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- 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
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