Chapter 6 Bandwidth Utilization: Multiplexing and Spreading 6.1 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
3-6 PERFORMANCE One important issue in networking is the performance of the network how good is it? We discuss quality of service, an overall measurement of network performance, in greater detail in Chapter 24. In this section, we introduce terms that we need for future chapters. opics discussed in this section: Bandwidth - capacity of the system Throughput - no. of bits that can be pushed through Latency (Delay) - delay incurred by a bit from start to finish Bandwidth-Delay Product 3.2
3.3 Note In networking, we use the term bandwidth in two contexts. The first, bandwidth in hertz, refers to the range of frequencies in a composite signal or the range of frequencies that a channel can pass. The second, bandwidth in bits per second, refers to the speed of bit transmission in a channel or link. Often referred to as Capacity.
Note Bandwidth utilization is the wise use of available bandwidth to achieve specific goals. Efficiency can be achieved by multiplexing; i.e., sharing of the bandwidth between multiple users. 6.4
6.5 6-1 MULTIPLEXING Whenever the bandwidth of a medium linking two devices is greater than the bandwidth needs of the devices, the link can be shared. Multiplexing is the set of techniques that allows the (simultaneous) transmission of multiple signals across a single data link. As data and telecommunications use increases, so does traffic. Topics discussed in this section: Frequency-Division Multiplexing Wavelength-Division Multiplexing Synchronous Time-Division Multiplexing Statistical Time-Division Multiplexing
6.6 Figure 6.1 Dividing a link into channels
6.7 Figure 6.2 Categories of multiplexing
6.8 Figure 6.3 Frequency-division multiplexing (FDM)
Note FDM is an analog multiplexing technique that combines analog signals. It uses the concept of modulation discussed in Ch 5. 6.9
6.10 Figure 6.4 FDM process
6.11 FM
6.12 Figure 6.5 FDM demultiplexing example
Example 6.1 Assume that a voice channel occupies a bandwidth of 4 khz. We need to combine three voice channels into a link with a bandwidth of 12 khz, from 20 to 32 khz. Show the configuration, using the frequency domain. Assume there are no guard bands. Solution We shift (modulate) each of the three voice channels to a different bandwidth, as shown in Figure 6.6. We use the 20- to 24-kHz bandwidth for the first channel, the 24- to 28-kHz bandwidth for the second channel, and the 28- to 32-kHz bandwidth for the third one. Then we combine them as shown in Figure 6.6. 6.13
Example 6.2 Five channels, each with a 100-kHz bandwidth, are to be multiplexed together. What is the minimum bandwidth of the link if there is a need for a guard band of 10 khz between the channels to prevent interference? Solution For five channels, we need at least four guard bands. This means that the required bandwidth is at least 5 100 + 4 10 = 540 khz, as shown in Figure 6.7. 6.14
6.15 Figure 6.7 Example 6.2
6.16 Figure 6.9 Analog hierarchy
Example 6.4 The Advanced Mobile Phone System (AMPS) uses two bands. The first band of 824 to 849 MHz is used for sending, and 869 to 894 MHz is used for receiving. Each user has a bandwidth of 30 khz in each direction. How many people can use their cellular phones simultaneously? Solution Each band is 25 MHz. If we divide 25 MHz by 30 khz, we get 833.33. In reality, the band is divided into 832 channels. Of these, 42 channels are used for control, which means only 790 channels are available for cellular phone users. 6.17
6.18 Figure 6.10 Wavelength-division multiplexing (WDM)
Note WDM is an analog multiplexing technique to combine optical signals. 6.19
6.20 Figure 6.11 Prisms in wavelength-division multiplexing and demultiplexing
6.21 Figure 6.12 Time Division Multiplexing (TDM)
Note TDM is a digital multiplexing technique for combining several low-rate digital channels into one high-rate one. 6.22
6.23 Figure 6.13 Synchronous time-division multiplexing
Note In synchronous TDM, the data rate of the link is n times faster, and the unit duration is n times shorter. 6.24
Example 6.5 In Figure 6.13, the data rate for each one of the 3 input connection is 1 kbps. If 1 bit at a time is multiplexed (a unit is 1 bit), what is the duration of (a) each input slot, (b) each output slot, and (c) each frame? Solution We can answer the questions as follows: a. The data rate of each input connection is 1 kbps. This means that the bit duration is 1/1000 s or 1 ms. The duration of the input time slot is 1 ms (same as bit duration). 6.25
Example 6.5 (continued) b. The duration of each output time slot is one-third of the input time slot. This means that the duration of the output time slot is 1/3 ms. c. Each frame carries three output time slots. So the duration of a frame is 3 1/3 ms, or 1 ms. Note: The duration of a frame is the same as the duration of an input unit. 6.26
Example 6.6 Figure 6.14 shows synchronous TDM with 4 1Mbps data stream inputs and one data stream for the output. The unit of data is 1 bit. Find (a) the input bit duration, (b) the output bit duration, (c) the output bit rate, and (d) the output frame rate. Solution We can answer the questions as follows: a. The input bit duration is the inverse of the bit rate: 1/1 Mbps = 1 μs. b. The output bit duration is one-fourth of the input bit duration, or ¼ μs. 6.27
Example 6.6 (continued) c. The output bit rate is the inverse of the output bit duration or 1/(4μs) or 4 Mbps. This can also be deduced from the fact that the output rate is 4 times as fast as any input rate; so the output rate = 4 1 Mbps = 4 Mbps. d. The frame rate is always the same as any input rate. So the frame rate is 1,000,000 frames per second. Because we are sending 4 bits in each frame, we can verify the result of the previous question by multiplying the frame rate by the number of bits per frame. 6.28
6.29 Figure 6.14 Example 6.6
Example 6.7 Four 1-kbps connections are multiplexed together. A unit is 1 bit. Find (a) the duration of 1 bit before multiplexing, (b) the transmission rate of the link, (c) the duration of a time slot, and (d) the duration of a frame. Solution We can answer the questions as follows: a. The duration of 1 bit before multiplexing is 1 / 1 kbps, or 0.001 s (1 ms). b. The rate of the link is 4 times the rate of a connection, or 4 kbps. 6.30
Example 6.7 (continued) c. The duration of each time slot is one-fourth of the duration of each bit before multiplexing, or 1/4 ms or 250 μs. Note that we can also calculate this from the data rate of the link, 4 kbps. The bit duration is the inverse of the data rate, or 1/4 kbps or 250 μs. d. The duration of a frame is always the same as the duration of a unit before multiplexing, or 1 ms. We can also calculate this in another way. Each frame in this case has four time slots. So the duration of a frame is 4 times 250 μs, or 1 ms. 6.31
Interleaving The process of taking a group of bits from each input line for multiplexing is called interleaving. We interleave bits (1 - n) from each input onto one output. 6.32
6.33 Figure 6.15 Interleaving
Example 6.8 Four channels are multiplexed using TDM. If each channel sends 100 bytes /s and we multiplex 1 byte per channel, show the frame traveling on the link, the size of the frame, the duration of a frame, the frame rate, and the bit rate for the link. Solution The multiplexer is shown in Figure 6.16. Each frame carries 1 byte from each channel; the size of each frame, therefore, is 4 bytes, or 32 bits. Because each channel is sending 100 bytes/s and a frame carries 1 byte from each channel, the frame rate must be 100 frames per second. The bit rate is 100 32, or 3200 bps. 6.34
6.35 Figure 6.16 Example 6.8
Example 6.9 A multiplexer combines four 100-kbps channels using a time slot of 2 bits. Show the output with four arbitrary inputs. What is the frame rate? What is the frame duration? What is the bit rate? What is the bit duration? Solution Figure 6.17 shows the output (4x100kbps) for four arbitrary inputs. The link carries 400K/(2x4)=50,000 2x4=8bit frames per second. The frame duration is therefore 1/50,000 s or 20 μs. The bit duration on the output link is 1/400,000 s, or 2.5 μs. 6.36
6.37 Figure 6.17 Example 6.9
6.38 Synchronization To ensure that the receiver correctly reads the incoming bits, i.e., knows the incoming bit boundaries to interpret a 1 and a 0, a known bit pattern is used between the frames. The receiver looks for the anticipated bit and starts counting bits till the end of the frame. Then it starts over again with the reception of another known bit. These bits (or bit patterns) are called synchronization bit(s). They are part of the overhead of transmission.
6.39 Figure 6.22 Framing bits
Example 6.10 We have four sources, each creating 250 8-bit characters per second. If the interleaved unit is a character and 1 synchronizing bit is added to each frame, find (a) the data rate of each source, (b) the duration of each character in each source, (c) the frame rate, (d) the duration of each frame, (e) the number of bits in each frame, and (f) the data rate of the link. Solution We can answer the questions as follows: a. The data rate of each source is 250 8 = 2000 bps = 2 kbps. 6.40
Example 6.10 (continued) b. Each source sends 250 characters per second; therefore, the duration of a character is 1/250 s, or 4 ms. c. Each frame has one character from each source, which means the link needs to send 250 frames per second to keep the transmission rate of each source. d. The duration of each frame is 1/250 s, or 4 ms. Note that the duration of each frame is the same as the duration of each character coming from each source. e. Each frame carries 4 characters and 1 extra synchronizing bit. This means that each frame is 4 8 + 1 = 33 bits. 6.41
Example 6.11 Two channels, one with a bit rate of 100 kbps and another with a bit rate of 200 kbps, are to be multiplexed. How this can be achieved? What is the frame rate? What is the frame duration? What is the bit rate of the link? Solution We can allocate one slot to the first channel and two slots to the second channel. Each frame carries 3 bits. The frame rate is 100,000 frames per second because it carries 1 bit from the first channel. The bit rate is 100,000 frames/s 3 bits per frame, or 300 kbps. 6.42
6.43 Figure 6.23 Digital hierarchy
6.44 Table 6.1 DS and T line rates
6.45 Figure 6.24 T-1 line for multiplexing telephone lines
6.46 Figure 6.25 T-1 frame structure
6.47 Table 6.2 E line rates
Inefficient use of Bandwidth Sometimes an input link may have no data to transmit. When that happens, one or more slots on the output link will go unused. That is wasteful of bandwidth. 6.48
6.49 Figure 6.18 Empty slots
6.50 Figure 6.26 TDM slot comparison
Chapter 4 Digital Transmission 4.51 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
4-1 DIGITAL-TO-DIGITAL CONVERSION 4.52 In this section, we see how we can represent digital data by using digital signals. The conversion involves three techniques: line coding, block coding, and scrambling. Line coding is always needed; block coding and scrambling may or may not be needed. opics discussed in this section: Line Coding Line Coding Schemes Block Coding Scrambling
Line Coding Converting a string of 1 s and 0 s (digital data) into a sequence of signals that denote the 1 s and 0 s. For example a high voltage level (+V) could represent a 1 and a low voltage level (0 or -V) could represent a 0. 4.53
4.54 Figure 4.1 Line coding and decoding
Mapping Data symbols onto Signal levels 4.55 A data symbol (or element) can consist of a number of data bits: 1, 0 or 11, 10, 01, A data symbol can be coded into a single signal element or multiple signal elements 1 -> +V, 0 -> -V 1 -> +V and -V, 0 -> -V and +V The ratio r is the number of data elements carried by a signal element.
Relationship between data rate and signal rate The data rate defines the number of bits sent per sec - bps. It is often referred to the bit rate. The signal rate is the number of signal elements sent in a second and is measured in bauds. It is also referred to as the modulation rate. Goal is to increase the data rate whilst reducing the baud rate. 4.56
4.57 Figure 4.2 Signal element versus data element
Example 4.1 A signal is carrying data in which one data element is encoded as one signal element ( r = 1). If the bit rate is 100 kbps, what is the average value of the baud rate if c is between 0 and 1? Solution We assume that the average value of c is 1/2. The baud rate is then 4.58
Note Although the actual bandwidth of a digital signal is infinite, the effective bandwidth is finite. 4.59
Example 4.2 The maximum data rate of a channel (see Chapter 3) is Nmax = 2 B log 2 L (defined by the Nyquist formula). Does this agree with the previous formula for N max? Solution A signal with L levels actually can carry log 2 L bits per level. If each level corresponds to one signal element and we assume the average case (c = 1/2), then we have 4.60
4.61 Considerations for choosing a good signal element referred to as line encoding Baseline wandering - a receiver will evaluate the average power of the received signal (called the baseline) and use that to determine the value of the incoming data elements. If the incoming signal does not vary over a long period of time, the baseline will drift and thus cause errors in detection of incoming data elements. A good line encoding scheme will prevent long runs of fixed amplitude.
Line encoding C/Cs DC components - when the voltage level remains constant for long periods of time, there is an increase in the low frequencies of the signal. Most channels are bandpass and may not support the low frequencies. This will require the removal of the dc component of a transmitted signal. 4.62
Line encoding C/Cs Self synchronization - the clocks at the sender and the receiver must have the same bit interval. If the receiver clock is faster or slower it will misinterpret the incoming bit stream. 4.63
4.64 Figure 4.3 Effect of lack of synchronization
Example 4.3 In a digital transmission, the receiver clock is 0.1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 kbps? How many if the data rate is 1 Mbps? Solution At 1 kbps, the receiver receives 1001 bps instead of 1000 bps. At 1 Mbps, the receiver receives 1,001,000 bps instead of 1,000,000 bps. 4.65
Line encoding C/Cs Error detection - errors occur during transmission due to line impairments. Some codes are constructed such that when an error occurs it can be detected. For example: a particular signal transition is not part of the code. When it occurs, the receiver will know that a symbol error has occurred. 4.66
Line encoding C/Cs Noise and interference - there are line encoding techniques that make the transmitted signal immune to noise and interference. This means that the signal cannot be corrupted, it is stronger than error detection. 4.67
Line encoding C/Cs Complexity - the more robust and resilient the code, the more complex it is to implement and the price is often paid in baud rate or required bandwidth. 4.68
4.69 Figure 4.4 Line coding schemes
Unipolar All signal levels are on one side of the time axis - either above or below NRZ - Non Return to Zero scheme is an example of this code. The signal level does not return to zero during a symbol transmission. Scheme is prone to baseline wandering and DC components. It has no synchronization or any error detection. It is simple but costly in power consumption. 4.70
4.71 Figure 4.5 Unipolar NRZ scheme
Polar - NRZ The voltages are on both sides of the time axis. Polar NRZ scheme can be implemented with two voltages. E.g. +V for 1 and -V for 0. There are two versions: NZR - Level (NRZ-L) - positive voltage for one symbol and negative for the other NRZ - Inversion (NRZ-I) - the change or lack of change in polarity determines the value of a symbol. E.g. a 1 symbol inverts the polarity a 0 does not. 4.72
4.73 Figure 4.6 Polar NRZ-L and NRZ-I schemes
Note In NRZ-L the level of the voltage determines the value of the bit. In NRZ-I the inversion or the lack of inversion determines the value of the bit. 4.74
Note NRZ-L and NRZ-I both have an average signal rate of N/2 Bd. 4.75
Note NRZ-L and NRZ-I both have a DC component problem and baseline wandering, it is worse for NRZ-L. Both have no self synchronization &no error detection. Both are relatively simple to implement. 4.76