数据通信基础：第3章 信号

1、数据通信基础第3章 (信号)20120305BitLiuJKPage 2 第3章内容提要第3章 信号3.1 模拟与数字3.2 模拟信号3.3 数字信号3.4 模拟与数字的比较3.5 数据速率的极限3.6 传输损伤3.7 信号的其它特性20120305BitLiuJKPage 3 Physical LayerPART II第二部分包括本书的第3-9章第二部分 物理层20120305BitLiuJKPage 4 Position of the physical layer物理层的作用和位置Duties for Physical Layer20120305Page 5 BitLiuJK物理层的功能比

2、特到信号的转换比特速率控制比特同步多路复用电路交换20120305BitLiuJKPage 6 Analog and Digital Data模拟与数字数据Analog and Digital Signals模拟与数字信号Periodic and Aperiodic Signals周期与非周期信号3.1 Analog and Digital20120305BitLiuJKPage 7 To be transmitted, data must be transformed to electromagnetic signals.Notes:Signals can be analog or digi

3、tal. Analog signals can have an infinite number of values in a range; digital signals can have only a limited number of values.Figure 3.1 Comparison of analog and digital signals20120305Page 8 BitLiuJK注意：区分模拟与数字信号，只关心纵轴（即幅度）；如果关心横轴，得到的是连续与离散信号。20120305BitLiuJKPage 9 Periodic & AperiodicPeriodic sign

4、al refers to a signal which consists of a continuously repeating pattern (cycle).Period Cycle （周期）Example Sine waveAperiodic signal refers to a signal without a definite period.20120305BitLiuJKPage 10 In data communication, we commonly use periodic analog signals and aperiodic digital signals.Note:2

5、0120305BitLiuJKPage 11 3.2 Analog SignalsSine WavePhaseExamples of Sine WavesTime and Frequency DomainsComposite SignalsBandwidth20120305BitLiuJKPage 12 Three characteristics of a sine wave include: amplitude幅度 frequnecy频率 phase相位。Note:Figure 3.3 Amplitude20120305Page 13 BitLiuJKFigure 3.4 Period an

6、d frequencyFrequency and period are inverses of each other. f = 1/T20120305Page 14 BitLiuJK20120305BitLiuJKPage 15 Table 3.1 Units of periods and frequenciesUnitEquivalentUnitEquivalentSeconds (s)1 shertz (Hz)1 Hzmilliseconds (ms)103 skilohertz (kHz)103 Hzmicroseconds (s)106 smegahertz (MHz)106 Hzna

7、noseconds (ns)109 sgigahertz (GHz)109 Hzpicoseconds (ps)1012 sterahertz (THz)1012 HzHertz?20120305BitLiuJKPage 16 More about UnitsPrefixSymbolFactorPrefixSymbolFactor十deka da101deci d 10-1分百hecto h102centi c 10-2厘千kilo k103milli m 10-3毫兆mega M106micro 10-6微吉giga G109nano n 10-9纳太tera T1012pico p 10-

8、12皮peta P1015femto f 10-15飞exa E1018atto a 10-18zetta Z1021zepto z 10-21yottaY1024yocto y 10-24Example 1Express a period of 100 ms in microseconds, and express the corresponding frequency in kilohertz. 将100毫秒的周期表示成微妙，并将其对应的频率表示成千赫。SolutionFrom Table 3.1 we find the equivalent of 1 ms. We make the fo

9、llowing substitutions:100 ms = 100 10-3 s = 100 10-3 106 ms = 105 ms Now we use the inverse relationship to find the frequency, changing hertz to kilohertz100 ms = 100 10-3 s = 10-1 s f = 1/10-1 Hz = 10 Hz = 10-2 kHz20120305Page 17 BitLiuJK20120305BitLiuJKPage 18 Frequency is the rate of change with

10、 respect to time. Change in a short span of time means high frequency（高频）. Change over a long span of time means low frequency（低频）.Note:Another meaning of frequency20120305BitLiuJKPage 19 If a signal does not change at all, its frequency is zero (Direct Curret,DC直流). If a signal changes instantaneou

11、sly, its frequency is infinite.Note:Figure 3.5 Relationships between different phasesPhase describes the position of the waveform relative to time zero.20120305Page 20 BitLiuJK20120305BitLiuJKPage 21 Example 2A sine wave is offset one-sixth of a cycle with respect to time zero. What is its phase in

12、degrees and radians?SolutionWe know that one complete cycle is 360 degrees. Therefore, 1/6 cycle is (1/6) 360 = 60 degrees = 60 x 2p /360 rad = 1.046 rad Figure 3.6 Sine wave examples20120305Page 22 BitLiuJKFigure 3.6 Sine wave examples (continued)20120305Page 23 BitLiuJK20120305BitLiuJKPage 24 An a

13、nalog signal is best represented in the frequency domain频域.Note:Figure 3.7 Time and frequency domains20120305Page 25 BitLiuJK20120305BitLiuJKPage 26 A single-frequency sine wave is not useful in data communications; we need to change one or more of its characteristics to make it useful. Notes:When w

14、e change one or more characteristics of a single-frequency signal, it becomes a composite signal made of many frequencies.20120305BitLiuJKPage 27 According to Fourier analysis, any composite signal can be represented as a combination of simple sine waves with different frequencies, phases, and ampli

15、tudes.Note:根据傅里叶分析，任何复合信号都可以表示成具有不同幅度、频率和相位的简单正弦波的组合叠加。Figure 3.9 Basic Square Wave20120305Page 28 BitLiuJKFigure 3.10 Adding first three harmonics20120305Page 29 BitLiuJKFigure 3.11 Frequency spectrum comparison20120305Page 30 BitLiuJKFigure 3.12 Signal distortion 信号失真20120305Page 31 BitLiuJK由于传输介质

16、的有限带宽导致的信号失真20120305BitLiuJKPage 32 The bandwidth带宽 is a property of a medium: It is the difference between the highest and the lowest frequencies that the medium can satisfactorily pass. Note:We will use the term bandwidth to refer to the property of a medium OR the width of a single spectrum频谱. Fi

17、gure 3.13 Bandwidth20120305Page 33 BitLiuJK20120305BitLiuJKPage 34 Example 3If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is the bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V.SolutionB = fh -f

18、l = 900 - 100 = 800 HzThe spectrum has only five spikes, at 100, 300, 500, 700, and 900 (see Figure 13.4 )Figure 3.14 Example 320120305Page 35 BitLiuJK20120305BitLiuJKPage 36 Example 4A signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz. What is the lowest frequency? Draw the spectrum i

19、f the signal contains all integral frequencies of the same amplitude.SolutionB = fh - fl20 = 60 - flfl = 60 - 20 = 40 HzFigure 3.15 Example 420120305Page 37 BitLiuJK20120305BitLiuJKPage 38 Example 5A signal has a spectrum with frequencies between 1000 and 2000 Hz (bandwidth of 1000 Hz). A medium can

20、 pass frequencies from 3000 to 4000 Hz (a bandwidth of 1000 Hz). Can this signal faithfully pass through this medium? SolutionThe answer is definitely no. Although the signal can have the same bandwidth (1000 Hz), the range does not overlap. The medium can only pass the frequencies between 3000 and

21、4000 Hz; the signal is totally lost.20120305BitLiuJKPage 39 3.3 Digital SignalsBit Interval and Bit RateAs a Composite Analog SignalThrough Wide-Bandwidth MediumThrough Band-Limited MediumVersus Analog BandwidthHigher Bit RateFigure 3.16 A digital signal20120305Page 40 BitLiuJK20120305BitLiuJKPage 4

22、1 Example 6A digital signal has a bit rate of 2000 bps. What is the duration of each bit (bit interval)SolutionThe bit interval is the inverse of the bit rate.Bit interval = 1/ 2000 s = 0.000500 s = 0.000500 x 106 ms = 500 msFigure 3.17 Bit rate and bit interval20120305Page 42 BitLiuJKFigure 3.18 Di

23、gital versus analog20120305Page 43 BitLiuJK20120305BitLiuJKPage 44 Analogue frequency = bit rateOR, for a channel of bandwidth, BThe max. binary bit rate (or channel capacity, C) it can support is:C = 2 * BNote:20120305BitLiuJKPage 45 The bit rate and the bandwidth are proportional to each other.Not

24、e:Example: A telephone line has an audio bandwidth of 3kHz its maximum binary bit rate is thus 6kbps20120305BitLiuJKPage 46 3.4 Analog versus DigitalLow-pass versus Band-passDigital TransmissionAnalog TransmissionFigure 3.19 Low-pass and band-pass 低通与带通20120305Page 47 BitLiuJK20120305BitLiuJKPage 48

25、 The analog bandwidth of a medium is expressed in hertz; the digital bandwidth, in bits per second.Note:20120305BitLiuJKPage 49 Digital transmission needs a low-pass channel.Notes:Analog transmission can use a band-pass channel.20120305BitLiuJKPage 50 3.5 Data Rate LimitNoiseless Channel: Nyquist Bi

26、t Rate无噪信道：奈奎斯特比特速率Noisy Channel: Shannon Capacity有噪信道：香农容量20120305BitLiuJKPage 51 Example 7Consider a noiseless channel with a bandwidth, B, of 3000 Hz transmitting a signal with two (binary) signal levels. The maximum bit rate can be calculated using the Nyquist bit rateNyquist Bit rate = 2 x B x

27、log2 Lwhere L is the number of signal levelsBit Rate = 2 3000 log2 2 = 6000 bpsIf we wish to squeeze more data down this channel, we have to transmit more than one bit at a given instant by having more than 2 signal levels20120305BitLiuJKPage 52 Example 8Consider the same noiseless channel, transmit

28、ting a signal with four signal levels (for each level, we can now send two bits). The maximum bit rate can be calculated as: Bit Rate = 2 x 3000 x log2 4 = 12,000 bps0V3V2V1V01111000By having 4 signal levels (L=4), we can now send 2 bits at a time20120305BitLiuJKPage 53 A real channel will have its

29、capacity limited by noise, which will limit the number of voltage levels which can be reliably distinguished.Shannon Capacity Formula:A noisy communications channel with bandwidth, B, has a maximum capacity, C, of:C = B x log2 (1 + SNR) bits/secWhere SNR is the signal to noise ratio, calculated as:

30、SNR = signal power/noise power20120305BitLiuJKPage 54 Example 9Consider an extremely noisy channel in which the value of the signal-to-noise ratio (SNR) is almost zero. In other words, the noise is so strong that the signal is faint. For this channel the capacity is calculated asC = B log2 (1 + SNR)

31、 = B log2 (1 + 0) = B log2 (1) = B 0 = 020120305BitLiuJKPage 55 Example 10We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal-to-noise ratio is usually around 3162 (35dB). For this channel

32、 the capacity is calculated asC = B log2 (1 + SNR) = 3000 log2 (1 + 3162) = 3000 log2 (3163)C = 3000 11.62 = 34,860 bpsExample 11We have a channel with a 1 MHz bandwidth. The SNR for this channel is 63; what is the appropriate bit rate and signal level?SolutionC = B log2 (1 + SNR) = 106 log2 (1 + 63

33、) = 106 log2 (64) = 6 MbpsThen we use the Nyquist formula to find the optimumnumber of signal levels. We reduce the required capacity to 4Mbps ( max of 6Mbps) to ensure good performance.4 Mbps = 2 1 MHz log2 L L = 4First, we use the Shannon formula to find our upper limit.20120305Page 56 BitLiuJK201

34、20305BitLiuJKPage 57 3.6 Transmission Impairment传输损伤Attenuation衰减Distortion失真Noise噪声Figure 3.20 Impairment typesTransmission Impairment20120305Page 58 BitLiuJKFigure 3.21 Attenuation20120305Page 59 BitLiuJK20120305BitLiuJKPage 60 Example 12Imagine a signal travels through a transmission medium and i

35、ts power is reduced to half. This means that P2 = 1/2 P1. In this case, the attenuation (loss of power) can be calculated asSolution 10 log10 (P2/P1) = 10 log10 (0.5P1/P1) = 10 log10 (0.5) = 10(0.3) = 3 dB20120305BitLiuJKPage 61 Example 13Imagine a signal travels through an amplifier and its power i

36、s increased ten times. This means that P2 = 10 x P1. In this case, the amplification (gain of power) can be calculated as 10 log10 (P2/P1) = 10 log10 (10P1/P1) = 10 log10 (10) = 10 (1) = 10 dB20120305BitLiuJKPage 62 Example 14One reason that engineers use the decibel to measure the changes in the strength of a signal is that decibel numbers can be added (or subtracted) when we are talking about several points instead of just two (cascading). In Figure 3.22 a signal travels a long dis