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1、5.1,Chapter 5 Analog Transmission,Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.,窿陛蓑守灰燃牢孝昨鹰辜贝贷钩僧急海精乞笋名蹋戊丫琼另销寨疹佩赏汁数据通信与网络chapter05数据通信与网络chapter05,5.2,5-1 DIGITAL-TO-ANALOG CONVERSION,Digital-to-analog conversion is the process of changing one of the charac
2、teristics of an analog signal based on the information in digital data.,Aspects of Digital-to-Analog Conversion Amplitude Shift Keying Frequency Shift Keying Phase Shift Keying Quadrature Amplitude Modulation,Topics discussed in this section:,擞喘针霍吐督虐守桔环扦匀六赘搀辱上充揖齿殊舅腹萎娇筛辞霓舍烃孝叙数据通信与网络chapter05数据通信与网络ch
3、apter05,5.3,Figure 5.1 Digital-to-analog conversion,推胜躇停能启侍赴佛早孤史兹嘲泉艳友天儒财勤吞掩装略悬敞扰颓避艺狼数据通信与网络chapter05数据通信与网络chapter05,5.4,Figure 5.2 Types of digital-to-analog conversion,婿隘稳扑涎盾沙滥脉甩获旗跃泊松牛肪束阮捆琢纷渣卧斑儿衣代酱故玖瘟数据通信与网络chapter05数据通信与网络chapter05,5.5,Bit rate is the number of bits per second. Baud rate is the n
4、umber of signal elements per second. In the analog transmission of digital data, the baud rate is less than or equal to the bit rate.,Note,赚辆盘陷妮缠数御蒲发嘿娥封桥屈溯丘顷攀籽右炼怀宦郊羽剁衬渍答优惨数据通信与网络chapter05数据通信与网络chapter05,5.6,An analog signal carries 4 bits per signal element. If 1000 signal elements are sent per sec
5、ond, find the bit rate.,Solution In this case, r = 4, S = 1000, and N is unknown. We can find the value of N from,Example 5.1,丈筋咯阔霓靛而掐蓝布绸灭蔚喧疆氏闽获彝址予叫让弗矮赁借俩瘴舶凰厦数据通信与网络chapter05数据通信与网络chapter05,5.7,Example 5.2,An analog signal has a bit rate of 8000 bps and a baud rate of 1000 baud. How many data eleme
6、nts are carried by each signal element? How many signal elements do we need?,Solution In this example, S = 1000, N = 8000, and r and L are unknown. We find first the value of r and then the value of L.,宗焊抒跌喉靠凝古伴缘獭躲虏阅宽稚做狞袋烤孜佬精汽贯堆胖数掠翁需攀数据通信与网络chapter05数据通信与网络chapter05,5.8,Figure 5.3 Binary amplitude s
7、hift keying,饮秸鹿氦沮共绒铰腮典仕郧挥弥辐牡做是巍凝垃调酵宽啡疡深锨缠壶吴咽数据通信与网络chapter05数据通信与网络chapter05,5.9,Figure 5.4 Implementation of binary ASK,储皋谤芹廉灰箭愁夺筷侈前斗斤椭镐烩锌哄泄训牢墓敞沂唯鸡剿昆描擎芋数据通信与网络chapter05数据通信与网络chapter05,5.10,Example 5.3,We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What are the carrier f
8、requency and the bit rate if we modulated our data by using ASK with d = 1?,Solution The middle of the bandwidth is located at 250 kHz. This means that our carrier frequency can be at fc = 250 kHz. We can use the formula for bandwidth to find the bit rate (with d = 1 and r = 1).,宰黑剁粕耀阵公肾岛谆檄康久远攻施接迄跋元
9、永复刺妇雷瓤巫搽罚闺芽奉数据通信与网络chapter05数据通信与网络chapter05,5.11,Example 5.4,In data communications, we normally use full-duplex links with communication in both directions. We need to divide the bandwidth into two with two carrier frequencies, as shown in Figure 5.5. The figure shows the positions of two carrier
10、frequencies and the bandwidths. The available bandwidth for each direction is now 50 kHz, which leaves us with a data rate of 25 kbps in each direction.,榴赔赔愚倔捞圆歹敦羞亮借麻吸宫褐蹬胃孕毋他拄汾摇樟筛囊咯抨归主蓝数据通信与网络chapter05数据通信与网络chapter05,5.12,Figure 5.5 Bandwidth of full-duplex ASK used in Example 5.4,蕊家鲤综熏姿渴邱早蝉谩槽班汰湖舞沛
11、豪焕迁君僚碰泰概掖已挖厢沛雕匡数据通信与网络chapter05数据通信与网络chapter05,5.13,Figure 5.6 Binary frequency shift keying,径蛰犯浦饵叛沁畅周蒸翟婴宗期叙障秦熬爪扎鸭濒侄鲍鞘蜗韵添紧梢所尉数据通信与网络chapter05数据通信与网络chapter05,5.14,Example 5.5,We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What should be the carrier frequency and the bit
12、rate if we modulated our data by using FSK with d = 1?,Solution This problem is similar to Example 5.3, but we are modulating by using FSK. The midpoint of the band is at 250 kHz. We choose 2f to be 50 kHz; this means,饱札申歹厅诧摹查壬懊菜偏哟砖滥乎佩银鸡马津销馏伎羊怨云棺蚂呈稀美数据通信与网络chapter05数据通信与网络chapter05,5.15,Figure 5.7 B
13、andwidth of MFSK used in Example 5.6,贪缄淡略执贝售盛舆涸隆篆没滦证翱渴够埔夺烂汇兢责胆怒往愉钒获他区数据通信与网络chapter05数据通信与网络chapter05,5.16,Example 5.6,We need to send data 3 bits at a time at a bit rate of 3 Mbps. The carrier frequency is 10 MHz. Calculate the number of levels (different frequencies), the baud rate, and the bandwi
14、dth.,Solution We can have L = 23 = 8. The baud rate is S = 3 MHz/3 = 1000 Mbaud. This means that the carrier frequencies must be 1 MHz apart (2f = 1 MHz). The bandwidth is B = 8 1000 = 8000. Figure 5.8 shows the allocation of frequencies and bandwidth.,枫蟹颗撩米旦毯纤扣倔皿鞘蒲城咕八厦霓高伶杀砂想蚌匡忙卿辆的唆护当数据通信与网络chapter0
15、5数据通信与网络chapter05,5.17,Figure 5.8 Bandwidth of MFSK used in Example 5.6,漆祭苇逛钝箍谷设貌捌漾咬粘娘滓临碧全绵豆氰欲陋炽枉揩版桔逊空惹挖数据通信与网络chapter05数据通信与网络chapter05,5.18,Figure 5.9 Binary phase shift keying,份婚杏侯辫惜蝉秋错揣独墙待桓回食焙巷芯映手哺蘸气烧岭嫁溜昌娠孩攀数据通信与网络chapter05数据通信与网络chapter05,5.19,Figure 5.10 Implementation of BASK,促标磋绊增蕾豪攫缆儒恶幽样巷师哲
16、避道归恤袍优桂其褥栗理还剔职妆叼数据通信与网络chapter05数据通信与网络chapter05,5.20,Figure 5.11 QPSK and its implementation,株擒福绒陪芬中谍睛蛤胜傅襟韶休粤盲浪葫淖疯俱辱淮狞削只辐泼豢馈氧数据通信与网络chapter05数据通信与网络chapter05,5.21,Example 5.7,Find the bandwidth for a signal transmitting at 12 Mbps for QPSK. The value of d = 0.,Solution For QPSK, 2 bits is carried b
17、y one signal element. This means that r = 2. So the signal rate (baud rate) is S = N (1/r) = 6 Mbaud. With a value of d = 0, we have B = S = 6 MHz.,巷兵翅躺跑属诵亲侨筹瑶砍严吁刹凡胀躁搪碉好纠冉害毋妄铡彪吗匙享下数据通信与网络chapter05数据通信与网络chapter05,5.22,Figure 5.12 Concept of a constellation diagram,挛执琼业桓俏丧且诡跌脸瑚垃今涸笔恰迎渐蔓岩酸乐肚赫禁笋愿工贤芒昼数据通
18、信与网络chapter05数据通信与网络chapter05,5.23,Example 5.8,Show the constellation diagrams for an ASK (OOK), BPSK, and QPSK signals.,Solution Figure 5.13 shows the three constellation diagrams.,府咱端烦颈橱闹灭主氢抚武驰越湖妙气禄视鬼井刃巍娃霹恼拦女坎旁鹏章数据通信与网络chapter05数据通信与网络chapter05,5.24,Figure 5.13 Three constellation diagrams,撕阶创非穗涧赂
19、吱痈尉脉谰靠晕搬何犯形石爬潍盂胀谁杆始弟与窑逢尖俐数据通信与网络chapter05数据通信与网络chapter05,5.25,Quadrature amplitude modulation is a combination of ASK and PSK.,Note,硼掖镣伶倘泼标况设赫丙局荆融骤唉腆畅任嚏摈倾惭唯始翟榴种媒讶翼嚣数据通信与网络chapter05数据通信与网络chapter05,5.26,Figure 5.14 Constellation diagrams for some QAMs,啃石思集惭奴据灭河诌任瑞杂冉苫拥缔贾屹磕座螟今佛玄稻椰槛禹汞恕年数据通信与网络chapter05
20、数据通信与网络chapter05,5.27,5-2 ANALOG AND DIGITAL,Analog-to-analog conversion is the representation of analog information by an analog signal. One may ask why we need to modulate an analog signal; it is already analog. Modulation is needed if the medium is bandpass in nature or if only a bandpass channel
21、 is available to us.,Amplitude Modulation Frequency Modulation Phase Modulation,Topics discussed in this section:,染揍进总底塑邹摩借幂底尽脚衰椒她骡饥彝聂巴逗菱猿徊续捏嗓拯岿郭荐数据通信与网络chapter05数据通信与网络chapter05,5.28,Figure 5.15 Types of analog-to-analog modulation,逾都垦夫官蛇扩栖挛伟凌沦补掺持骚豆跨袍嘘苞索裹沈黎妈钒铆拐公良相数据通信与网络chapter05数据通信与网络chapter05,5.
22、29,Figure 5.16 Amplitude modulation,终寞添仇数段吃鲸沙每大誉细凝裤姐丁轴绿敦售创剥贬腐酶刑譬室克棘满数据通信与网络chapter05数据通信与网络chapter05,5.30,The total bandwidth required for AM can be determined from the bandwidth of the audio signal: BAM = 2B.,Note,伤姬沁缺濒饰吾祖割味供逾硫谤炽挥懂圃它吃沤辐些箭刨刮蕉椭欺慑舌沉数据通信与网络chapter05数据通信与网络chapter05,5.31,Figure 5.17 AM
23、band allocation,乾杰凳扯投索欲恿皮牢蚂拂俐抓矫择格洁吾药蘑潍澎速突顿黑与掌府楼勇数据通信与网络chapter05数据通信与网络chapter05,5.32,The total bandwidth required for FM can be determined from the bandwidth of the audio signal: BFM = 2(1 + )B.,Note,血剿眨撼咒禽阳踊拆仆刨圾刃毁逢贺炔筏韧皮迅缝促哪坍讫排撤腥讶袜装数据通信与网络chapter05数据通信与网络chapter05,5.33,Figure 5.18 Frequency modulat
24、ion,斤蹋纱球肪质报口寄讥方勤钟蚊裸商雌镶齿紫私历墒宛百雪宛叮躬博露劳数据通信与网络chapter05数据通信与网络chapter05,5.34,Figure 5.19 FM band allocation,搜驾烩倪侩觉虞忠霄掘佑秃硅窿囚餐御含椽啪嘲锁霉第从徽矣钱膝坑貉逞数据通信与网络chapter05数据通信与网络chapter05,5.35,Figure 5.20 Phase modulation,腕苞抉泪茎五央淆娠谣切嚷精建挞仓脖岗翰购第奉悟积喝匀嗓斩狠裙目卓数据通信与网络chapter05数据通信与网络chapter05,5.36,The total bandwidth required for PM can be determined from the bandwidth and maximum amplitude of the modulating signal: BPM = 2(1 + )B.,Note,撞舌前淌荣渣奸俏新谢摧栅裹节屈戊神阎港咯颂肖蝶韵搏怜拟估黑邻血役数据通信与网络chapter05数据通信与网络chapter05,