Modulation and coding course- lecture 3

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Transforming the information source to a form compatible with a digital system -Sampling -Aliasing -Quantization -Uniform and non-uniform -Baseband modulation -Binary pulse modulation -M-ary pulse modulation -M-PAM (M-ay Pulse amplitude modulation)

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Nội dung Text: Modulation and coding course- lecture 3

Digital Communication I: Modulation and Coding Course Period 3 - 2007 Catharina Logothetis Lecture 3
Last time we talked about: Transforming the information source to a form compatible with a digital system Sampling Aliasing Quantization Uniform and non-uniform Baseband modulation Binary pulse modulation M-ary pulse modulation M-PAM (M-ay Pulse amplitude modulation) Lecture 3 2
Formatting and transmission of baseband signal Digital info. Bit stream Pulse waveforms (Data bits) (baseband signals) Textual Format source info. Pulse Analog Sample Quantize Encode modulate info. Sampling at rate Encoding each q. value to f s = 1 / Ts l = log 2 L bits (sampling time=Ts) (Data bit duration Tb=Ts/l) Quantizing each sampled Mapping every m = log 2 M data bits to a value to one of the symbol out of M symbols and transmitting L levels in quantizer. a baseband waveform with duration T Information (data) rate: Rb = 1 / Tb [bits/sec] Symbol rate : R = 1 / T [symbols/sec] For real time transmission: Rb = mR Lecture 3 3
Quantization example amplitude x(t) 111 3.1867 110 2.2762 Quant. levels 101 1.3657 100 0.4552 011 -0.4552 boundaries 010 -1.3657 001 -2.2762 x(nTs): sampled values xq(nTs): quantized values 000 -3.1867 Ts: sampling time PCM t codeword 110 110 111 110 100 010 011 100 100 011 PCM sequence Lecture 3 4
Example of M-ary PAM Assuming real time tr. and equal energy per tr. data bit for binary-PAM and 4-ary PAM: • 4-ary: T=2Tb and Binay: T=Tb • A2 = 10B 2 Binary PAM 4-ary PAM (rectangular pulse) (rectangular pulse) 3B A. ‘11’ ‘1’ B T T T ‘01’ T -B ‘00’ T T ‘0’ ‘10’ -A. -3B Lecture 3 5
Example of M-ary PAM … 0 Ts 2Ts 2.2762 V 1.3657 V 0 Tb 2Tb 3Tb 4Tb 5Tb 6Tb 1 1 0 1 0 1 Rb=1/Tb=3/Ts R=1/T=1/Tb=3/Ts 0 T 2T 3T 4T 5T 6T Rb=1/Tb=3/Ts R=1/T=1/2Tb=3/2Ts=1.5/Ts 0 T 2T 3T Lecture 3 6
Today we are going to talk about: Receiver structure Demodulation (and sampling) Detection First step for designing the receiver Matched filter receiver Correlator receiver Lecture 3 7
Demodulation and detection mi Pulse g i (t ) Bandpass si (t ) M-ary modulation Format modulate modulate i = 1, K, M channel transmitted symbol hc (t ) estimated symbol n(t ) Demod. Format Detect ˆ mi z (T ) & sample r (t ) Major sources of errors: Thermal noise (AWGN) disturbs the signal in an additive fashion (Additive) has flat spectral density for all frequencies of interest (White) is modeled by Gaussian random process (Gaussian Noise) Inter-Symbol Interference (ISI) Due to the filtering effect of transmitter, channel and receiver, symbols are “smeared”. Lecture 3 8
Example: Impact of the channel Lecture 3 9
Example: Channel impact … hc (t ) = δ (t ) − 0.5δ (t − 0.75T ) Lecture 3 10
Receiver job Demodulation and sampling: Waveform recovery and preparing the received signal for detection: Improving the signal power to the noise power (SNR) using matched filter Reducing ISI using equalizer Sampling the recovered waveform Detection: Estimate the transmitted symbol based on the received sample Lecture 3 11
Receiver structure Step 1 – waveform to sample transformation Step 2 – decision making Demodulate & Sample Detect z (T ) Threshold ˆ mi r (t ) Frequency Receiving Equalizing comparison down-conversion filter filter For bandpass signals Compensation for channel induced ISI Received waveform Baseband pulse Baseband pulse Sample (possibly distored) (test statistic) Lecture 3 12
Baseband and bandpass Bandpass model of detection process is equivalent to baseband model because: The received bandpass waveform is first transformed to a baseband waveform. Equivalence theorem: Performing bandpass linear signal processing followed by heterodying the signal to the baseband, yields the same results as heterodying the bandpass signal to the baseband , followed by a baseband linear signal processing. Lecture 3 13
Steps in designing the receiver Find optimum solution for receiver design with the following goals: 1. Maximize SNR 2. Minimize ISI Steps in design: Model the received signal Find separate solutions for each of the goals. First, we focus on designing a receiver which maximizes the SNR. Lecture 3 14
Design the receiver filter to maximize the SNR Model the received signal si (t ) hc (t ) r (t ) r (t ) = si (t ) ∗h c (t ) + n(t ) n(t ) AWGN Simplify the model: Received signal in AWGN Ideal channels hc (t ) = δ (t ) si (t ) r (t ) r (t ) = si (t ) + n(t ) n(t ) AWGN Lecture 3 15
Matched filter receiver Problem: Design the receiver filter h(t ) such that the SNR is maximized at the sampling time when si (t ), i = 1,..., M is transmitted. Solution: The optimum filter, is the Matched filter, given by h ( t ) = hopt ( t ) = s i (T − t ) * H ( f ) = H opt ( f ) = S i ( f ) exp( − j 2π fT ) * which is the time-reversed and delayed version of the conjugate of the transmitted signal si (t ) h(t ) = hopt (t ) 0 T t 0 T t Lecture 3 16
Example of matched filter y (t ) = si (t ) ∗h opt (t ) si (t ) h opt (t ) A2 A A T T T t T t 0 T 2T t y (t ) = si (t ) ∗h opt (t ) si (t ) h opt (t ) A2 A A T T T/2 T t T/2 T t 0 T/2 T 3T/2 2T t −A −A − A2 2 T T Lecture 3 17
Properties of the matched filter The Fourier transform of a matched filter output with the matched signal as input is, except for a time delay factor, proportional to the ESD of the input signal. Z ( f ) =| S ( f ) |2 exp(− j 2πfT ) 1. The output signal of a matched filter is proportional to a shifted version of the autocorrelation function of the input signal to which the filter is matched. z (t ) = Rs (t − T ) ⇒ z (T ) = Rs (0) = Es The output SNR of a matched filter depends only on the ratio of the signal energy to the PSD of the white noise at the filter input. ⎛S⎞ Es max⎜ ⎟ = ⎝ N ⎠T N 0 / 2 1. Two matching conditions in the matched-filtering operation: spectral phase matching that gives the desired output peak at time T. spectral amplitude matching that gives optimum SNR to the peak value. Lecture 3 18
Correlator receiver The matched filter output at the sampling time, can be realized as the correlator output. z (T ) = hopt (T ) ∗ r (T ) T = ∫ r (τ )si (τ )dτ =< r (t ), s (t ) > * 0 Lecture 3 19
Implementation of matched filter receiver Bank of M matched filters z1 (T ) s (T − t ) ⎡ z1 ⎤ * 1 Matched filter output: r (t ) ⎢ M ⎥=z z Observation ⎢ ⎥ vector ⎢zM ⎥ ⎣ ⎦ sM (T − t ) * zM (T ) zi = r (t ) ∗ s ∗i (T − t ) i = 1,..., M z = ( z1 (T ), z 2 (T ),..., z M (T )) = ( z1 , z 2 ,..., z M ) Lecture 3 20