# B - Optimum Receiver_ver1 by yaoyufang

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```									         OPTIMUM RECEIVER
 Simple demodulation technique (previous lectures),
center of bit period
 If the sampled value is  0 volts; assumed the
corresponding data is a 1
 If the sampled value is  0 volts; assumed the
corresponding data is a 0 (since a “0” from a source
would produce a negative pulse from the transmitter

Sample in center
of each bit period

signal                                                                 “1” if sample  0
Comparator
“0” if sample  0
f c = bandwidth of
transmitted signal
Threshold = 0V

Fig. 1: A simple PAM receiver

 The lowpass filter has a bandwidth as the bandwidth of
transmitted signal
 Errors due to channel (attenuation, noise) where the
transmitted bit is changed from “0” to “1” or vise versa
 This inaccuracy/error is quantify by measuring noise
margin

EEE377 – A: Optimum Receiver                                 1/36                            March 2009
 Another powerful tool is to use – probability of bit
error

Noise Margin

 The distance between a noiseless received signal and
the threshold at the instant of sampling

 Transmission over perfect channel (no attenuation) and
perfect synchronization the noise margin for the third
bit – 1 Volts
 Transmission over channel (50% attenuation) and
perfect synchronization the noise margin for the third
bit – 0.5 Volts

Probability of bit error

Let’s quantify accuracy by calculating the probability that
a particular bit – say the ith bit is demodulated in error.
For the PAM receiver in Fig. 1, the following notations
are defined;

EEE377 – A: Optimum Receiver     2/36                    March 2009
Let     bi  the ith transmitted bit bi  1, or bi  0 
Let     st   the transmitted signal
Let     nt   noise at the input of the receiver

Let r t   the received signal, which has been attenuated
and corrupted by noise
Let z t   the voltage at the input of the sampler in the
Let no t   noise at the output of the receiver’s LPF

information
signal st                     signal r t 
Noise nt 

Fig. 2: Block diagram of communication system

Sample

zt 
signal r t         LPF                                         Comparator
“1” if sample  0
“0” if sample  0
no t 

Threshold = 0V

The sampling time for ith bit (center of bit period);

Tb
t  i  1Tb 
2
EEE377 – A: Optimum Receiver                                        3/36                                           March 2009
The voltage output of sampler for ith bit;

           T 
zout t   z i  1Tb  b 
            2

We can derive the probability of bit error as follows;

P{ith bit =1, but the sample
The probability                              value at the receiver during
of the ith bit is                            the ith bit period < 0}
demodulated in                      =
+
error                                        P{ ith bit =0, but the sample
the ith bit period  0}
Thus
                        T      
P bi 1, and z i  1Tb  b   0
The                                                                   2     
probability of                                               
the ith bit is                    =                                            
           T 
P bi  0, and z i  1Tb  b   0
demodulated                                                            2     
in error

Note that
PX  a and event Z  Pevent ZPX  a | event Z

EEE377 – A: Optimum Receiver                         4/36                            March 2009
The                                                                 T                 
Pbi 1P  z i  1Tb  b   0 when bi 1
probability of                                                       2                

the ith bit is                      =
              T                  
Pbi  0P  z i  1Tb  b   0 when bi  0
(1a)
demodulated                                                           2                 
in error

The last two expressions;

1. P z i  1Tb  Tb   0 when bi 1
                                 
               2                
              T                  
2.    P  z i  1Tb  b   0 when bi  0
               2                 

are called “conditional probabilities”. We can use the
notation of “|” instead of “when”. Thus;

              T              
Pbi 1P  z i  1Tb  b   0 | bi 1
The probability                                                          2             
of the ith bit is                                                 
demodulated in                          =
              T               
error                                           Pbi  0P  z i  1Tb  b   0 | bi  0
               2              
(1b)

Assume that,
1
Pbi  0  Pbi  1 
2
(i.e. “0” and “1” are equiprobable and the value of each
bit is independent of the value of all other bits in the data
stream)
EEE377 – A: Optimum Receiver                             5/36                                  March 2009
Substituiting this value;

1                T              
The probability                            P  z i  1Tb  b   0 | bi 1
2                 2             
of the ith bit is
demodulated in                       =                      +                          (2)
1                T               
error                                       P  z i  1Tb  b   0 | bi  0
2                 2              

Let’s take a look at the input to sampler z(t);

z(t) = received signal after filtering

But before that, let’s examine the received signal (before
filtering or after the channel). Assume the channel has
sufficient bandwidth (the channel does not distort the
transmitted signal, although it can attenuate the signal);

Received signal = attenuated transmitted signal + noise

Symbolically;

r t   st   nt 
where
 - constant reflecting attenuation of the channel
nt  - noise at receiver input

           T                   T                 T 
r i  1Tb  b   s i  1Tb  b   ni  1Tb  b                     (3)
            2                   2                 2
EEE377 – A: Optimum Receiver                        6/36                              March 2009
Effect of filtering to the signal;

The filter has the same bandwidth of the transmitted
signal, thus st  will pass through the filter.

Noise has greater bandwidth than the filter, thus we use
the variable n0 t  to represent the noise the filter output.

The voltage at the sampler input (output of filter);

zt   st   no t 

So

           T                   T                   T 
z i  1Tb  b   s i  1Tb  b   no i  1Tb  b    (4)
            2                   2                   2

Let’s say the transmitted signal is binary PAM compose
of rectangular, sinc-shaped or raised cosine pulses with
peak amplitude of A if bi=1 and –A if bi=0.

Since there is no ISI in the exact center of bit period;

           T 
s i  1Tb  b   A when bi 1,
            2
and
           T 
s i  1Tb  b    A when bi  0
            2

EEE377 – A: Optimum Receiver                    7/36                       March 2009
Substuting into Eq 4, thus the expression is reduced to;

                     T           
A  no i  1Tb  b  if bi  1 
           Tb  
                      2          
z i  1Tb                                       
           2                          Tb 
 A  no i  1Tb   if bi  0

                       2         


Thus, we can express Eq. 2 as;

1                     T      
The probability                   P A  no i  1Tb  b   0
2                      2     
of the ith bit is
demodulated in               =                +                             (6)
1                       T      
error                             P  A  no i  1Tb  b   0
2                        2     

Manipulating Eq (6);

Pith bit in error 
1                T          1                 T           (7)
P no i  1Tb  b    A  P no i  1Tb  b   A
2                 2         2                  2      

NOTE
For the case of independent bits and equirobable “1”s and
“0”s, Eq. 7 tells that the ith bit is demodulated in error:–
If at the time of the ith sampling at the receiver, the noise
is stronger than the transmitted signal (attenuated) and is
of the opposite sign.
EEE377 – A: Optimum Receiver             8/36                          March 2009
2. Stochastic Mathematics

Although we cannot say certainly how much noise will
occur in the system at any particular time, but we can
determine the probability that the amount of noise at that
time will be less than a certain value.

           T 
We can’t determine in advance the value of       no i  1Tb  b  ,
            2
but we can determine the probability that it will be
  A volts.

NOTE
1. This is very different approach
(probabilistic/stochastic, mathematics) compared to
what we are used to (deterministic mathematic).

2. Deterministic mathematic says – “I can tell you with
certainty what the value of a particular variable will
be at any time t”.

3. Probabilistic or stochastic, mathematics says- “I don’t
know enough about a particular variable (like n(t)) to
tell you with certainty its value at particular time in
the future (like t  i  1Tb  Tb ), but I can tell you
2
something about the tendencies of n(t) at time
T
t  i  1Tb  b .
2

EEE377 – A: Optimum Receiver       9/36                        March 2009
A probabilistic description isn’t as good as a deterministic
description, but often it’s the best we can do and is better
than nothing.

A phenomena that we cannot describe deterministically,
but that we can describe probabilistically is called random

We don’t know exactly how it will behave at any
particular time, but through experimentation and analysis
of the physical occurrences that produce the random
event, we do know its tendencies.

Let’s develop a set of mathematic parameters that we can
use to describe the tendencies of a random phenomenon
such as noise.

Probability distribution function
As we saw in calculating probability of error, one
parameter we are interested is the probability that n(t) will
be less than a certain value  A at a certain time
( t  i  1Tb  Tb ).
2

We are interested in knowing the probability that the
result of a random phenomena (e.g. noise voltage) is less
than or equal to a certain value at certain time.

EEE377 – A: Optimum Receiver    10/36                   March 2009
Let’s us use the variable X to describe the voltage of noise
at a particular time (random variable).

The probability distribution function for X is symbolized
as
Fxa  PX  a            (8)

In other words, Fxa is the probability that the random
variable X is less than or equal to some specific value a.

The probability distribution function has two forms;
1. Discrete variable – tossing coin, rolling die

2. Continuous variable – lifetime of electronic

component

Probability distribution properties;

1.    Fx   0                          (9)
2.   Fx  1                             (10)
3.   If b  a, then Fx b  Fx a        (11)
4.   The probability that an outcome will have a
random variable value x, within the range a  x  b is
given by expression F b  F a .
X           X

EEE377 – A: Optimum Receiver            11/36                  March 2009
In our case, we need to determine the probability
distribution function for the noise.

The most prevalent type of noise in the channel (and front
end of the receiver too) is noise due to the random motion
of electrons within electronic devices – thermal noise, it
probability distribution function;
  x   n 2   
                                     
1             2 2           
F x a                   e             n      
dx
       2  n

where (mean) n  0 , and (variance)  n2 = average
normalized noise power.

Probability density function
It is symbolized as f X x and defined as

dFX  x 
f X x  
dx

It can be used to express the probability that the random
variable X lies between two values, say, a and b (b>a).

Pa  X  b PX  b  PX  a
 FX b   FX a 

  f X  x dx   f X  x dx
b                    a

                      

  f X  x dx
b

a

EEE377 – A: Optimum Receiver                                     12/36                    March 2009
Properties;
1. f X x 0 for    x  

2.  f X x  dx  1


3. FX a    f X  x dx
a



Combining property 2 and 3;

PX  c1  PX  c1  F X c 

1   f X  x dx
c



  f X  x  dx   f X  x dx
c

                

  f X  x dx
c

Thermal noise
Earlier the thermal noise is given as

  x   n 2   
                                 
1             2 2           
F x a               e             n      
dx
   2  n

where (mean) n  0 , and (variance)  n2 = average
normalized noise power.

Thus, the probability density function for thermal noise;

EEE377 – A: Optimum Receiver                                  13/36                    March 2009
  x   n 2   
                
1          2 2           
f X x              e             n      
2  n

This function usually called as –Gaussian probability
function, or Gaussian function.

Mean (expected value)
Defined as;

 X   xf X x dx


In other words,  X is the average value of random variable
X. Often it is also called as the expected value of X and
symbolized as EX ;

EX    X   xf X  x dx


Variance
The variance of random variable X is defined as
             2
    x   X  f X  x dx
2
X


Or it also can be expressed as
            2

X 
2
 x   X                            
f X  x dx  E  X   X 
2



Variance is a measure of how much the values of the
random variable X fluctuate (or vary) around its average.
Thus, variance is a measure of the unpredictability of X.

EEE377 – A: Optimum Receiver                                     14/36                 March 2009
NOTE
If a random variable has zero mean ( X  0) , then its
variance  n2 is equal to its average normalized power of
X.
               2                               2

X         x             f X  x dx      x  0       f X  x dx
2
X
                                  

       x 2 f X  x dx


Joint Probabilities
This is useful when determining how two random
variables may be related to each other.

The joint probability distribution function of two random
variables X and Y is defined as;

 2 FX ,Y  x, y 
f X , Y  x, y  
xy

Conditional Probabilities
The notation P X  a | event Z  to express “the probability
that random variable X is greater than a given that event Z
has occurred.”

NOTE (We always encounter)
PX  a and event Z  Pevent ZPX  a | event Z

EEE377 – A: Optimum Receiver                                 15/36                     March 2009
3. Examining Thermal Noise

This defines by random motion of electrons within
electronic devices.
Through experimentation and analysis of the physical
events that produce the noise (random motion of
electrons), we can determine the following tendencies,
or stochastic parameters, for the noise voltage;
1. Its mean (average value) is 0. n  0
2. It has a Gaussian probability density function,
  x   n 2   
                
1             2 2           
f X x              e             n      
2  n
and thus the probability that the noise voltage
nt  at time t o will be less than or equal to a
certain value, say  A , is

  x   n 2      
 A                                      
1             2 2              
Pnto    A  FX  A                                          e            n          
dx

2  n

And the probability that nt  is greater than a
certain value, say A , is;

  x   n 2   
                                 
1             2 2           
Pnto   A  1  FX A                                         e             n      
dx       (28)
   A
2  n

EEE377 – A: Optimum Receiver                                        16/36                                                    March 2009
Since the random variable has zero mean ( X  0) ,
and its variance  n2  is equal to the average
normalized power of X.

A typical Gaussian probability density function
is shown below ( X  0 and  n2  1) ;

3. The probabilistic properties of nt  do not change
with time ( nt  is said stationary). Thus,

Pnto   a  Pnt1   a  Pnt2   a .......... for any to , t1 and t2
..

This is not to say that nt  is the same value at
times to , t1, and t2 , but merely it has the same
probabilistic tendencies.

4. Thermal noise has the average normalized power
spectral density Gn(f) [Volts2/Hz] as constant from
dc to approximately 1012 Hz .
EEE377 – A: Optimum Receiver                    17/36                           March 2009
This power spectral density, shown in Fig.4-4,
means that thermal noise contains equal amount of
all frequency components in the band from dc to
1012 Hz .

As an analogy, consider the color white that is
composed of all equal component colors. Similarly,
thermal noise is composed of the sum of equal
strength components of all frequencies.
Thermal noise is often called as white noise or
more specifically additive white Gaussian noise.

Returning to our discussion on the receiver – we know the
stochastic properties of nt  - the thermal noise at the input
of the receiver, therefore we can find the stochastic
properties of output noise after LPF is no t  .
If the LPF has an ideal frequency response and cutoff
frequency of f c , then the average normalized power
spectrum of the noise at the filter output is as shown in
Fig. 4-5.
EEE377 – A: Optimum Receiver         18/36                 March 2009
If a random signal with a Gaussian probability density
function is passed through a linear, time-invariant filter,
then the output is a random signal that also has a Gaussian
probability density function.

Thus no t  is still has a Gaussian probability density
function. The mean of no t  is zero, but different value for
variance  n2 

Using Eq. 28, the probability that no t  is greater than
some constant a at time t  to  can be expressed as;

  x   o 2   
                                
1             2 2           
Pno to   a                e            o       
dx   (29)
a
2  o
Where o  0 and              2
o    average        normalized power [Volts2]
at the filter output.

As defined in lecture notes (Signal Analysis – Review)
where the average normalized power [Volts2] as Ps, we
can express  o2 as;
EEE377 – A: Optimum Receiver                            19/36                         March 2009
         fc    No
o
2
 Ps   G no  f  df          df  N o f c
         fc   2                    [Volts2]   (30)

NOTE
The area under Fig 4-5 which is  o2  N o f c (average
normalized power after filtering) is significantly smaller
than  n2 (area under Fig. 4-4 due to extra frequencies
after f c ) which signifies average normalized power before
filtering.

The smaller the value of o2 , the smaller the value of
integral in Eq. 29 - thus, the smaller the probability that
the noise will exceed a given threshold a.

Let’s come back to Eq. 7. Such a probability is related to
the probability that the transmitted signal is received in
error.

Pith bit in error
1                T          1                 T       
    P no i  1Tb  b    A  P no i  1Tb  b   A
2                 2         2                  2      
1                   1
 Pno to    A Pno to   A              (31)
2                   2

For any t o since the noise is stationary.

EEE377 – A: Optimum Receiver                20/36                     March 2009
From Eq. 29, we know that
  x   o 2   
                                         
1               2 2           
Pno to   A                             e            o       
dx
   A
2  o

Since the probability density function is continuous and
does not contain any discrete terms, we can write;
  x   o 2   
                                         
1               2 2           
Pno to   A                             e            o       
dx
   A
2  o

By inspecting Fig. 4-3, we see that the Gaussian
probability density function is symmetric about its mean,
so
  x   o 2                                         x   o 2   
 A                                                                                  
1             2 2                                     1          2 2           
Pno to    A                            e            o       
dx                        e            o       
dx

2  o                                      A
2  o
 Pno to   A

Substituting into Eq. 31;

EEE377 – A: Optimum Receiver                                      21/36                                                         March 2009
Pith bit in error
1                 T          1                 T       
  P no i  1Tb  b    A  P no i  1Tb  b   A
2                  2         2                  2      
1                     1
 Pno to    A Pno to   A
2                     2
  x   o 2   
                                   
1             2 2           
   
A
2  o
e            o       
dx

This is often called probability of bit error and denoted               Pb

EXAMPLE
(Probability of bit error for a simple PAM receiver)

A transmitter uses raised cosine pulse shaping with pulse
amplitudes of +3 and -3 volts. By the time the signal
been attenuated to half of the transmitted signal, and the
signal has been corrupted with additive white Gaussian
noise. The average normalized power at the output of the
receiver’s filter is 0.36 volts2 and the mean µo= 0. Find Pb
assuming perfect synchronization.

SOLUTION

EEE377 – A: Optimum Receiver                       22/36           March 2009
5. Gaussian Probability Density Function

As has been established the Gaussian probability Density
function is expressed as;
  x   2   
              
1         2 2         
f X x          e                  
(19R)
2 

Thus, if a random variable X can be described using the
Gaussian probability density function;
  x   o 2   
                                       
value of Gaussian                               1                2 2           
P
random var iable 
 
a       a
2  o
e            o       
dx   (29R)

Many random occurrences can be expressed using the
Gaussian function;
 thermal noise (zero mean  and variance  2 as the
average normalized power)
 IQ results (mean  = 100, and variance  2 of 256)

The Gaussian function is described by mean  and
standard deviation .
Changing mean  corresponds to shifting the graph along
the x-axis.
Changing  affects the “spread” of the function.

EEE377 – A: Optimum Receiver                                  23/36                                  March 2009
As seen in the previous example, solving the integral of
Gaussian function (Eq. 29R) requires numerical analysis
using computer program. Is there any easier solution?

1. We create a single table that calculate the value of
integral for various value of a with   0 and   1. This
has the following integral, or defines as Q(a);
  x  0 2   
                                       

x2
1           2 12                      1
Qa                e                  
dx           e        2 dx   (35a)
a
2 1                                 a
2

Fig. 4-7 shows the Q(a) integral graphically. Table 4-
1 shows the values for Q(a) for various values of a.

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EEE377 – A: Optimum Receiver   25/36   March 2009
NOTE
a. Q(0.53) = 0.2981
b. For a  3 correspond to Qa  0.0014
a2
1
Qa        e                   2     for a  3                     (36)
a 2

2. What if  is not zero?

Consider the integral;
                   x   2
1       

a
2
e             2        dx

Changing variable,
Let w  x   ; thus, dw  dx
Lower limit: w  x  ; when x  a; w  a  
Thus,
                   x   2                        w2
1                                       1 
        e             2        dx             e    2 dw    Qa             (38)
a
2                                a
2

Fig. 4-8a and 4-8b show the equivalency.

EEE377 – A: Optimum Receiver                                   26/36                       March 2009
Geometrically, Eq. 38 corresponds to sliding the
Gaussian probability density function to the left by  .
Table 4-1 is suitable to solve integral of a Gaussian
function with any value of a ,  , but only for   1.

3. What if  is not 1?

Consider the integral;
  x   2   
                            
1         2 2         
a
2 
e                  
dx

EEE377 – A: Optimum Receiver                               27/36   March 2009
Changing variable,
Let w  x   ; thus, dw  dx
Lower limit: w  x  ; when x  a; w  a  
Thus,
                       x   2                               w2
                                         
1                                          1

a
2 
e             2 2
dx     
a
2 
e          2 2
dw

Now let’s perform a second change in variable;
Let z  w ; thus, dz  dw
              
Lower limit: z  w  x    ; when x  a; z  a   
                                                

Thus,
                       x   2                               w2                            z2
                                                                       
1                                          1                                  1

a
2 
e             2 2
dx     
a
2 
e          2 2
dw     
a
2
e        2   dz



Or
                       x   2                          z2
1           
1                    a  

a
2 
e             2 2
dx     
a
2
e        2   dz  Q
  
                  (42)


Now our table 4-1 can be used to solve Eq. 28 for any
value of a ,  , and  .

Fig. 4-9a to Fig. 4-9c shows the equivalency after
changing the variables.

EEE377 – A: Optimum Receiver                                     28/36                                            March 2009
NOTE
The combination of expending the y axis by  and
compressing the x axis by  (Fig. 4-9c) produces no
net effect on the total shaded area under the curve.
EEE377 – A: Optimum Receiver       29/36                 March 2009
EXAMPLE (Revisit)
(Probability of bit error for a simple PAM receiver)

A transmitter uses raised cosine pulse shaping with pulse
amplitudes of +3 and -3 volts. By the time the signal
been attenuated to half of the transmitted signal, and the
signal has been corrupted with additive white Gaussian
noise. The average normalized power at the output of the
receiver’s filter is 0.36 volts2 and the mean µo= 0. Find Pb
assuming perfect synchronization.

SOLUTION

Simplifying the Expression of Bit Error
The probability of bit error for PAM receiver is given by;

Pith bit in error
1                T          1                 T       
    P no i  1Tb  b    A  P no i  1Tb  b   A
2                 2         2                  2      
1                   1
 Pno to    A Pno to   A
2                   2
  x   o 2   
                                     
1              2 2           
   
A
2  o
e            o       
dx

EEE377 – A: Optimum Receiver                         30/36     March 2009
Since µo=0, using the Q function we can simplify this
equation to;
  x   o 2   
                
                2 2                     A   o   
Pith bit in error   
1
e         o        dx    Q
 


A   2  o                                     o      
 A  0 
 Q
  

 o                                                                             (43)
 A 
 Q
 

 o

The physical interpretation;
 A 
Pith bit in error  Q 
 
 o
 received signal strength of sampled value 
 Q
                                           
                                  (44)
                     o                    
 received signal strength of sampled value 
 Q                                             
 ave normalized noise power at filter output 
                                             

Consider the following observations;
1. The larger the value of a, the smaller the value Q(a)
(see Figure 4-7, 4-8 etc). Thus, increasing the
argument of function Q, decreases the probability of
bit error.
2. The argument of Q is the ratio of received signal
strength versus noise  o . The larger this signal-to-
noise- ratio (SNR), the lower the probability bit error.
3. The LPF use in the receiver is to reduce the average
noise power at the receiver, while not affecting the
EEE377 – A: Optimum Receiver                       31/36                                      March 2009
strength of received signal (compatible to the
bandwidth of signal). Thus, increases the value of
SNR, and reducing the probability of bit error.

OBSERVE
1. Pb can be interpreted as the probability that a
particular transmitted bit is demodulated in error.
2. Pb can also be interpreted as the percentage of all
transmitted bits, which on average will be
demodulated in error.

EXAMPLE
A transmitter uses raised cosine pulse shaping with pulse
amplitudes of +1 and -1 volts. By the time the signal
been attenuated to half of the transmitted signal, and the
signal has been corrupted with additive white Gaussian
noise. The average normalized power at the output of the
receiver’s filter is 0.035 volts2 and the mean µo= 0. Find
Pb assuming perfect synchronization.

SOLUTION

A = 1 volts
      = fraction of the transmitted signal voltage = 0.5
EEE377 – A: Optimum Receiver      32/36                     March 2009
2
 o = average normalized power = 0.035 volts
2

 received signal strength sampled value 
Pb  Q
                                            

                    o                      
 A           0.51 
 Q          Q             Q2.67   0.0038
 0.035        0.035 

Thus, on average, 0.38% of the transmitted bits will be
received in error or using different interpretation, on
average 3.8 out of every thousand bits will be received in
error.

Sample

zt 
signal r t        LPF                                    Comparator
“1” if sample  0
“0” if sample  0
no t 

Let develop a receiver with better performance
Can we think about other than just sampling to detect the
signal (previous notes).

Consider a general block diagram for demodulation
process;
EEE377 – A: Optimum Receiver                                33/36                             March 2009
Restrict the “processing” part to be linear, thus it can be
represented as transfer function H(f) as shown below;

noise                         z t   r t  * ht 
r t   st   nt 
z iTb    Threshold
comparison   “1” if z iTb   0;
(Threshold
Hf                                                           “0” if z iTb   0;
z t 
Sample at                 = 0 volts)
t  iT

Fig. 4-11: Receiver for implementing general demodulation process

Let;
r t 
- the received signal which can be decomposed into a
component due to the attenuated signal st  , and a
component due to noise nt 

EEE377 – A: Optimum Receiver                                      34/36                                     March 2009
ht 
- the impulse response corresponding to the transfer
function H(f);
ht   1H  f 

zt    - the processed received signal
z t   r t  * ht   st   nt  * ht 
 st  * ht   nt  * ht 

OBSERVE
1. nt  has a Gaussian probability distribution, so nt * ht 
has also Gaussian probability distribution.
2. Processing of the received signal for the ith bit occurs
during the time interval i 1Tb  t  iTb . At the end of this
period, the processing produces a single value z(iTb ) ,
which is compared to a threshold to determine whether
the ith bit is demodulated as a “1” or “0”.

For a simple PAM receiver, the probability bit error
 A 
Pith bit in error  Q 
 
 o
 received signal strength of sampled value 
 Q                                             
 ave normalized noise power at filter output 
                                             

EEE377 – A: Optimum Receiver                          35/36   March 2009
Since nt * ht  (processed noise) exhibits a zero-mean
Gaussian probability distribution;
 received signal strength of sampled processed signal 
Pb , simple receiver  Q



       ave normalized noise power at filter output    
 siTb  * hiTb  
 Q                    
       p           
                    
(47)
Where  2 represents the average normalized power of the
p

processed noise (and the variance of its probability
distribution)
QUESTION
What type of processing will minimize the probability of
bit error for receiver Fig. 4-11?
This is equivalent of asking what expression for H(f) will