Generation of TDM-PAM signal (example) Input signals TDM-PAM signal f 2 f 1 f ( t 3 ) F 1 0 m F 2 F 3 is very complicated. 0 m Low-pass filter Impulse response Transmitted signal f4 = f3( t) hx F 4 = F3 H x ERG2310A: Principles of Communication Systems (2002-2003) 17
Reception of TDM-PAM signal (example) Use Receiver 1 (slide 12) Sampled output 1 0 m Sampled output 2 Outputs of LPF (cutoff frequency= ) Output of LPF 1 Output of LPF 2 f m f 1 f 2 F 1 F 2 0 m 0 m 0 m ERG2310A: Principles of Communication Systems (2002-2003) 18
Problem (Example 7.2.1): Channel 1 of a two-channel PAM system handles 0-8 khz signals; the second channel handles 0-10 khz signals. The two channels are sampled at equal intervals of time using very narrow pulses at the lowest frequency that is theoretically adequate. The sampled signals are time-multiplexed and passed through a lowpass filter before transmission. At the receiver that pulses in each of the two channels are passed through appropriate holding circuits (i.e. sample-and-hold) and low-pass filters. a) What is the minimum clock frequency of the PAM system? b) What is the minimum cutoff frequency of the LPF before transmission that will preserve the amplitude information on the output pulses? c) What would be the minimum bandwidth if these channels were frequency-multiplexed, using normal AM and SSB techniques? d) Assume the signal in channel 1 is sin 5000π t and that in channel 2 is sin10000π t. Sketch these signals; sketch the wave shapes at the input to the first LPF, at the filter output, and at the output of the sample-and-hold circuit ant output of the LPF in channel 2. ERG2310A: Principles of Communication Systems (2002-2003) 19
Solution: a) Signal bandwidth: fm1 = 8 khz (Ch. 1) and f m2 = 10 khz (Ch. 2) Nyquist sampling rate: f =16 N1 khz (Ch. 1) and f = 20 N 2 khz (Ch. 2) For two-channel PAM system, the minimum clock rate is 40 khz ( 2 f N 2 ). 1 1 b) Tx = sec cutoff frequency (bandwidth): Bx = 20 khz 40000 2Tx c) For AM, the minimum bandwidth is B AM = 2 (8 + 10) = 36 khz For SSB, the minimum bandwidth is = ( 8 + 10) = 18 khz d) B SSB ERG2310A: Principles of Communication Systems (2002-2003) 20
Solution: (continued) d) ERG2310A: Principles of Communication Systems (2002-2003) 21
Inter-symbol interference (ISI) Assumptions for attaining the zero-interference condition between adjacent pulses (slides 17-18): (1) impulse sampling (2) ideal low-pass filter (3) distortionless transmission (4) precise synchronization This assumptions are, however, physically unrealizable. In the case of multichannel pulse modulation, overlap between pulses gives rise to interchannel cross-talk, which in general is called inter-symbol interference (ISI). ISI can be decreased by purposely widening the transmission bandwidth. But, to minimize the ISI within as small a transmission bandwidth as possible, we need to design received signal waveforms and hence the transmission filters. pulse shaping ERG2310A: Principles of Communication Systems (2002-2003) 22
Pulse shaping A general received PAM signal: m : an integer; T : sampling period; x(t) : pulse waveform; a m : sampled values of the input signal f (t) ; n(t) : additive noise. t = kt y( kt ) = ak x(0) + amx( kt mt ) + n( kt ) Let, we have y = am x( t mt ) + n( t) th First term: the k sampled value of f (t) that has be transmitted. Second term: the overlap of other pulses adding to the desired th pulse a k x( t kt ) at the k sampling time. It is the ISI present. Third term: additive noise at the time t = kt. 1 m = k; x( kt mt ) = Nyquist 0 m k. waveform ISI can be eliminated if All Nyquist waveforms have uniformly spaced zeros at all multiples of a basic interval except one (at the center). m k m ERG2310A: Principles of Communication Systems (2002-2003) 23
Pulse shaping (continued) Raised-cosine waveform: a popular Nyquist waveform Impulse response x(t) 1 α : roll-off factor W = π T x( t) sinwt cosαwt Wt 1 = 2 ( 2αWt π ) α T T π X = 1 sin 2 2αW 0 ( W ) (1 α) W Larger faster decaying pulses synchronization will be less critical modest timing errors will not cause large amount of ISI. Expense: more required bandwidth for transmission: Since, we get 1 (2T ) B 1 T. 0 α 1 0 (1 α) W > (1 + α) W B = < (1 + α) W 1+ α 2T ERG2310A: Principles of Communication Systems (2002-2003) 24