数据通信与网络chapter05.ppt

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1、5.1Chapter 5Analog Transmission5.2Aspects of Digital-to-Analog ConversionAmplitude Shift KeyingFrequency Shift KeyingPhase Shift KeyingQuadrature Amplitude Modulation5.3Figure 5.1 Digital-to-analog conversion5.4Figure 5.2 Types of digital-to-analog conversion5.5Bit rate is the number of bits per sec

2、ond.Baud rate is the number of signalelements per second.In the analog transmission of digital data,the baud rate is less than or equal to the bit rate.Note5.6An analog signal carries 4 bits per signal element.If 1000 signal elements are sent per second,find the bit rate.SolutionIn this case,r=4,S=1

3、000,and N is unknown.We can find the value of N fromExample 5.15.7Example 5.2An analog signal has a bit rate of 8000 bps and a baud rate of 1000 baud.How many data elements are carried by each signal element?How many signal elements do we need?SolutionIn this example,S=1000,N=8000,and r and L are un

4、known.We find first the value of r and then the value of L.5.8Figure 5.3 Binary amplitude shift keying5.9Figure 5.4 Implementation of binary ASK5.10Example 5.3We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz.What are the carrier frequency and the bit rate if we modulated our

5、 data by using ASK with d=1?SolutionThe 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).5.11Example 5.4In data communications,we normally use full-duplex links with co

6、mmunication 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 frequencies and the bandwidths.The available bandwidth for each direction is now 50 kHz,which leaves us with a data rate of 25 kbp

7、s in each direction.5.12Figure 5.5 Bandwidth of full-duplex ASK used in Example 5.45.13Figure 5.6 Binary frequency shift keying5.14Example 5.5We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz.What should be the carrier frequency and the bit rate if we modulated our data by us

8、ing FSK with d=1?SolutionThis 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 means5.15Figure 5.7 Bandwidth of MFSK used in Example 5.65.16Example 5.6We need to send data 3 bits at a time at a bit rate of 3 Mb

9、ps.The carrier frequency is 10 MHz.Calculate the number of levels(different frequencies),the baud rate,and the bandwidth.SolutionWe 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 sh

10、ows the allocation of frequencies and bandwidth.5.17Figure 5.8 Bandwidth of MFSK used in Example 5.65.18Figure 5.9 Binary phase shift keying5.19Figure 5.10 Implementation of BASK5.20Figure 5.11 QPSK and its implementation5.21Example 5.7Find the bandwidth for a signal transmitting at 12 Mbps for QPSK

11、.The value of d=0.SolutionFor QPSK,2 bits is carried by 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.5.22Figure 5.12 Concept of a constellation diagram5.23Example 5.8Show the constellation diagrams for an ASK(OOK),BPSK

12、,and QPSK signals.SolutionFigure 5.13 shows the three constellation diagrams.5.24Figure 5.13 Three constellation diagrams5.25Quadrature amplitude modulation is a combination of ASK and PSK.Note5.26Figure 5.14 Constellation diagrams for some QAMs5.27Amplitude ModulationFrequency ModulationPhase Modul

13、ation5.28Figure 5.15 Types of analog-to-analog modulation5.29Figure 5.16 Amplitude modulation5.30The total bandwidth required for AM can be determinedfrom the bandwidth of the audio signal:BAM=2B.Note5.31Figure 5.17 AM band allocation5.32The total bandwidth required for FM can be determined from the bandwidth of the audio signal:BFM=2(1+)B.Note5.33Figure 5.18 Frequency modulation5.34Figure 5.19 FM band allocation5.35Figure 5.20 Phase modulation5.36The total bandwidth required for PM can be determined from the bandwidth and maximum amplitude of the modulating signal:BPM=2(1+)B.Note