Chapter 5 Analog Analog Transmission Transmission Analog Transmission Digital to Analog Encoding : is the representation of digital information by a analog signal Analog Transmission of Analog Data Analog to Analog Encoding : is the representation of analog information by a analog signal McGraw-Hill Analog Transmission of Digital Data Encoding means transform information into signal ©The McGraw-Hill Companies, Inc., 2004 Analog Analog Transmission Transmission of of Digital Digital Data Data is the representation of digital information by an analog signal. Ex : transmit data from one computer to another across a public access phone line, the data start as digital , but because telephone wires carry analog signals, the data must be converted. Analog Analog Transmission Transmission of of Analog Analog Data Data Digital / Analog encoding 1 5.1 Digital to Analog Conversion Digital-to-analog conversion • Aspects of Digital-to-Analog Conversion • Amplitude Shift Keying (ASK) • Frequency Shift Keying (FSK) • Phase Shift Keying (PSK) • Quadrature Amplitude Modulation McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004 Figure 5.2 Types of digital-to-analog conversion Note Bit rate is the number of bits per second. Baud rate is the number of signal elements per second. In the analog transmission of digital data, the baud rate is less than or equal to the bit rate. 2 Example 5.1 Relationship between data rate and signal rate An analog signal carries 4 bits per signal element. If 1000 signal elements are sent per second, find the bit rate. S : the signal rate (baud) N : the data rate (bps) r : ratio : the numbers of data elements carried in one signal element L : the type of signal element , not the level Solution In this case, r = 4, S = 1000, and N is unknown. We can find the value of N from Example 5.2 An 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? Carrier Signal In analog transmission the sending device produces a high-frequency signal that acts as a basis for the information signal. 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. This base signal is called the carrier signal or carrier frequency. Digital information is then encoded onto the carrier signal. 3 Binary amplitude shift keying (BASK) Figure 5.4 Implementation of binary ASK S : the number of signal element N : the data rate (bps) d : a factor related to the condition of a line Both frequency and phase remain constant while the amplitude changes. ASK transmission is highly susceptible to noise interference. Example 5.3 We 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 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). McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004 4 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 below figure . 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 kbps in each direction. Binary frequency shift keying (BFSK) S : the number of signal element N : the data rate (bps) d : a factor related to the condition of a line ∆f : the differences between two frequencies Example 5.5 the frequency of the signal is varied to represent binary 1 or 0. both peak amplitude and phase remain constant. FSK avoids most of the noise problems of ASK. The limiting factors of FSK are the physical capabilities of the carrier. We 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 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 2∆f to be 50 kHz; this means McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004 5 Binary phase shift keying (BPSK) Figure 5.10 Implementation of BPSK S : the number of signal element N : the data rate (bps) d : a factor related to the condition of a line Figure 5.9 phase refers to the direction in which the wave begins PSK constellation 2-PSK 4-PSK 8-PSK etc 2-PSK 0 o phase wave : the waves that start by moving up and to the right. 1o phase wave : the waves that start down and to the right. Every time there is a change in the binary value (0 or 1), there is a 180 o change in the phase McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004 McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004 6 Figure 5.11 The 4-PSK characteristics McGraw-Hill Figure 5.12 ©The McGraw-Hill Companies, Inc., 2004 McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004 Figure 5.15 The 4-QAM and 8-QAM constellations 4-QAM 1 Amplitudes , 4 Phases The 8-PSK characteristics Time domain for an 8-QAM signal 8-QAM 2 Amplitudes , 4 Phases Quadrature amplitude modulation is a combination of ASK and PSK so that a maximum contrast between each signal unit (bit, dibit, tribit, and so on) is achieved. McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004 McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004 7 Figure 5.16 16-QAM constellations Figure 5.17 McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004 Table 5.1 Bit and baud rate comparison Modulation Bits/Baud Baud rate Bit Rate Bit 1 N N 4-PSK, 44-QAM Dibit 2 N 2N 8-PSK, 88-QAM Tribit 3 N 3N 1616-QAM Quadbit 4 N 4N 3232-QAM Pentabit 5 N 5N 6464-QAM Hexabit 6 N 6N 128128-QAM Septabit 7 N 7N 256256-QAM Octabit 8 N 8N McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004 Example Units ASK, FSK, 22-PSK McGraw-Hill Bit and baud ©The McGraw-Hill Companies, Inc., 2004 A constellation diagram consists of eight equally spaced points on a circle. If the bit rate is 4800 bps, what is the baud rate? Solution The constellation indicates 8-PSK with the points 45 degrees apart. Since 23 = 8, 3 bits are transmitted with each signal unit. Therefore, the baud rate is 4800 / 3 = 1600 baud McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004 8 Example Example Compute the bit rate for a 1000-baud 16-QAM signal. Compute the baud rate for a 72,000-bps 64-QAM signal. Solution Solution A 16-QAM signal has 4 bits per signal unit since log216 = 4. Thus, (1000)(4) = 4000 bps A 64-QAM signal has 6 bits per signal unit since log2 64 = 6. Thus, 72000 / 6 = 12,000 baud McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004 McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004 Figure 5.24 Analog-to-analog modulation 5.4 Analog to Digital Conversion Amplitude Modulation (AM) Frequency Modulation (FM) Phase Modulation (PM) McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004 McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004 9 Figure 5.15 Types of analog-to-analog modulation McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004 Figure 5.18 Frequency modulation McGraw-Hill Figure 5.16 Amplitude modulation McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004 Figure 5.20 Phase modulation ©The McGraw-Hill Companies, Inc., 2004 McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004 10