【数据通信与网络】Lecture4-ModulationAndDemodulation分解.ppt

Data Communications and Networking;;4.1 Basic concepts;◆ Carrier （载波）;Bit Rate and Baud Rate;Nyquist Bit Rate ;Channel Capacity—Shannon’s Formula;Example: A noisy channel, the SNR is 20dB and the bandwidth is 3.1kHz. Calculate the channel capacity. Solution: B = 3100 Hz 10log10(S/N) = 20 dB then, S/N = 100 C = 3100 log2(1 + 100) = 20.6 kb/s Telephone line B: 3100-3500 Hz SNR: 33-39 dB C ≈ 33-45 kb/s (What about ADSL and 56K modems?);4.2 Basic Digital Modulation;(1) ASK; ASK is highly susceptible to noise interference, because noise usually affects the amplitude.; System Bandwidth requirement for ASK: BW=(1+d)Nbaud Nbaud —— the baud rate, d —— a factor related to the condition of the line ( d min = 0 );Example for ASK;Waveform of Modulation sample; The anti-jamming ability of FSK is better than ASK The bandwidth required for FSK is wider than ASK;Modulation Bandwidth for FSK;Example 1: Find the minimum bandwidth for an FSK signal transmitting at 2000bps.Transmission is in half-duplex mode and the carriers must be separated by 3000Hz.; The phase of the carrier signal is varied to represent binary 1 or 0. Both amplitude and frequency remain constant.;Performance for PSK;Relative Modulation; 4-PSK(QPSK): with 4 different phases, each phase represents 2 bits. 8-PSK: with 8 different phases, each phase represents 3 bits. 16-PSK: with 16 different phases, each phase represents 4 bits.;Waveform of 4-PSK;4-PSK Characteristics; The performance of anti-jamming of ASK is poor, so we cannot use N-ASK. The bit rate of N-PSK is several times greater than that of ASK given the same bandwidth.;Example 1: Find the minimum bandwidth for a 4-PSK signal at 9600 bps. transmission is in half-duplex mode.;—— ( Quadrature Amplitude Modulation, 正交幅度调制); —— Combining ASK and PSK in such a way that we have x variations in phase and y vari