+10x3y2+10x2y3+5xy4+y5and(x+y)6=x6+6x5y+15x4y2+20x3y3+15x2y4+6xy5+y6.5.We have(2x?y)7=7k=07k27?k(?1)k x7?k y k.6.The coe?cient of x5y13is35(?2)13 185.The coe?cient of x8y9is0since8+9=18./doc/6215673729.htmling the binomial theorem,3n=(1+2)n=nk=0nSimilarly,for any real number r,(1+r)n=nk=0nkr k./doc/6215673729.htmling the binomial theorem,2n=(3?1)n=nk=0(?1)knk3n?k.29.We haven k =0(?1)k nk 10k =(?1)n n k =0(?1)n ?k n k 10k =(?1)n (10?1)n =(?1)n 9n .The sum is 9n for n even and ?9n for n odd.10.Given integers 1≤k ≤n we showk n k =n n ?1k ?1.Let S denote the set of ordered pairs (x,y )such that x is a k -subset of {1,2,...,n }and yis an element of x .We compute |S |in two ways.(i)To obtain an element (x,y )of S there are n k choices for x ,and for each x there are k choices for y .Therefore |S |=k n k .(ii)Toobtain an element (x,y )of S there are n choices for y ,and for each y there are n ?1k ?1 choices for x .Therefore |S |=n n ?1k ?1.The result follows.11.Given integers n ≥3and 1≤k ≤n .We shown k ? n ?3k = n ?1k ?1 + n ?2k ?1 + n ?3k ?1.Let S denote the set of k -subsets of {1,2,...,n }.Let S 1consist of the elements in S thatcontain 1.Let S 2consist of the elements in S that contain 2but not 1.Let S 3consist of the elements in S that contain 3but not 1or 2.Let S 4consist of the elements in S that do|S |= n k ,|S 1|= n ?1k ?1 ,|S 2|= n ?2k ?1 ,|S 3|= n ?3k ?1 ,|S 4|= n ?3k .The result follows.12.We evaluate the sumnk =0(?1)k nk 2.First assume that n =2m +1is odd.Then for 0≤k ≤m the k -summand and the (n ?k )-summand are opposite.Therefore the sum equals 0.Next assume that n =2m is even.Toevaluate the sum in this case we compute in two ways the the coe?cient of x n in (1?x 2)n .(i)By the binomial theorem this coe?cient is (?1)m 2m m .(ii)Observe (1?x 2)=(1+x )(1?x ).We have(1+x )n =n k =0n k x k,(1?x )n =n k =0nk (?1)k x k .3By these comments the coe?cient of x n in(1?x2)n isn k=0nn?k(?1)knk=nk=0(?1)knk2.2=(?1)m2mm.13.We show that the given sum is equal ton+3k .The above binomial coe?cient is in row n+3of Pascal’s /doc/6215673729.htmling Pascal’s formula, write the above binomial coe?cient as a sum of two binomial coe?ents in row n+2of Pascal’s triangle.Write each of these as a sum of two binomial coe?ents in row n+1of Pascal’s triangle.Write each of these as a sum of two binomial coe?ents in row n of Pascal’s triangle.The resulting sum isn k+3nk?1+3nk?2+nk?3.14.Given a real number r and integer k such that r=k.We showr k=rr?kr?1k.First assume that k≤?1.Then each side is0.Next assume that k=0.Then each side is 1.Next assume that k≥1.ObserverP(r?1,k?1)k!,andr?1k=P(r?1,k)k!=(r?k)P(r?1,k?1)k!.The result follows.15.For a variable x consider(1?x)n=nk=0nk(?1)k x k.4Take the derivative with respect to x and obtain n(1x)n1=nk=0nk(?1)k kx k?1.Now set x=1to get(?1)k k.The result follows.16.For a variable x consider(1+x)n=nk=0nkx k.Integrate with respect to x and obtain(1+x)n+1 n+1=nk=0nkx k+1k+1+Cfor a constant C.Set x=0to?nd C=1/(n+1).Thus (1+x)n+1?1n+1=nk=0nkx k+1k+1.Now set x=1to get2n+1?1 n+1=k+1.17.Routine.18.For a variable x consider(x?1)n=nk=0nk(?1)n?k x k.Integrate with respect to x and obtain(x?1)n+1 n+1=nk=0nk(?1)n?kx k+1k+1+Cfor a constant C.Set x=0to?nd C=(?1)n+1/(n+1).Thus (x?1)n+1?(?1)n+1n+1=nk=0nk(?1)n?kx k+1k+1Now set x =1to get(?1)n n +1=n k =0n k(?1)n ?k 1k +1.Therefore1n +1=n k =0 n k (?1)k 1k +1 .19.One readily checks2 m 2 + m 1=m (m ?1)+m =m 2.Therefore n k =1k 2=nk =0k 2=2nk =0 k 2 +n k =0k1=2 n +13 +n +12 =(n +1)n (2n +1)6.20.One readily checksm 3=6 m 3 +6 m 2 + m1.Thereforen k =1k3=n=6nk =0 k3+6n k =0 k2 +n k =0k1 =6 n +14 +6 n +13 +n +12 =(n +1)2n 24= n +12 2.621.Given a real number r and an integer k .We showrk=(?1)kr +k ?1k .First assume that k <0.Then each side is zero.Next assume that k ≥0.Observe r k =(r )(r 1)···(r k +1)k !=(?1)kr (r +1)···(r +k ?1)k !=(?1)kr +k ?1k.22.Given a real number r and integers k,m .We showr m m k = r k r ?km ?k.First assume that mObserver m m k =r (r ?1)···(r ?m +1)m !m !k !(m ?k )!=r (r ?1)···(r ?k +1)k !(r ?k )(r ?k ?1)···(r ?m +1)(m ?k )!= r k r ?k m ?k .23.(a) 2410.(b) 94 156.(c) 949363.(d)94156949363.24.The number of walks of length 45is equal to the number of words of length 45involving10x ’s,15y ’s,and 20z ’s.This number is45!10!×15!×20!.725.Given integers m 1,m 2,n ≥0.Shown k =0m 1k m 2n ?k = m 1+m 2n .Let A denote a set with cardinality m 1+m 2.Partition A into subsets A 1,A 2with cardinalitiesm 1and m 2respectively.Let S denote the set of n -subsets of A .We compute |S |in two ways.(i)By construction|S |= m 1+m 2n .(ii)For 0≤k ≤n let the set S k consist of the elements in S whose intersection with A 1has cardinality k .The sets {S k }n k =0partition S ,so |S |= nk =0|S k |.For 0≤k ≤n we now compute |S k |.To do this we construct an element x ∈S k via the following 2-stage procedure: stage to do #choices 1pick x ∩A 1 m 1k2The number |S k |is the product of the entries in the right-most column above,which comes to m 1k m 2n ?k .By these comments |S |=n k =0m 1k m 2n ?k .The result follows.26.For an integer n ≥1shown k =1 n k n k ?1 =12 2n +2n +1 ? 2n n .Using Problem 25,n k =1 n k nk ?1 =n k =0n k n k ?1 =n k =0n k nn +1?k =2n n +1 =12 2n n ?1 +12 2n n +1.8It remains to show12 2nn ?1 +12 2n n +1 =12 2n +2n +1 ? 2n n.This holds since2n n ?1 +2 2n n + 2n n +1 = 2n +1n +2n +1n +1= 2n +2n +1.27.Given an integer n ≥1.We shown (n +1)2n ?2=nk =1Let S denote the set of 3-tuples (s,x,y )such that s is a nonempty subset of {1,2,...,n }and x,y are elements (not necessarily distinct)in s .We compute |S |in two ways.(i)Call an element (s,x,y )of S degenerate whenever x =y .Partition S into subsets S +,S ?with S +(resp.S ?)consisting of the degenerate (resp.nondegenerate)elements of S .So |S |=|S +|+|S ?|.We compute |S +|.To obtain an element (s,x,x )of S +there are n choices for x ,and given x there are 2n ?1choices for s .Therefore |S +|=n 2n ?1.We compute |S ?|.To obtain an element (s,x,y )of S ?there are n choices for x,and given x there are n ?1choices for y ,and given x,y there are 2n ?2choices for s .Therefore |S ?|=n (n ?1)2n ?2.By these comments|S |=n 2n ?1+n (n ?1)2n ?2=n (n +1)2n ?2.(ii)For 1≤k ≤n let S k denote the set of elements (s,x,y )in S such that |s |=k .Thesets {S k }nk =1give a partition of S ,so |S |= n k =1|S k |.For 1≤k ≤n we compute |S k |.To obtain an element (s,x,y )of S k there are n k choices for s ,and given s there are k 2ways to choose the pair x,y .Therefore |S k |=k 2 nk .By these comments|S |=n k =1k 2 n k .The result follows.28.Given an integer n ≥1.We shown k =1k n k 2=n 2n ?1n ?1 .Let S denote the set of ordered pairs (s,x )such that s is a subset of {±1,±2,...,±n }andx is a positive element of s .We compute |S |in two ways.(i)To obtain an element (s,x )of S There are n choices for x ,and given x there are 2n ?1n ?1 choices for s .Therefore|S |=n 2n ?1n ?1.9(ii)For1≤k≤n let S k denote the set of elements(s,x)in S such that s contains exactlyk positive elements.The sets{S k}nk=1partition S,so|S|=nk=1|S k|.For1≤k≤nwe compute|S k|.To obtain an element(s,x)of S k there are nkways to pick the positiveelements of s and nn?kways to pick the negative elements of s.Given s there are kways to pick x.Therefore|S k|=k nk2.By these comments |S|=nk=1knk2.The result follows.29.The given sum is equal tom2+m2+m3n .To see this,compute the coe?cient of x n in each side of(1+x)m1(1+x)m2(1+x)m3=(1+x)m1+m2+m3.In this computation use the binomial theorem.30,31,32.We refer to the proof of Theorem5.3.3in the text.Let A denote an antichain such that|A|=nn/2.For0≤k≤n letαk denote the number of elements in A that have size k.Sonk=0αk=|A|=nn/2.As shown in the proof of Theorem5.3.3,≤1,with equality if and only if each maximal chain contains an element of A.By the above commentsnk=0αknn/2nknk≤0,with equality if and only if each maximal chain contains an element of A.The above sum is nonpositive but each summand is nonnegative.Therefore each summand is zero and the sum is zero.Consequently(a)each maximal chain contains an element of A;(b)for0≤k≤n eitherαk is zero or its coe?cient is zero.We now consider two cases.10Case:n is even.We show that for0≤k≤n,αk=0if k=n/2.Observe that for0≤k≤n, if k=n/2then the coe?cient ofαk isnonzero,soαk=0.Case:n is odd.We show that for0≤k≤n,eitherαk=0if k=(n?1)/2orαk=0 if k=(n+1)/2.Observe that for0≤k≤n,if k=(n±1)/2then the coe?cient ofαk is nonzero,soαk=0.We now show thatαk=0for k=(n?1)/2or k=(n+1)/2. To do this,we assume thatαk=0for both k=(n±1)/2and get a contradiction.By assumption A contains an element x of size(n+1)/2and an element y of size(n?1)/2. De? ne s=|x∩y|.Choose x,y such that s is maximal.By construction0≤s≤(n?1)/2. Suppose s=(n?1)/2.Then y=x∩y?x,contradicting the fact that x,y are incomparable. So s≤(n?3)/2.Let y denote a subset of x that contains x∩y and has size(n?1)/2. Let x denote a subset of y ∪y that contains y and has size(n+1)/2.By construction |x ∩y|=s+1.Observe y is not in A since x,y are comparable.Also x is not in A by the maximality of s.By construction x covers y so they are together contained in a maximal chain.This chain does not contain an element of A,for a contradiction.33.De?ne a poset(X,≤)as follows.The set X consists of the subsets of{1,2,...,n}. For x,y∈X de?ne x≤y whenever x?y.Forn=3,4,5we display a symmetric chain decomposition of this poset.We use the inductive procedure from the text.For n=3,,1,12,1232,233,13.For n=4,,1,12,123,12344,14,1242,23,23424,For n=5,,1,12,123,1234,123455,15,125,12354,14,124,124545,1452,23,234,234525,23524,2453,13,134,134535,13534,345.1134.For 0≤k ≤ n/2 there are exactlyn kn k ?1symmetric chains of length n ?2k +1.35.Let S denote the set of 10jokes.Each night the talk show host picks a subset of S for his repertoire.It is required that these subsets form an antichain.By Corollary 5.3.2each antichain has size at most 105 ,which is equal to 252.Therefore the talk show host can continue for 252nights./doc/6215673729.htmlpute the coe?cient of x n in either side of(1+x )m 1(1+x )m 2=(1+x )m 1+m 2,In this computation use the binomial theorem.37.In the multinomial theorem (Theorem 5.4.1)set x i =1for 1≤i ≤t .38.(x 1+x 2+x 3)4is equal tox 41+x 42+x 43+4(x 31x 2+x 31x 3+x 1x 32+x 32x 3+x 1x 33+x 2x 33)+6(x 21x 22+x 21x 23+x 22x 23)+12(x 21x 2x 3+x 1x 22x 3+x 1x 2x 23).39.The coe?cient is10!3!×1!×4!×0!×2!which comes to 12600.40.The coe?cient is9!3!×3!×1!×2!41.One routinely obtains the multinomial theorem (Theorem 5.4.1)with t =3.42.Given an integer t ≥2and positive integers n 1,n 2,...,n t .De?ne n = ti =1n i .We shownn 1n 2···n t=t k =1n ?1n 1···n k ?1n k ?1n k +1···n t.Consider the multiset{n 1·x 1,n 2·x 2,...,n t ·x t }.Let P denote the set of permutations of this multiset.We compute |P |in two ways.(i)We saw earlier that |P |=n !n 1!×n 2!×···×n t != n n 1n 2···n t.12(ii)For1≤k≤t let P k denote the set of elements in P that have?rst coordinate x k.Thesets{P k}tk=1partition P,so|P|=tk=1|P k|.For1≤k≤t we compute|P k|.Observe that|P k|is the number of permutations of the multiset{n1·x1,...,n k?1·x k?1,(n k?1)·x k,n k+1·x k+1,...,n t·x t}. Therefore|P k|=n?1n1···n k?1n k?1n k+1···n t.By these comments|P|=tn1···n k?1n k?1n k+1···n t.The result follows.43.Given an integer n≥1.Show by induction on n that1 (1?z)n =∞k=0n+k?1kz k,|z|<1.The base case n=1is assumed to hold.We show that the above identity holds with n replaced by n+1,provided that it holds for n.Thus we show1(1?z)n+1=∞=0n+z ,|z|<1.Observe1(1?z)n+1=1(1?z)n11?z=∞k=0n+k?1kz k∞h=0z h=0c zwherec =n?1+n1+n+12+···+n+ ?1=n+.The result follows.1344.(Problem statement contains typo)The given sum is equal to (?3)n .Observe (?3)n =(?1?1?1)n=n 1+n 2+n 3=nnn 1n 2n 3(?1)n 1+n 2+n 3=n 1+n 2+n 3=nnn 1+n 2+n 3=nnn 1n 2n 3(?1)n 2.45.(Problem statement contains typo)The given sum is equal to (?4)n .Observe (?4)n =(?1?1?1?1)n=n 1+n 2+n 3+n 4=nnn 1n 2n 3n 4(?1)n 1+n 2+n 3+n 4=n 1+n 2+n 3+n 4=nnn 1n 2n 3n 4(?1)n 1?n 2+n 3?n 4.Also0=(1?1+1?1)n= n 1+n 2+n 3+n 4=nnn 1n 2n 3n 4(?1)n 2+n 4.46.Observe√30=5=5∞ k =01/2k z k.For n =0,1,2,...the n th approximation to √30isa n =5n k =0 1/2k 5?k.We have14n a n051 5.52 5.4753 5.47754 5.47718755 5.477231256 5.4772246887 5.4772257198 5.4772255519 5.477225579 47.Observe101/3=21081/3=2(1+z)1/3z=1/4,=2∞k=01/3kz k.For n=0,1,2,...the n th approximation to101/3isnk=01/3k4?k.We haven a n021 2.1666666672 2.1527777783 2.1547067904 2.1543852885 2.1544442306 2.1544327697 2.1544350898 2.1544346059 2.15443470848.We show that a poset with mn+1elements has a chain of size m+1or an antichain of size n+1.Our strategy is to assume the result is false,and get a contradiction.By assumption each chain has size at most m and each antichain has size at most n.Let r denote the size of the longest chain.So r≤m.By Theorem5.6.1the elements of the posetcan be partitioned into r antichains{A i}ri=1.We have|A i|≤n for1≤i≤r.Thereforemn+1=ri=1|A i|≤rn≤mn, 15for a contradiction.Therefore,the poset has a chain of size m+1or an antichain of size n+1.49.We are given a sequence of mn+1real numbers,denoted{a i}mni=0.Let X denote the setof ordered pairs{(i,a i)|0≤i≤mn}.Observe|X|=mn+1.De?ne a partial order≤on X as follows:for distinct x=(i,a i)and y=(j,a j)in X,declare xof{a i}mni=0,and the antichains correspond to the(strictly)decreasing subsequences of{a i}mni=0sequence{a i}mni=0has a(weakly)increasing subsequence of size m+1or a(strictly)decreasingsubsequence of size n+1.50.(i)Here is a chain of size four:1,2,4,8.Here is a partition of X into four antichains:8,124,6,9,102,3,5,7,111Therefore four is both the largest size of a chain,and the smallest number of antichains that partition X. (ii)Here is an antichain of size six:7,8,9,10,11,12.Here is a partition of X into six chains:1,2,4,83,6,1295,10711Therefore six is both the largest size of an antichain,and the smallest number of chains that partition X.51.There exists a chain x116。
外研版高中英语必修第三册Unit5 Section Ⅰ Understanding ideas讲义
核心单词initial，thus，zone，oxygen，border，confirm，crowd，attempt，failure，psychologist，port，unaware ，alcohol，rubber，cotton，wool，guideline，cloth，bleeding，liquid，injury，stretch，league，servant，charge，captain，ahead，permit，chapter，relate，nut，astonish，ray，consume，distinguish，gradually，surround，wrinkled，shore，carpet，intensity，depth，broad，beneath，resident，male重点短语good money，make money，succeed in(doing)sth．，take one's life，bring...into focus，figure out，refer to...as...，leave an impression upon/on sb．，account for，in broad daylight，make sense重点句式1．动词-ing形式作结果状语2．as if引导方式状语从句单元语法过去将来时主题写作写潜水经历Section ⅠUnderstanding ideas1．____ causing the delay 从而造成了拖延2．time ____ 时区3．take in ______ 吸收氧气4．across the ______ 穿越边界1．_______ adj．开始的，最初的→ initially ad v．最初，首先；开头2．_______ v．证实，证明→confirmation n．确认；证实；证明3．_____ n．人群→ crowded adj．拥挤的；塞满的4．_______ n．努力，尝试→attempted adj．企图的；未遂的5．fail v．失败→_______ n．失败6．psychology n．心理学→____________ n．心理学家1．____________ 寻找，寻求2．__________ 大笔的钱3．__________ 赚钱，挣钱4．_______________________ 成功(做)某事5．______________ 夺去某人的生命6．__________________ 使……成为焦点7．___________ 转身，转向反方向8．__________ 弄明白9．________________ 把……称为……10．_________ 代表1．[句型公式]动词-ing形式作结果状语For these people，climbing Qomolangma is an experience like no other，_____________________ and others，powerful．对于这些人来说，攀登珠穆朗玛峰是一项独一无二的体验，它让一些人感受到脆弱的同时也让另一些人感受到强大。
Nyquist-Shannon sampling theorem
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
研究生英语多维教程熟谙Unit5英语summary
研究生英语多维教程熟谙Unit5英语summaryThe summary of Unit5 Sandwich generationIn the modern society，there are a large group，They are the so-called Sandwich generation，people who are struggling to care for both their children and their elders，often while holding down a job as well.The Sandwich generation are living a hard life.The statistics show that the proportion of seniors living with their children is decreasing，those who do move in with their children enter households profoundly changed from previous generations.Because toda y′s seniors had fewer children than their predecessors，there are fewer family members to share the burden.They not only have some elder-care responsibilities，ranging from occasional help with groceries to fulltime care at home，but also care for their children.In fact，the parents of an average family now work 65 to 80 hours a week，up from 40 to 45 hours a week in the 1950s.To stay in the placethey are working twice as hard.It lead to that they have more stress，less job satisfaction and more absence than their colleagues.That change often is overlooked amid increasing public pressure to transfer some of the governments health-care burden to individual families.But families have changed，you cannot make the assumption the people are available.Even when an elderly parent lives independently，responsibilities can weigh heavily on their children，How do they break away from their job to see a sick mother?And then they fell guilty when they leave her alone.Even if their elderly parents are not living with them，there is the stress of juggling obligations at home，to their extended family and their employer.Fortunately，there are many house holds that cope well，especially when the senior is healthy or the younger family members have outside help in times of crisis.But for many families who bear a wide range of new responsibilities，elder care can take a physical，and emotional，sacrifice.T o the sandwich generation，all these responsibilities are a trouble to them，their burden will be more heavy than any generations before.。
Unit 5 课文原文——A Beautiful Mind
Unit 5 Leading Scientist and Pioneering WorkPassage A A Beautiful Mind[1]John Forbes Nash Jr., -mathematical genius, inventor of a theory of rational behavior, visionary of the thinking machine- had been sitting with his visitor, also a mathematician, for nearly half an hour. It was late on a weekday afternoon in the spring of 1959, and though it was only may uncomfortably warm, Nash was slumped in an armchair in one corner of the hospital lounge, carelessly dressed in a nylon shirt that hung limply over his unbelted trousers. His powerful frame was slack as a rag doll's his fine molded features expressionless. He had been staring dully at a spot immediately in front of the left foot of Harvard professor George Mackey, hardly moving except to brush his long, dark hair away from his forehead in a fitful, repetitive motion. His visitor sat upright, oppressed by the silence, acutely conscious that the doors to the room were locked. Mackey finally could contain himself no longer. His voice was slightly querulous, but he strained to be gentle. “How could you,” began Mackey, “how could you, a mathematician, a man devoted to reason and logical proof... how could you believe that extraterrestrials are sending you messages? How could you believe that you are being recruited by aliens from outer space to save the world? How could you...?”[2] Nash looked up at last and fixed Mackey with an unblinking stare as cool and dispassionate as that of any bird or snake. "Because," Nash said slowly in his soft, reasonable southern drawl, as if talking to himself, "the ideas I had about supernatural beings came to me the same way that my mathematical ideas did. So I took them seriously.[3] The young genius from Bluefield, West Virginia—handsome, arrogant, and highly eccentric—burst onto the mathematical scene in 1948. Over the next decade, a decade as notable for its supreme faith in human rationality as for its dark anxieties about mankind's survival, Nash proved himself, in the words of the eminent geometer Mikhail Gromov, "the most remarkable mathematician of the second half of the century." Games of strategy, economic rivalry, computer architecture, the shape of the universe, the geometry of imaginary spaces, the mystery of prime numbers—allengaged his wide-ranging imagination. His ideas were of the deep and wholly unanticipated kind that pushes scientific thinking in new directions.[4]Geniuses, the mathematician Paul Halmos wrote, " are of two kinds: the ones who are just like all of us, but very much more so, and the ones who, apparently, have an extra human spark. We can all run, and some of us can run the mile in less than 4 minutes; but there is nothing that most of us can do that compares with the creation of the Great:G-minor Fugue." Nash's genius was of that mysterious variety more often associated with music and art than with the oldest of all sciences: It wasn't merely/that his mind worked faster, that his memory was more retentive, or that his power of concentration was greater. The flashes of intuition were nonrational. Like other great mathematical intuitionists—Georg Friedrich Bernhard Riemann, Jules Henri Poincaré, Srinivasa Ramanujan—Nash saw the vision first; constructing the laborious proofs long afterward. But even after he'd try to explain some astonishing result, the actual route he had taken remained a mystery to others who tried to follow his reasoning. Donald Newman, a mathematician who knew Nash at MIT in the1950s, used to say about him that "everyone else would climb a peak by looking for a path somewhere on the mountain. Nash would climb another mountain altogether and from that distant peak would shine a searchlight back onto the first peak".[5] No one was more obsessed with originality, more disdainful of authority, or more jealous of his independence. As a young man he was surrounded by the high priests of twentieth-century science——Albert Einstein, John von Neumann, and Norbert Wiener—but he joined no school, became no one's disciple, got along: largely without guides or followers. In almost everything he did—from game theory to geometry—he thumbed his nose at the received wisdom, current fashions, established methods. He almost always worked alone, in his head, usually walking, often whistling Bach. Nash acquired his knowledge of mathematics not mainly from studying what other mathematicians had discovered, but by rediscovering their truths for himself. Eager to astound, he was always on the lookout for the really big problems. When he focused on some new puzzle, he saw dimensions that people who really knew the subject (he never did) initially dismissed as naive or wrong-headed.Even as a student, his indifference to otthers' skepticism, doubt, and ridicule was awesome.[6] Nash's faith in rationality and the power of pure thought was extreme, even for a very young mathematician and even for the new age of computers, space travel, and nuclear weapons. Einstein once chided him for wishing to amend relativity theory without studying physics. His heroes were solitary thinkers and supermen like Newton and Nietzsche. Computers and science fiction were his passions. He considered "thinking machines", as he called them superior in some ways to human beings. At one point, he became fascinated by the possibility that drugs could heighten physical and intellectual performance. He was beguiled by the idea of alien races of hyper-rational beings who had taught themselves to disregard all emotion. Compulsively rational, he wished to turn life's decisions-whether to take the first elevator or wait for the next one, where to bank his money, what job to accept, whether to marry-intocalculations of advantage and disadvantage, algorithms or mathematical rules divorced from emotion, convention, and tradition. Even the small act of saying an automatic hello to Nash in a hallway could elicit a furious "Why are you saying hello to me?"[7] His contemporaries, on the whole, found him immensely strange. They described him as “aloof”,“haughty”, “without affect”,“detached”, “spooky”“isolated”, and “queer”. Nash mingled rather than mixed with hispeers. Preoccupied with his own private reality, he seemed not to sharetheir mundane concerns. His manner-slightly cold, a bit superior, somewhat secretive-suggested something “mysterious and unnatural”. His remoteness was punctuated by flights of garrulousness about outer space and geopolitical trends, childish pranks, and unpredictable eruptions of anger. But these outburstswere, more often than not, as enigmatic as his silences. "He is not one of us' was a constant refrain. A mathematician at the Institute for Advanced Study remembers meeting Nash for the first time at a crowded student party at Princeton:I noticed him very definitely among a lot of other people who were there. He was sitting on the floor in a half-circle discussing something. He made me feel uneasy. Hegave me a peculiar feeling. I had a feeling of a certain strangeness. He was different in some way. I was not aware of the extent of his talent. I had no idea he would contribute as much as he really did.[8] But he did contribute, in a big way. The marvelous paradox was that the ideas themselves were not obscure. In1958, Fortune singled Nash out for his achievements in game theory, algebraic geometry, and nonlinear theory, calling him the most brilliant of the younger generation of new ambidextrous mathematicians who worked in both pure and applied mathematics. Nash's insight into the dynamics of human rivalry—his theory of rational conflict and cooperation—was to become one of the most influential ideas of the twentieth century, transforming the young science of economics the way that Mendel's ideas of genetic transmission, Darwin's model of natural selection, and Newton's celestial mechanics reshaped biology and physics in their day.。
Unit 5 Understanding Ideas(课件)——高中英语外研版(2019)必修第三册
Think and share Do you agree with their opinions? Why?
Read para 4~5 carefully and
answer the questions.
1. What is Type T personality?
2. Who developed it?
Climbing Qomolangma: Worth the risks
The risks
Para 1
climbers
The reasons Conclusion
scientific reasons
Para 6~7
Para 2~3 Para 4~5
Read para 1 carefully and answer the two questions. 1. What are the risks of
He __________ Qomolangma _______ and ___________
Name: George Mallory Reasons: What we get form this adventure is just _s_h_e_er_j_o_y____We _d_o_n_o_t _live to eat and make money. We eat and make money to _b_e_a_b_le__to_e_n_j_oy__li_fe____. That is _w__h_at_l_if_e_m__ea_n_s__ and __w_h_a_t _li_fe__is_f_or
If you were given the chance, would you like to climb Qomolangma? Why or why not?
Unit5WorkingtheLandReadingandThinking教学设计-高中英语人教版选
While-readingfora map 1.T invites Ss to read the title and the first sentences of each paragraph and asks Ss what type of text it is, where such a piece of writing can be found and what the text will include. Possible answers:(1)It’s a biography, which can be found from people column of the magazine.(2)It may contains one’s birth/death time, childhood/family background/personal life, character/quality, achievement/contribution/influence, hardship/effort/turning point, quote/comment and so on.2.T asks Ss to read through the text to find the structure of the passage and summarize the main idea according to the given map.Possible answers:Para1-C, Para2-D, Para3-A, Para4-E, Para5-F, Para6-B; Part1-Yuan’s brief introduction, Part2-Yuan’s devotion to hybrid rice, Part3-The evaluation of Yuan1.T asks each student to work out his /her own mind map withformind forto blackboard and the other Ss on their worksheet.(Independent Work)❖Level A：Ss can draw the mind map as they like as long as they can analyze the the reasons, process, results and influence of Yuan and his work.❖Level B : Ss can fill the blank below and use it as a tool for to sort out the right information.❖Level C: Ss answer the questions listed on the textbook and find the information sentences in the text.and own 1.Ss share their mind maps in groups and compare them with the ones on the blackboard. If necessary, they can add more key points or delete some needless points.(Group Work)2.T asks Ss to explain their own mind maps on the blackboard or works projected on the screen, and if there is something missing, T will enlighten Ss to make supplement.Possible answers:After cooperation, T shows Ss a video about Yuan and thefor Medal of the Republic and ask Ss to answer the following questions:1.Yuan Longping considers himself a farmer. What does the writer regard him as? What evidence can you find from the reading?Possible answers:(1) a pioneer for all people①While many people thought it was impossible to develop hybrid rice, he was convinced the idea and developed the first hybird rice through intense effort.②Yuan’s innovation has helped feed not just China, but many other countries that depend on rice as well.③Yuan is an example for all people----What impresses people most about Yuan is his ongoing ability to fulfill his dreams. Besides, he is still young at heart and full of vision.(2) a man of the soil①His slim but strong body is just like that of millions of Chinese farmers, to whom he has devoted his life.②Yuan cares little for celebrity or money. Instead, he makes large donations to support agricultural research.2.How do you understand the quote “ My lifelong pursuit is to keep all the people away from hunger” on the open page of this unit?One possible opinion:It expresses Yuan’s desire to prevent hunger and starvation in the world. (to help those who are suffering from the effects of famine.)3.How can we understand the title A Pioneer For All People? One possible opinion:First,a pioneer means a person who is the first to study and develop a particular area that other people then continue to develop. So,to be a pioneer, first you must be the first or one of the few to do something, and then there must be others who want to follow you. According to the main idea of each paragraph,we can find Yuan is really a pioneer.time.1.T invites the class to summarize what they have learned. (topic, language, text and its features, and the intention and value of this text, using mind map to introduce Yuan, etc.)2.Ss review the content of this class and do their best to consolidate what they have learned.3.T draws Ss’ attention back to the learning aims and see whether those aims have been achieved or not. Ss can assess themselves according to the chart.Homework。
信息科学与电子工程专业英语
5. The sampling rate limit the frequency range of the audio file; to reproduce a given frequency, the sampling rate must be at least twice that frequency. For example, if the audio contains audible frequencies as high as 8,000 Hz, your need a sample rate of 16,000 samples per second to reproduce this audio accurately in digital form. This calculation comes from the Nyquist Theorem, and the highest frequency that can be reproduced by a given sample rate is known as the Nyquist Frequency. CDs have a sampling rate of 44,100 samples per second that allows samples up to 22,050 Hz, which is higher than the limit of human hearing, 20,000Hz.
分子等等,导致一波高压进入空气; 高压力波通过空气移动,他们身后留下低压区域。当这 些压力低点和高点——或波动——到达我们,他们在我们的耳朵振动接收,我们听到时发 出的震动声。
高三英语教案：上学期Unit 5a2
人教修订版英语高三上Unit 5 Getting the messageReading教案Teaching aims:1.Improve the students’ reading ability.2.Get the students to know about the advertisement and people’sattitude towards ads.3.Let the students learn to protect themselves from misleading ads. Teaching important points:1.Master the important words and phrases.2.Improve the students’ reading ability.3.Enable the students to understand the passage better.Teaching difficult points:1.How to help the students understand the reading material exactly.2.How to help the students finish the concerned exercises afterreading.Teaching methods:1.Discussion method to talk about the advantages and disadvantagesof ads.2.Fast-reading to help the students get the general idea of the passage.3.Careful reading to help the students get the detailed informationabout the passage.4.Individual, pair or group work to make every student work in class.Teaching procedures:Step 1: Revision1.Check the homework.2.Get the students to show their own advertisements(this was left forthem the former period).(after showing)T: Which advertisement do you think is good? Why? Can you giveyour opinion?Step 2: Warming upT: With the development of market economy, advertisements have become a dominant feature in television, radio and newspaper industry. Ads are used to persuade us to buy things. They can be useful when we are trying to decide what we should buy. But some ads can be misleading when they give incorrect or false information.Are ads good or bad for people?Please make a list of advantages and disadvantages of ads. Work in pairs. You are given two minutes to prepare it.(two minutes later, collect the students’ answers on the blackboard) advantages: help to buy and sell goodswiden people’s knowledgemake people more experienceddisadvantages: always persuade people to buy their poorly made productsgive people some misinformationwaste people too much timeStep 3: Fast readingT: Today we’re going to read a passage about advertising. It will tell us the advantages and disadvantages of ads.Please read the passage as quickly as possible, and underline the advantages and disadvantages of ads mentioned in the text while reading, at the same time find out the main idea of each paragraph. (after reading, get the students to tell their answers)Para 1: Advertising is a highly developed industry.Para 2: People react to ads in different ways.Para 3: What is the basic principle of advertising?Para 4-6: Ads are helpful in many ways.Para 7: How can we spot bad ads?Para 8: We should take a critical attitude towards ads.Step 4: Careful readingT: Let’s read the paragraph carefully one by one and find out the following points(show on the screen)Para 1:1.Is the opinion about ads in this paragraph correct?2.Are the ads everywhere in your life?3.Where else can you find the ads?Para 2:1.How do people react to ads?2.What’s your opinion? Can you give some reasons?Para 3:1.What is the basic principle of advertising?2.How do you make a good ad?Para 4-6:1.How important are ads in today’s society?2.How do ads help companies and customers?3.What are good ads?Para 7:1.What are bad ads?2.Why should we keep an eye out for “hidden information” in ads? Para 8:1.How can we make use of good ads?2.How can we protect ourselves from false ads?3.How can we develop a critical attitude towards ads?Step 5: Listening and readingPlay the tape for the students to follow in a low voice, paying attention to their pronunciation and intonation. Then the students read the passage for a while aloud by themselves and try to understand the phrases learnt just now.Step 6: DiscussionT: Here are the two advertisements on Page 41. Work in pairs and analyse the ads. Talk about which of the techniques described in the reading are used to sell the product and which of the claims in the ads may not be truthful.I’ll give you three minutes to discuss them. Then I’ll ask some of you to talk about the ads.(Three minutes later, teacher asks the students to analyse the ads.)Step 7: SummaryT: in this class we’ve learned something about advertising. We’ve learned the advantages and disadvantages of ads, the basic principle of advertising and the effects on companies and customers. At the same time, we’ve learned some useful phrases. After class, try to make some sentences using these phrases.Step 8: Homework1.Prepare for the dictation (words & expressions)2.Write a brief introduction of advertisement.。
高二英语上学期unit5-grammar
森林具有三个层次：遗传多样性、物种多样性和生态系统多样性。森林包含了区域中生物种类的组合、生物与环境间相互作用的过程，以及经受干扰后的演变过程最为完整的记录。正如气候顶极类 型提供的当地植被完整的演变历史那样。这些生态过程，是从人为干预下生长时间较短的人工植被中无法获得的。或许，天然林和人工林是完全不同的两回事。
何况，在哈拉哈河两岸的森林里，枯朽的空洞老树，还是紫貂、青鼬、艾虎、花鼠、灰鼠、鼯鼠等兽类和原生蜜蜂栖居的巢穴。大空洞树是黑熊蹲仓冬眠的极好场所。猞猁也常常借助于大树窟窿而 栖身。
森林的奥秘，也许就藏在那些枯朽老树的树洞里。森林有自己的秩序和逻辑。当一种现象超过某种确定的界限，森林就会调整内部的结构关系，重新确定秩序——这就是森林法则。
森林就是森林。森林里没有多余的东西，更没有废物。即使森林中那些枯朽的老树也不是废物。只有父母儿孙的生存，而没有爷爷奶奶的存在，并不能算得上是一个完整的人类社会，而森林，同样 是一个老中青幼连结着的群体。正因为有枯朽老树的存在，才意味着一座森林的生长有着不同寻常的历史，才构成了完整的自然生态系统。
Unit 5：The Sampling Theorem
perceive：察觉，感知，感到
当我们以极快的速度来顺次观察这些瞬时画面时，感觉到的 就是对原来连续活动场景的一个准确再现。
Specialized English for Communication Engineering
Much of the importance of the sampling theorem also lies in its role as a bridge between continuous-time signals and discrete-time signals. continuous-time signals : 连续时间信号
sequence :次序, 顺序, 序列
Space：留间隔， 隔开 在一定的条件下，一个连续时间信号完全可以由该信号在时 间等间隔点上的瞬时值或样本值来表示，并且能用这些样本 值恢复出原信号来。 Specialized English for Communication Engineering
This somewhat surprising property follows from a basic result that is referred to as the sampling theorem. This theorem is extremely important and useful. somewhat :稍微, 有点, 有些
Specialized English for Communication Engineering
Sampling theorem can be stated as follows: Let x(t) be a band-limited signal with X(jω)=0 for |ω|>ωm. Then x(t) is uniquely determined by its samples x(nT), n=0, ±1, ±2, …, if ωm：被抽样信号的最高频率 ωs>2ωm where ωs ：抽样频率 T ：抽样间隔 ωs=2π/T
全新版大学英语第三册课件 Unit 5
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在美国，每年11月的最后一个星期四是举国同庆的感恩节，规定从这一 天起休假三天。即使远在异乡的美国人也都要在节日前赶回去与家人团聚， 共庆佳节。美国人对感恩节重视的程度，堪与传统的盛大节日－圣诞节相 比。 • 为什么叫感恩节呢？追根溯源，感恩节是和美洲古代的印第安人，特别是 和玉米的种植有十分密切的关系。1620年9月6日，一批英国清教徒难以忍 受宗教的迫害，搭乘“五月花号”木船驶往美洲。他们在疲劳、饥饿、寒 冷和疾病的袭击下在大西洋上漂泊了65天，最后到达北美殖民地的普利茅 斯。 • 当时正值冬季，气候严寒，田野寂寥。当地印第安人慷慨地拿出贮藏越冬 的玉米和土豆，送去猎获的野鸭和火鸡。春天来了，印第安人教他们种植 玉米和南瓜，饲养火鸡。白人移民和印第安人建立了亲密的友谊。这年秋 季玉米丰收，移民们举行了丰盛的感恩会，用烤火鸡和玉米糕点款待印第 安人。印第安人带着各种玉米制品、烤火鸡、南瓜馅饼、野葡萄以及玉米 酒浆等参加晚会，人们欢唱歌舞，通霄达旦。以后在每年玉米收获后的11 月底，定居在这里的白人移民都要举行感恩会，家家烤火鸡，烹制玉米食 品，款待印第安人。长此以往，这种感恩会就成为一种惯例。但是，当时 感恩节的时间并不固定，直至1863年，美国总统林肯宣布把感恩节定为全 国性的节日，号召人民同心同德，为美国的繁荣昌盛作出努力。为了纪念 感恩节，在普利茅斯城的萨摩特大街修建了一座巨大的移民先驱者纪念碑。 高达10米的乳白色先民雕像，搭着披肩，手指苍穹，在四块大理石浮雌上 描述了到普利茅斯未的第一批移民的故事和感恩节的渊源。现今，玉米制 品仍然是欢度感恩节的不可缺少的食品之
The Pilgrims The Pilgrims were the 102 English people who sailed to America on the Mayflower in 1620. Their group included 35 Puritans whose aim was to create a safe religious community in the New World.
unit 5大学英语精读3(课堂PPT)
well-paid positions and p_r_o_f_e_s_si_o_n_s_ once reserved almost exclusively for men, such as _m_e_d_i_c_in_e_ , law and management.
Most women work _b_e_ca_u__se__th__ei_r_f_a_m__il_ie_s_n__ee_d__t_h_e _m_o_n_e_y_ . With the price of food and fuel, and housing, transportation and education rising steadily, more and more families are depending upon a second _in_c_o_m__e_. Many women work also because they want to have careers of their own and enjoy full e_q_u_a_l_it_y_ with men.
3
Not until I became a mother did I understand How much my mother had sacrificed for me; Not until I became a mother did I feel How hurt my mother was when I disobeyed; Not until I became a mother did I know How proud my mother was when I achieved; Not until I became a mother did I realize How much my mother loves me.
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.。
外研版Unit 5 Understanding idea(Grammar) 课件 七年级英语上册
Understanding ideas (Grammar)
Grammar
Learning objectives
பைடு நூலகம்
通过本节课的学习，学生能够： 1. 掌握现在分词构成用法。 2. 掌握现在进行时态的结构和用法。 3. 运用现在进行时态描述正在发生的行为。
(3) Listen!They are singing.
(4)It's six a clock. We are having breakfast.
be(am,is,are) +V-ing be(am / is are)随着主语的变化而变化。
Grammar
Exploration of the Course
Exploration of the Course
（3）一般疑问句及回答:
问句, 直接把be提前, Be +主语+v-ing+其它? 肯定回答: Yes,主语 +be 否定回答: No,主语+be+not.
Grammar
Exploration of the Course
—Is she dancing ? —Yes, she is.
Grammar
Exploration of the Course
1.定义： （1）表示现在正在进行或发生的动作；
What is the monkey doing? It is eating a banana now.
Grammar
Exploration of the Course
（2）表示现在这一阶段正在进行的动作 What is he doing these days? He is working hard.
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——Specialized English
第一部分 专业英语基础知识
被动语态及其译法
科技英语中大量使用被动语态 科技英语所表达的是客观规律，要尽量避免使用主观色彩较 强的第一、第二人称； 被动语态主要的信息置于句首，更能突出主要论证、说明对 象，引人注目； 在很多情况下被动结构比主动结构更简短。
Specialized English for Communication Engineering
It is exploited, for example, in moving pictures, which consist of a sequence of individual frames, each of which represents an instantaneous view (i.e., a sample in time) of a continuously changing scene. exploit: 使用,开拓, 开发, 开采 moving pictures : 活动图象, 电影 individual :个别的, 单独的, 个人的 frame: 帧,画面,框架 instantaneous: 瞬间的, 即刻的, 即时的 例如，抽样定理在电影里得到了利用。电影由一组按时序排列的 单个画面所组成，其中每一个画面都代表着连续变化景象中的一 个瞬时画面（即时间样本）。
This somewhat surprising property follows from a basic result that is referred to as the sampling theorem. This theorem is extremely important and useful. somewhat :稍微, 有点, 有些 follow from：是从[根据]...得出的 saturate:饱和 这有点不可思议的性质是根据抽样定理这个基本结论得出的。 这个理论及其重要且非常有用
Specialized English on Communication Engineering
被动语态的译法
1.译成汉语被动句： 常用“被”、“受到”、“得到”、“让”、“给”、“把”、“使”、“由” 等措辞。 A digital image is composed of a finite number of elements, each of which has a particular location and value. 数字图像由有限个元素（像素）构成，每个像素都具有特定 的位置与数值。
Specialized English for Communication Engineering
This is due in large part to the dramatic development of digital technology over the past few decades, resulting in the availability of inexpensive, lightweight, programmable, and easily reproducible discrete-time signal, process the discretetime system. dramatic：引人注目的，戏剧性的, 生动的 lightweight ：重量轻的，轻质的 这在很大程度上是由于在过去几十年中数字技术的发展，导 致了便宜的、轻质的，处理可编程且易于再生的离散时间信 号的离散时间系统的实用化。 Specialized English for Communication Engineering
Specialized English for Communication Engineering
The fact that under certain conditions a continuous-time signal can be completely recovered from a sequence of its samples provides a mechanism for representing a continuoustime signal by a discrete-time signal. recover：重获，恢复，复原 sequence：序列，顺序, 次序 mechanism ：机理， 机构, 机制 在一定的条件下，可以用信号的时序样本值完全恢复出原连 续时间信号，这就提供了用一个离散时间信号来表示一个连 续时间信号的机理。 Specialized English for Communication Engineering
Specialized English for Communication Engineering
TEXT
Under certain conditions, a continuous signal can be completely represented by and recoverable from a sequence of its values, or samples, at points equally spaced in time. recoverable :可重获的 sequence :次序, 顺序, 序列 Space：留间隔， 隔开 在一定的条件下，一个连续时间信号完全可以由该信号在时 间等间隔点上的瞬时值或样本值来表示，并且能用这些样本 值恢复出原信号来。 Specialized English for Communication Engineering
Specialized English on Communication Engineering
被动语态的译法
2.译成汉语的主动句： 直接将原文的被动语态按照主动语态译成汉语的主动句。 The experiment will be finished in a month. 这项实验将在一个月后完成。 Using time multiplexing, a digital system can be used to process a number of digital signals. 利用时分复用技术，一个数字系统能处理多路数字信号。
Specialized English for Communication Engineering
Sampling theorem can be stated as follows: Let x(t) be a band-limited signal with X(jω)=0 for |ω|>ωm. Then x(t) is uniquely determined by its samples x(nT), n=0, ±1, ±2, …, if ωs>2ωm ωm：被抽样信号的最高频率 where ωs=2π/T ωs ：抽样频率 T ：抽样间隔 band-limited signal:频带有限的（低通）信号 is uniquely determined：被唯一地确定 抽样定理可以有下述关系表示：设x(t)是一个频带有限的（低 通）信号，对于|ω|>ωm， X(jω)=0 。若ωs>2ωm ，其中 ωs=2π/T，则x (t)可以被其抽样值x(nT)唯一地确定。 Specialized English for Communication Engineering
In many contexts, processing discrete-time signals is more flexible and is often preferable to processing continuous-time 更优越 在许多方面，处理离散时间信号要更加灵活些，因此往往比 处理连续时间更为可取。
Specialized English for Communication Engineering
Much of the importance of the sampling theorem also lies in its role as a bridge between continuous-time signals and discrete-time signals. continuous-time signals : 连续时间信号 discrete-time signals: 离散时间信号 抽样定理的重要性很大程度上体现在：它起着连续时间信号 和离散时间信号之间的桥梁的作用。
Specialized English for Communication Engineering
When these samples are viewed in sequence at a sufficiently fast rate, we perceive an accurate representation of the original continuously moving scene. in sequence：顺次, 挨次 perceive：察觉，感知，感到 当我们以极快的速度来顺次观察这些瞬时画面时，感觉到的 就是对原来连续活动场景的一个准确再现。
Specialized English on Communication Engineering
第二部分 通信技术
Unit 5： The Sampling Theorem ——抽样定理
NEW WORDS AND PHRASES
theorem n. 定理，定律，法则 exploit vt. 开拓，开发，利用 sequence n. 次序，顺序，序列 instantaneous adj. 瞬间的，即刻的，即时的 reproducible adj. 可再生的，可复制的 cut off n. 截止，中止，切掉 replication n. 复制，重现 perceive v. 感知，感到，认识到 aliasing n. 混淆现象，混叠 Nyquist rate n. 奈奎斯特（抽样）速率