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# Tele3113 wk9tue

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• 1. TELE3113 Analogue and Digital Communications – Multiplexing Wei Zhang w.zhang@unsw.edu.au School of Electrical Engineering and Telecommunications The University of New South WalesTELE3113 - Multiplexing 22 Sept. 2009 p. -1
• 2. Multiplexing Multiplexing: a number of independent signals are combined (without interfering each other) into a composite signal for transmission over a communication channel share the resources among users maximize the total bandwidth utilization of transmission channel Multiplexer: combine the multiple signals into one composite (multiplexed) signal Demultiplexer: separate individual signals from the multiplexed signalTELE3113 - Multiplexing 22 Sept. 2009 p. -2
• 3. Frequency Division Multiplexing (FDM)LPF are neededbefore mixingoperation to limitthe bandwidth ofthe input signalsTELE3113 - Multiplexing 22 Sept. 2009 p. -3
• 4. Time Division Multiplexing (TDM) Each information signal is a PCM signal (sampled, quantized and pulse-coded) Bit period Time frame N 2 1 N 2 1 N 2 1 N 2 1 time Sampling rate at each source: fs TDM frame rate: fs or TDM frame period: Ts=1/ fs T 1 1 Nf Bit period Tx = s = minimum required bandwidth BTDM ≥ = s N Nf 2Tx 2 p. -4TELE3113 - Multiplexing 22 Sept. 2009 s
• 5. Time Division Multiplexing (TDM) If each information source is a PAM signal: x1(t) x2(t) Ts time Considering a N-channel PAM system, each signal is band-limited to B Hz so that the sampling period Ts is 1 Ts = 2B Thus, time spacing Tx between adjacent samples in the time-multiplexed waveform is Ts Tx = N 1 The minimum bandwidth needed for the TDM signal transmission BTDM ≥ . 2TxTELE3113 - Multiplexing 22 Sept. 2009 p. -5
• 6. Time Division Multiplexing (TDM) Example: In a two-channel PAM system, channel 1 handles 0-8 kHz signals and channel 2 handles 0-10 kHz signals. The two channels are sampled at equal intervals of time using very narrow pulse at the lowest frequency that is theoretically adequate. The sampled signals are time-multiplexed before transmission. In order to sample channel 2 adequately, we must take samples at least 20-kHz rate. So, the TDM frame interval Ts is 1/(20kHz) = 50µs. As there are only two channels, each of them can at most occupy ½ of the frame interval. Thus, bit period Tx= Ts/2=25µs the clock rate of the TDM system is 1/Tx=40kHz. The minimum requirement of the channel bandwidth is 1/(2Tx)= 20kHz If these channels were frequency-multiplexed, instead of time-multiplexed The bandwidth required will be 2(10+8) = 36kHz (AM) or (10+8)=18kHz (SSB)TELE3113 - Multiplexing 22 Sept. 2009 p. -6
• 7. Time Division Multiplexing (TDM) TDM with analog and digital inputs:TELE3113 - Multiplexing 22 Sept. 2009 p. -7
• 8. Digital Multiplexing for Digital Telephone • Voice bandwidth ~4kHz sampling rate=8kHz • 8-bit per PCM word (voice channel) • Twenty-four 8-bit voice channels are time-multiplexed to give 192 bits per frame. • One extra bit is inserted to give frame synchronization, yielding 193 bits/frame. • Each channel is sampled at an 8-kHz rate resulting in a frame interval of 125 µsec.TELE3113 - Multiplexing 22 Sept. 2009 p. -8