Homework #8
1. Prob. 4.2.8
1.1 The problem is to determine mT  m1 m2  m N  so as to minimize
M
E   Pr H i  s i  m
2
i 1
where sTi  si1
si 2  siN  .
The average energy can be rewritten as




M
N
M N
E   Pr H i   sij  m j 2    PrH i  sij  m j 2 .
i 1
j 1
i 1 j 1
To determine m , we solve the N simultaneous equations given by


M
E
 2  PrH i  sij  m j  0 ; j  1,2,, N .
m j
i 1
Because the equations are uncoupled with respect to the components of m, it follows
readily that
M
M
m j  Pr H i    sij Pr H i  ; j  1,2,, N
i 1
i 1
M
m j   sij Pr H i  ; j  1,2,, N .
i 1
In vector form the solution is given by
M
m   s i Pr H i  .
i 1
1.2 The jth component of m is the expected value of the jth component of the signal
set. When the hypotheses are equally likely, m corresponds to the center of
gravity of the signal set.
1.3 For M orthogonal equal-energy signals the components can be chosen as
sij  E ij ; i  1,2,, M , j  1,2,, N
Assuming the hypotheses are equally likely,
1
Pr H i  
; i  1,2,  , M ,
M
1 M
E
m j 
; j  1,2,, N .
 E ij 
M i 1
M
M
Assuming one orthogonal axis to each signal, N = M. Since  s i  Mm , any one of
i 1
s i can be expressed as a linear combination of the others. Therefore, M simplex
signals span a space (M-1) dimensions.
M=2

s1  m T

s 2  m T
 0
E


E  E

2   2

E 
E
  
2   2
 E
0 
 2
 E
E 
 2

E

2 
E

2 
2
E
2

E
2
1
E
2

E
2
Observe that the two simplex signals lie on a straight line through the origin and are
E
equidistant from the original with energy .
2
M=3

s1  m T

s 2  m T
 0
s 3  m T
 0 0
E



 E
0 0 
 2
E

 E
0 
 2

 E
E 
 2
E
2
E  E

2   2
E
2
E 
E
  
2   2
E
2
E 
E
  
2   2

E
2

E

2 
E
2

E

2 

E
2
E

2 
The distance between any two simplex signals equals 2E . The three simplex signals
are the vertices of equilateral triangle, centered at the origin, where sides are length
2E . All three signals and the origin lie in a plane. Each signal is equidistant from
2
the origin with energy E
3
M=4
The simplex signals are the vertices of a tetrahedron centered at the origin. Each edge
of the tetrahedron is the length 2E . Thus, the tetrahedron is compressed for four
identical equilateral triangles. Each signal is equidistant from the origin with energy
3
E.
4
1.4
Ei  s i  m
2

E
 
 M

E
E
  
  E 

M 
M 

E

M
2
1 

 Ei  M  1
 E1  
2
 M
M

E
2
1 

 M  1
 E 1 

M2
 M M2 
E
E

M  1  M 2  2M  1 
M2 M
2
2
M
M
M 1
E
M
E M  1
Consequently, each simplex signal is at a distance
from the origin.
M
E




1.5 Since the spacing between the original signals is unaffected by the translation,
the probability of error for the simplex set equals that for the original set. The
energy reduction for each transmitted signal is
M 1
E
 M 1
E
E  1 
.
E 
M
M 
M

The reduction is largest for M = 2 (i.e., 3 dB improvement). For large value of M the
savings becomes negligible.
2. Prob. 4.2.9
2.1
T
 E ;i  j
E  si t s j t dt  
.

E
;
i

j

0
Using the Schwartz inequality,
2
2
T

T 2
 T 2







s
t
s
t
dt

s
t
dt
 i

 i
   s j t dt 
j
 0
 0

  0

For i  j , it follows that
T
 si t s j t dt    1 .
0
Now consider
TM
2
M M T



s
t
dt

   si t s j t dt
 i 
i 1 j 1 0

0 i 1
MT
M
M
T
   si2 t dt     si t s j t dt
i 1
0
0 
 i 1 j 1, j  i 
 

1



 M  M M 0
1   M  1  0
1

.
M 1
2
We conclude that

1
  1
M 1
2.2 Let 1 t ,2 t ,,M t  be a set of M orthogonal waveforms with unit
1
energy. Assume all hypotheses are equally likely. Thus, Pr H i  
M
th
; i  1,2,, M . The i signal in the simplex set is given by


1 M
Ei si t   i t  
  j t  E .
M j 1


The correlation coefficient is given by
T 

 
1 M
1 M
 k t   j t    l t dt E
  i t  
M k 1
M l 1
 

 
  0
;i  j
Ei E j
 

where E is the energy of each simplex signal. From problem 4.2.8
M 1
Ei  E j  E
.
M
Hence, we have
T
M 1
1 M T
   i t  j t dt 
   k t  j t dt
M
M
k
1 0
0
-
1 MT
1 M MT





t

t
dt

 i l
    k t l t dt
M l 10
M 2 k 1 l 10
Since i  j ,
M 1
1 1
M
1
  0 - 

2
M
M M M
M
  
1 M
1

M M 1
M 1
2.3 We first demonstrate that any set of equally-correlated signals can be
generated from an orthogonal set by a linear transformation of the form
M
si t   i t      j t 
j 1
where  is a constant and 1 t ,2 t ,,M t  is a set of M orthogonal waveforms
with energy Eorth . Let E denote the energy of the equally correlated signals. Thus,
T
T
M
M


E   si2 t dt   i t      k t  i t     l t  dt
k 1
l 1


0
0
T
M T
0
k 1 0
  i2 t dt     i t  k t dt
MT
M MT
l 10
k 1 l 10
    l t i t dt   2     k t l t dt




 1  2  M 2 Eorth .
In a similar fashion, assuming i  j , the correlation coefficient is given by
T
E   1  2  M 2 Eorth   si t s j t dt
0
T
M
M


  i t      k t   j t     l t  dt
k 1
l 1


0


  2  M 2 Eorth
It follows that


 1  2  M 2  2  M 2 ,
M 1    2  21       0 .
Solving for  , we have
21     41   2  4 M 1   
2M 1   
1 
M 

1  1 

M
1  
1 
1   M  1 

1 

M
1 

1
1
Observe that   
results in  
and we have the simplex set of part(2.2)
M 1
M
A linear translation does not affect the error probabilities. From the previous
paragraph,



E   1  2  M 2 Eorth


 E  1  2  M 2 Eorth


E  Eorth   2  M 2 Eorth .
Also,


E   2  M 2 Eorth  E  Eorth
 E 1     Eorth .
1
. Thus, probability of error for a simplex set
M 1
with signal energy E is the same as the probability of error for an orthogonal
set with signal energy
2.4 For a simplex set   
1 
M

Eorth  E 1     E 1 
E.

 M 1 M 1