MSE课件-数据通信ch08.ppt

Chapter 8Multiplexing Many to one/one to many Types of multiplexing Telephone system Example 6 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 the next Figure. Example 7 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 The next Figure shows the output for four arbitrary inputs. Framing Bits 帧定位比特：由于在同步时分复用中各帧内的时间片顺序不变，因而在帧头上只需要很少的额外开销，一个或若干各个同步比特，这些比特称为帧定位比特。 帧定位比特具有一定格式，并且每帧都有，以便于多路分解器根据输入流进行同步，从而可以精确地分离各时间片。 多数情况下由一比特构成，0、1交替出现。 Data Rate 假设：一条同步时分复用链路上由四个输入信源，传输流安字符交织，每个信源每秒产生250个字符，每帧容纳每个信源的一个字符，则传输通路每秒能容纳250帧。 若每个字符有8个比特，则每帧有33个比特。 则每个设备的数据产生率为2000bps，线路上的负载是8250bps(8000比特数据，250比特开销)。 Example 8 We have four sources, each creating 250 characters per second. If the interleaved unit is a character and 1 synchronizing bit is added to each frame, find (1) the data rate of each source, (2) the duration of each character in each source, (3) the frame rate, (4) the duration of each frame, (5) the number of bits in each frame, and (6) the data rate of the link. Solution 1. The data rate of each source is 2000 bps = 2 Kbps. 2. The duration of a character is 1/250 s, or 4 ms. 3. The link needs to send 250 frames per second. 4. The duration of each frame is 1/250 s, or 4 ms. 5. Each frame is 4 x 8 + 1 = 33 bits. 6. The data rate of the link is 250 x 33, or 8250 bps. Example 9 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 t