第3章课后习题参考答案
第三章连接课后习题参考答案

第三章连接课后习题参考答案第三章连接课后习题参考答案焊接连接参考答案一、概念题3.1 从功能上分类,连接有哪几种基本类型?3.2 焊缝有两种基本类型—对接坡口焊缝和贴角焊缝,二者在施工、受力、适用范围上各有哪些特点?3.3 对接接头连接需使用对接焊缝,角接接头连接需采用角焊缝,这么说对吗?3.4 h和lw相同时,吊车梁上的焊缝采用正f面角焊缝比采用侧面角焊缝承载力高?3.5 为何对角焊缝焊脚尺寸有最大和最小取值的限制?对侧面角焊缝的长度有何要求?为什么?【答】(1)最小焊脚尺寸:角焊缝的焊脚尺寸不能过小,否则焊接时产生的热量较小,致使施焊时冷却速度过快,导致母材开裂。
《规范》规定:h f≥1.5t,式中:t2——较厚焊件厚度,单2位为mm。
计算时,焊脚尺寸取整数。
自动焊熔深较大,所取最小焊脚尺寸可减小1mm;T形连接的单面角焊缝,应增加1mm;当焊件厚度小于或等于4mm时,则取与焊件厚度相同。
(2)最大焊脚尺寸:为了避免焊缝区的主体金属“过热”,减小焊件的焊接残余应力和残余变形,角焊缝的焊脚尺寸应满足12.1t h f式中: t 1——较薄焊件的厚度,单位为mm 。
(3)侧面角焊缝的最大计算长度侧面角焊缝在弹性阶段沿长度方向受力不均匀,两端大而中间小,可能首先在焊缝的两端破坏,故规定侧面角焊缝的计算长度l w ≤60h f 。
若内力沿侧面角焊缝全长分布,例如焊接梁翼缘与腹板的连接焊缝,可不受上述限制。
3.6 简述焊接残余应力产生的实质,其最大分布特点是什么? 3.7 画出焊接H 形截面和焊接箱形截面的焊接残余应力分布图。
3.8 贴角焊缝中,何为端焊缝?何为侧焊缝?二者破坏截面上的应力性质有何区别?3.9 规范规定:侧焊缝的计算长度不得大于焊脚尺寸的某个倍数,原因何在?规范同时有焊缝最小尺寸的规定,原因何在? 3.10 规范禁止3条相互垂直的焊缝相交,为什么。
3.11 举3~5例说明焊接设计中减小应力集中的构造措施。
人工智能课后习题第3章 参考答案

第3章确定性推理参考答案3.8 判断下列公式是否为可合一,若可合一,则求出其最一般合一。
(1) P(a, b), P(x, y)(2) P(f(x), b), P(y, z)(3) P(f(x), y), P(y, f(b))(4) P(f(y), y, x), P(x, f(a), f(b))(5) P(x, y), P(y, x)解:(1) 可合一,其最一般和一为:σ={a/x, b/y}。
(2) 可合一,其最一般和一为:σ={y/f(x), b/z}。
(3) 可合一,其最一般和一为:σ={ f(b)/y, b/x}。
(4) 不可合一。
(5) 可合一,其最一般和一为:σ={ y/x}。
3.11 把下列谓词公式化成子句集:(1)(∀x)(∀y)(P(x, y)∧Q(x, y))(2)(∀x)(∀y)(P(x, y)→Q(x, y))(3)(∀x)(∃y)(P(x, y)∨(Q(x, y)→R(x, y)))(4)(∀x) (∀y) (∃z)(P(x, y)→Q(x, y)∨R(x, z))解:(1) 由于(∀x)(∀y)(P(x, y)∧Q(x, y))已经是Skolem标准型,且P(x, y)∧Q(x, y)已经是合取范式,所以可直接消去全称量词、合取词,得{ P(x, y), Q(x, y)}再进行变元换名得子句集:S={ P(x, y), Q(u, v)}(2) 对谓词公式(∀x)(∀y)(P(x, y)→Q(x, y)),先消去连接词“→”得:(∀x)(∀y)(¬P(x, y)∨Q(x, y))此公式已为Skolem标准型。
再消去全称量词得子句集:S={¬P(x, y)∨Q(x, y)}(3) 对谓词公式(∀x)(∃y)(P(x, y)∨(Q(x, y)→R(x, y))),先消去连接词“→”得:(∀x)(∃y)(P(x, y)∨(¬Q(x, y)∨R(x, y)))此公式已为前束范式。
大学物理第三章课后习题答案

r3
, k 为常量。试求两粒子相距为 r 时的势能,设力为零的
r = a cos ωt i + b sin ωt j , r 式中 a , b , ω 是正值常数,且 a ≻ b 。
(1)说明这质点沿一椭圆运动,方程为
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x2 y 2 + = 1; a2 b2
(2)求质点在 A 点 (a ,0) 时和 B 点 (0, b ) 时的动能; (3)当质点从 A 点到 B 点,求力 F 所做的功,并求 F 的分力 Fx i 和 Fy j 所做的 功; (4) F 力是不是保守力? 12 . 如果物体从髙为 h 处静止下落,试求(1)时间为自变量; 12. (2)高度为自变量, 画出它的动能和势能图线,并证明两曲线中动能和势能之和相等。 . 一质量为 m 的地球卫星,沿半径为 3R e 的轨道运动, R e 为地球的半径,已知 13 13. 地球的质量为 M e ,求(1)卫星的动能; (2)卫星的引力势能; (3)卫星的机械 能。 . 如图所示, 14 14. 小球在外力作用下, 由静止开始从 A 点出发做匀加速运动,到达 B 点时撤消外力,小球 无摩擦的冲上竖直的半径为 R 的半圆环, 到达最高 点 C 时,恰能维持在圆环上做圆周运动,并以此速 度抛出而刚好落回到原来的出发点 A 处, 如图试求 小球在 AB 段运动的加速度为多大? . 如图所示,有一自动卸货矿车,满载时的质量 15 15. 为 M ,从与水平倾角 α = 30° 斜面上的点 A 由静 止下滑。设斜面对车的阻力为车重的 0.25 倍, 矿 车下滑距离 l 时,矿车与缓冲弹簧一道沿斜面运 动。当矿车使弹簧产生最大压缩形变时,矿车自 动卸货, 然后矿车借助弹簧的弹性力作用, 使之返回原位置 A 在装货。试问要完成这 一过程,空载时车的质量与满载时车的质 量之比应为多大? . 半径为 R 的光滑半球状圆塔的顶点 A 16 16. 上,有一木块 m ,今使木块获得水平速度
自动控制原理第三章课后习题 答案()

3-1 设系统的微分方程式如下:(1) )(2)(2.0t r t c= (2) )()()(24.0)(04.0t r t c t c t c=++ 试求系统闭环传递函数Φ(s),以及系统的单位脉冲响应g(t)和单位阶跃响应c(t)。
已知全部初始条件为零。
解:(1) 因为)(2)(2.0s R s sC =闭环传递函数ss R s C s 10)()()(==Φ 单位脉冲响应:s s C /10)(= 010)(≥=t t g单位阶跃响应c(t) 2/10)(s s C = 010)(≥=t t t c(2))()()124.004.0(2s R s C s s =++ 124.004.0)()(2++=s s s R s C `闭环传递函数124.004.01)()()(2++==s s s R s C s φ单位脉冲响应:124.004.01)(2++=s s s C t e t g t 4sin 325)(3-= 单位阶跃响应h(t) 16)3(61]16)3[(25)(22+++-=++=s s s s s s C t e t e t c t t 4sin 434cos 1)(33----=3-2 温度计的传递函数为11+Ts ,用其测量容器内的水温,1min 才能显示出该温度的98%的数值。
若加热容器使水温按10ºC/min 的速度匀速上升,问温度计的稳态指示误差有多大解法一 依题意,温度计闭环传递函数11)(+=ΦTs s 由一阶系统阶跃响应特性可知:o o T c 98)4(=,因此有 min 14=T ,得出 min 25.0=T 。
视温度计为单位反馈系统,则开环传递函数为Tss s s G 1)(1)()(=Φ-Φ=⎩⎨⎧==11v TK !用静态误差系数法,当t t r ⋅=10)( 时,C T Ke ss ︒===5.21010。
解法二 依题意,系统误差定义为 )()()(t c t r t e -=,应有 1111)()(1)()()(+=+-=-==ΦTs TsTs s R s C s R s E s e C T sTs Ts ss R s s e s e s ss ︒==⋅+=Φ=→→5.210101lim )()(lim 203-3 已知二阶系统的单位阶跃响应为)1.536.1sin(5.1210)(2.1o tt et c +-=-试求系统的超调量σ%、峰值时间tp 和调节时间ts 。
《统计学概论》第三章课后练习题答案

《统计学概论》第三章课后练习题答案一、思考题1.什么是统计整理,统计整理的对象是什么?P612.什么是统计分组,它可以分为哪几种形式?P633.简述编制变量数列的一般步骤。
P70-754.统计表分为哪几种?P785.什么是统计分布,它包括哪两个要素?P686.单项式分组和组距公式分组分别在什么情况下运用?P667.如何正确选择分组标志?P658.为什么要进行统计分组?其主要作用是什么?P63(2009.01)二、判断题1.统计整理只能对统计调查所得到的原始资料进行加工整理。
(×)P61【解析】统计整理分为两情况:一种是对原始资料进行整理,另一种是对次级资料即已加工过的现成资料进行在整理。
2.对一个既定总体而言,合理的分组标志只有一个。
(×)P67【解析】复合分组就是对同一总体选择两个或两个以上标志进行的分组。
3.在异距数列中,计算次数密度主要是为了消除组距因素对次数分布的影响。
(√)P74 4.组中值是指各组上限和下限之中点数值,故在任何情况下它都能代表各组的一般水平。
(×)P72【解析】当组内标志值分布均匀时,组中值能代表各组的一般水平(平均水平),当组内标志值分布不均匀时,组中值不能代表各组的一般水平(平均水平)。
5.在变量数列中,组数等于全距除以组距。
(×)(2010.01)P71【解析】变量数列的分组可分为等距分组和异距分组,只有在等距分组的情况下,组数等于全距除以组距。
6.统计分组的关键问题是确定组数和组距。
(×)(2009.10)P65【解析】统计分组的关键问题是选择恰当的分组标志。
7.按数量标志分组的目的,就是要区分各组在数量上的差别。
(×)P66【解析】按数量标志分组的目的,并不是单纯确定各组在数量上的差别,而是要通过数量上的变化来区分各组的不同类型和性质。
8.连续型变量可以作单项式分组或组距式分组,而离散型变量只能作组距式分组。
利息理论第三章课后答案

《金融数学》课后习题参考答案第三章 收益率1、某现金流为:元,元,元,元,求该现金流的收益率。
解:由题意得:2、某投资者第一年末投资7000元,第二年末投资1000元,而在第一、三年末分别收回4000元和5500元,计算利率为及时的现金流现值,并计算该现金流的内部收益率。
解:由题意得:当时,当时,令3、某项贷款1000元,每年计息4次的年名义利率为12%,若第一年后还款400元,第5年后还款800元,余下部分在第7年后还清,计算最后一次还款额。
解:由题意得:4、甲获得100000元保险金,若他用这笔保险金购买10年期期末付年金,每年可得15380元,若购买20年期期末付年金,则每年可得10720元,这两种年金基于相同的利率,计算。
解:由题意得: 08688.010720153802010=⇒=i a a i i5、某投资基金按积累,,在时刻0基金中有10万元,在时刻1基金中有11万元,一年中只有2次现金流,第一次在时刻时投入15000元,第二次在时刻时收回2万元,计算k。
解:由题意得:6、某投资业务中,直接投资的利率为8%,投资所得利息的再投资利率为4%,某人为在第10年末获得本息和1万元,采取每年末投资相等的一笔款项,共10年,求证每年投资的款项为:。
证明:7.某投资人每年初在银行存款1000元,共5年,存款利率为5%,存款所得利息的再投资利率为4%,证明:V(11)=1250(。
V(11)=1000[5(1++(Is)8.甲年初投资2000元,年利率为 17%,每年末收回利息,各年收回的利息按某一利率又投资出去,至第10 年末,共得投资本息和元。
乙每年末投资150元,年利率14%,共20年,每年收回的利息按甲的再投资利率投资。
计算乙在第20年末的投资本息和。
9.某投资基金年初有投资2万元,年收益率为12%,3月末又投入资金5000元,9月末赎回资金8000元,假设1-t it=(1-t)i 计算年末基金的资金量。
(完整版)钢结构课后习题第三章

(完整版)钢结构课后习题第三章第三章部分习题参考答案3.8 已知A3F 钢板截⾯mm mm 20500?⽤对接直焊缝拼接,采⽤⼿⼯焊焊条E43型,⽤引弧板,按Ⅲ级焊缝质量检验,试求焊缝所能承受的最⼤轴⼼拉⼒设计值。
解:焊缝质量等级为Ⅱ级,抗拉的强度设计值20.85182.75/w f f f N mm == 采⽤引弧板,故焊缝长度500w l b mm ==承受的最⼤轴⼼拉⼒设计值3500*20*182.75*101827.5N btf kN -===3.9 焊接⼯字形截⾯梁,在腹板上设⼀道拼接的对接焊缝(如图3-66),拼接处作⽤荷载设计值:弯矩M=1122kN ·mm ,剪⼒V=374kN ,钢材为Q235B ,焊条为E43型,半⾃动焊,三级检验标准,试验算该焊缝的强度。
解:(1)焊缝截⾯的⼏何特性惯性矩3341(28102.827.2100)26820612x I cm =-= ⼀块翼缘板对x 轴的⾯积矩3128 1.4(507)2234.4X S cm =??+=半个截⾯对x 轴的⾯积矩31500.8253234.4X X S S cm =+??=(2)焊缝强度验算焊缝下端的剪应⼒33214374102234.41038.9/268206108x x w VS N mm I t τ===?? 焊缝下端的拉应62max4112210500209/0.852********x M h N mm f I σ??=?==>? 所以,该焊缝不满⾜强度要求(建议将焊缝等级质量提为⼆级)则 max σ2209/N mm =<215f =2/N mm下端点处的折算应2222max 3219.6/ 1.1236.5/N mm f N mm στ+=<=且焊缝中点处剪应⼒33224374103234.41056.3/125/268206108w x v x w VS N mm f N mm I t τ===<=??3.10 试设计如图3-67所⽰双⾓钢和节点板间的⾓焊缝连接。
组合数学第三章课后习题答案

3.1题(宗传玉)某甲参加一种会议,会上有6位朋友,某甲和其中每人在会上各相遇12次,每二人各相遇6次,每三人各相遇3次,每五人各相遇2次,每六人各相遇一次,1人也没有遇见的有5次,问某甲共参加了几次会议解:设A i为甲与第i个朋友相遇的会议集,i=1,…,6.则故甲参加的会议数为:28+5=33.3.2题(宗传玉)求从1到500的整数中被3和5整除但不被7整除的数的个数.解:设A3:被3整除的数的集合A5:被5整除的数的集合A7:被7整除的数的集合所以3.3.题(宗传玉)n个代表参加会议,试证其中至少有2人各自的朋友数相等。
解:每个人的朋友数只能取0,1,…,n-1.但若有人的朋友数为0,即此人和其他人都不认识,则其他人的最大取数不超过n-2.故这n个人的朋友数的实际取数只有n-1种可能.,所以至少有2人的朋友数相等.3.4题(宗传玉)试给出下列等式的组合意义.解:(a) 从n 个元素中取k 个元素的组合,总含有指定的m 个元素的组合数为)()(kn mn m k m n --=--。
设这m 个元素为a 1,a 2,…,a m ,Ai 为不含a i 的组合(子集),i=1,…,m.()∑∑∑==∈⊄==⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛-=-+⎪⎪⎭⎫ ⎝⎛==⎪⎪⎭⎫⎝⎛--⎪⎪⎭⎫⎝⎛-=ml l m l l m i i lj i lk l n k m A k n k n m n k l n l j 01),(),...,(1m1i i i i i 1)1(A A A A 111213.5题(宗传玉)设有三个7位的二进制数:a1a2a3a4a5a6a7,b1b2b3b4b5b6b7,c1c2c3c4c5c6c7.试证存在整数i 和j,1≤i≤j≤7,使得下列之一必定成立:a i=a j=b i=b j,a i=a j=c i=c j,b i=b j=c i=c j.证:显然,每列中必有两数字相同,共有种模式,有0或1两种选择.故共有·2种选择.·2=6.现有7列,.即必有2列在相同的两行选择相同的数字,即有一矩形,四角的数字相等.3.6题(宗传玉)在边长为1的正方形内任取5个点试证其中至少有两点,其间距离小于证:把1×1正方形分成四个(1/2)×.则这5点中必有两点落在同一个小正方形内.而小正方形内的任两点的距离都小于.3.7题(王星)在边长为1的等边三角形内任取5个点试证其中至少有两点,期间距离小于1/2.证:把边长为1的三角形分成四个边长为1/2的三角形,如上图:则这5点中必有两点落在同一个小三角形中.小三角形中任意两点间的距离都小于1/2.3.8题(王星)任取11个整数,求证其中至少有两个数它们的差是10的倍数。
计算机组成原理第三章课后题参考答案

计算机组成原理第三章课后题参考答案第三章课后习题参考答案1.有⼀个具有20位地址和32位字长的存储器,问:(1)该存储器能存储多少个字节的信息(2)如果存储器由512K×8位SRAM芯⽚组成,需要多少芯⽚(3)需要多少位地址作芯⽚选择解:(1)∵ 220= 1M,∴该存储器能存储的信息为:1M×32/8=4MB (2)(1024K/512K)×(32/8)= 8(⽚)(3)需要1位地址作为芯⽚选择。
3.⽤16K×8位的DRAM芯⽚组成64K×32位存储器,要求:(1) 画出该存储器的组成逻辑框图。
(2) 设DRAM芯⽚存储体结构为128⾏,每⾏为128×8个存储元。
如单元刷新间隔不超过2ms,存储器读/写周期为µS, CPU在1µS内⾄少要访问⼀次。
试问采⽤哪种刷新⽅式⽐较合理两次刷新的最⼤时间间隔是多少对全部存储单元刷新⼀遍所需的实际刷新时间是多少解:(1)组成64K×32位存储器需存储芯⽚数为N=(64K/16K)×(32位/8位)=16(⽚)每4⽚组成16K×32位的存储区,有A13-A0作为⽚内地址,⽤A15 A14经2:4译码器产⽣⽚选信号,逻辑框图如下所⽰:(2)根据已知条件,CPU在1us内⾄少访存⼀次,⽽整个存储器的平均读/写周期为,如果采⽤集中刷新,有64us的死时间,肯定不⾏;所以采⽤分散式刷新⽅式:设16K×8位存储芯⽚的阵列结构为128⾏×128列,按⾏刷新,刷新周期T=2ms,则分散式刷新的间隔时间为:t=2ms/128=(s) 取存储周期的整数倍s的整数倍)则两次刷新的最⼤时间间隔发⽣的⽰意图如下可见,两次刷新的最⼤时间间隔为tMAXt MAX=×2-=(µS)对全部存储单元刷新⼀遍所需时间为tRt R=×128=64 (µS)4.有⼀个1024K×32位的存储器,由128K×8位DRAM芯⽚构成。
3. 第三章课后习题及答案

第三章1. (Q1) Suppose the network layer provides the following service. The network layer in the source host accepts a segment of maximum size 1,200 bytes and a destination host address from the transport layer. The network layer then guarantees to deliver the segment to the transport layer at the destination host. Suppose many network application processes can be running at the destination host.a. Design the simplest possible transport-layer protocol that will get application data to thedesired process at the destination host. Assume the operating system in the destination host has assigned a 4-byte port number to each running application process.b. Modify this protocol so that it provides a “return address” to the destination process.c. In your protocols, does the transport layer “have to do anything” in the core of the computernetwork.Answer:a. Call this protocol Simple Transport Protocol (STP). At the sender side, STP accepts from thesending process a chunk of data not exceeding 1196 bytes, a destination host address, and a destination port number. STP adds a four-byte header to each chunk and puts the port number of the destination process in this header. STP then gives the destination host address and the resulting segment to the network layer. The network layer delivers the segment to STP at the destination host. STP then examines the port number in the segment, extracts the data from the segment, and passes the data to the process identified by the port number.b. The segment now has two header fields: a source port field and destination port field. At thesender side, STP accepts a chunk of data not exceeding 1192 bytes, a destination host address,a source port number, and a destination port number. STP creates a segment which contains theapplication data, source port number, and destination port number. It then gives the segment and the destination host address to the network layer. After receiving the segment, STP at the receiving host gives the application process the application data and the source port number.c. No, the transport layer does not have to do anything in the core; the transport layer “lives” inthe end systems.2. (Q2) Consider a planet where everyone belongs to a family of six, every family lives in its own house, each house has a unique address, and each person in a given house has a unique name. Suppose this planet has a mail service that delivers letters form source house to destination house. The mail service requires that (i) the letter be in an envelope and that (ii) the address of the destination house (and nothing more ) be clearly written on the envelope. Suppose each family has a delegate family member who collects and distributes letters for the other family members. The letters do not necessarily provide any indication of the recipients of the letters.a. Using the solution to Problem Q1 above as inspiration, describe a protocol that thedelegates can use to deliver letters from a sending family member to a receiving family member.b. In your protocol, does the mail service ever have to open the envelope and examine theletter in order to provide its service.Answer:a.For sending a letter, the family member is required to give the delegate the letter itself, theaddress of the destination house, and the name of the recipient. The delegate clearly writes the recipient’s name on the top of the letter. The delegate then puts the letter in an e nvelope and writes the address of the destination house on the envelope. The delegate then gives the letter to the planet’s mail service. At the receiving side, the delegate receives the letter from the mail service, takes the letter out of the envelope, and takes note of the recipient name written at the top of the letter. The delegate than gives the letter to the family member with this name.b.No, the mail service does not have to open the envelope; it only examines the address on theenvelope.3. (Q3) Describe why an application developer might choose to run an application over UDP rather than TCP.Answer:An application developer may not want its application to use TCP’s congestion control, which can throttle the application’s sending rate at times of congestion. Often, designers of IP telephony and IP videoconference applications choose to run their applications over UDP because they want to avoid TCP’s congestion control. Also, some applications do not need the reliable data transfer provided by TCP.4. (P1) Suppose Client A initiates a Telnet session with Server S. At about the same time, Client B also initiates a Telnet session with Server S. Provide possible source and destination port numbers fora. The segment sent from A to B.b. The segment sent from B to S.c. The segment sent from S to A.d. The segment sent from S to B.e. If A and B are different hosts, is it possible that the source port number in the segment fromA to S is the same as that fromB to S?f. How about if they are the same host?Yes.f No.5. (P2) Consider Figure 3.5 What are the source and destination port values in the segmentsflowing form the server back to the clients’ processes? What are the IP addresses in the network-layer datagrams carrying the transport-layer segments?Answer:Suppose the IP addresses of the hosts A, B, and C are a, b, c, respectively. (Note that a,b,c aredistinct.)To host A: Source port =80, source IP address = b, dest port = 26145, dest IP address = a To host C, left process: Source port =80, source IP address = b, dest port = 7532, dest IP address = cTo host C, right process: Source port =80, source IP address = b, dest port = 26145, dest IP address = c6. (P3) UDP and TCP use 1s complement for their checksums. Suppose you have the followingthree 8-bit bytes: 01101010, 01001111, 01110011. What is the 1s complement of the sum of these 8-bit bytes? (Note that although UDP and TCP use 16-bit words in computing the checksum, for this problem you are being asked to consider 8-bit sums.) Show all work. Why is it that UDP takes the 1s complement of the sum; that is , why not just sue the sum? With the 1s complement scheme, how does the receiver detect errors? Is it possible that a 1-bit error will go undetected? How about a 2-bit error?Answer:One's complement = 1 1 1 0 1 1 1 0.To detect errors, the receiver adds the four words (the three original words and the checksum). If the sum contains a zero, the receiver knows there has been an error. All one-bit errors will be detected, but two-bit errors can be undetected (e.g., if the last digit of the first word is converted to a 0 and the last digit of the second word is converted to a 1).7. (P4) Suppose that the UDP receiver computes the Internet checksum for the received UDPsegment and finds that it matches the value carried in the checksum field. Can the receiver be absolutely certain that no bit errors have occurred? Explain.Answer:No, the receiver cannot be absolutely certain that no bit errors have occurred. This is because of the manner in which the checksum for the packet is calculated. If the corresponding bits (that would be added together) of two 16-bit words in the packet were 0 and 1 then even if these get flipped to 1 and 0 respectively, the sum still remains the same. Hence, the 1s complement the receiver calculates will also be the same. This means the checksum will verify even if there was transmission error.8. (P5) a. Suppose you have the following 2 bytes: 01001010 and 01111001. What is the 1scomplement of sum of these 2 bytes?b. Suppose you have the following 2 bytes: 11110101 and 01101110. What is the 1s complement of sum of these 2 bytes?c. For the bytes in part (a), give an example where one bit is flipped in each of the 2 bytesand yet the 1s complement doesn’t change.0 1 0 1 0 1 0 1 + 0 1 1 1 0 0 0 0 1 1 0 0 0 1 0 1 1 1 0 0 0 1 0 1 + 0 1 0 0 1 1 0 0 0 0 0 1 0 0 0 1Answer:a. Adding the two bytes gives 10011101. Taking the one’s complement gives 01100010b. Adding the two bytes gives 00011110; the one’s complement gives 11100001.c. first byte = 00110101 ; second byte = 01101000.9. (P6) Consider our motivation for correcting protocol rdt2.1. Show that the receiver, shown inthe figure on the following page, when operating with the sender show in Figure 3.11, can lead the sender and receiver to enter into a deadlock state, where each is waiting for an event that will never occur.Answer:Suppose the sender is in state “Wait for call 1 from above” and the receiver (the receiver shown in the homework problem) is in state “Wait for 1 from below.” The sender sends a packet with sequence number 1, and transitions to “Wait for ACK or NAK 1,” waiting for an ACK or NAK. Suppose now the receiver receives the packet with sequence number 1 correctly, sends an ACK, and transitions to state “Wait for 0 from below,” waiting for a data packet with sequence number 0. However, the ACK is corrupted. When the rdt2.1 sender gets the corrupted ACK, it resends the packet with sequence number 1. However, the receiver is waiting for a packet with sequence number 0 and (as shown in the home work problem) always sends a NAK when it doesn't get a packet with sequence number 0. Hence the sender will always be sending a packet with sequence number 1, and the receiver will always be NAKing that packet. Neither will progress forward from that state.10. (P7) Draw the FSM for the receiver side of protocol rdt3.0Answer:The sender side of protocol rdt3.0 differs from the sender side of protocol 2.2 in that timeouts have been added. We have seen that the introduction of timeouts adds the possibility of duplicate packets into the sender-to-receiver data stream. However, the receiver in protocol rdt.2.2 can already handle duplicate packets. (Receiver-side duplicates in rdt 2.2 would arise if the receiver sent an ACK that was lost, and the sender then retransmitted the old data). Hence the receiver in protocol rdt2.2 will also work as the receiver in protocol rdt 3.0.11. (P8) In protocol rdt3.0, the ACK packets flowing from the receiver to the sender do not havesequence numbers (although they do have an ACK field that contains the sequence number of the packet they are acknowledging). Why is it that our ACK packets do not require sequence numbers?Answer:To best Answer this question, consider why we needed sequence numbers in the first place. We saw that the sender needs sequence numbers so that the receiver can tell if a data packet is a duplicate of an already received data packet. In the case of ACKs, the sender does not need this info (i.e., a sequence number on an ACK) to tell detect a duplicate ACK. A duplicate ACK is obvious to the rdt3.0 receiver, since when it has received the original ACK it transitioned to the next state. The duplicate ACK is not the ACK that the sender needs and hence is ignored by the rdt3.0 sender.12. (P9) Give a trace of the operation of protocol rdt3.0 when data packets and acknowledgmentpackets are garbled. Your trace should be similar to that used in Figure 3.16Answer:Suppose the protocol has been in operation for some time. The sender is in state “Wait for call fro m above” (top left hand corner) and the receiver is in state “Wait for 0 from below”. The scenarios for corrupted data and corrupted ACK are shown in Figure 1.13. (P10) Consider a channel that can lose packets but has a maximum delay that is known.Modify protocol rdt2.1 to include sender timeout and retransmit. Informally argue whyyour protocol can communicate correctly over this channel.Answer:Here, we add a timer, whose value is greater than the known round-trip propagation delay. We add a timeout event to the “Wait for ACK or NAK0” and “Wait for ACK or NAK1” states. If the timeout event occurs, the most recently transmitted packet is retransmitted. Let us see why this protocol will still work with the rdt2.1 receiver.• Suppose the timeout is caused by a lost data packet, i.e., a packet on the senderto- receiver channel. In this case, the receiver never received the previous transmission and, from the receiver's viewpoint, if the timeout retransmission is received, it look exactly the same as if the original transmission is being received.• Suppose now that an ACK is lost. The receiver will eventually retransmit the packet on atimeout. But a retransmission is exactly the same action that is take if an ACK is garbled. Thus the sender's reaction is the same with a loss, as with a garbled ACK. The rdt 2.1 receiver can already handle the case of a garbled ACK.14. (P11) Consider the rdt3.0 protocol. Draw a diagram showing that if the network connectionbetween the sender and receiver can reorder messages (that is, that two messagespropagating in the medium between the sender and receiver can be reordered), thenthe alternating-bit protocol will not work correctly (make sure you clearly identify thesense in which it will not work correctly). Your diagram should have the sender on theleft and the receiver on the right, with the time axis running down the page, showingdata (D) and acknowledgement (A) message exchange. Make sure you indicate thesequence number associated with any data or acknowledgement segment.Answer:15. (P12) The sender side of rdt3.0 simply ignores (that is, takes no action on) all received packetsthat are either in error or have the wrong value in the ack-num field of anacknowledgement packet. Suppose that in such circumstances, rdt3.0 were simply toretransmit the current data packet . Would the protocol still work? (hint: Consider whatwould happen if there were only bit errors; there are no packet losses but prematuretimeout can occur. Consider how many times the nth packet is sent, in the limit as napproaches infinity.)Answer:The protocol would still work, since a retransmission would be what would happen if the packet received with errors has actually been lost (and from the receiver standpoint, it never knows which of these events, if either, will occur). To get at the more subtle issue behind this question, one has to allow for premature timeouts to occur. In this case, if each extra copy of the packet is ACKed and each received extra ACK causes another extra copy of the current packet to be sent, the number of times packet n is sent will increase without bound as n approaches infinity.16. (P13) Consider a reliable data transfer protocol that uses only negative acknowledgements.Suppose the sender sends data only infrequently. Would a NAK-only protocol bepreferable to a protocol that uses ACKs? Why? Now suppose the sender has a lot ofdata to send and the end to end connection experiences few losses. In this second case ,would a NAK-only protocol be preferable to a protocol that uses ACKs? Why?Answer:In a NAK only protocol, the loss of packet x is only detected by the receiver when packetx+1 is received. That is, the receivers receives x-1 and then x+1, only when x+1 is received does the receiver realize that x was missed. If there is a long delay between the transmission of x and the transmission of x+1, then it will be a long time until x can be recovered, under a NAK only protocol.On the other hand, if data is being sent often, then recovery under a NAK-only scheme could happen quickly. Moreover, if errors are infrequent, then NAKs are only occasionally sent (when needed), and ACK are never sent – a significant reduction in feedback in the NAK-only case over the ACK-only case.17. (P14) Consider the cross-country example shown in Figure 3.17. How big would the windowsize have to be for the channel utilization to be greater than 80 percent?Answer:It takes 8 microseconds (or 0.008 milliseconds) to send a packet. in order for the sender to be busy 90 percent of the time, we must have util = 0.9 = (0.008n) / 30.016 or n approximately 3377 packets.18. (P15) Consider a scenario in which Host A wants to simultaneously send packets to Host Band C. A is connected to B and C via a broadcast channel—a packet sent by A is carriedby the channel to both B and C. Suppose that the broadcast channel connecting A, B,and C can independently lose and corrupt packets (and so, for example, a packet sentfrom A might be correctly received by B, but not by C). Design a stop-and-wait-likeerror-control protocol for reliable transferring packets from A to B and C, such that Awill not get new data from the upper layer until it knows that B and C have correctlyreceived the current packet. Give FSM descriptions of A and C. (Hint: The FSM for Bshould be essentially be same as for C.) Also, give a description of the packet format(s)used.Answer:In our solution, the sender will wait until it receives an ACK for a pair of messages (seqnum and seqnum+1) before moving on to the next pair of messages. Data packets have a data field and carry a two-bit sequence number. That is, the valid sequence numbers are 0, 1, 2, and 3. (Note: you should think about why a 1-bit sequence number space of 0, 1 only would not work in the solution below.) ACK messages carry the sequence number of the data packet they are acknowledging.The FSM for the sender and receiver are shown in Figure 2. Note that the sender state records whether (i) no ACKs have been received for the current pair, (ii) an ACK for seqnum (only) has been received, or an ACK for seqnum+1 (only) has been received. In this figure, we assume that theseqnum is initially 0, and that the sender has sent the first two data messages (to get things going).A timeline trace for the sender and receiver recovering from a lost packet is shown below:Sender Receivermake pair (0,1)send packet 0Packet 0 dropssend packet 1receive packet 1buffer packet 1send ACK 1receive ACK 1(timeout)resend packet 0receive packet 0deliver pair (0,1)send ACK 0receive ACK 019. (P16) Consider a scenario in which Host A and Host B want to send messages to Host C. HostsA and C are connected by a channel that can lose and corrupt (but not reorder)message.Hosts B and C are connected by another channel (independent of the channelconnecting A and C) with the same properties. The transport layer at Host C shouldalternate in delivering messages from A and B to the layer above (that is, it should firstdeliver the data from a packet from A, then the data from a packet from B, and so on).Design a stop-and-wait-like error-control protocol for reliable transferring packets fromA toB and C, with alternating delivery atC as described above. Give FSM descriptionsof A and C. (Hint: The FSM for B should be essentially be same as for A.) Also, give adescription of the packet format(s) used.Answer:This problem is a variation on the simple stop and wait protocol (rdt3.0). Because the channel may lose messages and because the sender may resend a message that one of the receivers has already received (either because of a premature timeout or because the other receiver has yet to receive the data correctly), sequence numbers are needed. As in rdt3.0, a 0-bit sequence number will suffice here.The sender and receiver FSM are shown in Figure 3. In this problem, the sender state indicates whether the sender has received an ACK from B (only), from C (only) or from neither C nor B. The receiver state indicates which sequence number the receiver is waiting for.20. (P17) In the generic SR protocol that we studied in Section 3.4.4, the sender transmits amessage as soon as it is available (if it is in the window) without waiting for anacknowledgment. Suppose now that we want an SR protocol that sends messages twoat a time. That is , the sender will send a pair of messages and will send the next pairof messages only when it knows that both messages in the first pair have been receivercorrectly.Suppose that the channel may lose messages but will not corrupt or reorder messages.Design an error-control protocol for the unidirectional reliable transfer of messages.Give an FSM description of the sender and receiver. Describe the format of the packetssent between sender and receiver, and vice versa. If you use any procedure calls otherthan those in Section 3.4(for example, udt_send(), start_timer(), rdt_rcv(), and soon) ,clearly state their actions. Give an example (a timeline trace of sender and receiver)showing how your protocol recovers from a lost packet.Answer:21. (P18) Consider the GBN protocol with a sender window size of 3 and a sequence numberrange of 1024. Suppose that at time t, the next in-order packet that the receiver isexpecting has a sequence number of k. Assume that the medium does not reordermessages. Answer the following questions:a. What are the possible sets of sequence number inside the sender’s window at timet? Justify your Answer.b .What are all possible values of the ACK field in all possible messages currentlypropagating back to the sender at time t? Justify your Answer.Answer:a.Here we have a window size of N=3. Suppose the receiver has received packet k-1, and hasACKed that and all other preceeding packets. If all of these ACK's have been received by sender, then sender's window is [k, k+N-1]. Suppose next that none of the ACKs have been received at the sender. In this second case, the sender's window contains k-1 and the N packets up to and including k-1. The sender's window is thus [k- N,k-1]. By these arguments, the senders window is of size 3 and begins somewhere in the range [k-N,k].b.If the receiver is waiting for packet k, then it has received (and ACKed) packet k-1 and the N-1packets before that. If none of those N ACKs have been yet received by the sender, then ACKmessages with values of [k-N,k-1] may still be propagating back. Because the sender has sent packets [k-N, k-1], it must be the case that the sender has already received an ACK for k-N-1.Once the receiver has sent an ACK for k-N-1 it will never send an ACK that is less that k-N-1.Thus the range of in-flight ACK values can range from k-N-1 to k-1.22. (P19) Answer true or false to the following questions and briefly justify your Answer.a. With the SR protocol, it is possible for the sender to receive an ACK for a packet thatfalls outside of its current window.b. With CBN, it is possible for the sender to receiver an ACK for a packet that fallsoutside of its current window.c. The alternating-bit protocol is the same as the SR protocol with a sender and receiverwindow size of 1.d. The alternating-bit protocol is the same as the GBN protocol with a sender andreceiver window size of 1.Answer:a.True. Suppose the sender has a window size of 3 and sends packets 1, 2, 3 at t0 . At t1 (t1 > t0)the receiver ACKS 1, 2, 3. At t2 (t2 > t1) the sender times out and resends 1, 2, 3. At t3 the receiver receives the duplicates and re-acknowledges 1, 2, 3. At t4 the sender receives the ACKs that the receiver sent at t1 and advances its window to 4, 5, 6. At t5 the sender receives the ACKs 1, 2, 3 the receiver sent at t2 . These ACKs are outside its window.b.True. By essentially the same scenario as in (a).c.True.d.True. Note that with a window size of 1, SR, GBN, and the alternating bit protocol arefunctionally equivalent. The window size of 1 precludes the possibility of out-of-order packets (within the window). A cumulative ACK is just an ordinary ACK in this situation, since it can only refer to the single packet within the window.23. (Q4) Why is it that voice and video traffic is often sent over TCP rather than UDP in today’sInternet. (Hint: The Answer we are looking for has nothing to do with TCP’s congestion-control mechanism. )Answer:Since most firewalls are configured to block UDP traffic, using TCP for video and voice traffic lets the traffic though the firewalls24. (Q5) Is it possible for an application to enjoy reliable data transfer even when the applicationruns over UDP? If so, how?Answer:Yes. The application developer can put reliable data transfer into the application layer protocol. This would require a significant amount of work and debugging, however.25. (Q6) Consider a TCP connection between Host A and Host B. Suppose that the TCP segmentstraveling form Host A to Host B have source port number x and destination portnumber y. What are the source and destination port number for the segments travelingform Host B to Host A?Answer:Source port number y and destination port number x.26. (P20) Suppose we have two network entities, A and B. B has a supply of data messages thatwill be sent to A according to the following conventions. When A gets a request fromthe layer above to get the next data (D) message from B, A must send a request (R)message to B on the A-to-B channel. Only when B receives an R message can it send adata (D) message back to A on the B-to-A channel. A should deliver exactly one copy ofeach D message to the layer above. R message can be lost (but not corrupted) in the A-to-B channel; D messages, once sent, are always delivered correctly. The delay alongboth channels is unknown and variable.Design(give an FSM description of) a protocol that incorporates the appropriatemechanisms to compensate for the loss-prone A-to-B channel and implementsmessage passing to the layer above at entity A, as discussed above. Use only thosemechanisms that are absolutely necessary.Answer:Because the A-to-B channel can lose request messages, A will need to timeout and retransmit its request messages (to be able to recover from loss). Because the channel delays are variable and unknown, it is possible that A will send duplicate requests (i.e., resend a request message that has already been received by B). To be able to detect duplicate request messages, the protocol will use sequence numbers. A 1-bit sequence number will suffice for a stop-and-wait type of request/response protocol.A (the requestor) has 4 states:• “Wait for Request 0 from above.” Here the requestor is waiting for a call from above to request a unit of data. When it receives a request from above, it sends a request message, R0, to B, starts a timer and make s a transition to the “Wait for D0” state. When in the “Wait for Request 0 from above” state, A ign ores anything it receives from B.• “Wait for D0”. Here the requestor is waiting for a D0 data message from B. A timer is always running in this state. If the timer expires, A sends another R0 message, restarts the timer and remains in this state. If a D0 message is received from B, A stops the time and transits to the “Wait for Request 1 from above” state. If A receives a D1 data message while in this state, it is ignored.• “Wait for Request 1 from above.” Here the requestor is again waiting for a call from above to request a unit of data. When it receives a request from above, it sends a request message, R1, to B, starts a timer and makes a transition to the “Wait for D1” state. When in the “Wait for Request 1 from above” state, A ignores anything it receives from B.• “Wait for D1”. Here the requestor is waiting for a D1 data message from B. A timer is always running in this state. If the timer expires, A sends another R1 message, restarts the timer and remains in this state. If a D1 message is received from B, A stops the timer and transits to the “Wait for Request 0 from above” state. If A receives a D0 data message while in this state, it is ignored.The data supplier (B) has only two states:。
机械制造技术基础(第2版)第三章课后习题答案

机械制造技术基础(第2版)第三章课后习题答案《机械制造技术基础》部分习题参考解答第三章机械制造中的加工方法及装备3-1表面发生线的形成方法有哪几种?答:(p69-70)表面发生线的形成方法有轨迹法、成形法、相切法、展成法。
具体参见第二版教材p69图3-2。
3-2试以外圆磨床为例分析机床的哪些运动是主运动,哪些运动是进给运动?答:如图3-20(p87),外圆磨削砂轮旋转nc是主运动,工件旋转nw、砂轮的横向移动fr、工作台往复运动fa均为进给运动。
3-3机床有哪些基本组成部分?试分析其主要功用。
答:(p70-71)基本组成部分动力源、运动执行机构、传动机构、控制系统和伺服系统、支承系统。
3-5试分析提高车削生产率的途径和方法。
答:(p76)提高切削速度;采用强力切削,提高f、ap;采用多刀加工的方法。
3-6车刀有哪几种?试简述各种车刀的结构特征及加工范围。
答:(p77)外圆车刀(左、右偏刀、弯头车刀、直头车刀等),内、外螺纹车刀,切断刀或切槽刀,内孔车刀(通孔、盲孔车刀、)端面车刀、成形车刀等。
顾名思义,外圆车刀主要是切削外圆表面;螺纹车刀用于切削各种螺纹;切断或切槽车刀用于切断或切槽;内孔车刀用于车削内孔;端面车刀切断面;成形车刀用于加工成形表面。
3-7试述CA6140型卧式车床主传动链的传动路线。
答:(p82)CA6140型卧式车床主传动链的传动路线:3-8CA6140型卧式车床中主轴在主轴箱中是如何支承的三爪自定心卡盘是怎样装到车床主轴上去的?答:(p83-84)3-9CA6140型卧式车床是怎样通过双向多片摩擦离合器实现主轴正传、反转和制动的?答:如教材图3-17和3-18所示,操纵手柄向上,通过连杆、扇形齿块和齿条带动滑套8向右滑移,拨动摆杆10使拉杆向左,压紧左边正向旋转摩擦片,主轴实现正转;若操纵手柄向下,通过连杆、扇形齿块和齿条带动滑套8向左滑移,拨动摆杆10使拉杆向右,压紧右边反向旋转摩擦片,主轴反转。
(完整版)计算机操作系统(第四版)课后习题答案第三章

第三章处理机调度与死锁1,高级调度与低级调度的主要任务是什么?为什么要引入中级调度?【解】(1)高级调度主要任务是用于决定把外存上处于后备队列中的那些作业调入内存,并为它们创建进程,分配必要的资源,然后再将新创建的进程排在就绪队列上,准备执行。
(2)低级调度主要任务是决定就绪队列中的哪个进程将获得处理机,然后由分派程序执行把处理机分配给该进程的操作。
(3)引入中级调度的主要目的是为了提高内存的利用率和系统吞吐量。
为此,应使那些暂时不能运行的进程不再占用宝贵的内存空间,而将它们调至外存上去等待,称此时的进程状态为就绪驻外存状态或挂起状态。
当这些进程重又具备运行条件,且内存又稍有空闲时,由中级调度决定,将外存上的那些重又具备运行条件的就绪进程重新调入内存,并修改其状态为就绪状态,挂在就绪队列上,等待进程调度。
3、何谓作业、作业步和作业流?【解】作业包含通常的程序和数据,还配有作业说明书。
系统根据该说明书对程序的运行进行控制。
批处理系统中是以作业为基本单位从外存调入内存。
作业步是指每个作业运行期间都必须经过若干个相对独立相互关联的顺序加工的步骤。
作业流是指若干个作业进入系统后依次存放在外存上形成的输入作业流;在操作系统的控制下,逐个作业进程处理,于是形成了处理作业流。
4、在什么情冴下需要使用作业控制块JCB?其中包含了哪些内容?【解】每当作业进入系统时,系统便为每个作业建立一个作业控制块JCB,根据作业类型将它插入到相应的后备队列中。
JCB 包含的内容通常有:1) 作业标识2)用户名称3)用户账户4)作业类型(CPU 繁忙型、I/O芳名型、批量型、终端型)5)作业状态6)调度信息(优先级、作业已运行)7)资源要求8)进入系统时间9) 开始处理时间10) 作业完成时间11) 作业退出时间12) 资源使用情况等5.在作业调度中应如何确定接纳多少个作业和接纳哪些作业?【解】作业调度每次接纳进入内存的作业数,取决于多道程序度。
税收筹划(第3版)课后习题答案第3章

第三章税收筹划的发展前沿(习题及参考答案)复习思考题1.企业战略管理与税收筹划有何关系?请分析说明。
参考答案:①战略管理指导税收筹划战略管理是首要的,税收筹划是为之服务的。
税收筹划只有在战略管理的框架下才能充分发挥作用,税收筹划对企业战略管理能够起到一定作用,但绝不可能起决定性作用。
②战略管理重视税收筹划战略管理是从宏观角度看待问题的,它关注的是企业整体,战略管理不仅考虑税收对企业的影响,还考虑其他非税收因素对企业的影响。
税收筹划是战略管理要考虑的一个重要因素。
在企业管理活动中,企业战略管理决定税收筹划的范围、目标和过程,而企业税收筹划对企业战略的制定和实施起着决策支持和促进作用。
战略管理和税收筹划有着共同的目的,即从企业整体绩效出发,为股东创造最大价值。
在实现这一目标过程中,税收筹划也明显带有长期战略的痕迹。
2.“税收链”思想有何价值?请举例说明。
参考答案:对于一个企业来说,所从事的经营活动是多种多样的,因此,在生产经营过程中面临的税收问题也是多种多样的,既有增值税等流转性质的税种,又有企业所得税和个人所得税等所得性质的税种,还有诸如房产税、资源税等其他性质的税种。
如果从企业的生产经营过程来看,企业先采购原材料,然后进行生产加工,再销售产品。
销售的产品,有的成了最终的消费品,有的则继续作为中间产品为后面的企业所利用。
所有这些环节都形成或实现价值,因此从总体来看就构成了产品的价值链。
在这条产品的价值链中,有一部分价值将以税收的形式上缴到国家财政。
这些上缴的税款,也形成了一个链条,这就是税收链。
“税收链”思想要求企业从利益相关者视角看问题,不仅要考虑企业自身的税收情况,还要从企业与利益相关者的交易在整个价值链上的位置来考虑,照顾到利益相关者的纳税要求,这样才能真正利用“税收链”彻底解决与利益相关者因交易活动产生的税收问题。
例如,对于一个小规模纳税人来说,即使其销售产品的价格(这里指含税价)比一般纳税人的要低,只要不低到一定的程度,别人就不一定会购买。
概率论与数理统计第三章课后习题及参考答案

概率论与数理统计第三章课后习题及参考答案1.设二维随机变量),(Y X 只能取下列数组中的值:)0,0(,)1,1(-,31,1(-及)0,2(,且取这几组值的概率依次为61,31,121和125,求二维随机变量),(Y X 的联合分布律.解:由二维离散型随机变量分布律的定义知,),(Y X 的联合分布律为2.某高校学生会有8名委员,其中来自理科的2名,来自工科和文科的各3名.现从8名委员中随机地指定3名担任学生会主席.设X ,Y 分别为主席来自理科、工科的人数,求:(1)),(Y X 的联合分布律;(2)X 和Y 的边缘分布律.解:(1)由题意,X 的可能取值为0,1,2,Y 的可能取值为0,1,2,3,则561)0,0(3833====C C Y X P ,569)1,0(381323====C C C Y X P ,569)2,0(382313====C C C Y X P ,561)3,0(3833====C C Y X P ,283)0,1(382312====C C C Y X P ,289)1,1(38131312====C C C C Y X P ,283)2,1(382312====C C C Y X P ,0)3,1(===Y X P ,563)0,2(381322====C C C Y X P ,563)1,2(381322====C C C Y X P ,0)2,2(===Y X P ,0)3,2(===Y X P .),(Y X 的联合分布律为:(2)X 的边缘分布律为X 012P1452815283Y 的边缘分布律为Y 0123P285281528155613.设随机变量),(Y X 的概率密度为⎩⎨⎧<<<<--=其他.,0,42,20),6(),(y x y x k y x f 求:(1)常数k ;(2))3,1(<<Y X P ;(3))5.1(<Y P ;(4))4(≤+Y X P .解:方法1:(1)⎰⎰⎰⎰--==+∞∞-+∞∞-422d d )6(d d ),(1yx y x k y x y x f ⎰--=42202d |)216(y yx x x k k y y k 8d )210(42=-=⎰,∴81=k .(2)⎰⎰∞-∞-=<<31d d ),()3,1(y x y x f Y X P ⎰⎰--=32102d d )216(yx yx x x ⎰--=32102d |)216(81y yx x x 83|)21211(81322=-=y y .(3)),5.1()5.1(+∞<<=<Y X P X P ⎰⎰+∞∞-∞---=5.1d d )6(81yx y x ⎰⎰--=425.10d d )6(81y x y x y yx x x d )216(81422⎰--=3227|)43863(81422=-=y y .(4)⎰⎰≤+=≤+4d d ),()4(y x y x y x f Y X P ⎰⎰---=2042d )6(d 81x y y x x ⎰+-⋅=202d )812(2181x x x 32|)31412(1612032=+-=x x x .方法2:(1)同方法1.(2)20<<x ,42<<y 时,⎰⎰∞-∞-=yxv u v u f y x F d d ),(),(⎰⎰--=y xv u v u 20d d )6(81⎰--=y xv uv u u 202d |)216(81⎰--=y v xv x x 22d )216(81y xv v x xv 222|)21216(81--=)1021216(81222x xy y x xy +---=,其他,0),,(=y x F ,∴⎪⎩⎪⎨⎧<<<<+---=其他.,0,42,20),1021216(81),(222y x x x xy y x xy y x F 83)3,1()3,1(==<<F Y X P .(3))42,5.1(),5.1()5.1(<<<=+∞<<=<Y X P Y X P X P )2,5.1()4,5.1(<<-<<=Y X P Y X P 3227)2,5.1()4,5.1(=-=F F .(4)同方法1.4.设随机变量),(Y X 的概率密度为⎩⎨⎧>>=--其他.,0,0,0,e ),(2y x A y x f y x 求:(1)常数A ;(2)),(Y X 的联合分布函数.解:(1)⎰⎰⎰⎰+∞+∞--+∞∞-+∞∞-==02d d e d d ),(1yx A y x y x f y x ⎰⎰+∞+∞--=002d e d e y x A y x2|)e 21(|)e (020A A y x =-⋅-=∞+-∞+-,∴2=A .(2)0>x ,0>y 时,⎰⎰∞-∞-=y xv u v u f y x F d d ),(),(⎰⎰--=yxv u vu 02d d e 2yv x u 020|)e 21(|)e (2---⋅-=)e 1)(e 1(2y x ----=,其他,0),(=y x F ,∴⎩⎨⎧>>--=--其他.,0,0,0),e 1)(e 1(),(2y x y x F y x .5.设随机变量),(Y X 的概率密度为⎩⎨⎧≤≤≤≤=其他.,0,10,10,),(y x Axy y x f 求:(1)常数A ;(2)),(Y X 的联合分布函数.解:(1)2121d d d d ),(11010⋅⋅===⎰⎰⎰⎰+∞∞-+∞∞-A y y x x A y x y x f ,∴4=A .(2)10≤≤x ,10≤≤y 时,⎰⎰∞-∞-=y xv u v u f y x F d d ),(),(⎰⎰=yxv u uv 0d d 4220202||y x v u yx =⋅=,10≤≤x ,1>y 时,⎰⎰∞-∞-=yx v u v u f y x F d d ),(),(⎰⎰=100d d 4xv u uv 210202||x v u x =⋅=,10≤≤y ,1>x 时,⎰⎰∞-∞-=yx v u v u f y x F d d ),(),(⎰⎰=100d d 4yu v uv 202102||y v u y =⋅=,1>x ,1>y 时,⎰⎰∞-∞-=yx v u v u f y x F d d ),(),(⎰⎰=101d d 4v u uv 1||102102=⋅=v u,其他,0),(=y x F ,∴⎪⎪⎪⎩⎪⎪⎪⎨⎧>>≤≤>>≤≤≤≤≤≤=其他.,0,1,1,1,10,1,,1,10,,10,10,),(2222y x y x y y x x y x y x y x F .6.把一枚均匀硬币掷3次,设X 为3次抛掷中正面出现的次数,Y 表示3次抛掷中正面出现次数与反面出现次数之差的绝对值,求:(1)),(Y X 的联合分布律;(2)X 和Y 的边缘分布律.解:由题意知,X 的可能取值为0,1,2,3;Y 的可能取值为1,3.易知0)1,0(===Y X P ,81)3,0(===Y X P ,83)1,1(===Y X P ,0)3,1(===Y X P 83)1,2(===Y X P ,0)3,2(===Y X P ,0)1,3(===Y X P ,81)3,3(===Y X P 故),(Y X 得联合分布律和边缘分布律为:7.在汽车厂,一辆汽车有两道工序是由机器人完成的:一是紧固3只螺栓;二是焊接2处焊点,以X 表示由机器人紧固的螺栓紧固得不牢的数目,以Y 表示由机器人焊接的不良焊点的数目,且),(Y X 具有联合分布律如下表:求:(1)在1=Y 的条件下,X 的条件分布律;(2)在2=X 的条件下,Y 的条件分布律.解:(1)因为)3,3()1,2()1,1()1,0()1(==+==+==+====Y X P Y X P Y X P Y X P Y P 08.0002.0008.001.006.0=+++=,所以43)1()1,0()1|0(=======Y P Y X P Y X P ,81)1()1,1()1|1(=======Y P Y X P Y X P ,101)1()1,2()1|2(=======Y P Y X P Y X P ,401)1()1,3()1|3(=======Y P Y X P Y X P ,故在1=Y 的条件下,X 的条件分布律为X 0123P4381101401(2)因为)2,2()1,2()0,2()2(==+==+====Y X P Y X P Y X P X P 032.0004.0008.002.0=++=,所以85)2()0,2()2,0(=======X P Y X P X Y P ,4)2()1,2()2,1(=======X P Y X P X Y P ,81)2()2,2()2,2(=======X P Y X P X Y P ,故在2=X 的条件下,Y 的分布律为:Y 012P8541818.设二维随机变量),(Y X 的概率密度函数为⎩⎨⎧>>=+-其他.,0,0,0,e ),()2(y x c y x f y x 求:(1)常数c ;(2)X 的边缘概率密度函数;(3))2(<+Y X P ;(4)条件概率密度函数)|(|y x f Y X ,)|(|x y f X Y .解:(1)⎰⎰⎰⎰+∞+∞+-+∞∞-+∞∞-==0)2(d d e d d ),(1yx c y x y x f y x⎰⎰+∞+∞--=002d e d ey x c y x2|)e (|)e 21(002c c y x =-⋅-=∞+-∞+-,∴2=c .(2)0>x 时,⎰+∞∞-=y y x f x f X d ),()(⎰+∞+-=0)2(d e 2y y x x y x 202e 2|)e (e 2-+∞--=-=,0≤x 时,0)(=x f X ,∴⎩⎨⎧≤>=-.0,0,0,e 2)(2x x x f x X ,同理⎩⎨⎧≤>=-.0,0,0,e )(y y y f y Y .(3)⎰⎰<+=<+2d d ),()2(y x y x y x f Y X P ⎰⎰---=2202d d e 2xy x yx 422202e e 21d e d e 2-----+-==⎰⎰xy x y x .(4)由条件概率密度公式,得,当0>y 时,有⎩⎨⎧>=⎪⎩⎪⎨⎧>==----其他.其他.,0,0,e 2,0,0,e e 2)(),()|(22|x x y f y x f y x f xy y x Y Y X ,0≤y 时,0)|(|=y x f Y X ,所以⎩⎨⎧>>=-其他.,0,0,0,e 2)|(2|y x y x f x Y X ;同理,当0>x 时,有⎩⎨⎧>=⎪⎩⎪⎨⎧>==----其他.其他.,0,0,e ,0,0,2e e 2)(),()|(22|y y x f y x f x y f yx y x X X Y 0≤x 时,0)|(|=x y f X Y ,所以⎩⎨⎧>>=-其他.,0,0,0,e )|(|y x x y f y X Y .9.设二维随机变量),(Y X 的概率密度函数为⎩⎨⎧<<<<=其他.,0,0,10,3),(x y x x y x f求:(1)关于X 、Y 的边缘概率密度函数;(2)条件概率密度函数)|(|y x f Y X ,)|(|x y f X Y .解:(1)10<<x 时,⎰+∞∞-=y y x f x f X d ),()(203d 3x y x x==⎰,其他,0)(=x f X ,∴⎩⎨⎧<<=其他.,0,10,3)(2x x x f X ,密度函数的非零区域为}1,10|),{(}0,10|),{(<<<<=<<<<x y y y x x y x y x ,∴10<<y 时,⎰+∞∞-=x y x f y f Y d ),()()1(23d 321y x x y-==⎰,其他,0)(=y f Y ,∴⎪⎩⎪⎨⎧<<-=其他.,0,10),1(23)(2y y y f Y .(2)当10<<y 时,有⎪⎩⎪⎨⎧<<-=⎪⎪⎩⎪⎪⎨⎧<<-==其他.其他.,0,1,12,0,1,)1(233)(),()|(22|x y y x x y y xy f y x f y x f Y Y X ,其他,0)|(|=y x f Y X ,故⎪⎩⎪⎨⎧<<<<-=其他.,0,10,1,12)|(2|y x y y xy x f Y X .当10<<x 时,有⎪⎩⎪⎨⎧<<=⎪⎩⎪⎨⎧<<==其他.其他.,0,0,1,0,0,33)(),()|(2|x y x x y x x x f y x f x y f X X Y ,其他,0)|(|=x y f X Y ,故⎪⎩⎪⎨⎧<<<<=其他.,0,10,0,1)|(|x x y x x y f X Y .10.设条件密度函数为⎪⎩⎪⎨⎧<<<=其他.,0,10,3)|(32|y x yx y x f Y X Y 的概率密度函数为⎩⎨⎧<<=其他.,0,10,5)(4y y y f Y 求21(>X P .解:⎩⎨⎧<<<==其他.,0,10,15)|()(),(2|y x y x y x f y f y x f Y X Y ,则6447d )(215d d 15d d ),(21(121421211221=-===>⎰⎰⎰⎰⎰>x x x x y y x y x y x f X P xx .11.设二维随机变量),(Y X 的概率密度为⎪⎩⎪⎨⎧<<<<+=其他.,0,20,10,3),(2y x xyx y x f 求:(1)),(Y X 的边缘概率密度;(2)X 与Y 是否独立;(3))),((D Y X P ∈,其中D 为曲线22x y =与x y 2=所围区域.解:(1)10<<x 时,x x y xy x y y x f x f X 322d )3(d ),()(222+=+==⎰⎰+∞∞-,其他,0)(=x f X ,∴⎪⎩⎪⎨⎧<<+=其他.,0,10,322)(2x x x x f X ,20<<y 时,⎰+∞∞-=x y x f y f Y d ),()(316)d 3(12+=+=⎰y x xy x ,其他,0)(=y f Y ,∴⎪⎩⎪⎨⎧<<+=其他.,0,20,316)(y y y f Y .(2)),()()(y x f y f x f Y X ≠,∴X 与Y 不独立.(3)}22,10|),{(2x y x x y x D ≤≤<<=,∴⎰⎰+=∈102222d d )3()),((x xx y xy x D Y X P 457d )32238(10543=--=⎰x x x x .12.设二维随机变量),(Y X 的概率密度为⎪⎩⎪⎨⎧>>+=-其他.,0,0,0,e )1(),(2y x y x y x f x试讨论X ,Y 的独立性.解:当0>x 时,xx x X x yx y y x y y x f x f -∞+-∞+-∞+∞-=+-=+==⎰⎰e |11e d )1(e d ),()(002,当0≤x 时,0)(=x f X ,故⎩⎨⎧≤>=-.0,0,0,e )(x x x x f x X ,同理,可得⎪⎩⎪⎨⎧≤>+=.0,0,0,)1(1)(2y y y y f Y ,因为)()(),(y f x f y x f Y X =,所以X 与Y 相互独立.13.设随机变量),(Y X 在区域}|),{(a y x y x g ≤+=上服从均匀分布,求X 与Y 的边缘概率密度,并判断X 与Y 是否相互独立.解:由题可知),(Y X 的联合概率密度函数为⎪⎩⎪⎨⎧≤+=其他.,0,,21),(2a y x a y x f ,当0<<-x a 时,有)(1d 21d ),()(2)(2x a ay a y y x f x f xa x a X +===⎰⎰++-+∞∞-,当a x <≤0时,有)(1d 21d ),()(2)(2x a a y a y y x f x f x a x a X -===⎰⎰---+∞∞-,当a x ≥时,0d ),()(==⎰+∞∞-y y x f x f X ,故⎪⎩⎪⎨⎧≥<-=.a x a x x a a x f X ,0,),(1)(2,同理,由轮换对称性,可得⎪⎩⎪⎨⎧≥<-=.a y a y y a a y f Y ,0,),(1)(2,显然)()(),(y f x f y x f Y X ≠,所以X 与Y 不相互独立.14.设X 和Y 时两个相互独立的随机变量,X 在)1,0(上服从均匀分布,Y 的概率密度为⎪⎩⎪⎨⎧≤>=-.0,0,0,e 21)(2y y y f yY (1)求X 和Y 的联合概率密度;(2)设含有a 的二次方程为022=++Y aX a ,试求a 有实根的概率.解:(1)由题可知X 的概率密度函数为⎩⎨⎧<<=其他.,0,10,1)(x x f X ,因为X 与Y 相互独立,所以),(Y X 的联合概率密度函数为⎪⎩⎪⎨⎧><<==-其他.,0,0,10,e 21)()(),(2y x y f x f y x f yY X ,(2)题设方程有实根等价于}|),{(2X Y Y X ≤,记为D ,即}|),{(2X Y Y X D ≤=,设=A {a 有实根},则⎰⎰=∈=Dy x y x f D Y X P A P d d ),()),(()(⎰⎰⎰---==1021002d )e 1(d d e 2122xx y x x y⎰--=102d e12x x ⎰--=12e 21212x x ππππ23413.01)]0()1([21-=Φ-Φ-=.15.设i X ~)4.0,1(b ,4,3,2,1=i ,且1X ,2X ,3X ,4X 相互独立,求行列式4321X X X X X =的分布律.解:由i X ~)4.0,1(b ,4,3,2,1=i ,且1X ,2X ,3X ,4X 相互独立,易知41X X ~)84.0,16.0(b ,32X X ~)84.0,16.0(b .因为1X ,2X ,3X ,4X 相互独立,所以41X X 与32X X 也相互独立,又32414321X X X X X X X X X -==,则X 的所有可能取值为1-,0,1,有)1()0()1,0()1(32413241======-=X X P X X P X X X X P X P 1344.016.084.0=⨯=,)1,1()0,0()0(32413241==+====X X X X P X X X X P X P )1()1()0()0(32413241==+===X X P X X P X X P X X P 7312.016.016.084.084.0=⨯+⨯=,)0()1()0,1()1(32413241=======X X P X X P X X X X P X P 1344.084.016.0=⨯=,故X 的分布律为X 1-01P1344.07312.01344.016.设二维随机变量),(Y X 的概率密度为⎩⎨⎧>>=+-其他.,0,0,0,e 2),()2(y x y x f y x 求Y X Z 2+=的分布函数及概率密度函数.解:0≤z 时,若0≤x ,则0),(=y x f ;若0>x ,则0<-=x z y ,也有0),(=y x f ,即0≤z 时,0),(=y x f ,此时,0d d ),()2()()(2==≤+=≤=⎰⎰≤+zy x Z y x y x f z Y X P z Z P z F .0>z 时,若0≤x ,则0),(=y x f ;只有当z x ≤<0且02>-=xz y 时,0),(≠y x f ,此时,⎰⎰≤+=≤+=≤=zy x Z yx y x f z Y X P z Z P z F 2d d ),()2()()(⎰⎰-+-=zx z y x y x 020)2(d e 2d z z z ----=e e 1.综上⎩⎨⎧≤>--=--.0,0,0,e e 1)(z z z z F z z Z ,所以⎩⎨⎧≤<='=-.0,0,0,e )()(z z z z F z f z Z Z .17.设X ,Y 是相互独立的随机变量,其概率密度分别为⎩⎨⎧≤≤=其他.,0,10,1)(x x f X ,⎩⎨⎧≤>=-.0,0,0,e )(y y y f y Y 求Y X Z +=的概率密度.解:0<z 时,若0<x ,则0)(=x f X ;若0≥x ,则0<-=x z y ,0)(=-x z f Y ,即0<z 时,0)()(=-x z f x f Y X ,此时,0d )()()(=-=⎰+∞∞-x x z f x f z f Y X Z .10≤≤z 时,若0<x ,则0)(=x f X ;只有当z x ≤≤0且0>-=x z y 时0)()(≠-x z f x f Y X ,此时,z zx z Y X Z x x x z f x f z f ---+∞∞--==-=⎰⎰e 1d e d )()()(0)(.1>z 时,若0<x ,0)(=x f X ;若1>x ,0)(=x f X ;若10≤≤x ,则0>-=x z y ,此时,0)()(≠-x z f x f Y X ,z x z Y X Z x x x z f x f z f ---+∞∞--==-=⎰⎰e )1e (d e d )()()(1)(.综上,⎪⎩⎪⎨⎧<>-≤≤-=--.0,0,1,e )1e (,10,e 1)(z z z z f z z Z .18.设随机变量),(Y X 的概率密度为⎪⎩⎪⎨⎧>>+=+-其他.,0,0,0,e)(21),()(y x y x y x f y x (1)X 和Y 是否相互独立?(2)求Y X Z +=的概率密度.解:(1)),()()(y x f y f x f Y X ≠,∴X 与Y 不独立.(2)0≤z 时,若0≤x ,则0)(=x f X ;若0>x ,则0<-=x z y ,0),(=y x f ,此时,0d ),()(=-=⎰+∞∞-x x z x f z f Z .0≥z 时,若0≤x ,则0)(=x f X ;只有当z x <<0且0>-=x z y 时0),(≠y x f ,此时,⎰+∞∞--=x x z x f z f Z d ),()(⎰+-+=zy x x y x 0)(d e )(21⎰-=z z x z 0d e 21z z -=e 212,所以⎪⎩⎪⎨⎧≤>=-.0,0,0,e 21)(2z z z z f zZ .19.设X 和Y 时相互独立的随机变量,它们都服从正态分布),0(2σN .证明:随机变量22Y X Z +=具有概率密度函数⎪⎩⎪⎨⎧<≥=-.0,0,0,e )(2222z z z z f z Z σσ.解:因为X 与Y 相互独立,均服从正态分布),0(2σN ,所以其联合密度函数为2222)(2e 121),(σσπy x y xf +-⋅=,(+∞<<∞-y x ,)当0≥z 时,有⎰⎰≤+=≤+=≤=zy x Z yx y x f z Y X P z Z P z F 22d d ),()()()(22⎰⎰≤++-⋅=zy x y x y x 22222d e 1212)(2σσπ⎰⎰-⋅=πσθσπ2022d ed 12122zr r r ⎰-=zr r r 022d e122σσ,此时,2222e)(σσz Z z z f -=;当0<z 时,=≤+}{22z Y X ∅,所以0)()()(22=≤+=≤=z Y X P z Z P z F Z ,此时,0)(=z f Z ,综上,⎪⎩⎪⎨⎧<≥=-.0,0,0,e )(2222z z z z f z Z σσ.20.设),(Y X 在矩形区域}10,10|),{(≤≤≤≤=y x Y X G 上服从均匀分布,求},min{Y X Z =的概率密度.解:由题可知),(Y X 的联合概率密度函数为⎪⎩⎪⎨⎧≤≤≤≤=其他.,0,20,10,21),(y x y x f ,易证,X ~]1,0[U ,Y ~]2,0[U ,且X 与Y 相互独立,⎪⎩⎪⎨⎧≥<≤<=.1,1,10,,0,0)(x x x x x F X ,⎪⎪⎩⎪⎪⎨⎧≥<≤<=.2,1,20,2,0,0)(y y yy y F Y ,可得)](1)][(1[1)(z F z F z F Y X Z ---=)()()()(z F z F z F z F Y X Y X -+=⎪⎪⎩⎪⎪⎨⎧≥<≤-<=.1,1,10,223,0,02z z z z z ,求导,得⎪⎩⎪⎨⎧<<-=其他.,0,10,23)(z z z f Z .21.设随机变量),(Y X 的概率密度为⎩⎨⎧+∞<<<<=+-其他.,0,0,10,e ),()(y x b y x f y x (1)试确定常数b ;(2)求边缘概率密度)(x f X 及)(y f Y ;(3)求函数},max{Y X U =的分布函数.解:(1)⎰⎰⎰⎰+∞+-+∞∞-+∞∞-==01)(d d e d d ),(1yx b y x y x f y x ⎰⎰+∞--=10d e d e y x b y x)e 1(|)e(|)e (10102-+∞---=-⋅=b b y x ,∴1e11--=b .(2)10<<x 时,1)(1e1e d e e 11d ),()(--∞++--∞+∞--=-==⎰⎰x y x X y y y x f x f ,其他,0)(=x f X ,∴⎪⎩⎪⎨⎧<<-=--其他.,0,10,e 1e )(1x x f xX ,0>y 时,⎰+∞∞-=x y x f y f Y d ),()(yy x x -+--=-=⎰e d e e 1110)(1,0≤y 时,0)(=y f Y ,∴⎩⎨⎧≤>=-.0,0,0,e )(y y y f y Y .(3)0≤x 时,0)(=x F X ,10<<x 时,101e1e 1d e 1e d )()(----∞---=-==⎰⎰xxt xX X t t t f x F ,1≥x 时,1)(=x F X ,∴⎪⎪⎩⎪⎪⎨⎧≥<<--≤=--.1,1,10,e 1e1,0,0)(1x x x x F x X ;0≤y 时,0)(=y F Y ,0>y 时,y yv y Y Y v v v f y F --∞--===⎰⎰e 1d e d )()(0,∴⎩⎨⎧≤>-=-.0,0,0,e 1)(y y y F y Y ,故有)()()(y F x F u F Y X U =⎪⎪⎩⎪⎪⎨⎧≥-<≤--<=---.1,e 1,10,e 1e1,0,01u u u uu .。
课后习题及答案_第3章离散傅里叶变换--习题

课后习题及答案_第3章离散傅里叶变换--习题数字信号处理第三版第3章离散傅里叶变换(DFT)习题1.计算以下序列的N点DFT,在变换区间0≤n≤N-1内,序列定义为(1) x(n)=1(2) x(n)=δ(n)(3) x(n)=δ(n-n0) 0n0N(4) x(n)=Rm(n) 0mN(5) n ) jNmn N x(=e,0 mπ 2 (6) n ) x(=cos mn ,0mN2π(7) x(n)=ejω0nRN(n)(8) x(n)=sin(ω0n)RN(n)(9) x(n)=cos(ω0n)RN(N)(10) x(n)=nRN(n)2.已知下列X(k),求x(n)=IDFT[X(k)]Njθ 2e N(1)X (k)= e jθ20 N k=m k=N m其它kNjθ j2e N jθ(2)X (k)= je 2 0 k=m k=N m 其它k其中,m为正整数,0mN/2, N为变换区间长度。
3.已知长度为N=10的两个有限长序列:做图表示x1(n)、x2(n)和y(n)=x1(n) * x2(n),循环卷积区间长度L=10。
,4.证明DFT的对称定理,即假设X(k)=DFT[x(n)]数字信号处理第三版证明DFT[X(n)]=Nx(N-k)5.如果X(k)=DFT[x(n)],证明DFT的初值定理1x(0)=N∑X(k)k=0N 16.设x(n)的长度为N,且X(k)=DFT[x(n)]0≤k≤N-1令h(n)=x((n))NRmN(n) m为自然数H(k)=DFT[h(n)]mN 0≤k≤mN-1求H(k)与X(k)的关系式。
7.证明: 若x(n)为实序列,X(k)=DFT[x(n)]N,则X(k)为共轭对称序列,即X(k)=__(N-k);若x(n)实偶对称,即x(n)=x(N-n),则X(k)也实偶对称;若x(n)实奇对称,即x(n)=-x(N-n),则X(k)为纯虚函数并奇对称。
概率论课后习题第3章答案

第三章 多维随机向量及其概率分布(一)基本题答案1、设X 和Y 的可能取值分别为.2,1,0;3,2,1,0,==j i j i 则与因盒子里有3种球,在这3种球中任取4个,其中黑球和红球的个数之和必不超过4.另一方面,因白球只有2个,任取的4个球中,黑球和红球个数之和最小为2个,故有j i 与ٛ且,42≤+≤j i ./),(474223C C C C j Y i X p j i j i −−===因而 或0),(===j Y i X P 2).2,1,0;3,2,1,0,4(<+j i ==>+j i j i于是 ,0)0,0(1111======y Y x X P P ,0)0,0(2112======y Y x X P p.35/1/)0,0(472212033113=======C C C C y Y x X P p即 2、X 和. ⎥⎦⎤⎢⎣⎡04.032.064.0210~X ⎥⎦⎤⎢⎣⎡25.05.025.0210~Y 由独立性知,X 和Y 的联合分布为3、Y 的分布函数为显知有四个可能值:).0(0)(),0(1)(≤=>−=−y y F y e y F y ),(21X X }{{}{}11−=e ,2,10,0).1,1(),0,1(),1,0(),0,0(121−≤=≤≤===Y P Y Y P X X P 易知{}{}{}{}{},221−−−=e e 12<=P ,10,1,02,11,02121≤≤>====>≤===Y Y Y P X X P Y Y P X X P{}{}{},212,10,12121−=≤<=≤>===e e Y P Y Y P X X P {}−− {}{}.22,11,1221−=>=>>===e Y P Y Y P X X P于是,可将X 1和X 24、∑=====nm m n P n X P 0),()(ηζ∑=−−−−=nm mn m n e m n m p p 0)!(!)1(λλ()[]).,2,1,0(!1!)1()!(!!!==−+=−−=−−−=−∑n n e p p n e p p m n m n n e n n n mn m nm n λλλλλλ即X 是服从参数为λ的泊松分布.∑∑∞=−−∞=−−−−−=−−==mn mn m n mn m m mn m n m n p m e p em n m p p m Y P )!()1(!)!(!)1()(λλλλλ).,2,1,0(,!)(!)()1( ==⋅=−−−−m m ep e e m ep pmp mλλλλλλ即Y 是服从参数为λp 的泊松分布.5、由定义F (y x ,)=P {}∫∫∞−∞−=≤≤x y dxdy y x y Y x X .),(,ϕ因为ϕ(y x ,)是分段函数,要正确计算出F (y x ,;1>y ),必须对积分区域进行适当分块:等5个部分.10,10,1;1,1;10,100≤≤≤≤>>>≤≤<x y x y x y y x 或;0<≤≤x (1)对于 有 F (,00<<y x 或y x ,)=P{X ≤,x Y ≤y}=0; (2)对于 有 ;,10,10≤≤≤≤y x 2204),(y x vdudv u y x F x y ==∫∫(3)对于, 有 10,1≤≤>y x {};,1),(2y y Y X P y x F =≤≤= (4)对于, 有 10,1≤≤>x y {}21,),(x Y x X P y x F =≤≤=; (5)对于 有 ,1,1>>y x 1),(=y x F .故X 和Y 的联合分布函数⎪⎪⎪⎩⎪⎪⎪⎨⎧<<≤≤<<≤≤≤≤≤≤<<=.1,1,.1,10,1,,1,10,,10,10,,00,0),(2222y x y x y y x x y x y x y x y x F 或6、(1) ,0,0;0),(,00>>=≤≤y x y x F y x 或),(y x F =∫∫+−x y t s dsdt ze)2())(())((200202yt x s y t x se e dt e ds e−−−−−−==∫∫=)1)(1(2y x e e −−−−即⎩⎨⎧>>−−=−−.,0,0,0),1)(1(),(2其它y x e e y x F y x (2)P ()()220(),22x x y x yxy xY X f x y dxdy dx e dy e e d +∞+∞−−−−<≤===−∫∫∫∫∫x∫∫∞+−−−∞+−−=−−=03220)(2)1(2dx e e dx e e x x x x .312131(2)2131(2023=−−=−=∞+−−x x e e7、(1)时,0>x ,0)(,0;)(=≤==∫∞+−−x f x e dy e x f X Xx y X 时 即 ⎩⎨⎧≤>=−.0,0,0,)(x x e x f x X (2){}2/111210121),(1−−≤+−−−+===≤+∫∫∫∫e e dy e dxdxdy y x f Y X P y x x xy8、(1)(i )时,,;),()(计算根据公式∫∞+∞−=dy y x f x f X 0≤x 当10;0)(<<=x x f X 当时()();24.224.2)2(8.4)(202x x x y dy x y x f xx X −=−=−=∫0)(,1=≥x f x X 时当即⎩⎨⎧<<−=.,0;10),2(4.2)(2其它x x x x f X (ii ) 利用公式计算. 当∫∞+∞−=dx y x f y f Y ),()(;0)(,0=≤y f y Y 时,10时当<<y112)22(8.4)2(8.4)(y y Y x x y dx x y y f ∫−=−=⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛−−⎟⎠⎞⎜⎝⎛−=222128.42y y y );43(4.2)2223(8.422y y y y y y +−=+−=当时,1≥y .0)(=y f Y 即⎩⎨⎧<<+−=.0;10),43(4.2)(2其它y y y y y f Y 121111222211111(2)((1(,1(,)1.22222P X Y P X Y f x y dxdy dx dxdy +∞+∞⎧⎫<<=−≥≥=−=−=⎨⎬⎩⎭∫∫∫∫∪58、47809、本题先求出关于x 的边缘概率密度,再求出其在2=x 之值. 由于平面区域D 的面积为)2(X f ,2121=dx =∫x S e D 故(X,Y )的联合概率密度为⎪⎩⎪⎨⎧∈=.,0;),(,21),其它D y x y x (f易知,X 的概率密度为∫∞+∞−⎪⎩⎪⎨⎧<<==,,0,1,21),()(2其它e x xdy y x f x f X 故.41221)2(=×=X f 10、(1)有放回抽取:当第一次抽取到第个数字时,第二次可抽取到该数字仍有十种可能机会,即为 k {}).9, ,1,0(101====i k Y i X P (2)不放回抽取:(i )当第一次抽取第)90(≤≤k k 个数时,则第二次抽到此(第个)数是不可能的,故 k {}.)9,,1,0,; =k i k (0====i k Y i X P(ii )当第一次抽取第个数时,而第二次抽到其他数字(非k )的机会为,知)90(≤≤k k 9/1{}.)9,,1,0,; =k i k (9/1≠===i k Y i X P 11、(1)因∫−=−=12,)1(12)1(24)(yy y ydx x y f η.,0)(;10其它=≤≤y f y n 故在0≤y ≤1时,⎩⎨⎧≤≤−−=;1)1/()1(2)(2其它x y y x y x f ηξ因()∫−=−=x y x ydy x x f 022,)1(12124)(ξ.,0)(;10其它=≤≤x f x ξ故在0≤x ≤1时,⎩⎨⎧≤≤=.0,0/2)(2其它x y x y x y f ξη(2)因;1,121)(2/12∞≤≤==∫x x nxdy y x X f x x ξ;,0)(其它=x f ξ故在1≤x<时,∞⎪⎩⎪⎨⎧<<=.,1121)(其它x y xnxy x y f ξη因 ⎪⎪⎪⎩⎪⎪⎪⎨⎧∞<<=≤<==∫∫∞∞,002121102121)(22/12其它y y dx y x y dx y x y f y y η 故在10≤<y 时,⎪⎩⎪⎨⎧∞<<=;011)(2其它x y y x x y f ξη 而在,1时∞<<y ⎪⎩⎪⎨⎧∞<<=.0)(2其它x y x yx y f ξη(3)在x >0,.0,0)(;0,)(≤=>==∫∞−−x x f x e dy e x f x xy ξξ⎪⎩⎪⎨⎧>=−.0,)(其它x y e x y f y x ξη ;0,)(0>==∫−−y ye dx e y f y yy η .故在y>0时,0,0)(≤=y y f η⎪⎩⎪⎨⎧<<=.0,01)(其它y x y y x f ηξ12、1(1)(2)2(),0(1)(1)X n n n n n f x dy x x y x ∞−−−−==+++∫>,故12(1)(2)0,(/1)0.n nY X n y y f y −⎧−+>=⎨⎩其它 13、X 和Y 是否独立,可用分布函数或概率密度函数验证.方法一:X 的分布函数的分布函数分别为 Y x F X 和)()(y F Y ⎩⎨⎧<≥−=+∞=−,0001),()(5.0x x e x F x F x X ⎩⎨⎧<≥−=+∞=−.0001),()(5.0y y e y F y F yY 由于独立.Y X y F x F y x F Y X 和知),()(),(={}{}{}[][]1.005.005.0)1.0(1)1.0(11.01.01.0,1.0−−−=⋅=−⋅−=>⋅>=>>=e e e F F Y P X P Y X P Y X αY X Y X x f x f y x f Y X 和分别表示和),,()()(),,(方法二:以的概率密度,可知 ⎩⎨⎧≥≥=∂∂∂=+−.00,025.0),(),()(5.02其它y x e y x y x F y x f y x ∫∞+∞−−⎩⎨⎧<≥==,0005.0),()(5.0x x e dy y x f x f x X ∫∞+∞−−⎩⎨⎧<≥==.00,05.0),()(5.0y y e dx y x f y f yY ∫∫∞+∞+−+−==>>==1.01.01.0)(5.0.25.0}1.0,1.0{.),()(),(e dxdy e Y X P a Y X y f x f y x f y x Y X 独立和知由于)()(),(j i j i y Y P x x P y Y x X P =⋅====14、因知X 与Y 相互独立,即有 . )3,2,1,2,1(==j i 首先,根据边缘分布的定义知 .2418161),(11=−===y Y x X P 又根据独立性有),(61)()(},{2411111i x X p y Y p x X p y Y x X p ===⋅===== 解得41)(==i x X P ,从而有 1218124141),(31=−−===y Y x X P 又由 )()(),(2121y Y P x X P y Y x X P =⋅====, 可得 ),(41812y Y P == 即有21)(2==y Y P , 从而 838121),(22=−===y Y x X P .类似地,由),()(),(3131y Y P x X P y Y x X P ===== 有),(411213y Y P ==得31)(3==y Y P ,从而,.111),(31=−===y Y x X P 最后=)(2x X P =1+3+1=3. 将上述数值填入表中有1x1/24 1/8 1/12 1/4 2x1/8 3/8 1/4 3/4 {}j P y X P j ⋅==1/6 1/2 1/3115、本题的关键是由题设P{X 1X 2=0}=1,可推出P{X 1X 2≠0}=0;再利用边缘分布的定义即可列出概率分布表.(1)由P{X 1X 2=0}=1,可见易见,0}1,1{}1,1{2121=====−=X X P X X P 25.0}1{}0,1{121=−===−=X P X X P 5.0}1{}1,0{221=====X P X X P 25.0}1{}0,1{121=====X P X X P 0}0,0{21===X X P121212.16、(1) ⎩⎨⎧<<=,,0,10,1)(其他x x f X ⎪⎩⎪⎨⎧≤>=−.0,0,021)(2y y ey f yY 因为X ,Y 独立,对任何y x ,都有 ).,()()y x f y f x Y =⋅(f X ⎪⎩⎪⎨⎧><<=−.,0,0,10,21),(2其他所以有y x e y x f y(2)二次方程 有实根,△ t Y Xt t 中022=++,04)2(2≥−=Y X ,02≥−Y X 即,2X Y ≤ 故=)(有实根t P dydx e dydx y x f X Y P yx y x 2122221),(}{−≤∫∫∫∫==≤∫−−=1022)(dx ex y=dx edx edx x x x 2101010222221211)21(−−∫∫−=−=−πππ21−=[∫∫∞−∞−−−−1022222121dx edx exx ππ].1445.08555.01]5.08413.0[21)]0()1([21=−≈−−≈Φ−Φ−=ππ17、(1)因为X ,Y 独立,所以 .⎩⎨⎧>>==+−.,0,0,0,)()(),()(其他y x e y f x f y x f uy x Y X λλμ(2)根据Z 的定义,有 P{z=1}=P{Y ≥X}∫∫∫∫∞+∞−+−≥==)(),(xy x xy dydx e dydx y x f μλλμ∫∫∞+∞+−−=)(dx dy e e xy x μλμλ ),0u dx ee x x +=⋅=∫∞+−−λλλμλ{}{110=−==Z P Z P Z 的分布函数为⎪⎪⎩⎪⎪⎨⎧≥<≤+<=.1,1,10,,0,0)(z z z z F Z μλμ18、∵X 、Y 分别仅取0,1两个数值,∴Z 亦只取0,1两个数值. 又∵X 与Y 相互独立,∴{}{}{}{}==========00)0,0(0),max(0Y P X P Y X P Y X P Z P 1/2×1/2=1/4, 故{}{}.4/34/110111=−==−===Z P Z P 19、 X 由2×2阶行列式表示,仍是一随机变量,且X=X 1X 4--X 2X 3,根据X 1,X 2,X 3,X 4的地位是等价且相互独立的,X 1X 4与X 2X 3也是独立同分布的,因此可先求出X 1X 4和X 2X 3的分布律,再求X 的分布律. ,则X=Y 1--Y 2.随机变量Y 1和Y 2独立同分布:322411,X X Y X X Y ==记}{}{}{{}.84.016.01}0{0112121=−========Y P Y Y P Y P 16.01,132===P X X P 显见, 随机变量X=Y 1--Y 2有三个可能值--1,0,1.易见 P{X=--1}=P{Y 1=0,Y 2=1}=0.84×0.16= 0.1344, P{X=1}=P{Y 1=1,Y 2=0}=0.16×0.84=0.1344, P{X=0}=1--2×0.1344=0.7312. 于是,行列式的概率分布为 4321X X X X X =~ ⎥⎦⎤⎢⎣⎡−1344.07312.01344.010120、因为{Z=i }={X+Y=i }={X=0,Y=i }}.0,{}1,1{==−==Y i X i Y X ∪ ∪∪ 由于上述各事件互不相容,且注意到X 与Y 相与独立,则有 ∑∑==−===−====i k ik k i Y P k X P k i Y k X P i Z P 00}{}{},{}{∑=+−−−−−=−−=iik ki n ki k i nkn kk n P p pC P p c 022111()1()1∑=−−+ik k i n k n in n C Cp 02121)(,,1,0,)1(212121n n i p p C i n n i i n n+=−=−++).,(~21p n n B Y X Z ++=故注:在上述计算过程中,已约定:当r>n 时,用到了公式 并,0=rnC .12121∑=+−=ik i n n k i n k n C C C21、X 和Y 的概率分布密度为},2)(exp{21)(22σσπy x x f X −−=);(+∞<<−∞x ⎩⎨⎧≤≤−=.,0,),2/(1)(其它πππy y f Y 因X 和Y 独立,考虑到 )仅在[)(y f Y ππ,−]上才有非零值,故由卷积公式知Z 的概率密度为.221)()()(222)(dy edy y f y z f z f a y z Y X Z ∫∫−−−−∞+∞−=−=ππμσππ令σμ−−=y z t ,则上式右端等于.(2122122⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛−−Φ−−+Φ=∫−+−−−σμπσμππππσμπσμπz z dt e z z t 22、(1)由题设知 {}y X X P y M P y F n M ≤=≤=),,max()()(1),,(1y X y X P n ≤≤= )()()()()(121y F y F y X P y X P y X P Xn X n =≤≤≤=.∵),1(],0[~:,,1n i U X X X i n ≤≤θ独立且同分布 ∴⎪⎩⎪⎨⎧><<≤=,0,1,0,,0,0)(x x x x x F i X θθ∴⎪⎪⎩⎪⎪⎨⎧≥<<≤=.,1,0,,0,0)(θθθy y y y y F n n M 故⎪⎩⎪⎨⎧<<=−.,0,0,)(1其它θθy ny y f n n M(2){}y X X P y N P y N P y F n N >−=>−=≤=),,min(1)(1)()(1()y X P y X P y X P y X y X y X P n n >>>−=>>>−= )()(1,,,12121()[])(11)(11y F y X P i X i ni −−=>Π−==故 ⎪⎩⎪⎨⎧<<−=⎪⎩⎪⎨⎧<<−−−=−−其它其它,0,00,)(,001(1()(11y y n y y n y f n n n N θθθθθ 23、由题设容易得出随机变量(X ,Y )的概率密度,本题相当于求随机变量X 、Y 的函数S=XY 的概率密度,可用分布函数微分法求之.依题设,知二维随机变量(X ,Y )的概率密度为()()()⎩⎨⎧∉∈=G y x Gy x y x f ,,0,2/1,若若 设为S 的分布函数,则 当{s S P s F ≤=)(}0≤s 时,()0=s F ; 当时, .2≥s ()1=s F 现设0<s<2. 曲线s xy =与矩形G 的上边交于点(s,1);位于曲线s xy =上方的点满足s xy >,位于下方的点满足s xy <. 故(){}{}{}).ln 2ln 1(2211211121s sdy dx dxdy S XY P s XY P s S P s F s x s sxy −+=−=−=>−=≤=≤=∫∫∫∫>于是,⎩⎨⎧≥≤<<−=.20,0,20,2/)ln 2(ln )(s s s s s f 或若若(二)、补充题答案1.由于即{},0)(),,min(,,max =<==Y X P Y X 故知ηξηξ{}{}{}03,23,12,1=========Y X P Y X P Y X P ;又易知{}{}{}{},9/1111,11,1==⋅=======ηξηξP P P Y X P{}{},9/12,22,2======ηξP Y X P {}{},9/13,33,3======ηξP Y X P {}{}{},9/29/19/11,22,11,2=+===+=====ηξηξP P Y X P{}{}{},9/22,33,22,3===+=====ηξηξP P Y X P {}.9/29/711,3=−===Y X P 所以2.(1)x{}.,2,1,0,0,)1( =≤≤−===n n m P P C n X m Y P m n {}(2){}{}n X P n X m Y P m Y n X P ======,.,2,1,0,0,!)1( =≤≤⋅⋅−=−−n n m e P P C n m n mm n λλ3.22)1()1()1()0()0()1(p p Y P X P Y P X P z P +−===+====)1(2)0()1()1()0()0(p p Y P X P Y P X P z P −===+====而,由2)1,1()1,1(p Y X P Z X P ======),1()1()1,1(=====Z P X P Z X P 得. 2/1=p 5.:设随机变量ξ和η相互独立,都服从分 )1,0(N 布.则⎭⎬⎫⎩⎨⎧+−⋅=)(21exp 21),(22y x y x p π.显然, ,),(),(∫∫∫∫<SGdxdy y x p dxdy y x p,其中 G 和S 分别是如图所示的矩形ABCD 和圆.22/)21(),(2∫∫∫−−=a ax Gdx e dxdy y x p π,令,sin ,cos ϕγϕγ==y x 则 ∫∫∫∫=ππ20221),(a aSdxdy y x p 所以221212/a aaxe dx e −−−−<∫π.6.设这类电子管的寿命为ξ,则(1)三个管子均不要替换的概率为;(2)三个管子均要替换的概率为 .∫∞+==>1502.3/2)/(100)150(dx x P ξ21(−27/8)3/2(3=27/1)3/3=7.假设总体X 的密度函数为,分布函数为,第次的观察值为,独立同分布,其联合密度函数)(x f ,(1x f )(x F )()2x f i (n x )1(n i X i ≤≤i X )(),1n f x f x =.依题意,所求的概率为{}∫∫∫∫∫∫∞+∞−∞−∞−∞−−−−=−==>>><n n n nx i x x x x n n nn nn n i n n n n dx x f dx x f dx x f dx x f dx dx xx f X X X X X X P 112211111,...,2,1121)(...)()()(),,(.,...,,∫∫∞+∞−∞+∞−−−==)()()()(11n n n n n n n x dF x F dx x f x F.1)(1n x F nn n=∞−∞+=8.)(),()(21211211n P n k P n k P =+=+===+=ξξξξξξξξ)()()(2121n P k n P k P =+−===ξξξξ.由普哇松分布的可加性,知服从参数为的普哇松分布,所以 21ξξ+21λλ+)(21212112121!)()!(!)(λλλλλλλλξξξ+−−−−+−⋅==+=e n e k n ek n k P n k n k.1211211kn kk n −⎟⎟⎠⎞⎜⎜⎝⎛+−⎟⎟⎠⎞⎜⎜⎝⎛+⎟⎟⎠⎞⎜⎜⎝⎛=λλλλλλ9.当,0≤z (),0)(=≤=z Z P z F z ,0>z 当()z Z P z F z ≤=)(∫∫−+−=20)2(02xz y x z dy e dx∫∫−−−−−−−==202012x z z z y z x ze e dy e dxe ,所以 Y X z 2+=的分布函数为 ⎩⎨⎧>+−≤=−.0,)1(1,0,0),(z e z z y x F z10.由条件知X 和Y 的联合密度为⎪⎩⎪⎨⎧≤≤≤≤=其他若,0,31,31,41),(y x y x p以表示随机{})()(∞<<−∞≤=u u U P u F 变量U 的分布函数.显然,当0≤u 时, 0)(=u F ;当时,; 2≥u 1)(=u F 当,则20<<u []∫∫∫∫≤−uy x y x p ||,(≤−−−=−−===uy x u u dxdy dxdy u F ||2)2(411)2(44141))(2u−于是,随机变量的密度为⎪⎩⎪⎨⎧<<−=其他,0;20),2(21)(u u u p .11.记为这3个元件无故障工作的时间,则的分布函数321,,X X X ),,min(321X X X T ={}[][].)(1),,min(1(31321t X P t X X X P t F T −=>−(11)13X P t ≤−−=>)()t T P =≤=⎩⎨⎧≤>−=∴⎩⎨⎧=≤>−=−−,0,0,0,1)()3,2,1(,0,0,0,1)(~3t t e t F i t t e t F X t T t i λλ∵ 故 ⎪⎩⎪⎨⎧≤>==−.0,0,0,3)(')(3t t e t F t f t T T λλ。
利息理论第三章课后答案

利息理论第三章课后答案《金融数学》课后习题参考答案第三章 收益率1、某现金流为:3000o o =元,11000o =元,12000I =元,24000I =元,求该现金流的收益率。
解:由题意得:2001122()()()0O I O I v O I v -+-+-=23000100040000v v --=4133v i ⇒=⇒=2、某投资者第一年末投资7000元,第二年末投资1000元,而在第一、三年末分别收回4000元和5500元,计算利率为0.09及0.1时的现金流现值,并计算该现金流的内部收益率。
解:由题意得:23(0)[(47) 5.5]1000V v v v =--+⨯ 当0.09i =时,(0)75.05V =当0.1i =时,(0)57.85V =-令(0)00.8350.198V v i =⇒=⇒=3、某项贷款1000元,每年计息4次的年名义利率为12%,若第一年后还款400元,第5年后还款800元,余下部分在第7年后还清,计算最后一次还款额。
解:由题意得:40.121(1)0.88854i v +=+⇒=571000400800657.86v pv p =++⇒= 4、甲获得100000元保险金,若他用这笔保险金购买10年期期末付年金,每年可得15380元,若购买20年期期末付年金,则每年可得10720元,这两种年金基于相同的利率i ,计算i 。
解:由题意得: 08688.010720153802010=⇒=i a a i i5、某投资基金按1(1)t k t k δ=+-积累,01t ≤≤,在时刻0基金中有10万元,在时刻1基金中有11万元,一年中只有2次现金流,第一次在时刻0.25时投入15000元,第二次在时刻0.75时收回2万元,计算k 。
解:由题意得:101(1)1k dt t k ek +-⎰=+ 10.251(1)10.75k dt t k ek +-⎰=+ 10.751(1)10.25kdt t k e k +-⎰=+ ⇒10000(1)15000(10.75)20000(10.25)1100000.141176k k k k +++-+=⇒=6、某投资业务中,直接投资的利率为8%,投资所得利息的再投资利率为4%,某人为在第10年末获得本息和1万元,采取每年末投资相等的一笔款项,共10年,求证每年投资的款项为:100.0410000210s -。
西方经济学-微观部分-第三章课后习题答案

第三章效用论1。
已知一件衬衫的价格为80元,一份肯德基快餐的价格为20元,在某消费者关于这两种商品的效用最大化的均衡点上,一份肯德基快餐对衬衫的边际替代率MRS是多少?解答:按照两商品的边际替代率MRS的定义公式,可以将一份肯德基快餐对衬衫的边际替代率写成:MRS XY=-错误!其中,X表示肯德基快餐的份数;Y表示衬衫的件数;MRS XY表示在维持效用水平不变的前提下,消费者增加一份肯德基快餐消费时所需要放弃的衬衫的消费数量。
在该消费者实现关于这两种商品的效用最大化时,在均衡点上有MRS XY=错误!即有MRS XY=错误!=0.25它表明,在效用最大化的均衡点上,该消费者关于一份肯德基快餐对衬衫的边际替代率MRS为0。
25。
2。
假设某消费者的均衡如图3-1(即教材中第96页的图3—22)所示。
其中,横轴OX1和纵轴OX2分别表示商品1和商品2的数量,线段AB为消费者的预算线,曲线图3-1某消费者的均衡U为消费者的无差异曲线,E点为效用最大化的均衡点.已知商品1的价格P1=2元。
(1)求消费者的收入;(2)求商品2的价格P2;(3)写出预算线方程;(4)求预算线的斜率;(5)求E点的MRS12的值。
解答:(1)图中的横截距表示消费者的收入全部购买商品1的数量为30单位,且已知P1=2元,所以,消费者的收入M=2元×30=60元.(2)图中的纵截距表示消费者的收入全部购买商品2的数量为20单位,且由(1)已知收入M=60元,所以,商品2的价格P2=错误!=错误!=3元.(3)由于预算线方程的一般形式为P1X1+P2X2=M所以,由(1)、(2)可将预算线方程具体写为:2X1+3X2=60。
(4)将(3)中的预算线方程进一步整理为X2=-错误!X1+20.很清楚,预算线的斜率为-23。
(5)在消费者效用最大化的均衡点E上,有MRS12=错误!,即无差异曲线斜率的绝对值即MRS等于预算线斜率的绝对值错误!。
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第三章模拟集成放大器及其应用1.判断对错,在括号中画√或×。
(1)若放大电路的A<0,则接人的反馈一定是负反馈。
( ×)(2)若A>0,则接人的一定是正反馈。
( ×)(3)由于接人负反馈,则反馈放大电路的A一定是负值。
( ×)(4)接人正反馈后,A一定是正值。
( ×)2.判断对错,在括号中画√或×。
(1)直流负反馈是指:①只存在于直接藕合电路,而阻容耦合电路中不存在的反馈。
( ×)②直流通路中的负反馈。
( √)③只在放大直流信号时才有的反馈。
( ×)(2)交流负反馈是指:①只存在于阻容耦合电路而直接耦合电路中没有的反馈。
( ×)②交流通路中的负反馈。
( √)③只在放大正弦波信号时才有的反馈。
( ×)3.填空:(1) 反相比例运算电路的输入电流基本上等于流过反馈电阻的电流,而___同相___比例运算电路的输电流几乎等于零。
(同相,反相)(2)反相比例放大电路的输入电阻较_低,同相比例放大电路的韵输入电阻较_高_。
(高,低)(3)在进行反相比例放大时,集成放大器两个输入端的共模信号Uic=__ 0_;若同相输入端接Ui,由集成放大器组成的比例放大电路的共模信号Uic=__Ui_。
(0,Ui)4.一个放大电路具有直流电流负反馈和交流电流并联负反馈,试问这些反馈有哪些作用?(判断下列答案是否正确,在括号中画√或×)(1)稳定输出电压( ×)(2)稳定输出电流( √)(3)增加输出电阻( √)(4)减小输出电阻( ×)(5)增大输入电阻( ×)(6)减小输入电阻( √)(7)增大工作点电流( ×)(8)稳定工作点电流。
(√)5.选择合适的答案填空为了实现以下目的,应引入:A.直流负反馈;B.交流负反馈。
(1)为稳定静态工作点;( A)(2)为稳定电压放大倍数( B )(3)为扩展频带( B)(4)为改变输入电阻或输出电阻( B)(5)为抑制温漂。
( A )6.对于下面的要求应分别选:A.电压串联;B.电压并联;C.电流串联;D.电流并联的负反馈形式。
(1)某仪表放大电路,要求Ri大,输出电流稳定,应选( C)。
(2)某传感器产生的是电压信号(几乎不能提供电流),经放大后希望输出电压与信号成正比,这放大电路应选( A)。
(a ) (b )(c )(a )(d )(b )(3)要得到一个由电流控制的电流源,应选( D )。
(4)要得到一个由电流控制的电压源,应选( B )。
(5)需要一个阻抗变换电路,Ri 大,Ro 小,应选( A )。
(6)要得到一个阻抗变换电路,Ri 小,Ro 大,应选( D )。
7.试分析图所示放大电路,指出反馈电路、反馈性质(正反馈还是负反馈,是直流反馈还是交流反馈),判断反馈方式为何种组态(串联还是并联,电压还是电流)。
解:(a )图:R f 、C f ;;电压串联交流负反馈; (b )图:R f 、C f ;电压串联交流负反馈。
8.试分析图所示电路,指出反馈通路、反馈性质,判断电路中各交流负反馈分别是什么组态。
解:(a 图)反馈通路:R L 、R 3;电流串联负反馈(b 图)反馈通路:R L 、R 6;电流串联正反馈 (c 图)反馈通路:R 2、R 3、C ;负反馈;电压串联; (d 图)反馈通路:A 、R 2、;负反馈;电压串联。
9.在负反馈运算放大电路的三种输入方式(A .反相输入;B .同相输入;C .差分输人)中,选择合适的方式填空。
(1)为给集成放大器引入电压串联负友馈,应采用( B )方式。
(2)要求引入电压并联负反馈,应采用( A )方式。
(3)在多个输入信号情况下,要求各输入信号互不影响,应采用( A )方式。
(4)要求向输入信号电压源索取的电流尽量小,应采用( B )方式。
(5)要求能放大两个信号的差值,又能抑制共模信号,应采用( C )方式。
10.判断下列说法是否正确(在括号中画√ 或×)。
(1)处于线性工作状态下的集成放大器,反相输入端可按“虚地”来处理。
(× )(2)反相比例运算电路属于电压串联负反馈,( × )同相比例运算电路属于电压并联负反馈。
( × )(3)处于线性工作状态的实际集成放大器,在实现信号运算时,两个输入端对地的直流电阻必须相等,才能防止输入偏置电流带来运算误差( √ )(4)在反相求和电路中,集成放大器的反相输入端为虚地点,流过反馈电阻的电流基本上等于各输入电流之代数和。
(√ )(5)同相求和电路和同相比例电路一样,各输入信号的电流几乎等于0( × ) 11.在图中各集成运放都是理想的,试写出各输出电压Uo 的值。
解: (a)9V .04.0)100501(100503.050100U O =⨯++⨯⨯=(b)2V )10201()3(3020)2(1020U O -=+⨯-⨯+-⨯-=(c) 2V(d)2V(c )(d )(b )(a )12.在图P3所示电路中,A 为理想集成放大器。
设Rl=R2=RL=10K,R3=Rf=100K ,Ui=0.55V ,求输出电压Uo 、负载电流IL 及集成放大器输出电流Io 。
解:5V .555.0)101001(10100100U )R1Rf 1(R3R2R3U i O =⨯++=++=55mA .0105.5RLU i O L ===U O =5.5V ; I O =0.6mA ;05mA .055.01010)(100100U R1)R3R2(R3i i f =⨯⨯+=+=6mA .005.055.0i O =+=13.设图中的A1,A2均为理想集成放大器。
(1)写出Uo 与Ui 的关系式;(2)若R1=10k ,R2=100k ,R=10K , Ui=:0.25V ,求Uo 。
解: (1)i o U R R U 122=(2)U O = 5V ;14.写出图所示运算电路中输出与输入的关系式。
设A1,A2均为理想集成放大器。
解:325210i i i o U U U U --=15.在图所示电路中,假设各集成放大器都是理想的。
(1)指出A1、A2、A3各组成什么电路; (2)写出Uo1、Uo2、Uo3的表达式。
解:(1)A 1组成电压跟随器;A 2组成反相输入加法器;A 3组成电流串联负反馈电路;(2)i o U U =1; i o U U 22-=;i o U U 43-=16.在图中,A1、A2、A3均为理想运算放大器,求瞬时值uo 与输入瞬时值uil ,ui2的关系。
-1V(a ) (b )解:216i i o U U U --=17.积分运算电路如图所示,已知初始状态ui=0,uo=0。
若ui 突加1V 直流电压,求1s 后的输出电压值。
设A 为理想集成放大器,R1=100K ,C=2uF 。
解:⎰-=-=tV dt uCR U Io 51118.已知图(a )所示电路中A 为理想集成放大器,R1=50K ,C=1uF ,输入电压ui 的波形为如图(b )所示的方波,试画出输出电压uo 的波形。
设t=0时,uo=0。
解:输出波形如图(c )所示。
19.在图所示电路中。
Al 、A2、A3均为理想集成放大器。
(1)写出uo1与ui1的关系表达式; (2)写出uo 与ui1,ui2的函数关系式。
解: (1)⎰-=dt uCR u i o 1111;(2)2241134i i o u R R dt u CR R R u -=⎰;(b ) (a ) (a ) (b ) (c )20.设图(a )所示基本微分电路中的电容C=0.0022uF ,电阻R=3k ,输入电压的波形如图(b )所示,试画出理想情况下输出电压的波形。
解:输出波形如图(c )所示。
21.两级理想集成放大器如图所示,为了稳定输出电压,且输入电阻较小,拟通过电阻Rf 引入一个级间反馈支路,问: (1)Rf 应如何连接?(2)这是什么类型的反馈?对输出电阻有何影响?(3)如果希望接入反馈后,不仅输出电阻小,而且输入电阻高,则上述电路应如何改接? 解:(1)R f 接在A 1反相输入端和A 2输出端之间;(2)属于电压并联负反馈;减小输出电阻;(3)R f 接在A 1同相输入端和A 2输出端之间;属于电压串联负反馈,提高输入电阻,减小输出电组;22.判断图中各电路所引反馈的极性及交流反馈的组态,并分别说明对输入电阻和输出电阻的影响。
解:(a)电流串联负反馈;提高输入电阻和输出电阻;(b)电流并联负反馈;减小输入电阻,提高输出电阻;(c)电压并联负反馈;减小输入电阻和输出电阻;。