Title: BASEBAND DATA TRANSMISSION
1BASEBAND DATA TRANSMISSION
by Dr. Uri Mahlab
2Communication system
3Block diagram of an Binary/M-ary signaling scheme
Channel noise
4Block diagram Description
dk1,1,1,1,0,0,1,1,0,0,0,1,1,1
For
Tb
Tb
For
5 Block diagram Description (Continue - 1)
dk1,1,1,1,0,0,1,1,0,0,0,1,1,1
6 Block diagram Description (Continue - 2)
dk1,1,1,1,0,0,1,1,0,0,0,1,1,1
100110
7Block diagram Description (Continue - 3)
dk1,1,1,1,0,0,1,1,0,0,0,1,1,1
Timing
Tb
100110
t
t
8Block diagram Description (Continue - 4)
Timing
HT(f)
HR(f)
(X(t
Information source
Pulse generator
Trans filter
Receiver filter
Channel noise n(t)
Tb
2Tb
5Tb
6Tb
t
3Tb
4Tb
t
t
9Block diagram Description (Continue - 5)
Tb
2Tb
5Tb
6Tb
t
3Tb
4Tb
t
t
10Block diagram of an Binary/M-ary signaling scheme
Timing
11Block diagram Description
Tb
2Tb
5Tb
6Tb
t
3Tb
4Tb
t
t
t
1 0 0 0 1 0
1 0 0 1 1 0
12Typical waveforms in a binary PAM system
13Block diagram of an Binary/M-ary signaling scheme
Timing
14Explanation of Pr(t)
15The element of a baseband binary PAM system
16Analysis and Design of Binary Signal
17The input to the A/D converter is
For and is the
total time delay in the system, we get.
t
t
t2
t3
t
tm
t1
18The output of the A/D converter at the sampling
time
tm mTbtd
Tb
2Tb
5Tb
6Tb
t
3Tb
4Tb
t2
t3
t
tm
t1
19ISI - Inter Symbol Interference
t2
t3
t
tm
t1
20Explanation of ISI
t
t
f
f
21Explanation of ISI - Continue
t
t
f
f
Tb
2Tb
5Tb
6Tb
t
3Tb
4Tb
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23-The pulse generator output is a pulse waveform
If kth input bit is 1 if kth input bit is 0
-The A/D converter input Y(t)
24 255.2 BASEBAND BINARY PAM SYSTEMS
- minimize the combined effects of inter symbol
interference and noise in order to achieve
minimum probability of error for given data rate.
265.2.1 Baseband pulse shaping
The ISI can be eliminated by proper choice of
received pulse shape pr (t).
Does not Uniquely Specify Pr(t) for all values
of t.
27To meet the constraint, Fourier Transform Pr(f)
of Pr(t), should satisfy a simple condition given
by the following theorem
Theorem
Proof
28Which verify that the Pr(t) with a transform
Pr(f) Satisfy ______________
29The condition for removal of ISI given in the
theorem is called Nyquist (Pulse Shaping)
Criterion
1
Tb
2Tb
-Tb
-2Tb
30The Theorem gives a condition for the removal of
ISI using a Pr(f) with a bandwidth larger then
rb/2/. ISI cant be removed if the bandwidth of
Pr(f) is less then rb/2.
Tb
2Tb
5Tb
6Tb
t
3Tb
4Tb
31Particular choice of Pr(t) for a given
application
p
(t)
r
32A Pr(f) with a smooth roll - off characteristics
is preferable over one with arbitrarily sharp
cut off characteristics.
Pr(f)
Pr(f)
33In practical systems where the bandwidth
available for transmitting data at a rate of rb
bits\sec is between rb\2 to rb Hz, a class of
pr(t) with a raised cosine frequency
characteristic is most commonly used. A raise
Cosine Frequency spectrum consist of a flat
amplitude portion and a roll off portion that has
a sinusoidal form.
34raised cosine frequency characteristic
35Summary
The BW occupied by the pulse spectrum is
Brb/2b. The minimum value of B is rb/2 and the
maximum value is rb. Larger values of b imply
that more bandwidth is required for a given bit
rate, however it lead for faster decaying pulses,
which means that synchronization will be less
critical and will not cause large ISI. b rb/2
leads to a pulse shape with two convenient
properties. The half amplitude pulse width is
equal to Tb, and there are zero crossings at
t3/2Tb, 5/2Tb. In addition to the zero
crossing at Tb, 2Tb, 3Tb,...
365.2.2 Optimum transmitting and receiving filters
The transmitting and receiving filters are chosen
to provide a proper
37-One of design constraints that we have for
selecting the filters is the relationship
between the Fourier transform of pr(t) and pg(t).
Where td, is the time delay Kc normalizing
constant.
In order to design optimum filter Ht(f) Hr(f),
we will assume that Pr(f), Hc(f) and Pg(f) are
known.
Portion of a baseband PAM system
38 If we choose Pr(t) Pr(f) to produce Zero ISI
we are left only to be concerned with noise
immunity, that is will choose
39Noise Immunity
Problem definition
- For a given
- Data Rate -
- Transmission power -
- Noise power Spectral Density -
- Channel transfer function -
- Raised cosine pulse -
Choose
40Error probability Calculations
At the m-th sampling time the input to the A/D is
We decide
41AaKc
The noise is assumed to be zero mean Gaussian at
the receiver input then the output should also be
Zero mean Gaussian with variance No given by
42b
430
44-A
A
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46Q(u)
dz
U
47Perror decreases as
increase
Hence we need to maximize the signal to noise
Ratio
Thus for maximum noise immunity the filter
transfer functions HT(f) and HR(f) must be xhosen
to maximize the SNR
48Optimum filters design calculations
We will express the SNR in terms of HT(f) and
HR(f)
We will start with the signal
The PSD of the transmitted signal is given by
49And the average transmitted power ST is
The average output noise power of n0(t) is given
by
50The SNR we need to maximize is
Or we need to minimize
51Using Schwartzs inequality
The minimum of the left side equaity is reached
when V(f)ConstW(f)
If we choose
52The filter should have a linear phase response in
a total time delay of td
53Finally we obtain the maximum value of the SNR to
be
54For AWGN with and pg(f) is chosen such that it
does not change much over the bandwidth of
interest we get.
Rectangular pulse can be used at the input of
HT(f).
555.2.3 Design procedure and Example
The steps involved in the design procedure.
ExampleDesign a binary baseband PAM system to
transmit data at a a bit rate of 3600 bits/sec
with a bit error probability less than
The channel response is given by
The noise spectral density is
56Solution
If we choose a braised cosine pulse spectrum with
57We choose a pg(t)
We choose
58Plots of Pg(f),Hc(f),HT(f),HR(f),and Pr(f).
59 To maintain a
For Pr(f) with raised cosine shape
And hence
Which completes the design.
60DUOBINARY BASEBAND PAM SYSTEM
5.3
In order to transmit data at a rate of rb
bits/sec with zero ISI PAM data transmission
system requires a bandwidth of at least rb /2
HZ.
1.such filters are physically unrealizable 2.
Any system with this filters will be extremely
sensitive to perturbations.
204
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62The duobinary scheme utilizes controlled amounts
of ISI for transmitting data at a rate of rb /2
HZ.
The shaping filters for duobinary are easier to
realize than the ideal rectangular filters.
The duobinary signaling schemes use pulse spectra
Pr(f)
63use of controlled ISI in duobinary signaling
scheme
5.3.1
The output Y(t) of the receive filter can be
written as
If the output is sampled at tm mTb/2 td than
it is obvious that in the absence of noise
The Am' s can assume one of two values /-A
depending on whether the m th input bit is 1 or
0. Since Y(tm) depends on Am Am-1 assuming no
noise
2A if the m th and (m-1st)bits are
both 1's 0 if
the m th and (m-1st)bits are different
-2A if the m th and
(m-1st)bits are both zero
Y(tm)
64Transmitting and receiving filters for optimum
performance
5.3.2
The receiving levels at the input to the A/D
converter are 2A, 0, and -2A with probabilities
1/2, 1/4.
The probability of bit error pe is given by
Since no is a zero mean Gaussian random variable
with a variance No we can write pe as
For the direct binary PAM case
65Where ?/2Gn(f) is the noise power spectral
density the probability of error is
The integral can be evaluated as
66M-ARY SIGNALING SCHEMES
5.4
In baseband binary PAM we use pulses with one of
2 possible amplitude,
In M-ary baseband PAM system we allowed M
possible levels (Mgt2) and there M distinct input
symbols.
The signal pulse noise passes through the
receiving filter and is sampled by the A/D
converter at an appropriate rate and phase.
the M-ary PAM scheme operating with the preceding
constraints can transmit data at a bit rate of
rs log2M bit/sec and require a minimum bandwidth
of rs/2 HZ.
675.7 MISCELLANEOUS TOPICS
5.7.1 Eye Diagram
The performance of baseband PAM systems depends
on the amount of ISI and channel noise.
The received waveform with no noise and no
distortion is shown in Figure 5.20a the open
eye pattern results
Figure 5.20b shows a distorted version of the
waveform the corresponding eye pattern.
Figure 5.20c shows a noise distorted version of
the received waveform and the corresponding eye
pattern.
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69In the figure 5.21 we see typical eye patterns of
a duobinary signal
70If the signal-to-noise ratio at the receiver is
high then the following observations can be made
from the eye pattern shown simplified in Figure
5.22
711. The best time to sample the received waveform
is when the eye is opening is largest. 2. The
maximum distortion is indicated by the vertical
width of the two branches at sampling time. 3.
The noise margin or immunity to noise is
proportional to the width of the eye opening. 4.
The sensitivity of the system to timing errors is
revealed by the rate of closing of the eye as
sampling time is varied. 5. The sampling time is
midway between zero crossing. 6. Asymmetries in
the eye pattern indicate nonlinearities in the
channel.
72synchronization
5.7.2
Three methods in which this synchronization can
be obtained are 1. Derivation of clock
information from a primary or secondary
standard. 2. Transmitting a synchronizing clock
signal. 3. Derivation of the clock signal from
the received waveform itself.
An example of a system used to derive a clock
signal from the received waveform is shown in
figure 5.23.
To illustrate the operating of the phase
comparator network let us look at the timing
diagram shown in figure 5.23b
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74Scrambler and unscrambler
5.7.3
Scrambler The scrambler shown in figure 5.24a
consists a feedback shift register.
UnscramblerThe matching unscrambler have a feed
forward shift register structure.
In both the scrambler and unscrambler the outputs
of several stages of of shift register are added
together modulo-2 and the added to the data
stream again in modulo-2 arithmetic
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76Scrambler affects the error performance of the
communication system in that a signal channel
error may cause multiple error at the output of
the unscrambler.
The error propagation effect lasts over only a
finite and small number of bits.
In each isolated error bit causes three errors in
the final output it must also be pointed out that
some random bit patterns might be scrambled to
the errors or all ones.