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Data communication and networking by william stallings pdf

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William Stallings. Upper Saddle River Copies of figures from the book in PDF format Data Communications and Networking for Today's Enterprise communications over long-distance networks. covers the material in the Computer Communication Networks course of the book in PDF (Adobe Acrobat ) format, and sign-up information for the book's . William Stallings. —Computer, terminal, phone, etc. • A collection of nodes and connections is a communications network. • Data routed by being switched from node to node.


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Stallings, William. Data and computer communications/William Stallings.—Tenth edition. Data Communications and Networking for Today's Enterprise 9. William Stallings Data and Computer Communications 7th Edition. 83 Pages· · data communication and networking by behrouz a. forouzan 4th edition David Evans†, Paul Gruba, Justin Zobel · Download PDF Chapter. Page. pdf. Data and Communication Network by -william stallings 5th. Pages transparency masters of figures in the book in PDF (Adobe Acrobat) format, and.

Each multiplexer also acts as a buffer. The input of cells to the VCC is monitored by the network to ensure that the negotiated parameters are not violated. Forum specifications. Checksum The start bit is the synchronization event, but it must be recognizable.

When transmitting video information, compression ratios can range as high as Interlacing is a method of reducing the bandwidth requirements for video transmissions. When using analog communications, channel capacity is measured in hertz Hz where 1 Hz equals oscillations per second.

Common examples of data include text and numerical information. Raster graphics involves the use of binary codes to represent object type, size, and orientation. Which of the following represents a digital form of information? Which of the following represents the most basic unit of digital information? Standard voice telephone lines, such as those found in residences, limit bandwidth to: A stereo compact disc typically requires the bandwidth for each channel to be: When using data communications with 8-bit codes, the number of alphabetic symbols A.

The time interval between when a user presses a key and when the result of that action arrives at his or her workstation is called the: In digital systems, the information rate and the capacity of a digital channel are measured in: Analog information sources include: UTF-8 B.

IRA C. Morse code D. The number of different characters that can be represented in the International Reference Alphabet text code is: IRA D. UCST RGB B. Color B. Black and white C. Pixilated D. Grayscale While an antenna this size is impractical, the U.

Defense Department has considered using large parts of Wisconsin and Michigan to make an antenna many kilometers in diameter. The received power will increase by a factor of 4 4. Thus, the higher the frequency, the higher power is needed to obtain a given SNR. Power is much more readily available at earth stations than at satellites. Therefore, it makes more sense to put the higher power requirements on the earth stations than on the satellites.

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Using Equation 4. The available received signal power is 20 — A lack of high-frequency components means that less bandwidth is required for transmission.

In addition, lack of a direct-current dc component means that ac coupling via transformer is possible. The magnitude of the effects of signal distortion and interference depend on the spectral properties of the transmitted signal.

Encoding can be used to synchronize the transmitter and receiver. Error detection: It is useful to have some error detection capability built into the physical signaling encoding scheme. Signal interference and noise immunity: Certain codes exhibit superior performance in the presence of noise.

Cost and complexity: The higher the signaling rate to achieve a given data rate, the greater the cost. Some codes require a signaling rate that is in fact greater than the actual data rate. The data themselves are encoded as the presence or absence of a signal transition at the beginning of the bit time.

A transition low to high or high to low at the beginning of a bit time denotes a binary 1 for that bit time; no transition indicates a binary 0.

The binary 1 pulses must alternate in polarity. For pseudoternary, a binary 1 a is represented by the absence of a line signal, and a binary 0 by alternating positive and negative pulses. In the Manchester code, there is a transition at the middle of each bit period; a low-to-high transition represents a 1, and a high-to-low transition represents a 0. In differential Manchester, the midbit transition is used only to provide clocking. The encoding of a 0 is represented by the presence of a transition at the beginning of a bit period, and a 1 is represented by the absence of a transition at the beginning of a bit period.

The filling sequence must be recognized by the receiver and replaced with the original data sequence. The filling sequence is the same length as the original sequence, so there is no data rate penalty. This approach is susceptible to sudden gain changes and is rather inefficient.

For PM, the phase is proportional to the modulating signal. For FM, the derivative of the phase is proportional to the modulating signal. Map this level as indicated by the definition for 1 and 0 for each of the other codes. Second, under worst case, E-NRZ provides a minimum of one transition for every 14 bits, reducing the synchronization problem.

Third, the parity bit provides an error check.

The disadvantages of E-NRZ are added complexity and the overhead of the extra parity bit. Bipolar-AMI 5. For AMI, positive and negative pulses are used alternately for binary 1.

The pulse in position 1 represents the third binary 1 in the data stream and should have a positive value. First consider NRZ-L. For the remaining codes, one must first determine the average number of pulses per bit.

For example, for Biphase-M, there is an average of 1. These higher components cause the signal to change more rapidly over time. Hence, DM will suffer from a high level of slope overload noise.

The demodulator portion of a modem expects to receive a very specific type of waveform e. Thus, it would not function as the coder portion of a codec. The case against using a codec in place of a modem is less easily explained, but the following intuitive argument is offered.

If the decoder portion of a codec is used in place of the modulator portion of a modem, it must accept an arbitrary bit pattern, interpret groups of bits as a sample, and produce an analog output. Some very wide value swings are to be expected, resulting in a strange-looking waveform.

Given the effects of noise and attenuation, the digital output produced at the receiving end by the coder portion of the codec will probably contain many errors. The actual step size, in volts, is: Thus the actual maximum quantized voltage is: The normalized step size is 2—8.

The maximum error that can occur is one-half the step size. Therefore, the normalized resolution is: For frequency deviation, recognize that the change in frequency is determined by the derivative of the phase: A stop binary one follows the character. One side transmitter or receiver pulses the line regularly with one short pulse per bit time. The other side uses these regular pulses as a clock. Another alternative is to embed the clocking information in the data signal.

For digital signals, this can be accomplished with Manchester or differential Manchester encoding. For analog signals, a number of techniques can be used; for example, the carrier frequency itself can be used to synchronize the receiver based on the phase of the carrier. That is, it provides more information that can be used to detect errors.

You could design a code in which all codewords are at least a distance of 3 from all other codewords, allowing all single-bit errors to be corrected. Suppose that some but not all codewords in this code are at least a distance of 5 from all other codewords. Then for those particular codewords, but not the others, a double- bit error could be corrected.

For 10, characters, there are 20, extra bits. The file takes 10 frames or additional bits. Ten times as many extra bits and ten times as long for both. Then the maximum effective data rate R is: There are 7 data bits, 1 start bit, 1. Write down a few dozen characters. Since some 1's will intervene before you find that zero, you will have moved the starting point of the framing process. Eventually, you will achieve proper framing.

The stop bit is needed so that the start bit can be recognized as such. The start bit is the synchronization event, but it must be recognizable. The start bit is always a 0, and the stop bit is always a 1, which is also the idle state of the line.

When a start bit occurs, it is guaranteed to be different from the current state of the line. Then a frame is 12T long. Let a clock period be T'. The last bit bit 12 is sampled at For a fast running clock, the condition to satisfy is T The resultant accuracy is 2 minutes in 1 year or: There are more bits that can be in error since the parity bit is now included.

The parity bit may be in error when there are no errors in the corresponding data bits. Therefore, the inclusion of a parity bit with each character would change the probability of receiving a correct message. The modulo 2 scheme is easy to implement in circuitry. It also yields a remainder one bit smaller than binary arithmetic.

We have: Each 1 bit will merge with a 1 bit exclusive-or to produce a 0; each 0 bit will merge with a 0 bit to produce a zero. The CRC bits are The string is sent. The errors are detected. The errors are not detected. The HDLC standard provides the following explanation. The addition of XK L X corresponds to a value of all ones. This addition protects against the obliteration of leading flags, which may be non-detectable if the initial remainder is zero.

The addition of L X to R X ensures that the received, error- free message will result in a unique, non-zero remainder at the receiver. The non-zero remainder protects against the potential non-detectability of the obliteration of trailing flags. The implementation is the same as that shown in Solution 6. At both transmitter and receiver, the initial content of the register is preset to all ones.

The final remainder, if there are no errors, will be For a codeword w to be decoded as another codeword w', the received sequence must be at least as close to w' as to w. Therefore all errors involving t or fewer digits are correctable. Data transmitted by one side are received by the other. In order to operate a synchronous data link without a modem, clock signals need to be supplied. The Transmitter and Receive Timing leads are cross-connected for this purpose.

The beginning and end of each frame must be recognizable. Flow control: The sending station must not send frames at a rate faster than the receiving station can absorb them.

Error control: Bit errors introduced by the transmission system should be corrected. On a multipoint line, such as a local area network LAN , the identity of the two stations involved in a transmission must be specified.

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Control and data on same link: The receiver must be able to distinguish control information from the data being transmitted. Link management: The initiation, maintenance, and termination of a sustained data exchange require a fair amount of coordination and cooperation among stations.

Procedures for the management of this exchange are required.

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With smaller frames, errors are detected sooner, and a smaller amount of data needs to be retransmitted. The window changes dynamically to allow additional packets to be sent. The sliding window flow control technique can send multiple frames before waiting for an acknowledgment.

Efficiency can be greatly improved by allowing multiple frames to be in transit at the same time. Based on stop-and-wait flow control.

A station retransmits on receipt of a duplicate acknowledgment or as a result of a timeout. Go-back-N ARQ: Based on sliding-window flow control. When an error is detected, the frame in question is retransmitted, as well as all subsequent frames that have been previously transmitted. Selective-reject ARQ. When an error is detected, only the frame in question is retransmitted.

Responsible for controlling the operation of the link. Frames issued by the primary are called commands. Secondary station: Operates under the control of the primary station.

Frames issued by a secondary are called responses. The primary maintains a separate logical link with each secondary station on the line. Combined station: Combines the features of primary and secondary. A combined station may issue both commands and responses. Used with an unbalanced configuration.

The primary may initiate data transfer to a secondary, but a secondary may only transmit data in response to a command from the primary. Asynchronous balanced mode ABM: Used with a balanced configuration. Either combined station may initiate transmission without receiving permission from the other combined station. Asynchronous response mode ARM: The secondary may initiate transmission without explicit permission of the primary.

The primary still retains responsibility for the line, including initialization, error recovery, and logical disconnection. This is achieved by bit stuffing.

Data and Computer Communications 10th Edition William Stallings Test Bank

Additionally, flow and error control data, using the ARQ mechanism, are piggybacked on an information frame. Supervisory frames S-frames provide the ARQ mechanism when piggybacking is not used. Unnumbered frames U-frames provide supplemental link control functions. Because only one frame can be sent at a time, and transmission must stop until an acknowledgment is received, there is little effect in increasing the size of the message if the frame size remains the same.

All that this would affect is connect and disconnect time. This would lower line efficiency, because the propagation time is unchanged but more acknowledgments would be needed.

Data and Computer Communications (Eighth Edition)

For a given message size, increasing the frame size decreases the number of frames. This is the reverse of b. Then, using Equation 7. Using Equation 7. The first frame takes 10 msec to transmit; the last bit of the first frame arrives at B 20 msec after it was transmitted, and therefore 30 msec after the frame transmission began.

It will take an additional 20 msec for B's acknowledgment to return to A. Thus, A can transmit 3 frames in 50 msec.

B can transmit one frame to C at a time. Thus, the total number of frames transmitted without an ACK is: The REJ improves efficiency by informing the sender of a bad frame as early as possible.

Station A sends frames 0, 1, 2 to station B. Station B receives all three frames and cumulatively acknowledges with RR 3. Because of a noise burst, the RR 3 is lost. A times out and retransmits frame 0.

B has already advanced its receive window to accept frames 3, 0, 1, 2. Thus it assumes that frame 3 has been lost and that this is a new frame 0, which it accepts. The sender never knows that the frame was not received, unless the receiver times out and retransmits the SREJ. Also from the standard: This would contradict the intent of the SREJ frame or frames.

From the beginning of the transmission of the first frame, the time to receive the acknowledgment of that frame is: However, for simplicity, bit stuffing is used on this field.

When a flag is used as both an ending and starting flag that is, one 8-bit pattern serves to mark the end of one frame and the beginning of the next , then a single-bit error in that flag alters the bit pattern so that the receiver does not recognize the flag.

Accordingly, the received assumes that this is a single frame. If a bit error somewhere in a frame between its two flags results in the pattern , then this octet is recognized as a flag that delimits the end of one frame and the start of the next frame. Any discrepancies result in discarding the frame. Bit-stuffing at least eliminates the possibility of a long string of 1's. This is the number of the next frame that the secondary station expects to receive.

The same frame format as for LAPB is used, with one additional field: The LAPB control field includes, as usual, a sequence number unique to that link. The MLC field performs two functions. First, LAPB frames sent out over different links may arrive in a different order from that in which they were first constructed by the sending MLP. Second, if repeated attempts to transmit a frame over one link fails, the DTE or DCE will send the frame over one or more other links.

The MLP sequence number is needed for duplicate detection in this case. In essence, a transmitter must subtract the echo of its own transmission from the incoming signal to recover the signal sent by the other side. This explains the basic difference between the 1. A scheme such as depicted in Figure 8. Each Hz signal can be sampled at a rate of 1 kHz. If 4-bit samples are used, then each signal requires 4 kbps, for a total data rate of 16 kbps.

This scheme will work only if the line can support a data rate of 16 kbps in a bandwidth of Hz. In time-division multiplexing, the entire channel is assigned to the source for a fraction of the time. If there is spare bandwidth, then the incremental cost of the transmission can be negligible.

The new station pair is simply added to an unused subchannel. If there is no unused subchannel it may be possible to redivide the existing subchannels creating more subchannels with less bandwidth.

If, on the other hand, a new pair causes a complete new line to be added, then the incremental cost is large indeed. What the multiplexer receives from attached stations are several bit streams from different sources. What the multiplexer sends over the multiplexed transmission line is a bit stream from the multiplexer. As long as the multiplexer sends what can be interpreted as a bit stream to the demultiplexer at the other end, the system will work.

The multiplexer, for example, may use a self-clocking signal. The incoming stream may be, on the other hand, encoded in some other format. The multiplexer receives and understands the incoming bits and sends out its equivalent set of multiplexed bits.

In synchronous TDM, using character interleaving, the character is placed in a time slot that is one character wide. The character is delimited by the bounds of the time slot, which are defined by the synchronous transmission scheme. Thus, no further delimiters are needed. When the character arrives at its destination, the start and stop bits can be added back if the receiver requires these.

TDM's focus is on the medium rather than the information that travels on the medium. Its services should be transparent to the user.

It offers no flow or error control. These must be provided on an individual-channel basis by a link control protocol. The actual bit pattern is If a receiver gets out of synchronization it can scan for this pattern and resynchronize. This pattern would be unlikely to occur in digital data.

Analog sources cannot generate this pattern. It corresponds to a sine wave at 4, Hz and would be filtered out from a voice channel that is band limited. One SYN character, followed by 20 bit terminal characters, followed by stuff bits. The available capacity is 1. This is a practical limit based on the performance characteristics of a statistical multiplexer.

If the receiver is on the framing pattern no searching , the minimum reframe time is 12 frame times the algorithm takes 12 frames to decide it is "in frame". Hence it must search the maximum number of bits 55 to find it. Each search takes 12T f. Assuming the system is random, the reframing is equally to start on any bit position. Hence on the average it starts in the middle or halfway between the best and worst cases.

Therefore, the channel cost will be only one-fourth, since one channel rather than four is now needed. The same reasoning applies to termination charges. The present solution requires eight low speed modems four pairs of modems.

The new solution requires two higher-speed modems and two multiplexers. The reliability of the multiplexed solution may be somewhat less. The new system does not have the redundancy of the old system.

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A failure anywhere except at the terminals will cause a complete loss of the system. Each multiplexer also acts as a buffer. It can accept bits in asynchronous form, buffer them and transmit them in synchronous form, and vice versa. Assume a continuous stream of STDM frames. If a delimiter is used, bit or character-stuffing may be needed. Only a recipient who knows the spreading code can recover the encoded information. A receiver, hopping between frequencies in synchronization with the transmitter, picks up the message.

Each user uses a different spreading code. The receiver picks out one signal by matching the spreading code. Thus, to achieve the desired SNR, the signal must be spread so that 56 KHz is carried in very large bandwidths. Thus a far higher SNR is required without spread spectrum. Period of the PN sequence is 15 b. MFSK c. Same as for Problem 9. This is from the example in Section 6. We need three more sets of 8 frequencies.

The second set can start at kHz, with 8 frequencies separated by 50 kHz each. The third set can start at kHz, and the fourth set at kHz. The first generator yields the sequence: The second generator yields the sequence: