Chapter 4: Baseband Pulse Transmission
CHAPTER 4
BASEBAND PULSE
TRANSMISSION
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Chapter 4: Baseband Pulse Transmission
Outline
•
•
•
•
4.1 Introduction
4.2. Matched Filter
4.3 Error Due to Noise
4.4 Intersymbol Interference (ISI)
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Chapter 4: Baseband Pulse Transmission
4.1 Introduction
• In Ch.3 – methods for digital transmission of analog
information bearing signals
• In Ch.4 – methods for digital transmission of digital
information using baseband channel
• Digital data – broad spectrum; low-frequency components;
• Transmission channel bandwidth – should accommodate
the essential frequency content of the data stream
• Channel is dispersive
– channel is noisy – control over additive white noise (old problem..)
– received signal pulses are affected by adjacent symbols (new
problem) – intersymbol interference (ISI); major source of
interference;
– Distorted pulse shape (new problem) - channel requires control
over pulse shape
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Chapter 4: Baseband Pulse Transmission
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Chapter 4: Baseband Pulse Transmission
• Main issue to be discussed:
– Detection of digital pulses corrupted by the effect
of the channel
• We know the shape of the transmitted pulse
– Device to be used – matched filter
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Chapter 4: Baseband Pulse Transmission
Outline
•
•
•
•
4.1 Introduction
4.2. Matched Filter
4.3 Error Due to Noise
4.4 Intersymbol Interference (ISI)
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Chapter 4: Baseband Pulse Transmission
4.2 Matched Filter
• Basic task – detecting transmitted pulses at the
front end of the receiver (corrupted by noise)
• Receiver model
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Chapter 4: Baseband Pulse Transmission
Details
• The filter input x(t) is:
• where
x(t )  g  t   w(t ),
0t T
(4.1)
– T is an arbitrary observation interval
– g(t) is a binary symbol 1 or 0
– w(t) is a sample function of white noise,
zero mean, psd N0/2
The function of the receiver is to detect the pulse g(t) in an
optimum manner, providing that the shape of the pulse is
known and the distortion is due to effects of noise =
To optimize the design of a filter so as to minimize the effects
of noise at the filter output in some statistical sense.
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Chapter 4: Baseband Pulse Transmission
Designing the filter
• Since we assume the
filter is linear its output
can be described as:
• where
– g0(t) is the recovered
signal
– n(t) produced noise
• This is equal to
maximizing the peak
signal-to-noise ratio:
y(t )  go (t )  n(t )
(4.2)
instantaneous power in
the output signal
2
| go (T ) |

(4.3)
2
E[n (t )]
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noise
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Chapter 4: Baseband Pulse Transmission
So,
• we have to define the impulse response of the
filter h(t) in such a way that the signal-to-noise
ratio (4.3) is maximized.
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Chapter 4: Baseband Pulse Transmission
Let us assume that:
- G(f) - FT of the signal g(t);
- H(f) – frequency response of the filter
then: FT of the output signal g0(t)= H(f).G(f), or

go (t )   H ( f )G( f ) exp( j 2 ft )df
(4.4)

sampled at time t=T and no noise
| go (T ) |2 



H ( f )G( f ) exp( j 2 ft )df
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(4.5)
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Chapter 4: Baseband Pulse Transmission
Next step is to add the noise.
What we know is that the power spectral density of white noise is:
N0
2
SN ( f ) 
| H( f )|
2
(4.6)
Thus the average power of the output noise n(t) is:

E[n (t )]   S N ( f )df
2

N0

2



2
| H ( f ) | df
(4.7)
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Chapter 4: Baseband Pulse Transmission
Substituting,




H ( f )G( f ) exp( j 2 ft )df
N0
2



2
(4.8)
| H ( f ) |2 df
So, given the function G(f), the problem is reduced to finding
such an H(f) that would maximize η.
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Chapter 4: Baseband Pulse Transmission
We use Schwartz inequality which states that
for two complex functions, satisfying the
conditions:




| 2 ( x) | dx  
2

| 1 ( x) | dx  
2

the following is true:



2


1 ( x)2 ( x)dx   | 1 ( x) | dx  | 2 ( x) |2 dx

and equality holds iff:
2

1 ( x)  k ( x)
*
2
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(4.10)
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Chapter 4: Baseband Pulse Transmission
In our case this inequality will have the form:



H ( f )G ( f ) exp( j 2 fT )df
2

  | H ( f ) | df
2




| G ( f ) |2 df
(4.11)
and we can re-write the equation for the peak signal-to-noise
ratio as:
2

N0



2
| G( f ) | df
(4.12)
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Chapter 4: Baseband Pulse Transmission
Remarks:
• The right hand side of this equation does not depend on H(f).
• It depends only on:
– signal energy
– noise power spectral density
• Max value is for:
2

N0



2
| G( f ) | df
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Chapter 4: Baseband Pulse Transmission
• Let us denote the optimum value of H(f) by Hopt(f).
To find it we use equation 4.10:
Hopt ( f )  kG ( f )exp( j 2 fT )
*
(4.14)
The result: Except for a scaling coefficient k exp(2πfT), the frequency response of the optimum filter is
the same as the complex conjugate of the FT of the
input signal.
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Chapter 4: Baseband Pulse Transmission
Definition of the filter functions:
• In the frequency domain:
– knowing the input signal we can define the
frequency response of the filter (in the
frequency domain) as the FT of its complex
conjugate.
• In the time domain…
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Chapter 4: Baseband Pulse Transmission
Take inverse FT of Hopt(f):

hopt (t )  k  G ( f ) exp[ j 2 f (T  t )]df
*

(4.15)
and keeping in mind that for real signals G*(f) = G(-f):

hopt (t )  k  G( f ) exp[ j 2 f (T  t )]df


 k  G ( f ) exp[ j 2 f (T  t )]df

 kg (T  t )
(4.16)
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Chapter 4: Baseband Pulse Transmission
• So,
– in the time domain it turns out that the impulse
response of the filter, except for a scaling factor k,
is a time-reversed and delayed function of the
input signal
• This means it is “matched” to the input signal, that is
why this type of time-invariant linear filters is known
as “matched filter”
• NOTE: The only assumption for the channel noise
was that it is stationary, white, with psd N0/2.
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Chapter 4: Baseband Pulse Transmission
Properties of Matched Filters
Property 1:
• A filter matched to a pulse signal g(t) of
duration T is characterized by an impulse
response that is time-reversed and delayed
version of the input g(t):
Time domain:
hopt(t) = k . g(T-t)
Frequency Domain: Hopt(f) = kG*(f)exp(j2πfT)
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Chapter 4: Baseband Pulse Transmission
Property 2:
• A matched filter is uniquely defined by the
waveform of the pulse but for the:
- time delay T
- scaling factor k
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Chapter 4: Baseband Pulse Transmission
Property 3:
Go ( f )  H opt ( f )G ( f )
 kG* ( f )G ( f ) exp( j 2 fT )
• The peak signal-to-noise
 k | G ( f ) |2 exp( j 2 fT )
ratio of the matched
filter depends only on
(4.17)
the ratio of the signal
energy to the power
using the inverse FT
spectral density of the

g o (T )   G0 ( f ) exp( j 2 fT )df
white noise at the filter

input.

k
2
|
G
(
f
)
|
df


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Chapter 4: Baseband Pulse Transmission
• The integral of the squared


magnitude spectrum of a
2
2
E

g
(
t
)
dt

G
(
f
)
|
df
pulse signal with respect to


frequency is equal to the
so g 0 (T )  k E
(4.18)
signal energy E (Rayleigh
Theorem) so substituting in
the previous formula we get:
*


H opt ( f )  kG ( f ) exp( j 2 fT )
• After substitution of (4.14)
into (4.7) we get the
expression for the average
output noise power as:
(4.14)

E[n (t )]   S N ( f )df
2

N0

2



| H ( f ) |2 df
(4.7)
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Chapter 4: Baseband Pulse Transmission
we finally get the following expression:
2
k N0 
2
E[n (t )] 
| G( f ) | df

2 
2
 k N0 E / 2
(4.19)
2
and
max
2
(kE )
2E
 2

(k N0 E / 2) N0
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(4.20)
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Chapter 4: Baseband Pulse Transmission
Conclusion:
• From (4.20) we see that the dependence of the peak
SNR on the input waveform g(t) has been completely
removed by the matched filter.
• So, in evaluating the ability of a matched-filter
receiver to overcome/remove additive white noise we
see that all signals with equal energy are equally
effective.
• We call the ratio E/N0 signal_energy-to-noise ratio
(dimensionless)
• The matched filter is the optimum detector of a pulse
of known shape in additive white noise.
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Chapter 4: Baseband Pulse Transmission
Outline
•
•
•
•
4.1 Introduction
4.2. Matched Filter
4.3 Error Due to Noise
4.4 Intersymbol Interference (ISI)
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Chapter 4: Baseband Pulse Transmission
4.3 Error Rate Due to Noise
• In this section we will derive some quantitative
results for the performance of binary PCM
systems, based on results for the matched
filter.
• We consider the BER performance for a
rectangular baseband pulse using and integrate
and dump filter for detection.
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Chapter 4: Baseband Pulse Transmission
The model:
• We assume
– binary PCM system based on polar NRZ signaling
(encoding)
– symbols 1 and 0 are equal amplitude and equal duration
– channel noise is modeled as AWGN w(t) with zero mean
and psd N0/2
– the received signal, in the interval 0 ≤ t ≤ Tb can be
represented as:
white
noise
 A  w(t ),

x(t )  
 A  w(t ),

amplitude
symbol 1 was sent
(4.21)
symbol 0 was sent
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Chapter 4: Baseband Pulse Transmission
• also it is assumed that the receiver has accurate
knowledge of the starting and ending times of each
pulse (perfect synchronization);
• the receiver has to make a decision whether the pulse
is a 1 or a 0.
• the structure of the receiver is as follows:
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Chapter 4: Baseband Pulse Transmission
Details:
• The filter is matched to a rectangular pulse of
amplitude A and duration Tb.
• The resulting matched filter output is sampled
at the end of each signaling interval.
• The channel noise adds randomness to the
matched filter output.
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Chapter 4: Baseband Pulse Transmission
• For the channel model we are discussing
(AWGN), integrating the noise over a period T
is equal to creating noise with Gaussian
distribution with variance σ02 = N0 T/2, where
N0 is the noise power in Watts/Hz.
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Chapter 4: Baseband Pulse Transmission
• If we consider a single symbol s(t) of voltage V
passing through the detector with additive noise
n(t) then the output of the integrator y(t) will be:
T
T
0
0
y(t )   {s(t )  n(t )}dt  V .T   n(t )dt
symbol contribution
noise contribution
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Chapter 4: Baseband Pulse Transmission
• Then the probability density of the integrated noise at the
sampling point is:
Pd [n0(t ) ] 
e
 n02 (T ) / 2 02
2
2
0

e
 n02 (T ). N oT
 N 0T
• A detection error will occur if the noise sample exceeds - VT/2.
• The probability for such an event is calculated from (1)
VT / 2
Ps 


VT / 2
Pd [no (t )]dno (t ) 


e
 n02 (T ). N oT
 N 0T
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Chapter 4: Baseband Pulse Transmission
x  n0 (T ) N0T
• substituting
we can write the expression for the probability of error as:

1 2
Ps 
2 
v T
2 No

x 
e
 x2
V
1
dx  erfc 
2
2
T
N0



complementary
error function
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Chapter 4: Baseband Pulse Transmission
• The probability of error can also be expressed in
terms of the signal energy as follows:
V
1
Ps  erfc 
2
2
T
N0
 1
1V T 
  erfc 

 4 N0 
 2
2
1/ 2
1/ 2
 1 Es 
1
 erfc 

2
 4 N0 
• As 0 and 1 are equiprobable the result for receiving
a logic 0 in error will be the same only in this case
the noise sample will be exceeding +VT/2.
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Chapter 4: Baseband Pulse Transmission
• For the special case we are discussing
unipolar NRZ the average symbol energy is
half that of the logic 1. So the symbol error
probability will be:
1/ 2
PSunipolar 
 1 Es 
1
erfc 

2
 2 N0 
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Chapter 4: Baseband Pulse Transmission
Figure 4.5
Noise analysis of PCM system.
(a) Probability density function of random variable Y at
matched filter output when 0 is transmitted.
(b) Probability density function of Y when 1 is transmitted.
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Chapter 4: Baseband Pulse Transmission
Unipolar vs Bipolar Symbols
• For unipolar symbols as we said the average energy is
equal to E0/2.
• For bipolar symbols (logic 1 is conveyed by +V, logic 0 –
by –V volts) it can be proved that
(An example of this is RS-232, where "one" is −5V to −12V
and "zero" is +5 to +12V)
1/ 2
PSunipolar 
 Es 
1
erfc 

2
 N0 
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Chapter 4: Baseband Pulse Transmission
Outline
•
•
•
•
4.1 Introduction
4.2. Matched Filter
4.3 Error Due to Noise
4.4 Intersymbol Interference (ISI)
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Chapter 4: Baseband Pulse Transmission
4.4 PAM and Intersymbol
Interference (ISI)
• ISI is due to the fact that the communication
channel is dispersive – some frequencies of
the received pulse are delayed which causes
pulse distortion (change in shape and delay).
• Most efficient method for baseband
transmission – both in terms of power and
bandwidth - is PAM.
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Chapter 4: Baseband Pulse Transmission
Considered model:
• baseband binary PAM system
• incoming binary sequence b0 consists of 1 and 0
symbols of duration Tb
Figure 4.7 Baseband binary data transmission system.
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Chapter 4: Baseband Pulse Transmission
• PAM changes this sequence into a new sequence of
short pulses each with amplitude ak, represented in
polar form as:
 1,

ak  
1,

if symbol bk is 1
(4.42)
if symbol bk is 0
applied to a transmit filter of impulse response g(t):
s(t )   ak g (t  kTb )
(4.43)
k
transmitted signal
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Chapter 4: Baseband Pulse Transmission
• s(t) is modified by a channel with channel impulse
response h(t)
• random white noise is added (AWGN channel model)
• x(t) is the channel output, the noisy signal arriving at
the receiver front end
• receiver has a receive filter with impulse response c(t)
and output y(t)
• y(t) is sampled synchronously with the transmitter
(clock signal is extracted from the receive filter
output)
• reconstructed samples are compared to a threshold
• decision is taken as for 1 or 0
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Chapter 4: Baseband Pulse Transmission
Receive filter output is:
y(t )    ak p(t  kTb )  n(t )
k
(4.44)
should have a constant delay t0 here set to 0
The scaled pulse µp(t) can be expressed as double
convolution: the impulse response of the transmit filter g(t),
the impulse response of the channel filter h(t) (channel) and
the impulse response of the receive filter c(t):
 p(t )  g (t ) h(t ) c(t )
(4.45)
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Chapter 4: Baseband Pulse Transmission
• In the frequency domain we have:
 P( f )  G( f ) H ( f )C ( f )
(4.47)
where P(f), G(f), H(f) and C(f) are FT of the respective p(t), g(t), h(t) and c(t)
The received signal is sampled at times ti= iTb which, taking
(4.44) in mind and the norm. condition p(0) = 1 (µ is a scaling
factor to account for amplitude changes), can be expressed as:

y (ti )    ak p[(i  k )Tb ]  n(ti )
k 

  ai    ak p[(i  k )Tb ]  n(ti )
residual effect due
to other transmitted
pulses
(4.48)
k 
k i
contribution of the
ith pulse
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Chapter 4: Baseband Pulse Transmission
Conclusion:
• In the absence of noise and ISI we get and
ideal pulse (from (4.48) ) y(ti)=µai
• The ISI can be controlled (reduced) by the
proper design of the transmit and receive filter
• Subject discussed further in the following
sections….
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