Data and computer communications eighth edition william stallings ppt

This would lower line efficiency, because the propagation time is unchanged but more acknowledgments would be needed. 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. 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. This would contradict the intent of the SREJ frame or frames. 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 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. 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: 1, 6, 10, 8, 9, 2, 12, 7, 3, 5, 4, 11, 1,. The second generator yields the sequence: 1, 7, 10, 5, 9, 11, 12, 6, 3, 8, 4, 2, 1,. Because of the patterns evident in the second half of the latter sequence, most people would consider it to be less random than the first sequence. See [KNUT98], page 13 for a discussion.

As discussed in the answer to Problem 9. Often, a and c are chosen to create a sequence of alternating even and odd integers. The simulation depends on counting the number of pairs of integers whose greatest common divisor is 1. With truly random integers, one-fourth of the pairs should consist of two even integers, which of course have a gcd greater than 1. This never occurs with sequences that alternate between even and odd integers.

For a further discussion, see Danilowicz, R. Subscriber line: the link between the subscriber and the network. Exchanges: the switching centers in the network. Trunks: the branches between exchanges. Two stations of different data rates can exchange packets because each connects to its node at its proper data rate. On a packet-switching network, packets are still accepted, but delivery delay increases.

Thus, if a node has a number of packets queued for transmission, it can transmit the higher- priority packets first. These packets will therefore experience less delay than lower-priority packets. In the virtual circuit approach, a preplanned route is established before any packets are sent. Once the route is established, all the packets between a pair of communicating parties follow this same route through the network.

As a smaller packet size is used, there is a more efficient "pipelining" effect, as shown in Figure However, if the packet size becomes too small, then the transmission is less efficient, as shown in Figure Help Preferences Sign up Log in. View by Category Toggle navigation. Products Sold on our sister site CrystalGraphics. Tags: communications computer data stallings william.

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