Chapter 5 Analog Transmission
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1 5-1 DIGITAL-TO-ANALOG CONVERSION Chapter 5 Analog Transmission Digital-to-analog conversion is the process of changing one of the characteristics of an analog signal depending on the information in digital data. From Chapter 5, we will study 5.1. Topics discussed in this section: Aspects of Digital-to-Analog Conversion Binary Amplitude Shift Keying Binary Frequency Shift Keying Phase Shift Keying Quadrature Amplitude Modulation Figure 5.1 Digital-to-analog conversion Figure 5.2 Types of digital-to-analog conversion
2 Aspects of Digital-to-Analog Conversion Aspects of Digital-to-Analog Conversion Bit rate is the number of bits per second. Baud rate is the number of signal elements per second. Data element versus Signal element Data rate (N) versus signal rate(s) In the analog transmission of digital data, the baud rate is less than or equal to the bit rate. Baud Rate = bit rate /number of bits per signal unit Bandwidth Carrier Signal Example 1 Example 2 An analog signal carries 4 bits per signal element. If 1000 signal elements are sent per second, find the bit rate. baud rate = 1000 baud/s Number of bits per signal= 4 An analog signal has a bit rate of 3000 bps and each signal unit carries 6 bits. What is the baud rate? Baud Rate = bit rate /number of bits per signal unit = 3000/6 = 500 baud/s Bit Rate
3 Carrier signal A carrier signal is a transmitted electromagnetic pulse or wave at a steady base frequency of alternation on which information can be imposed by increasing signal strength, varying the base frequency, phase, or amplitude. This modification is called modulation. Digital Modulation Techniques There are three major classes of digital modulation techniques used for transmission of digitally represented data: Amplitude shift keying (ASK). Frequency shift keying (FSK). Phase shift keying (PSK) Amplitude Shift Keying (ASK) Amplitude Shift Keying (ASK) ASK is a form of modulation that represents digital data as variations in the amplitude of a carrier wave. The amplitude of the carrier signal varies according to the bit stream (modulating signal), keeping frequency and phase constant
4 Amplitude Shift Keying (ASK) Bandwidth for ASK Advantages Very simple modulation and demodulation Disadvantages High sensitivity to noise Low bandwidth efficiency Bandwidth = (1+d) N baud N baud is the baud rate d is the modulation factor with minimum value =0 In BASK baud rate = bit rate Minimum BW = N baud ASK on/off keying (OOK) Example 5.4 We have an available bandwidth of 10 khz (1000 to 11,000) In data communications, we normally use full-duplex links with communication in both directions. Draw a diagram of the system? Frequency shift keying (FSK) FSK is a method of transmitting digital signals. The two binary states, logic 0 (low) and 1 (high), are represented by an analog waveform. Logic 0 is represented by a wave at a specific frequency, and logic 1 is represented by a wave at a different ,000 frequency. BW = 10000/2 = 5000 Hz f c (forward) = /2= 3500Hz f c (backward) = /2= 8500Hz Peak amplitude and phase are constant
5 Frequency shift keying (FSK) Frequency shift keying (FSK) FSK rejects unwanted signals (noise) that are weaker than the desired signal. It can ignore spikes. Better bandwidth efficiency than ASK Simple modulation and demodulation Disadvantages: FSK is limited Bandwidth of FSK Bandwidth of FSK FSK is a combination of two ASK. BW= f 2 -f 1 +N baud BW= f 2-f 1+N baud Example 7 Find the maximum bit rate for FSK signal if the bandwidth is Hz, and the carriers are separated by 2000 Hz using fullduplex links. Example 5.5 Find the minimum bandwidth for FSK signal transmitting at 2000 bps using half-duplex mode, and the carriers are separated by 3000 Hz. BW= bit rate + (f 2 -f 1 )= = 5000 Hz BW= baud rate + (f 2 -f 1 ) Baud rate = BW (f 2 -f 1 ) Baud rate = = 4000 baud/s Bite rate = 4000 bps
6 Phase shift keying Phase-shift keying (PSK): the phase of a transmitted signal is varied to convey information. There are several methods that can be used to accomplish PSK. Phase shift keying The simplest PSK technique is called 2-PSK or BPSK. It uses two opposite signal phases (0 o and 180 o ). The state of each bit is determined according to the state of the preceding bit. If the phase of the wave does not change, then the signal state stays the same (0 or 1). If the phase of the wave reverses that is changes by 180 o then the signal state is flipped Phase shift keying 4-Phase shift keying PSK minimum bandwidth = ASK minimum bandwidth. Advantages of PSK: Not susceptible to noise. No bandwidth limitation Disadvantages: Distinguishing small difference in phase depending on the equipment used
7 2-PSK Constellation Diagram 8-PSK Constellation Diagram Bit Phase Bits 4-PSK Constellation Diagram Bit Phase Dibit Bit Phase Tribits Example 8 Example 9 Find the bandwidth for a signal transmitting at 2 kbps for 4-PSK using half-duplex mode. What are the baud rate and bit rate giving a bandwidth of 5000 Hz for an 8-PSK signal. For 4-PSK, 2 bits is carried by one signal element. This means that r = 2. So the signal rate (baud rate) is = N (1/r) = 1 kbaud/s. In PSK bandwidth = baud rate. Bandwidth = 1000 Hz. In PSK bandwidth = baud rate. Baud rate = 5000 baud/s. For 8-PSK, 3 bits is carried by one signal element. This means that r = 3. So the bit rate is = 3 * baud rate= 15,000 bps
8 Figure 5.12 Concept of a constellation diagram Quadrature amplitude modulation is a combination of ASK and PSK. To achieve the maximum contrast between each signal unit. In QAM the number of phase shift is always larger than the number of amplitude shifts. Bandwidth of QAM = ASK=PSK Constellation diagrams for some QAMs Constellation diagrams for some QAMs
9 Constellation diagrams for some QAMs Bit rate for different modulation techniques Modulation Units Bits/Baud Baud rate Bit Rate ASK, FSK, 2-PSK Bit 1 N N 4-PSK, 4-QAM Dibit 2 N 2N 8-PSK, 8-QAM Tribit 3 N 3N 16-QAM Quadbit 4 N 4N 32-QAM Pentabit 5 N 5N 64-QAM Hexabit 6 N 6N 128-QAM Septabit 7 N 7N 256-QAM Octabit 8 N 8N Example 10 Bit rate and Baud rate A constellation diagram consists of eight equally spaced points on a circle. If the bit rate is 4800 bps, what is the baud rate? For 8-PSK with 45 o apart, 3 bits is carried by one signal element. This means that r = 3. So the signal rate (baud rate) is = 4800 (1/r) = 1600 baud/s
10 Example 11 Compute the bit rate for 1000 baud 16-QAM signal? For 16-QAM, 4 bits is carried by one signal element. This means that r = 4. So the bite rate is = = 4000 bps. Example 12 Compute the baud rate for bps 64-QAM signal? For 64-QAM, 6 bits is carried by one signal element. This means that r = 6. So the baud rate is = /6= baud. Telephone Modems 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 is available to us. Chapter 9 section 2 pages Topics discussed in this section: Telephone Modems Modem Standards Telephone line bandwidth A telephone line has a bandwidth of almost 2400 Hz for data transmission. 10
11 Modulation/demodulation Modem stands for modulator/demodulator Modem Standards V.32 and V.32bis The V.32 uses trellis coded modulation. Trellis is 32-QAM plus a redundant bit 5bits is transmitted instead of 4bits. Bit rate = 2400 * 4 = 9600 bps. The V.32bis constellation and bandwidth The V.32bis is a ITU-T standard support 14,400 bps. It uses 128-QAM. Fall-back and fall-forward. V.90 and V.92 Traditional Modem has maximum data rate of 33.6Kbps. 56K Modem which is asymmetric in downloading rate such as V90 V.90 V.92 upload rate of 48 Kbps and download rate of 56 Kbps. Calles + internet access. ( 7 bits/baud 6 for data + 1 for error control) Bit rate = 2400 * 6 = 14,400 bps. 11
12 Traditional modems 56K modems 12
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