TUTORIAL FOR CHAPTER 8

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1 TUTORIAL FOR CHAPTER 8 PROBLEM 1) The informaiton in four analog signals is to be multiplexed and transmitted over a telephone channel that has a 400 to 3100 Hz bandpass. Each of the analog baseband signals is bandlimited to 500 Hz. Design a communication system (block diagram) that will allow the transmission of these four sources over the telephone channel using frequency division multiplexing with SSB (Single Sideband) subcarriers. Show the block diagram of the complete system, including the transmission channel,and reception portions. Include the bandwidths of the signals at the various points in the system. The available bandwidth is. M 1 (t) S 1 (t) M 2 (t) S 2 (t) M b (t) M 3 (t) S 3 (t) FDM Signal Transmitter f e S(t) M 4 (t) S 4 (t) (a) Transmitter S 1 (t) M 1 (t) Bandpass filter f 1 Demodulator f 1 S 2 (t) M 2 (t) Bandpass filter f 2 Demodulator f 2 S(t) M b (t) Main Receiver FDM S 3 (t) M 3 (t) Bandpass filter f 3 Demodulator f 3 Bandpass filter f 4 S 4 (t) Demodulator f 4 M 4 (t) (b) Receiver

2 f 100 Hz 500 Hz 800 Hz 1300 Hz 1500 Hz 2000 Hz 2200Hz 2700 Hz f 1 f 2 f 3 f 4 PROBLEM 2) Twenty four voice signals are to be multiplexed and transmitted over twisted pair. What is the bandwidth requires for FDM? Assuming a bandwidth efficiency (ratio od data rate to transmission bandwidth) of 1 bps/hz, what is the bandwidth required for TDM using PCM? Assuming 4 KHz per voice signal, the required bandwidth for FDM is With PCM, each voice signal requires a data rate of 64 Kbps, for a total data rate of. At 1 bps/hz, this requires a bandwidth of MHz. PROBLEM 3) Assume that you are to design a TDM carrier, say DS-Y89, to support 30 voice channels using 6 bit samples and a structure similar to DS-1. Determine the required bit rate. ; 6 bits/sample. Thus 30 voice channels : 1 synchronous bit/channel: 1 synchronous bit/frame: Total = 1688 Kbps

3 PROBLEM 4) Find the number of the following devices that could be accomodated by a T1-type TDM line if 1% of the T1 line capacity is reserved for synchronization purposes. a) computer terminals. b) 9600 bps computer output parts. How would these numbers change if each of the sources were transmitting on an average of 10% of the time and a statistical multiplexer was used? We know that the capacity of the T1 line is 1544 Mbps. The available capacity is a) AC/300=5.095 computer terminals. b) AC/9600=159 computer output ports. If the sources were active only 10% of the time, a statistical multiplexer could be used to boost the number of devices by a factor of about seven or eight in each case. This is a practical limit based on the performance characteristics of a statistical multiplexer. PROBLEM 5) Given multiplexing configuration and input data streams. Determine the multiplexed data stream. Use the Data Link Control on TDM channels. Input 1 Output 1 Input 2 Output 2 Input 3 Output 3 Input 4 Output 4 Input 1...F 1 f 1 f 1 d 1 d 1 d 1 C 1 A 1 F 1 Input 2...F 2 f 2 d 2 d 2 d 2 d 2 C 2 A 2 F 2 Input 3...F 3 f 3 f 3 f 3 d 3 d 3 C 3 A 3 F 3 Input data streams. Input 4...F 4 f 4 d 4 d 4 d 4 d 4 d 4 A 4 F 4 F 1 F 2 F 3 F 4 f 1 f 2 f 3 f 4 f 1 d 2 f 3 d 4 d 1 d 2 f 3 d 4 d 1 d 2 d 3 d 4 d 1 d 2 d 3 d 4 C 1 C 2 C 3 d 4 A 1 A 2 A 3 A 4 F 1 F 2 F 3 F 4 : Multiplexed data stream.

4 PROBLEM 6) Using the DS-1 format, what id the control sşgna data rate for each voice channel? There is one control bşt per channel per six frames. Each frame lasts 125 µsec. PROBLEM 7) Ten 9600 bps lines are to be multiplexed using TDM. Ignoring overhead bits in the TDM frame, what is the total capacity required for synchronous TDM? Assuming that we wish to limit average link utilization of 0,8, and assuing that each link is busy 50% of the time, what is the capacity required to statistical TDM? Synchronous TDM: Statistical TDM: PROBLEM 8) A character-interleaved time, division multiplexer is used to combine the data streams of a number of 110 bps asynchronous terminals for data transmission over a 2400 bps digital line. Each terminal sends asynchronous characters consisting of 7 data bits, 1 parity bit, 1 start bit, and 2 stop bits. Assume that one synchronization character is sent every 10 data characters and, in additioni at least 3% of the line capacity is reserved for pulse to 3% accomodate speed variations from the various terminals. a) Determine the number of bits per character. b) Determine the number of terminals that can be accomodated by the multiplexer. c) Sketch a possible framing pattern for the multiplexer. a) bits/character b) Available capacity = bps If we use 20 terminals sending one character at a time in TDM plus a synchronouzation character,the total capacity used is: c) One SYN character, followed by stuff bits.

5 PROBLEM 9) Draw a block diagram for TDM PCM system that will accommodate for, synchronous, digital inputs and one analog input with a bandwidth of 500 Hz.Assume that the analog samples will be encoded into 4-bit PCM words. The structure is with one analog signal and four digital signals. The 500 Hz analog signal is converted into a PAM signal at 1 KHz; with 4-bits encoding, this becomes a 4 Kbps PCM digital bit stream. A simple multiplexing technique is to use a 260 bit frame, with 200 bits for the analog signal and 15 bits for each digital signa, transmitted at a rate of 5.2 Kbps or 20 frames per second. Thus the PCM source transmits an (20 frames/sec) x (15 bits/frame)=. Analog: 4 bit From source 1 Khz TDM PAM Signal A Hz TDM PAM Signal 4 Kbps Digital: From S.2 Pulse 5.2 Kbps (200 bits x 20)=4000 bps From S.3 Pulse 5.2 Kbps Scan operator TDM PCM 260 bit/frame From S.4 Pulse 5.2 Kbps From S.5 Pulse 5.2 Kbps

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