Answers of Exercise 2
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Answers of Exercise 2 's Theorem to determine the maximum rate in bits per second at which data across a transmission system that has a bandwidth of 4000 Hz and use four oltage to encode information. What's the maximum rate when encoding the with 16 values of voltage? n M=4, D=2x4000xlog 2 4 =16000bps=16kbps When M=16, D=2x4000xlog 2 16=32000bps=32kbps ble to increase a number of the encoded values without limit in order to inc on speed of system? Why? Assume a bandwidth of a system is 4000 Hz and -noise ratio S/N=1023, What's the maximum rate of the system? due to existence of noise. C=4000xlog 2 (1+1023)=40kbps ) A digital modulator using ASK, PSK or QAM is a digital-to-digital system. lse! A digital modulator is a digital-to-analogy system. bit rate of 4-PSK signal is 2400bps, what’s its baud rate? Answer: 1200 baud baud rate of 256-QAM is 2400 baud, what’s its bit rate? Answer: 19.2kbps te of one digital telephone channel is 64Kbps. If the transmission capacity ical fiber is 2Gbps, how many telephone channel can be multiplexed to the fi is used. 10 9 /(64x10 3 )=31250 telephone channels
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Answers of Exercise 2. 1. Use Nyquist's Theorem to determine the maximum rate in bits per second at which data can be send across a transmission system that has a bandwidth of 4000 Hz and use four values of voltage to encode information. What's the maximum rate when encoding the - PowerPoint PPT Presentation
Transcript of Answers of Exercise 2
Answers of Exercise 2Answers of Exercise 2
1. Use Nyquist's Theorem to determine the maximum rate in bits per second at which data can be send across a transmission system that has a bandwidth of 4000 Hz and use four values of voltage to encode information. What's the maximum rate when encoding the information with 16 values of voltage? Answer: When M=4, D=2x4000xlog24 =16000bps=16kbps When M=16, D=2x4000xlog216=32000bps=32kbps
2. Is it possible to increase a number of the encoded values without limit in order to increase transmission speed of system? Why? Assume a bandwidth of a system is 4000 Hz and a signal-to-noise ratio S/N=1023, What's the maximum rate of the system? Answer: No, due to existence of noise. C=4000xlog2(1+1023)=40kbps
3. (True/false) A digital modulator using ASK, PSK or QAM is a digital-to-digital system. Answer: False! A digital modulator is a digital-to-analogy system.
4. (1) If the bit rate of 4-PSK signal is 2400bps, what’s its baud rate? Answer: 1200 baud (2) If the baud rate of 256-QAM is 2400 baud, what’s its bit rate? Answer: 19.2kbps
5. The bite rate of one digital telephone channel is 64Kbps. If the transmission capacity of a single optical fiber is 2Gbps, how many telephone channel can be multiplexed to the fiber? Assume TDM is used. Answer: 2x109/(64x103)=31250 telephone channels
