电子科技大学离散期末试题13年
电子科技大学839自动控制原理_2013年真题及答案
电⼦科技⼤学839⾃动控制原理_2013年真题及答案2013年试题1. (共15分) 某控制系统状态⽅程为X AX ?=,其中，A 是22?常数⽅阵。
当1(0)1X ??=??-??时，22t t e X e --??=??-??当2(0)1X ??=??-??时，2t t e X e --??=??-??，试求该系统的状态转移矩阵()t φ2. (共20分) 某负反馈系统结构框图如图1所⽰，其中1()G s 环节的单位阶跃响应为58(1)5t e -- 1) 当()201(),()0r t t f t =?=时，试求系统响应的超调量，调节时间及输出稳态值。
2) 当()201(),()1()r t t f t t =?=时，试求系统的稳态误差。
其中，系统误差()E s 如图1所⽰。
图13. (共15分) 某负反馈系统的特征⽅程为2(10)(1)0s s K s +++=1) 试绘制该系统的根轨迹图。
2) 试确定使系统暂态响应分别为衰减振荡及单调变化时K 的取值范围。
4. (共15分) 某受控对象()G s 的单位脉冲响应为0.54()(12)t tg t K ee --=+-，将()G s 构成单位负反馈系统，如图2所⽰。
1) 试绘制该单位负反馈系统的开环幅相曲线(Nyquist 曲线) 2) 试求出该系统处于临界稳定时的开环增益和振荡频率值。
(r图25. (共20分) 已知某单位负反馈最⼩相位系统，其被控对象0()G s 的开环对数幅频特性0()L ω和串联校正装置()c G s 的对数幅频特性()c L ω如图3所⽰。
1) 试写出校正后系统的开环传递函数0()()()c G s G s G s =。
2) 试分析图中所采⽤的校正装置的类型，并讨论校正装置对系统相⾓裕度γ、剪切频率c ω及抗⾼频⼲扰等指标的影响。
图36. (共15分) 某离散控制系统如图4所⽰，T 为采样周期，0.25T s = 1) 求使得系统稳定的K 的取值范围。
2012-2013-2《数字逻辑设计及应用》期末考试题-A参考解答
电子科技大学2012 -2013学年第二学期期末考试 A 卷课程名称：_数字逻辑设计及应用__ 考试形式：闭卷考试日期：20 13 年07 月05 日考试时长：_120___分钟课程成绩构成：平时30 %，期中30 %，实验0 %，期末40 %本试卷试题由___七__部分构成，共__7___页。
I. Fill out your answers in the blanks (3’ X 10=30’)1. If a 74x138 binary decoder has 110 on its inputs CBA, the active LOW output Y5 should be ( 1 or high ).2. If the next state of the unused states are marked as “don’t-cares” when designing a finite state machine, this approach is called minimal ( cost ) approach.3.The RCO_L of 4-bit counter 74x169 is ( 0 or low ) when counting to 0000 in decreasing order.4. To design a "001010" serial sequence generator by shift registers, the shift register should need ( 4 ) bit at least.5. One state transition equation is Q*=JQ’+K’Q. If we use T flip-flop with enable to complete the equation，the enable input of T flip-flop should have the function EN=( JQ’+KQ ).6. A 4-bit Binary counter can have ( 16 ) normal states at most, 4-bit Johnson counter with no self-correction can have ( 8 ) normal states, 4-bit linear feedback shift-register (LFSR) counter with self-correction can have ( 16 ) normal states.7. If we use a ROM, whose capacity is 16 × 4 bits, to construct a 4-bit binary code to gray code converter, when the address inputs are 1001, ( 1101 ) will be the output.8. When the input is 10000000 of an 8 bit DAC, the corresponding output voltage is 2V. The output voltage is ( 3.98 ) V when the input is 11111111.II. Please select the only one correct answer in the following questions.(2’ X 5=10’)1. If a 74x85 magnitude comparator has ALTBIN=1, AGTBIN=0, AEQBIN=0, A3A2A1A0=1101, B3B2B1B0=0111 on its inputs, the outputs are ( D ).A) ALTBOUT=0, AEQBOUT=0, AGTBOUT=0 B) ALTBOUT=1, AEQBOUT=0, AGTBOUT=0C) ALTBOUT=1, AEQBOUT=0, AGTBOUT=1 D) ALTBOUT=0, AEQBOUT=0, AGTBOUT=12. As shown in Figure 1, what would the outputs of the 4-bit adder 74x283 be ( B ) when A3A2A1A0=0100, B3B2B1B0=1110 and S/A=1.A) C4=1, S3S2S1S0=0010 B) C4=0, S3S2S1S0=0110 C) C4=0, S3S2S1S0=1010D) C4=0, S3S2S1S0=1110Figure 13. Which of the following statements is INCORRECT? ( A )A) A D latch is edge triggered and it will follow the input as long as the control input C is activelow.B) A D flip flop is edge triggered and its output will not change until the edge of the controllingCLK signal.C) An S-R latch may go into metastable state if both S and R are changing from 11 to 00simultaneously.D) The pulse applying to any input of an S -R latch must meet the minimum pulse width requirement.4. The capacity of a memory that has 13 bits address bus and can store 8 bits at each address is ( B ).A) 8192 B) 65536 C) 104 D) 2565. Which state in Figure 2 is NOT ambiguous ( C ).A) A B) B C) C and D D) CABCD WX W+Y ZZ ’X ’+YYZ1X ’Z ’Figure 2III. Analyze the sequential-circuit as shown in Figure 3, D Flip-Flop with asynchronous presetand clear inputs. [15’]1.Write out the excitation equations, transition equations and output equation. [5’]2.Assume the initial state Q 2Q 1=00, complete the timing diagram for Q 2 ,Q 1 and Z. [10’]Figure 3参考答案：激励方程: D 1=Q 2/，D 2= Q 1转移方程：Q 1 *= D 1=Q 2/，Q 2 *=D 2= Q 1 输出方程：Z= (CLK+Q 2)/参考评分标准：1. 5个方程正确得5分；每错一个扣1分，扣完5分为止；2. Q 1、Q 2、Z 的波形边沿判断正确，得3分，错一个，扣1分，扣完3分为止；每个上升沿和下降沿各0.5分，错1处扣0.5分，扣完7分为止。
电子科技大学微机原理2012-2013期末A卷及答案
电子科技大学2012-2013学年第 1 学期期 末 考试 A 卷课程名称：微处理器系统结构与嵌入式系统设计 考试形式：一本书开卷 考试日期：2013年1月16日 考试时长：120分钟平行班课程成绩构成：平时 10 %， 期中 15 %， 实验 15 %， 期末 60 % 英才班课程成绩构成：平时 30 %， 期中 0 %， 实验 20 %， 期末 50 % 本试卷试题由 3 部分构成，共 4 页。
注意：请将第一、二题答案填入指定位置。
一、单选题答案（共30分，共30空，每空1分）二、填空题答案（共28分，共28空，每空1分）1. ① 存储器 ② I/O 端口 （可交换顺序）2. ① 一条机器指令由一段微程序来解释执行3. ① 指令 ② 总线4. ① IRQ ② FIQ （可交换顺序）5. ① 1.78 ② 44.94 ③ 2.886. ① 1100 0011 0000 1010 ② 1010 0010 1001 1010 ③ 1 ④ 1 ⑤ 1 ⑥ 0 （说明：该题评阅时按上述答案给分，但实际ARM 是32位CPU ，基本没有同学考虑到）7. ① 非流水线执行时间相对流水线执行时间之比8. ① 一段时间内，计算机工作时所需的指令和数据总是集中存放在临近地址的存储单元 9. ① 复位 ② 电源 ③ 时钟 ④ 存储系统 ⑤ 调试接口 （可交换顺序） 10. ① 异常 （或中断） 11. ① BIC ② ORR12. ① =ULCON0(或=0x50000000) ② #0x2B （或#0xAB ）一、单选题（共30分，共30空，每空1分）1.以下常用总线标准中，不属于片内总线的是（）。
A、Core ConnectB、AMBAC、AvalonD、SATA2.计算机系统中，以下不属于“异常”的是（）。
A、系统复位B、软件中断C、未定义指令陷阱D、函数调用3.一般地，微机接口电路一定是（）。
A．可编程的B．可寻址的C．可中断的D．可定时的4.某减法定时/计数器的输入时钟周期为Ti，若计数初值为N，则定时时间为（）。
#西安电子科技大学2013年上学期期末考试机电一体化原理与应用试卷
西安电子科技大学2013上学期期末考试机电一体化原理与应用试卷姓名 专业 学号 题 号 一二 三 四 五 总 分 分 数一、单项选择题(本大题共10小题，每小题1分，共10分) 在每小题列出的四个备选项中只有一个是符合题目要求的，请将其代码填写在题后的括号内。
错选、多选或未选均无分。
1.系统中用于提供驱动力改变系统运行状态，产生所希望运动的模块称为 【 】A ．驱动模块B ．接口模块C .微计算机模块D ．测量模块2． 根据传动轴的不同特点，可选用不同类型的传动部件。
要求传动轴相交的传动部件为 【 】A ．直齿圆柱齿轮B ．斜齿圆柱齿轮C ．直齿锥齿轮D ．蜗轮蜗杆3． 齿轮副的间隙会造成齿轮传动的回差，属于柔性消隙法的结构是 【 】A ．调整中心距法B ．选择装配法C ．带锥度齿轮法D ．压簧消隙结构4． 某光栅的条纹密度是l00条mm ，要用它测出1 m 的位移，应采用的细分电路是【 】A ．四倍频B ．八倍频C ．十倍频D ．十六倍频5． 下列电动机中，没有绕组和磁极的电动机是 【 】A ．直流电动机B ．超声波电动机C ．永磁同步电动机D ．交流感应电动机6 下列操作中．可以使直流伺服电动机的理想空载转速升高的是 【 】A.增大电枢电阻B.减小电枢电压 C ．增大电枢电压 D.减小电枢电阻7．采样一数据系统中，若考虑系统的抑制干扰能力时，采样速率应为闭环系统通频带的 【 】A .10倍以上B ．5倍C .2倍 D.(0.1～1)倍 8. 自动导引车在平面上运动，一般具有的运动自由度是 【 】A ．2个B ．3个C .4个D ．6个9 . 属于模块化小型PLC ，并能满足中等性能应用要求的PLC 系统是 【 】 A ．SIMAT IC S7-200PLC B SIMATIC S7-300PLC C SIMATIC S7-400PLC D. SIMATIC NET10．载入一个动合触点的指令是 【 】A ．OR0.00B ．LDNOT0.00C ．AND 0.00 D .LD 0.00二、简答题(本大题共8小题，每小题4分，共32分) 11机电一体化的所渭“4A 革命”是指什么?12.机电一体化设计中，“增强机械系统”的设计思想是什么?得分 评卷人得分 评卷人13试说明滚珠丝杠螺母副传动装置的优缺点。
电子科技大学《通信原理》期末考试(A卷)含答案
学院_______________________ 系别____________ 班次_____________ 学号__________ 姓名________________………….……密…..……….封……..……线………..…以………..…内………....答…………...题…………..无…….….效…..………………..电子科技大学《通信原理》期末考试（A 卷）1. 填空:1) PCM 编码的三个基本步骤 是__抽样_, ___量化__, __编码___. 2) 通信系统包含哪三个子系统: __发射机__ , 信道 and _接收机__. 3) 设模拟信号为 )1000042sin()(t t t s ππ+⨯=, 其最小抽样频率为 _20kHz_.4) 若数据速率是 100kbps,单极性 NRZ 信号, 双极性 NRZ 信号及Manchester NRZ 信号 的第一零点带宽分别为__100kHz___, __100kHz _, ___200kHz __.5) 对于无 ISI 得情形, 若升余弦滚降滤波器的6dB 带宽, f 0, 为 50kHz, 则通信系统的符号速率是 _100k _ baud.6) 若 AM 信号表达式为)100002cos()]42cos(5.01[5)(t t t s ⨯⨯+=ππ, 则该信号的复包络是_)]42cos(5.01[5)(t t g ⨯+=π_.7) 对于非均匀量化两个主要的压扩律 是: __u __律, ___A__律. 8) 设输入信号为 0()[cos][()-(-)] s t t t U t U t T . 当输入噪声为白噪声, 匹配滤波器的冲击响应为h(t)=___)]()())[(cos()(00000T t t U t t U t t t t +-----ω ___，最佳采样时刻为 =0t __T __.9) 若调制信号的带宽为 50kHz, 则AM 、DSB-SC 、SSB 信号的传输带宽分别为 _100kHz__, __100kHz__, __200kHz__.10) 若 s(t)的PSD 为 []22222)4(12cos 16)(f T fT T A f b b b c g -=ππp ,其第一零点带宽是 2. 设()m t 是 单极性 NRZ 信号，数据速率为 1000 bits/s (其峰值为 1V ,二进制 1 码或 0码 发生的概率均为 1/2). OOK 信号表达式为 ()()cos c c s t A m t t ω=, 其中1c A V =. (a) 计算 OOK 信号的PSD. 设载波频率为 10 KHz. (b) 画出OOK 信号的 PSD 并给出第一零点带宽的值. (c) 计算 OOK 信号的频谱效率.Solution:(a) ()()cos c c s t A m t t ω= , m (t )为单极性 ()()c g t A m t =()sin ()()b b b b fT t f t F f T T fT ππ⎛⎫=↔= ⎪⎝⎭∏22:01,2210,2c n n kn For k A a a a and I +=⎧⎪⎪===⎨⎪⎪⎩依概率依概率222111(0)()0222cn n i i c i AR a a P A ===⨯+⨯=∑:0For k ≠,()[][][]24cn n k n n k AR k E a a E a E a ++===()22,02,04c c A k R k A k ⎧=⎪⎪=⎨⎪≠⎪⎩ ()()()222221()1()4111()()444sin sin sin c b g n bb c b bbA T n bP f f T T bA T T b b f f T bbfT fT fT fT fT fT δδδππππππ∞=-∞⎡⎤=+-⎢⎥⎣⎦⎡⎤=+=+⎢⎥⎣⎦⎛⎫∑⎪⎝⎭⎛⎫⎛⎫⎪ ⎪⎝⎭⎝⎭3311/1010b T R ms === ()()221()[()()]41()()16sin ()sin ()()()s g g c c b bc cP f P f f P f f c b c b T T f f f f c b c b f f T f f T f f T f f T δδππππ=-+--⎡⎤⎢⎥=++-++⎢⎥⎢⎥⎣⎦-+⎛⎫⎛⎫ ⎪ ⎪-+⎝⎭⎝⎭(b)2null B KHz =(c) 1/1//22null R Kb s b s Hz B KHz η===3. 某模拟信号，其带宽为3400Hz ，要在一个 PCM 系统中传输。
电子科技大学微机原理2012-2013期末A卷及答案
………密………封………线………以………内………答………题………无………效……电子科技大学2012-2013学年第 1 学期期 末 考试 A 卷课程名称：微处理器系统结构与嵌入式系统设计 考试形式：一本书开卷 考试日期：2013年1月16日 考试时长：120分钟平行班课程成绩构成：平时 10 %， 期中 15 %， 实验 15 %， 期末 60 % 英才班课程成绩构成：平时 30 %， 期中 0 %， 实验 20 %， 期末 50 % 本试卷试题由 3 部分构成，共 4 页。
注意：请将第一、二题答案填入指定位置。
一、单选题答案（共30分，共30空，每空1分）二、填空题答案（共28分，共28空，每空1分）1. ① 存储器 ② I/O 端口 （可交换顺序）2. ① 一条机器指令由一段微程序来解释执行3. ① 指令 ② 总线4. ① IRQ ② FIQ （可交换顺序）5. ① 1.78 ② 44.94 ③ 2.886. ① 1100 0011 0000 1010 ② 1010 0010 1001 1010 ③ 1 ④ 1 ⑤ 1 ⑥ 0 （说明：该题评阅时按上述答案给分，但实际ARM 是32位CPU ，基本没有同学考虑到）7. ① 非流水线执行时间相对流水线执行时间之比8. ① 一段时间内，计算机工作时所需的指令和数据总是集中存放在临近地址的存储单元 9. ① 复位 ② 电源 ③ 时钟 ④ 存储系统 ⑤ 调试接口 （可交换顺序） 10. ① 异常 （或中断） 11. ① BIC ② ORR12. ① =ULCON0(或=0x50000000) ② #0x2B （或#0xAB ）………密………封………线………以………内………答………题………无………效……一、单选题（共30分，共30空，每空1分）1.以下常用总线标准中，不属于片内总线的是（）。
A、Core ConnectB、AMBAC、AvalonD、SATA2.计算机系统中，以下不属于“异常”的是（）。
电子科技大学2012-2013学年第 一 学期期 末 考试 A 卷答案
………密………封………线………以………内………答………题………无………效……电子科技大学2012-2013学年第 一 学期期 末 考试 A 卷答题卡一．判断题二．选择题一、判断题（正确的打“√”，错误的打“×”，共10分，共10小题，每小题1分） 1.考虑某消费者消费价格均为正的良好总商品。
如果其中一种商品的价格下降了，而收入和另一种商品价格保持不变，那么预算集范围缩小了。
2. 效用函数 U(x, y) = 10 + y2 + x 代表凸偏好。
3.具有{}y x y x u ,max ),(=效用函数的消费者认为商品x 和y 为完全互补品。
4.如果消费者不具有良性偏好，无差异曲线与预算线的切点仍可能为最优选择。
5.对于一个消费者来说，它的需求曲线在所有价格水平下都向上倾斜是不可能的。
6.某消费者预算保持不变，如果随着价格的改变他的状况变好了，那么可以判断在旧的价格下新的消费束的花费比旧的消费束要多。
7.等成本线上的产量最大化选择点也是平均成本的最低点。
8.在古诺模型中，每家厂商的反应函数是在假定其竞争对手价格保持不变的前提下进行利润最大化决策所得出的。
9. 在一个纳什均衡策略组合中，每个博弈者的策略一定都是占优策略。
10..根据瓦尔拉斯法则，从初始配置出发，市场最后一定能达到竞争性均衡。
二、单项选择题（共30分，共 15小题，每小题2 分）1.小王把他全部收入花在香蕉和西瓜上，他能够消费11只香蕉和4只西瓜，也能偶消费3只香蕉和8只西瓜。
香蕉每只6元，请问小王的收入是多少？………密………封………线………以………内………答………题………无………效……A.115元B.105元C.114元D.119元2.小张消费商品x和y，他的无差异曲线可以用方程y=k/(x+7)表示，更大的k值表示更偏好的无差异曲线。
下列哪项是正确的？A.小张喜欢商品x，讨厌yB.小张偏好（12,16）于（16,12）C.小张偏好（8,5）于（5,8）D.小张喜欢商品y，讨厌商品x3.小王的效用函数为{}xmin2),+(==。
杭电离散数学期末总复习_离散数学期末考试
m3 m2 P∧Q P∧Q F F F F F T T F
m1 m0 P∧Q P∧Q F T T F F F F F
6)大项及其性质. M0 M1 M2 M3 P Q P∨Q P∨Q P∨Q P∨Q 00 F F F T T T 01 F T T F T T 10 T F T T F T 11 T T T T T F
((A∧B)C)∧D∧(C∨D) A∨B 3.反证法: ⑴ (A∨B) P(假设前提) ⑵A∧B T⑴ E9 ⑶(A∧B)C P ⑷C T ⑵ ⑸ I11 ⑸ D P ⑹C∨D P ⑺C T ⑻⑼ I10 ⑻C∧C T ⑷ ⑺ I9
第二章 谓词逻辑 1.准确掌握有关概念. 2.会命题符号化.(如P60题(2)) 3.掌握常用的等价公式和永真蕴涵式.包括: 带量词的公式在论域内展开式,量词否定,量词辖域扩充, 量词分配公式. 4.会用等价公式求谓词公式的真值.(如P66题(3)) *5.会写前束范式 6.熟练掌握谓词逻辑推理.
第四章 二元关系 1.关系的概念,表示方法. 2.二元关系的 性质的定义, 熟练掌握性质的判断及证明. 3.掌握关系的复合,求逆及闭包运算(计算方法及有关性质) 4.掌握等价关系的判断,证明,求等价类和商集. *4.掌握相容关系定义,简化图和简化矩阵,相容类,最大相 容类,完全覆盖. 5.偏序关系的判断,会画Hasse图,会求一个子集的极小(大) 元,最小(大)元,上界与下界,最小上界及最大下界. 第六章 函数 1.函数的定义. 2.函数的类型, 会判断,会证明. 3.会计算函数的复合(左复合),求逆函数.知道有关性质. *4.了解集合的特征函数,了解集合的基数,可数集合.
5.等价公式的证明,记住常用的公式. 方法1.列真值表. 方法2.用公式的等价变换. 例如:证明 P(QR)(P∧Q)R P(QR)P(QR) (PQ)R (PQ)∨R (P∧Q)R
双语离散数学期末考试_2012年春季_试卷A
电子科技大学2011 -2012学年第 2学期期 末 考试 A 卷课程名称： 离散数学 考试形式： 闭卷 考试日期： 2012 年 6 月 日 考试时长：120分钟 课程成绩构成：平时 10 %， 期中 20 %， 实验 0 %， 期末 70 % 本试卷试题由____ _部分构成，共_____页。
I.Multiple Choice (15%)1. (⌝p ∧q)→(p ∨q) is logically equivalent toa) T b) p ∨q c) F d) ⌝ p ∧q （ ） 2. If P(A) is the power set of A, and A = , what is |P(P(P(A)))|?a) 4 b) 24 c) 28 d) 216（ ） 3. Which of these statements is NOT a proposition?a) Tomorrow will be Friday. b) 2+3=4.c) There is a dog. d) Go and play with me.（ ）4. The notation K n denotes the complete graph on n vertices. K n is the simple graph thatcontains exactly one edge between each pair of distinct vertices. How many edges comprise a K 20?a) 190 b) 40 c) 95 d) 380（ ）5. Suppose | A | = 5 and | B | = 9. The number of 1-1 functions f : A → B isa) 45 b) P (9,5). c) 59 d) 95（ ）6. Let R be a relation on the positive integers where xRy if x divides y . Whichof the following lists of properties best describes the relation R ?a) reflexive, symmetric, transitive b) reflexive, antisymmetric, transitive c) reflexive, symmetric, antisymmetric d) symmetric, transitive （ ）7. Which of the following are partitions of }8,7,6,5,4,3,2,1{=U ?a) }8,7,6,5,4,3{},3,2,1{},1{ b) }8,7,6,5,4,3{},3,2{},1{c) }8,6,5{},3,2{},7,4,1{ d) }8,7,6,5,4{},3,2{},2,1{（ ） 8. The function f(x)=3x 2log(x 3+21) is big-O of which of the following functions? a) x 3 b) x 2(logx)3 c) x 2logx d) xlogx （ ） 9.In the graph that follows, give an explanation for why there is no path from a back to a that passes through each edge exactly once.a) There are vertices of odd degree, namely {B,D}. b) There are vertices of even degree, namely {A,C}. c) There are vertices of even degree, namely {B,D}. d) There are vertices of odd degree, namely {A,C}.（ ） 10. Which of the followings is a function from Z to R ?a) )1()(-±=n n f . ` b) 1)(2+=x x f . c) x x f =)( d) 11)(2-=n n fII. True or False (10%)（ ） 1. If 3 < 2, then 7 = 6. （ ） 2. p ∧ (q ∨ r)≡ (p ∧ q) ∨ r（ ） 3. If A , B , and C are sets, then (A -C )-(B -C )=A -B . （ ） 4. Suppose A = {a ,b ,c }, then {{a }} ⊆ P (A ).（ ） 5. ()100h x x =+is defined as a function with domain R and codomain R.（ ） 6. Suppose g : A → B and f : B → C , where f g is 1-1 and f is 1-1. g must be 1-1? （ ） 7. If p and q are primes (> 2), then p + q is composite .（ ） 8.If the relation R is defined on the set Z where aRb means that ab > 0, then R is an equivalence relation on Z .（ ） 9. Every Hamilton circuit for W n has length n .（ ） 10. There exists a simple graph with 8 vertices, whose degrees are 0, 1, 2, 3, 4, 5, 6, 7.III. Fill in the Blanks (20%)1. Let p and q be the propositions “I am a criminal” and “I rob banks”. Express in simple English the propositi on “if p then q”: .2. P (x ,y ) means “x + 2y = xy ”, where x and y are integers. The truth value of ∃x ∀yP (x ,y ) is .3. T he negation of the statement “No tests are easy.” is .4. If 11{|}i A x x R x i i =∈∧-≤≤ then 1i i A +∞=is .5. Suppose A = {x , y }. Then ()P A is .6. Suppose g : A →A and f :A →A where A ={1,2,3,4},g = {(1, 4), (2,1), (3,1), (4,2)} andf ={(1,3),(2,2),(3,4),(4,2)}.Then fg = . 7.The sum of 2 + 4 + 8 + 16 + 32 + ... + 210 is .8. The expression of gcd(45, 12) as a linear combination of 12 and 45 is .9.There are permutations of the seven letters A,B ,C ,D ,E ,F have A immediately to the left of E .10. If G is a planar connected graph with 18 vertices, each of degree 3, then G has _ __regions. IV. Answer the Questions (32%)：1. Determine whether the following argument is valid: p → r q → r q ∨ ⌝r ________∴ ⌝p2. S uppose you wish to prove a theorem of the form “if p then q ”. (a) If you give a direct proof, what do you assume and what do you prove? (b) If you give an indirect proof, what do you assume and what do you prove? (c) If you give a proof by contradiction, what do you assume and what do you prove?3. Prove that A B A B ⋂=⋃ by giving a proof using logical equivalence.4.Suppose f:R→R where f(x) =⎣x/2⎦.(a) If S={x| 1 ≤x≤ 6}, find f(S).(b) If T={3,4,5}, find f-1(T).e the definition of big-oh to prove that5264473n nn+--is O(n3).6.Solve the linear congruence 5x≡ 3 (mod 11).e the Principle of Mathematical Induction to prove that131 1392732nn+-++++...+=for alln≥ 0.8.Draw the directed graph for the relation defined by the matrix1111 0111 0011 0001⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦.V. (6%) Without using the truth table, show that the following are tautologiesa) [⌝p ∧(p ∨q)]→q b) [p ∧(p →q)]→qVI. (6%) Devise an algorithm which will find the minimum of n integers. What is the worst casetime complexity of this algorithm?VII. (5%) Give the definition of a transitive relation, and Prove or disprove that the union oftwo transitive relations is transitive.VIII.(6%) The pseudo-code of Prim’s algorithm is given as following:Procedure Prim(G: connected weighted undirected graph with n vertices)T := a minimum-weight edgefor i := 1 to n 2begine := an edge of minimum weight incident to a vertex in T and notforming a simple circuit in T if added to TT := T with e addedPrint eend {T is a minimum spanning tree of G}(a)Find a minimum spanning tree using Prim’s algorithm given above. For every iterative in for-loop, list theresult for “Print e” statement.(b)Compute the total weight of the spanning tree.。
电子科技大学《通信原理》期末试卷(C卷)含答案
学院_______________________ 系别____________ 班次_____________ 学号__________ 姓名________________………….……密…..……….封……..……线………..…以………..…内………....答…………...题…………..无…….….效…..………………..电子科技大学《通信原理》期末试卷（C 卷）1. Blank Filling:1) The three basic operations of the generation for PCM signal are __sampling_, ___quantizing__, __coding___.2) All communication systems involve three main subsystems: __Transmitter __ , channel and _Receiver __.3) If the analog signal is )1000042sin()(t t t s ππ+⨯=, the minimum sampling rate is _20kHz_.4) If the data rate is 100kbps, the first null bandwidth for the unipolar NRZ signal, the polar NRZ signal and the Manchester NRZ signal are__100kHz___, __100kHz _, ___200kHz __ separately.5) For no ISI, and the 6 dB bandwidth of the raised cosine-rolloff filter, f 0, is designed to be 50kHz, the symbol rate of communication system is _100k _ baud.6) If the AM signal is )100002cos()]42cos(5.01[5)(t t t s ⨯⨯+=ππ, its complex envelope signal is _)]42cos(5.01[5)(t t g ⨯+=π_.7) Two main compression laws for non-uniform quantizing are: __u __-law, ___A__-law. 8) Assume that the input signals is()[cos][U(t)-U(t-T)] s t t t . When the input noise is white, the impulseresponse of the matched filter h(t)=___)]()())[(cos()(00000T t t U t t U t t t t +-----ω ___and the optimum sampling time =0t __T __.9) If the bandwidth of modulating signal is 50kHz, the transmission bandwidth for AM 、DSB-SC 、SSB 、signals are _100kHz__, __100kHz__, __200kHz__ separately.10) If the PSD of s(t) is []22222)4(12cos 16)(f T fT T A f b b b c g -=ππp2. Assume()m t is a unipolar NRZ signal occurring at a rate of 1000 bits/s (The peak values of which is1V , and the probability of obtaining either a binary 1 or 0 is 1/2). The OOK signal is()()cos c c s t A m t t ω=, where 1c A V =.(a) Evaluate the PSD of the OOK signal. Assume a carrier frequency of 10 KHz.(b) Plot the PSD of the OOK signal and indicate the value of the first null-to-null bandwidth.(c) Evaluate the spectral efficiency for the OOK signal.Solution:(a) ()()cos c c s t A m t t ω= , m (t )为单极性 ()()c g t A m t =()sin ()()b b b b fT t f t F f T T fT ππ⎛⎫=↔= ⎪⎝⎭∏22:01,2210,2c n n kn For k A a a a and I +=⎧⎪⎪===⎨⎪⎪⎩依概率依概率222111(0)()0222cn n i i c i AR a a P A ===⨯+⨯=∑:0For k ≠,()[][][]24cn n k n n k AR k E a a E a E a ++===()22,02,04c c A k R k A k ⎧=⎪⎪=⎨⎪≠⎪⎩ ()()()222221()1()4111()()444sin sin sin c b g n bb c b bbA T n bP f f T T bA T T b b f f T bbfT fT fT fT fT fT δδδππππππ∞=-∞⎡⎤=+-⎢⎥⎣⎦⎡⎤=+=+⎢⎥⎣⎦⎛⎫∑⎪⎝⎭⎛⎫⎛⎫⎪ ⎪⎝⎭⎝⎭3311/1010b T R ms === ()()221()[()()]41()()16sin ()sin ()()()s g g c c b bc c P f P f f P f f c b c b T T f f f f c b c b f f T f f T f f T f f T δδππππ=-+--⎡⎤⎢⎥=++-++⎢⎥⎢⎥⎣⎦-+⎛⎫⎛⎫⎪ ⎪-+⎝⎭⎝⎭(b)2null B KHz =(c) 1/1//22null R Kb s b s Hz B KHz η===3. An analog signal with the bandwidth of 3400Hz is to be transmitted over a PCM system. The number of quantizing steps is 64. Assume the channel bandwidth is 15KHz. (a) Find the minimum sampling rate.(b) Determine the minimum bit rate required in the PCM signal.(c) Assume the overall equivalent system transfer function is of the raised cosine-rolloff type. When we use 4-level signal for sending data, is there ISI ？Solutiona.2234006800s f B Hz ==⨯=b.6466680040.8/s M n R nf Kb s=→===⨯=c.20.42302channel RD KBd B KHz ==<= ()230115110.471220.4T T D B B r KHz r D =+=→=-=-=So, we can realizing zero ISI transmission.4. An modulated signal is described by the question ]2sin 102cos[100)(t f t f t s m c ππ+= where fc=10MHz, fm=1000Hz.(a) Evaluate the percentage of AM.(b) Assume that s(t) is an FM signal, find the frequency modulation index and the transmission bandwidth.(c) Assume that s(t) is a PM signal, find the peak phase deviation and the phase modulation index. (d) Assume that s(t) is fed into a 50-Ωload, evaluate the average power in the load.解：(a) 0(b) Hz dtt d F tf t m 10000})(max{212sin 10)(==∆=θππθ kHz f B f F m f T m f 221000)110(2)1(210100010000=⨯+=+===∆=ββ(c) 1010=∆==∆θβθP rad(d) W A P c 10050210050222=⨯=⨯=5. If the binary baseband signal is passed through a raised cosine-rolloff filter with 50% rolloff factor and is then modulated onto a carrier, the data rate is 32 kbits/s. Evaluate (a) The absolute bandwidth of resulting QPSK signal.(b) The approximate bandwidth of a resulting FSK signal when the mark frequency is 50 kHz and the space frequency is 60kHz. Solution:(a)(1)(1)22basebandD RB r r l=+=+ where 2=l for QPSK 320002(1)(10.5)242T basebandR B B r kHz l ==+=+=(b)1222()(1)T baseband B F B f f R r =∆+=-++(6000050000)32000(10.5)58kHz =-+⨯+=6. A binary data stream {1011010010100001001011} is transmitted using π/4 QPSK signal , and the first bit on the left is first passed into the multilevel system, the data rate is R. Assume the phase shift π/4 for 00, -π/4 for 01, +3π/4 for 10, -3π/4 for 11. Evaluate (a) calculate the carrier phase shift.(b) what is the minimum transmission bandwidth?(c) Assume the baseband signal is pass through a raised cosine-rolloff filter with a 50% rolloff factor first, what is the transmission bandwidth? (d) What is the spectrum efficiency? Solution:(a) The carrier phase shift isFrom the description of π/4 QPSK,use the table shown at the following:So the carrier phase is, assume the initial phase is 0(b) The minimum transmission bandwidth is:When the waveshape is sinc(x) type function, we can get the minimum transmission bandwidth02(1)24T r D RDR B r ===+=(c) Assume the baseband signal is pass through a raised cosine-rolloff filter with a 50% rolloff factor first,the transmission bandwidth is:0.523(1)28T r D RDR B r ===+= (d) The spectrum efficiency isFor (b) : 44TR RR B η===For (c) : 8338T R R R B η===7. The communication system is shown in the following figure, find the (S/N) for the receiver output.( ()cos 210m t π=⨯()out m t 10cos 210t ⨯n(t)Noise PSD 1mW/HzSolution:()10cos cos c s t t t ω=Ω23310/425221040002()210400082252 6.2588in T in T out inS WB HzN f B WG S S G dB N N ϕ-===⨯⨯==⨯⨯=⨯⨯==⎛⎫⎛⎫==⨯=≈ ⎪ ⎪⎝⎭⎝⎭8.The digital receiver is shown in the following figure. Gaussian noise having a PSD of 1mW/Hz plus a polar signal with a peak level of A=10V is present at the receiver input. Find the expression for BER as a function Q( ).Digital outputSolution:()010230010 102000 22200010425e s V s V N B Hz B W P Q Q Q σ-==-===⨯⨯====9.Suppose that the PSD of five digital transmission systems with bit rate of 1200bps are as following figures(a) to (e). Please answer the questions in the form below.Spectral efficiency Figures Baseband or bandpass? M-level? B Null(Hz)(bps/Hz)Baseband 4-ary 600 2Baseband Binary 1200 1Baseband Binary 900 4/3bandpass Binary 2400 0.5。
电子科技大学-信息论课件及历年考题
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电子科技大学离散期末试题13年
电⼦科技⼤学离散期末试题13年电⼦科技⼤学2012 -2013学年第 2学期期末考试 A 卷课程名称：离散数学考试形式：闭卷考试⽇期： 2013 年⽉⽇考试时长：120分钟课程成绩构成：平时 10 %，期中 20 %，实验 0 %，期末 70 % 本试卷试题由____ _部分构成，共_____页。
I.Multiple Choice (15%, 10 questions, 1.5 points each)（） 1.Which of these propositions is not logically equivalent to the other three? a) (p → q) ∧ (r → q) b) (p ∨ r) → q c) (p ∧ r) → q d) ?q → (?p ∧?r) （） 2. Suppose A = {x ,y } and B = {x ,{x }}, then we don ’t havea) x ∈ B b)? ∈ P (B ). c) {x } ? A - B . d)| P (A ) | = 4.（） 3.Suppose the variable x represents students, F (x ) means “xis a freshman”, and M (x ) means “x is a math major”. Match the statement “??x (M (x ) ∧ ?F (x ))” with one of the English statements below:A. Some freshmen are math majors.B. Every math major is a freshman.C. No math major is a freshman.D. Some freshmen are not math majors. （） 4.The two's complement of -13 isA. 1 0011.B. 0 1101 a)C. 1 0010D. 0 1100（） 5.The chromatic number of a graph is the least number of colors needed for a coloring of this graph. The chromatic number of the graph G isa) 2 b)3c)4 d) 5（） 6. The function f(x)=x 2log(x 3+100) is big-O of which of the following functions? a) x 2 b)x 2logx c) x(logx)3 d) xlogx（） 7.Which of the following complete graphs is planar?a) K 5 b) K 3,3 c) K 6 d) K 4（） 8.Which of the following set is uncountable ?a) The set of real numbers between 172 and 173. b) The set of integersc) The set of integers not divisible by 3. d) The union of two countable sets. （） 9.How many numbers must be selected from the set {2,4,6,8,10,12,14,16,18,20} in order to guarantee that at least one pair adds up to 22?a) 5 b) 6 c) 7 d) 8（） 10. Which of the following is false?a) {x}?{x} b) {x}∈{x, {x}} c) {x}?P({x}), where P({x}) is the power set of {x} d) {x}?{x, {x}}II. True or False (10%, 10 questions, 1 point each)（） 1. The proposition ((p → q ) ∧ ?p ) → ?q is a tautology. （） 2. If pigs can fly, then it will be raining tomorrow. （） 3. Suppose A = {a ,b ,c }, then {{a }} ? P (A ).（） 4. “My daughter visited Europe last week” implies the conclusion “Someone visited Europe last week”.（） 5. For all integers a ,b ,c ,d , if a | b and c | d , then (a + c )|(b + d ). （） 6. For all real numbers x and y , ?x - y ? = ?x ? -y .（）7. ()h x =is defined as a function with domain R and codomain R.（） 8. There exists a simple graph with 6 vertices, whose degrees are 2,2,2,3,4,4.（） 9.If G is a simple graph with n ≥3 vertices such that the degree of every vertex in G is at least n/2, then G has a Hamilton circuit.（） 10. Let P (m ,n ) be the statement “m|n ,” where the u.d . of m and n is the set of positive integer.Then ),(n m mP n ??holds.III. Fill in the Blanks (20%, 10 questions, 2 points each)1. Suppose A = {x | x ∈ Z and x 2 < 10}. Then ()P A is .2.If 11{|}i A x x R x i i =∈∧-≤≤ then 1i i A +∞= is .3.Give a relation on {a ,b ,c } that is reflexive and transitive, but not antisymmetric..4.Suppose g : A → B and f : B → C where A = B = C = {1,2,3,4}, g = {(1,4), (2,1), (3,1), (4,2)} and f = {(1,3),(2,2),(3,4),(4,2)}. Then fg =. 5.W rite the negation of the statement “No tests are easy ” in good English:. 6. The expression of GCD(45,12) as a linear combination of 12 and 45is . 7.There are permutations of 7letters A ,B ,C ,D ,E ,F ,G have A immediately to the left of E .8. If f (n ) = f (n - 1) / f (n - 2), f (0) = 2, f (1) = 5, Then f (2) = . 9.The negation of the statement ?x ?y (xy = 0) is.10. Let }|),{(},|),{(2221b a b a R b a b a R ≠?∈=≤?∈=Then 21R R ? is.IV. Answer the Questions (35%,7 questions, 5 points each)：1. Write the truth table for the proposition ：?(r → ?q ) ∨ (p ∧ ?r )2. Suppose f : R → R where f (x ) = ?x /2?.(a) If S = {x | 1 ≤ x ≤ 6}, find f (S ). (b) If T = {3,4,5}, find f -1(T ).3. Find the matrix that represents the relation of R on {1,2,3,4} where aRb means | a - b | ≤ 1. Use elements in the order given to determine rows and columns of the matrix.4.Prove that is a tautology using propositional equivalence and the laws of logic.5.Encrypt the message “HELP” by translating the letters into numb ers, applying the encryption functionf (p) = (3p+ 7) mod 26, and then translating the numbers back into letters.6.Solve the linear congruence 5x≡ 3 (mod 11).7.Determine whether these two graphs are isomorphic. If they are isomorphic, give a one-to-one and onto function f from the one on the left to the one on the right.A = {(x, y) | x, y ∈R}?{(0, 0)}.Define a relation on A by the rule: (a, b)R(c, d) (a, b) and (c, d) lie on the same line through the origin.(a) Prove that R is an equivalence relation.(b) Describe the equivalence classes arising from the equivalence relation R in part (a).(c) If A is replaced by the entire plane, is R an equivalence relation?VI. (7%) Determine whether this argument is valid: Lynn works part time or full time.If Lynn does not play on the team, then she does not work part time. If Lynn plays on the team, she is busy. Lynn does not work full time. Therefore, Lynn is busy.VII. (7%) The pseudo-code of Dijstra ’s algorithm is given as following: Procedure Dijkstra (G : weighted connected simple graph) for i :=1 to n L (v i ):=∞ ; L (a ):=0 ; S :=Φ;While z S Beginu := a vertex not in S with L (u ) minimal; S :=S ∪{u }; Print S for all vertices v not in S if L (u )+w (u,v )(a) Find a shortest path between b and f using Dijstra ’s algorithm given above. For every iterative inwhile-loop, list S ’s elements for “Print S ” statement. (b) Compute the length of the shortest path.。
电子科技大学离散考题13年B
电子科技大学2012 -2013学年第 2学期期 末 考试 B 卷课程名称： 离散数学 考试形式： 闭卷 考试日期： 2013 年 月 日 考试时长：120分钟 课程成绩构成：平时 10 %， 期中 20 %， 实验 0 %， 期末 70 % 本试卷试题由____ _部分构成，共_____页。
I.Multiple Choice (15%, 10 questions, 1.5 points each)（ ） 1.Which of these propositions is not logically equivalent to the other three? a) (p → q) ∧ (r → q) b) (p ∨ r) → qc) (p ∧r) → q d) The contrapositive of ¬q → (¬p ^ ¬r) （ ） 2. Suppose A = {a ,b ,c } , then we don ’t havea) {b ,c } ∈ P (A ) b) {{a }} ⊆ P (A ) c) ∅ ⊆ A d) {a ,b } ∈ A ⨯ A .（ ） 3.If 1111011100110001R ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦M , then R is not (a) reflexive (b) antisymmetric (c) transitive. (d) symmetric（ ） 4. Suppose R 1 and R 2 be transitive on A. Which of the following is transitive?a) a) R 1∪R 2 b) R 1oR 2 c) R 2oR 1 d) R 1∩R 2 （ ） 5.The chromatic number of a graph is the least number of colors needed for a coloring of this graph. The chromatic number of the graph H isa) 2 b)3 c)4 d) 5（ ） 6.If all sets are finite, which of the following must be true?a) If a function is bijective, its domain and co-domain have the same cardinality. b) If a function is one-to-one, its domain and co-domain have the same cardinality. c) If a function is onto, its domain and co-domain have the same cardinality.a) d) If a function is neither one-to-one nor onto, its domain and co-domain do not havethe same cardinality.（ ） 7.Which of the following set is uncountable?a) The set of real numbers between 172 and 173. b) The set of integers.c) The set of integers not divisible by 3. d) The union of two countable sets.（ ） 8.Which of these propositions is false (the domain is the set of real numbers)? a) ∀x ∃y(x ≠= 0 →x · y = 1) b) ∃y ∀x(x + y = x) c) ∀x ∀y[(x ≠= y) → ∃z(x < z < y ∨ y < z < x)] d) ∀x ∀y ∃z(x < z < y)（ ） 9.Which of the following does NOT belong to S ? S is a collection of strings of symbols. It is recursively defined by 1) a and b belong to S ; 2) if string x belongs to S , so does Xb . a) abbb b) bba c) bb d) ab（ ） 10. Which of the following complete graphs is planar?a) K 5 b) K 3,3 c) K 4 d) K 6II. True or False (10%, 10 questions, 1 point each)（ ） 1. The following sentence is a proposition: “ x+ 1 =9.” （ ） 2. If 1 + 2 = 3 or 1 + 2 = 5 then 2 + 2 =4 and 2 +3 = 6. （ ） 3. A ⋃ (B ⋂ C ) ⊇ (A ⋃ B ) ⋂ C .（ ） 4. The premises "No juniors left campus for the weekend" and "Some math majors are not juniors" imply the conclusion "Some math majors left campus for the weekend." （ ） 5. For all integers a ,b ,c ,d , if a | b and c | d , then (ac ) | (b + d ).（ ） 6. For all real numbers x and y , ⎣x + ⎣x ⎦ + 0.5⎦ = ⎣2x +0.5⎦. （ ） 7. h : R → Ris a function, where ()h x =（ ） 8. There exists a simple graph with 8 vertices, whose degrees are 0,1,2,3,4,5,6,7.（ ） 9.Suppose g : A → B and f : B → C , where f g is 1-1 and f is 1-1. g must be 1-1?（ ） 10. Let P (m ,n ) be the statement “m|n ,” where the u.d . of m and n is the set of positive integer.Then ),(n m nP m ∀∃holds.III. Fill in the Blanks (20%, 10 questions, 2 points each) 1. The size of {x | x ∈ N and 9x 2 - 1 = 0} is .2. 61((2)2).iii =--∑= .3. Give a relation on {1,2} that is symmetric and transitive, but not reflexive. .4.Suppose g : A → B and f : B → C where A = B = C = {1,2,3,4}, g = {(1,4),(2,1),(3,1),(4,2)} and f = {(1,3),(2,2),(3,4),(4,2)}. Then g f . =.5.Write the negation of the statement “Roses are red and violets are blue ” in good English: .6.Suppose the variable x represents people, and F (x ): x is friendly; T (x ): x is tall. Write thestatement “All tall people are friendly” using these predicates and any needed quantifiers: . 7. The best big-oh function for the function g (n ) =4332423n n n n--- is . 8. If f (n ) = f (n - 1) / f (n - 2), f (0) = 2, f (1) = 5, Then f (3) = . 9.The negation of the statement ∃x ∀y (xy = 0) is.10. Let }|),{(},|),{(2221b a b a R b a b a R ≥ℜ∈=>ℜ∈= Then 12R R -is: . IV. Answer the Questions (32%)：1. Prove that (q ∧ (p → ⌝q )) → ⌝p is a tautology using propositional equivalence and the laws of logic.2. Suppose f : R → R where f (x ) = ⎣x /2⎦. (a) If S = {x | 1 ≤ x ≤ 6}, find f (S ). (b) If T = {3,4,5}, find f -1(T ).3.On the island of knights and knaves you encounter two people. A and B. Person A says, "B is a knave." Person B says, "At least one of us is a knight." Determine whether each person is a knight or a knave.⋂=⋃by giving a containment proof (that is, prove that the left side is a 4.Prove that A B A Bsubset of the right side and that the right side is a subset of the left side).5.Encrypt the message “stop” by tran slating the letters into numbers, applying the encryption function f (p)= (p+ 6) mod 26, and then translating the numbers back into letters.6.Express gcd(450,120) as a linear combination of 120 and 450.7.Determine whether these two graphs are isomorphic, and explain the reason.defined as R-1= {(b, a) | (a, b) ∈R}, is also an equivalence relation on A? Prove youranswer.VI. (7%) Use powers of the adjacency matrix to find the following numbers of paths in thisdigraph:(a) paths from e to c of length 3. (b) paths from e to e of length 4. (c) paths from f to b of length 6.VII. (7%) Suppose we have:“Every student in this class is a Junior.”“Every Junior in this class passed the final exam.” “Allen is a student in this class.”Explain why we can draw the conclusion “Allen passed the final exam.”。
电子科技大学2011至2012 学年第2学期通信技术与系统期末考试
………密………封………线………以………内………答………题………无………效……电子科技大学2011 至2012 学年第2 学期期末考试通信技术与系统课程考试题 A 卷（120分钟）考试形式：一页纸开卷考试日期200 8 年6月日课程成绩构成：平时10 分，期中分，实验分，期末90 分1．为什么要调制？至少说出三个理由。
答：通过调制可以将基带信号转换成适合信道传输的频带信号；实现信道复用；减少干扰，提高系统的抗干扰能力；实现传输带宽与信噪比之间的互换。
2．通信系统中的可靠性和有效性指的是什么？对于模拟和数字通信系统分别用什么指标来度量？答：可靠性是指接收信息的准确程度，也就是传输的质量；有效性是指在给定的信道内所传输的信息内容的多少，也就是传输的速度。
模拟通信的有效性用传输频带来度量，可靠性用信噪比来度量；数字通信的有效性用传输速率来度量，可靠性用差错率来度量。
3．什么是码间干扰？简述码间干扰形成的原因。
答：接收信号除本码以外的其他波形在当前抽样时刻上的影响称为码间干扰。
码间干扰是由相临脉冲的拖尾叠加产生，本质上是由信道带限特性引起。
4．设二进制原代码为10000101000001，（设第一个信码为高电平，前一个破坏脉冲为高电平），写出此代码的HDB3码。
答：当第一个信码为高电平，前一个破坏脉冲为高电平时，原代码为：1 0 0 0 0 1 0 1 0 0 0 0 0 1HDB3码：B+ C- 0 0 V- B+ 0 B-C+0 0 V+0 B-对应电平：+1 -1 0 0 -1 +1 0 -1 +1 0 0 +1 0 -1（其中：B为信码；C为补信码；V为破坏点；±为电平极性。
）5．简述相干解调和非相干解调的基本原理，试比较两者的特点？。
答：相干解调的基本原理是：在接收端提供一个与接收的已调载波严格同步的本地载波，它与接收信号相乘后取低通分量，得到原始的调制信号；非相干解调的基本原理是：在接收端不需要产生本地的同步载波，而是利用接收的信号本身特点进行包络检波；相干解调主要特点是没有解调门限效应，而非相干解调主要特点是有解调门限效应。
电子科技大学 计算机系统结构2013- 2014年期末考试题目回忆版本
电子科技大学计算机系统结构2013- 2014年期末考试题目回忆版本题型：选择十分每个一分简答30分综合题目60分选择题：系列机是向后兼容的。
评价：四个选项向前向后向上向下尼玛有区别吗？提高组相联的联合度主要是为了解决冲突失效的问题。
RISC体系结构是寄存器-寄存器型的体系。
一个存储系统为每个处理机提供相同的访问时间，是集中共享存储器的特点，这样的结构也称为均匀存储器访问UMA结构（尼玛竟然是第七章的内容）MISD实际代表何种计算机，目前不知道，就是不存在了。
多核芯片组成的处理机系统结构属于MIMD。
（尼玛这个知识点竟然考了两遍）S：add r1 r2,r3S+1: sub r1 r2,r4属于写后写相关。
前瞻执行的最后一步是确认（第四章内容）还有就记不得了。
简答题：机器语言物理机和虚拟机的概念。
（尼玛第一章的内容）TLB快表的作用,TLB快表的内容（两个）。
（第五章）BTB表格的两个内容。
（第四章）RISC的两个重要特点。
（第二章，尼玛我真的不知道那么多特点哪两个重要的）试分析采用哪种设计方案实现求浮点数除法FPMUL对系统性能提高更大。
假定FPMUL 操作占整个测试程序执行时间的12%。
一种设计方案是增加专门的FPMUL硬件，可以将FPMUL操作的速度加快到16倍；另一种设计方案是提高所有FP运算指令的执行速度，使得FP指令的执行速度加快为原来的1.6倍，设FP运算指令在总执行时间中占50%。
（6分）（这个题目就是期中考试题目改了一下）写出三级Cache的平均访问时间的公式。
（题目当然不是这样的，但是这个原理懂了之后，那个小小的计算题就不是问题了，就是给了两级Cache的相关参数，求L1的失效时间，知道公式往里面带入就行了）大题：一题是：如图所示，题目给出了这个图，下面是9分的填空和5分的简答。
简答题：在wb级的时候这一级的作用和各个信号的值作用。
题目中给出的指令是load指令，我这个图是随便在ppt上找的。
西安电子科技大学信号与系统期末真题2
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©西安电子科技大学电路与系统教研中心
信号与系统 电子教案 12、如题12图所示电路系统，以电 感电流iL(t)为输出，则该系统的冲 激响应h(t) = 。
iS(t) 2Ω iL(t) 1H
题12图
13、频谱函数F(jω) = 1 + f(t) = 。
|H(jω )| π ω -5 5 0 -5 (b) 题5图
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信号与系统 电子教案 ___ 6、若f1(t)←→F1(jω)，则F2(jω) = 1 F (j ) e 的原 1 2 2 函数f2(t)等于 (A) f1(2 t +5) (B) f1(2 t -5) (C) f1(-2 t+5) (D) f1[2(t-5)]
1 2
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信号与系统 电子教案
___ 4、对信号f(t) = cos(πt +30o) +2sin(4πt +45o)，当取 样间隔 T 至多为何值时，f(t)就能唯一地由均匀取样样 本f (kT) (k = 0，1，2，…)确定。 (A) 0.25 s (B) 0.5s (C) 1s (D) 2s
h1(t) ∑ f(t) h2(t) (a) 题16图
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©西安电子科技大学电路与系统教研中心
信号与系统 电子教案 （8分）17、周期信号f(t) = 4 + 2cos(
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电子科技大学2012 -2013学年第 2学期期 末 考试 A 卷课程名称： 离散数学 考试形式： 闭卷 考试日期： 2013 年 月 日 考试时长：120分钟 课程成绩构成：平时 10 %， 期中 20 %， 实验 0 %， 期末 70 % 本试卷试题由____ _部分构成，共_____页。
I.Multiple Choice (15%, 10 questions, 1.5 points each)（ ） 1.Which of these propositions is not logically equivalent to the other three? a) (p → q) ∧ (r → q) b) (p ∨ r) → q c) (p ∧ r) → q d) ¬q → (¬p ∧¬r) （ ） 2. Suppose A = {x ,y } and B = {x ,{x }}, then we don ’t havea) x ∈ B b)∅ ∈ P (B ). c) {x } ⊆ A - B . d)| P (A ) | = 4.（ ） 3.Suppose the variable x represents students, F (x ) means “xis a freshman”, and M (x ) means “x is a math major”. Match the statement “⌝∃x (M (x ) ∧ ⌝F (x ))” with one of the English statements below:A. Some freshmen are math majors.B. Every math major is a freshman.C. No math major is a freshman.D. Some freshmen are not math majors. （ ） 4.The two's complement of -13 isA. 1 0011.B. 0 1101 a)C. 1 0010D. 0 1100（ ） 5.The chromatic number of a graph is the least number of colors needed for a coloring of this graph. The chromatic number of the graph G isa) 2 b)3c)4 d) 5（ ） 6. The function f(x)=x 2log(x 3+100) is big-O of which of the following functions? a) x 2 b)x 2logx c) x(logx)3 d) xlogx（ ） 7.Which of the following complete graphs is planar?a) K 5 b) K 3,3 c) K 6 d) K 4（ ） 8.Which of the following set is uncountable ?a) The set of real numbers between 172 and 173. b) The set of integersc) The set of integers not divisible by 3. d) The union of two countable sets. （ ） 9.How many numbers must be selected from the set {2,4,6,8,10,12,14,16,18,20} in order to guarantee that at least one pair adds up to 22?a) 5 b) 6 c) 7 d) 8（ ） 10. Which of the following is false?a) {x}⊆{x} b) {x}∈{x, {x}} c) {x}⊆P({x}), where P({x}) is the power set of {x} d) {x}⊆{x, {x}}II. True or False (10%, 10 questions, 1 point each)（ ） 1. The proposition ((p → q ) ∧ ⌝p ) → ⌝q is a tautology. （ ） 2. If pigs can fly, then it will be raining tomorrow. （ ） 3. Suppose A = {a ,b ,c }, then {{a }} ⊆ P (A ).（ ） 4. “My daughter visited Europe last week” implies the conclusion “Someone visited Europe last week”.（ ） 5. For all integers a ,b ,c ,d , if a | b and c | d , then (a + c )|(b + d ). （ ） 6. For all real numbers x and y , ⎣x - y ⎦ = ⎣x ⎦ - ⎣y ⎦.（ ）7. ()h x =is defined as a function with domain R and codomain R.（ ） 8. There exists a simple graph with 6 vertices, whose degrees are 2,2,2,3,4,4.（ ） 9.If G is a simple graph with n ≥3 vertices such that the degree of every vertex in G is at least n/2, then G has a Hamilton circuit.（ ） 10. Let P (m ,n ) be the statement “m|n ,” where the u.d . of m and n is the set of positive integer.Then ),(n m mP n ∀∃holds.III. Fill in the Blanks (20%, 10 questions, 2 points each)1. Suppose A = {x | x ∈ Z and x 2 < 10}. Then ()P A is .2.If 11{|}i A x x R x i i =∈∧-≤≤ then 1i i A +∞= is .3.Give a relation on {a ,b ,c } that is reflexive and transitive, but not antisymmetric..4.Suppose g : A → B and f : B → C where A = B = C = {1,2,3,4}, g = {(1,4), (2,1), (3,1), (4,2)} and f = {(1,3),(2,2),(3,4),(4,2)}. Then f g =. 5.W rite the negation of the statement “No tests are easy ” in good English:. 6. The expression of GCD(45,12) as a linear combination of 12 and 45is . 7.There are permutations of 7letters A ,B ,C ,D ,E ,F ,G have A immediately to the left of E .8. If f (n ) = f (n - 1) / f (n - 2), f (0) = 2, f (1) = 5, Then f (2) = . 9.The negation of the statement ∃x ∀y (xy = 0) is.10. Let }|),{(},|),{(2221b a b a R b a b a R ≠ℜ∈=≤ℜ∈=Then 21R R ⋂ is.IV. Answer the Questions (35%,7 questions, 5 points each)：1. Write the truth table for the proposition ：⌝(r → ⌝q ) ∨ (p ∧ ⌝r )2. Suppose f : R → R where f (x ) = ⎣x /2⎦.(a) If S = {x | 1 ≤ x ≤ 6}, find f (S ). (b) If T = {3,4,5}, find f -1(T ).3. Find the matrix that represents the relation of R on {1,2,3,4} where aRb means | a - b | ≤ 1. Useelements in the order given to determine rows and columns of the matrix.4.Prove that is a tautology using propositional equivalence and the laws of logic.5.Encrypt the message “HELP” by translating the letters into numb ers, applying the encryption functionf (p) = (3p+ 7) mod 26, and then translating the numbers back into letters.6.Solve the linear congruence 5x≡ 3 (mod 11).7.Determine whether these two graphs are isomorphic. If they are isomorphic, give a one-to-one and ontofunction f from the one on the left to the one on the right.A = {(x, y) | x, y ∈R}−{(0, 0)}.Define a relation on A by the rule: (a, b)R(c, d) (a, b) and (c, d) lie on the same line through the origin.(a) Prove that R is an equivalence relation.(b) Describe the equivalence classes arising from the equivalence relation R in part (a).(c) If A is replaced by the entire plane, is R an equivalence relation?VI. (7%) Determine whether this argument is valid: Lynn works part time or full time.If Lynn does not play on the team, then she does not work part time. If Lynn plays on the team, she is busy. Lynn does not work full time. Therefore, Lynn is busy.VII. (7%) The pseudo-code of Dijstra ’s algorithm is given as following: Procedure Dijkstra (G : weighted connected simple graph) for i :=1 to n L (v i ):=∞ ; L (a ):=0 ; S :=Φ;While z S Beginu := a vertex not in S with L (u ) minimal; S :=S ∪{u }; Print S for all vertices v not in S if L (u )+w (u,v )<L (v ) then L (v ):=L(u )+w (u,v ) End(a) Find a shortest path between b and f using Dijstra ’s algorithm given above. For every iterative inwhile-loop, list S ’s elements for “Print S ” statement. (b) Compute the length of the shortest path.。