![The bandwidth of a communication
channel is 12.5 kHz. The S/N ratio is
25 dB. Calculate (a) the maximum
theoretical data rate in bits per
second, (b) the maximum
theoretical channel capacity, and (c)
the number of coding levels N
needed to achieve the maximum
speed. [For part (c), use the yx key
on a scientific calculator.]](https://image.slidesharecdn.com/week-3-230517064636-97701927/85/Week-3-pptx-29-320.jpg)
Week-3.pptx
- 1. Principles of Data Communication ENGR. CHRISTINE T. RASIMO
- 2. Serial Transmission Parallel data transmission is not practical for long-distance communication. Data transfers in long-distance communication systems are made serially; each bit of a word is transmitted one after another
- 3. Expressing the Serial Data Rate The speed of data transfer is usually indicated as number of bits per second (bps or b/s). The speed of serial transmission is, of course, related to the bit time of the serial data. The speed in bits per second, denoted by bps, is the reciprocal of the bit time t, or bps = 1/t. Assume a bit time of 104.17 μs. The speed is bps = 1/104.17 x 10-6 = 9600 bps.
- 4. Expressing the Serial Data Rate If the speed in bits per second is known, the bit time can be found by rearranging the formula: t = 1/bps. For example, the bit time at 230.4 kbps (230,400 bps) is t = 1/230,400 = 4.34 x 10-6 = 4.34 μs.
- 5. Baud Rate Another term used to express the data speed in digital communication systems is baud rate. Baud rate is the number of signaling elements or symbols that occur in a given unit of time, such as 1 s. A signaling element is simply some change in the binary signal transmitted. In many cases, it is a binary logic voltage level change, either 0 or 1, in which case, the baud rate is equal to the data rate in bits per second. In summary, the baud rate is the reciprocal of the smallest signaling interval.
- 6. Baud Rate Bit rate = baud rate x bit per symbol Bit rate = baud rate x log2 S where S = number of states per symbol
- 7. Assume, e.g., a system that represents 2 bits of data as different voltage levels. With 2 bits, there are 22 = 4 possible levels, and a discrete voltage is assigned to each. 00 0V 01 1V 10 2V 11 3V In this system, sometimes called pulse-amplitude modulation (PAM), each of the four symbols is one of four different voltage levels. Each level is a different symbol representing 2 bits.
- 8. Baud Rate Assume, e.g., that it is desired to transmit the decimal number 201, which is binary 11001001. The number can be transmitted serially as a sequence of equally spaced pulses that are either on or off. If each bit interval is 1 μs, the bit rate is 1/1 x 106-6 = 1,000,000 bps (1 Mbps). The baud rate is also 1 million bps. Using the four-level system, we could also divide the word to be transmitted into 2-bit groups and transmit the appropriate voltage level representing each. The number 11001001 would be divided into these groups: 11 00 10 01. Thus the transmitted signal would be voltage levels of 3, 0, 2 and 1 V, each occurring for a fixed interval of, say, 1 μs.
- 9. Based on the formula, Bit rate = baud rate x bit per symbol Bit rate = baud rate x log2 S Bit rate = 1 x 106 bps x 2 bits per symbol = 2 x 106 bps
- 11. A good example of modern PAM is the system used in U.S. digital high-definition television (HDTV). The video and audio to be transmitted are put into serial digital format, then converted to 8- level PAM. Each level can represent one of eight 3-bit combinations from 000 to 111. The symbols occur at a 10,800 rate, producing a net bit rate of 10,800 x 3 = 32.4 Mbps. The PAM signal is then used to amplitude-modulate the transmitter carrier. Part of the lower sideband is suppressed to save spectrum.
- 12. Asynchronous Transmission In asynchronous transmission each data word is accompanied by start and stop bits that indicate the beginning and ending of the word. When no information is being transmitted, the communication line is usually high, or binary 1. In data communication terminology, this is referred to as a mark. To signal the beginning of a word, a start bit, a binary 0 or space, as shown in the figure, is transmitted. The start bit has the same duration as all other bits in the data word. The transmission from mark to space indicates the beginning of the word and allows the receiving circuits to prepare themselves for the reception of the remainder of the bits.
- 13. After the start bit, the individual bits of the word are transmitted. In this case, the 7-bit ASCII code for the letter U, 1010101, is transmitted. Once the last code bit has been transmitted, a stop bit is included. The stop bit may be the same duration as all other bits and again is a binary 1 or mark. In some systems, 2 stop bits are transmitted, one after the other, to signal the end of the word.
- 14. Sequential words transmitted asynchronously There can also be time gaps between characters or groups of characters, as the illustration shows, and thus the stop “bit’’ may be of some indefinite length. The primary disadvantage of asynchronous communication is that the extra start and stop bits effectively slow down data transmission. This is not a problem in low-speed applications such as those involving certain printers and plotters. But when huge volumes of information must be transmitted, the start and stop bits represent a signii cant percentage of the bits transmitted. We call that percentage the overhead of transmission.
- 15. Synchronous Transmission The technique of transmitting each data word one after another without start and stop bits, usually in multiword blocks, is referred to as synchronous data transmission. To maintain synchronization between transmitter and receiver, a group of synchronization bits is placed at the beginning of the block and the end of the block. Each block of data can represent hundreds or even thousands of 1-byte characters. At the beginning of each block is a unique series of bits that identifies the beginning of the block.
- 16. Synchronous Transmission Once the receiving equipment finds these characters, it begins to receive the continuous data, the block of sequential 8-bit words or bytes. At the end of the block, another special ASCII code character, ETX, signals the end of transmission. The receiving equipment looks for the ETX code; detection of this code is how the receiving circuit recognizes the end of the transmission. An error detection code usually appears at the very end of the transmission.
- 17. How is the receiving station keeps track of the individual bits and bytes, especially when the signal is noisy, since there is no clear separation between them?
- 18. How is the receiving station keeps track of the individual bits and bytes, especially when the signal is noisy, since there is no clear separation between them? This is done by transmitting the data at a fixed, known, precise clock rate. Then the number of bits can be counted to keep track of the number of bytes or characters transmitted. For every 8 bits counted, 1 byte is received. The number of received bytes is also counted. Synchronous transmission assumes that the receiver knows or has a clock frequency identical to that of the transmitter clock. Usually, the clock at the receiver is derived from the received signal, so that it is precisely the same frequency as, and in synchronization with, the transmitter clock.
- 19. A block of 256 sequential 12-bit data words is transmitted serially in 0.016 s. Calculate (a) the time duration of 1 word, (b) the time duration of 1 bit, and (c) the speed of transmission in bits per second. 𝑎. 𝑡𝑤𝑜𝑟𝑑 = 0.016 256 = 0.000625 = 625𝜇𝑠 b. 𝑡𝑏𝑖𝑡 = 625𝜇𝑠 12 𝑏𝑖𝑡𝑠 = 52.0833 𝜇𝑠 c. bps = 1 𝑡 = 1 52.0833 𝑥 10−6 = 19.2 𝑘𝑏𝑝𝑠
- 20. Transmission Efficiency - the accuracy and speed with which information, whether it is voice or video, analog or digital, is sent and received over communication media. Hartley’s Law The amount of information that can be sent in a given transmission is dependent on the bandwidth of the communication channel and the duration of transmission.
- 21. Hartley’s Law Stated mathematically, Hartley’s law is C = 2B Here C is the channel capacity expressed in bits per second and B is the channel bandwidth. It is also assumed that there is a total absence of noise in the channel. When noise becomes an issue, Hartley’s law is expressed as 𝐶 = (𝐵 log 2)(1 + 𝑆 𝑁 ) where S/N is the signal-to-noise ratio in power.
- 22. Hartley’s Law The greater the bandwidth of a channel, the greater the amount of information that can be transmitted in a given time. It is possible to transmit the same amount of information over a narrower channel, but it must be done over a longer time. This general concept is known as Hartley’s law, and the principles of Hartley’s law also apply to the transmission of binary data. The greater the number of bits transmitted in a given time, the greater the amount of information that is conveyed. But the higher the bit rate, the wider the bandwidth needed to pass the signal with minimum distortion
- 23. Transmission Media and Bandwidth Coaxial cables have a wide usable bandwidth, ranging from 200 to 300 MHz for smaller cables to 500 MHz to 50 GHz for larger cables. The bandwidth decreases drastically with length. Twisted-pair cable has a narrower bandwidth, from a few kilohertz to over 800 MHz.
- 24. Find the bandwidth of an NRZ with bit interval of 100 ns. Bit rate = 1/t = 1 / 100 ns = 10 Mbps C = 2B B = C / 2 = 10 Mbps / 2 = 5 MHz
- 25. Multiple Coding Levels C = 2B log2 N where N is the number of different encoding levels per time interval. The implication is that for a given bandwidth, the channel capacity, in bits per second, will be greater if more than two encoding levels are used per time interval.
- 26. Two levels or symbols (0 or 1) were used in transmitting the binary signal. The bit or symbol time is 1 μs. The bandwidth needed to transmit this 1,000,000-bps signal can be computed from C = 2B, or B = C/2. Thus a minimum bandwidth of 1,000,000 bps/2 = 500,000 Hz (500 kHz) is needed. Same with, C = 2B log2 N C = 2(500x 103) log2 (2) = 1 x 106 Mbps
- 27. Shannon-Hartley theorem: Another important aspect of information theory is the impact of noise on a signal. As discussed in earlier chapters, increasing bandwidth increases the rate of transmission but also allows more noise to pass, and so the choice of a bandwidth is a tradeoff. The relationship between channel capacity, bandwidth, and noise is summarized in what is known as the Shannon-Hartley theorem: 𝐶 = (𝐵 log 2)(1 + 𝑆 𝑁 ) where C = channel capacity, bps B = bandwidth, Hz S/N = signal-to-noise ratio
- 28. Assume, e.g., that the maximum channel capacity of a voice-grade telephone line with a bandwidth of 3100 Hz and an S/N of 30 dB is to be calculated. First, 30 dB is converted to a power ratio. If dB = 10 log P, where P is the power ratio, then P = antilog (dB/10). Antilogs are easily computed on a scientific calculator. A 30-dB S/N ratio translates to a power ratio of
- 29. The bandwidth of a communication channel is 12.5 kHz. The S/N ratio is 25 dB. Calculate (a) the maximum theoretical data rate in bits per second, (b) the maximum theoretical channel capacity, and (c) the number of coding levels N needed to achieve the maximum speed. [For part (c), use the yx key on a scientific calculator.]
Editor's Notes
- #22: The greater the bandwidth of a channel, the greater the amount of information that can be transmitted in a given time.