or since N =2
ν
D
total
=
x
2
max
3 · 4
ν
(1+4p
b
(4
ν
− 1)) =
x
2
max
3N
2
(1+4p
b
(N
2
− 1))
4)
SNR =
E[X
2
]
D

total
=
E[X
2
]3N
2
x
2
max
(1+4p
b
(N
2
− 1))
If we let
˘
X =
X
x
max
, then
E[X
2
]
x
2
max
= E[
˘
X

2
]=
˘
X
2
. Hence,
SNR =
3N
2
˘
X
2
1+4p
b
(N
2
− 1)
=
3 · 4
ν
˘
X
2
1+4p
b
(4
ν
− 1)
Problem 6.57
1)

g(x)=
log(1 + µ
|x|
x
max
)
log(1 + µ)
sgn(x)
Differentiating the previous using natural logarithms, we obtain
g

(x)=
1
ln(1 + µ)
µ/x
max
(1 + µ
|x|
x
max
)
sgn
2
(x)
Since, for the µ-law compander y
max
= g(x
max
) = 1, we obtain
D ≈

y
2
max
3 × 4
ν

∞
−∞
f
X
(x)
[g

(x)]
2
dx
=
x
2
max
[ln(1 + µ)]
2
3 × 4
ν
µ
2

∞
−∞


1+µ
2
|x|
2
x
2
max
+2µ
|x|
x
max

f
X
(x)dx
=
x
2
max
[ln(1 + µ)]
2
3 × 4
ν
µ
2

1+µ
2
E[
˘

X
2
]+2µE[|
˘
X|]

=
x
2
max
[ln(1 + µ)]
2
3 × N
2
µ
2

1+µ
2
E[
˘
X
2
]+2µE[|
˘
X|]

where N
2
=4

ν
and
˘
X = X/x
max
.
2)
SQNR =
E[X
2
]
D
=
E[X
2
]
x
2
max
µ
2
3 · N
2
[ln(1 + µ)]
2
(µ
2
E[
˘
X

2
]+2µE[|
˘
X|]+1)
=
3µ
2
N
2
E[
˘
X
2
]
[ln(1 + µ)]
2
(µ
2
E[
˘
X
2
]+2µE[|
˘
X|]+1)
3) Since SQNR
unif
=3·N
2
E[

˘
X
2
], we have
SQNR
µ
law
= SQNR
unif
µ
2
[ln(1 + µ)]
2
(µ
2
E[
˘
X
2
]+2µE[|
˘
X|]+1)
= SQNR
unif
G(µ,
˘
X)
158
where we identify
G(µ,

˘
X)=
µ
2
[ln(1 + µ)]
2
(µ
2
E[
˘
X
2
]+2µE[|
˘
X|]+1)
3) The truncated Gaussian distribution has a PDF given by
f
Y
(y)=
K
√
2πσ
x
e
−
x
2
2σ
2
x

where the constant K is such that
K

4σ
x
−4σ
x
1
√
2πσ
x
e
−
x
2
2σ
2
x
dx =1=⇒ K =
1
1 − 2Q(4)
=1.0001
Hence,
E[|
˘
X|]=
K
√
2πσ
x


4σ
x
−4σ
x
|x|
4σ
x
e
−
x
2
2σ
2
x
dx
=
2K
4
√
2πσ
2
x

4σ
x
0
xe
−
x

2
2σ
2
x
dx
=
K
2
√
2πσ
2
x

−σ
2
x
e
−
x
2
2σ
2
x

4σ
x
0
=
K
2

√
2π
(1 − e
−2
)=0.1725
In the next figure we plot 10 log
10
SQNR
unif
and 10 log
10
SQNR
mu−law
vs. 10 log
10
E[
˘
X
2
] when the
latter varies from −100 to 100 db. As it is observed the µ-law compressor is insensitive to the
dynamic range of the input signal for E[
˘
X
2
] > 1.
-50
0
50
100

150
200
-100
-80 -60 -40 -20 0 20 40 60 80 100
mu-law
uniform
E[X^2] db
SQNR (db)
Problem 6.58
The optimal compressor has the form
g(x)=y
max


2

x
−∞
[f
X
(v)]
1
3
dv

∞
−∞
[f
X
(v)]

1
3
dv
−


where y
max
= g(x
max
)=g(1).

∞
−∞
[f
X
(v)]
1
3
dv =

1
−1
[f
X
(v)]
1
3
dv =


0
−1
(v +1)
1
3
dv +

1
0
(−v +1)
1
3
dv
=2

1
0
x
1
3
dx =
3
2
159
If x ≤ 0, then

x
−∞
[f
X

(v)]
1
3
dv =

x
−1
(v +1)
1
3
dv =

x+1
0
z
1
3
dz =
3
4
z
4
3




x+1
0
=

3
4
(x +1)
4
3
If x>0, then

x
−∞
[f
X
(v)]
1
3
dv =

0
−1
(v +1)
1
3
dv +

x
0
(−v +1)
1
3
dv =
3

4
+

1
1−x
z
1
3
dz
=
3
4
+
3
4

1 − (1 − x)
4
3

Hence,
g(x)=



g(1)

(x +1)
4
3

− 1

−1 ≤ x<0
g(1)

1 − (1 − x)
4
3

0 ≤ x ≤ 1
The next figure depicts g(x) for g(1) = 1. Since the resulting distortion is (see Equation 6.6.17)
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-1
-0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
g(x)
x
D =
1
12 × 4
ν



∞
−infty
[f
X
(x)]
1
3
dx

3
=
1
12 × 4
ν

3
2

3
we have
SQNR =
E[X
2
]
D
=
32
9

× 4
ν
E[X
2
]=
32
9
× 4
ν
·
1
6
=
16
27
4
ν
Problem 6.59
The sampling rate is f
s
= 44100 meaning that we take 44100 samples per second. Each sample is
quantized using 16 bits so the total number of bits per second is 44100 × 16. For a music piece of
duration 50 min = 3000 sec the resulting number of bits per channel (left and right) is
44100 × 16 × 3000 = 2.1168 × 10
9
and the overall number of bits is
2.1168 × 10
9
× 2=4.2336 × 10
9

160
Chapter 7
Problem 7.1
The amplitudes A
m
take the values
A
m
=(2m − 1 − M)
d
2
,m=1, M
Hence, the average energy is
E
av
=
1
M
M

m=1
s
2
m
=
d
2
4M
E
g

M

m=1
(2m − 1 − M)
2
=
d
2
4M
E
g
M

m=1
[4m
2
+(M +1)
2
− 4m(M + 1)]
=
d
2
4M
E
g

4
M

m=1

m
2
+ M(M +1)
2
− 4(M +1)
M

m=1
m

=
d
2
4M
E
g

4
M(M + 1)(2M +1)
6
+ M(M +1)
2
− 4(M +1)
M(M +1)
2

=
M
2
− 1

3
d
2
4
E
g
Problem 7.2
The correlation coefficient between the m
th
and the n
th
signal points is
γ
mn
=
s
m
· s
n
|s
m
||s
n
|
where s
m
=(s
m1
,s
m2

, ,s
mN
) and s
mj
= ±

E
s
N
. Two adjacent signal points differ in only one
coordinate, for which s
mk
and s
nk
have opposite signs. Hence,
s
m
· s
n
=
N

j=1
s
mj
s
nj
=

j=k

s
mj
s
nj
+ s
mk
s
nk
=(N − 1)
E
s
N
−
E
s
N
=
N − 2
N
E
s
Furthermore, |s
m
| = |s
n
| =(E
s
)
1
2

so that
γ
mn
=
N − 2
N
The Euclidean distance between the two adjacent signal points is
d =

|s
m
− s
n
|
2
=





±2

E
s
/N





2
=

4
E
s
N
=2

E
s
N
161
Problem 7.3
a) To show that the waveforms ψ
n
(t), n =1, ,3 are orthogonal we have to prove that

∞
−∞
ψ
m
(t)ψ
n
(t)dt =0,m= n
Clearly,
c
12
=


∞
−∞
ψ
1
(t)ψ
2
(t)dt =

4
0
ψ
1
(t)ψ
2
(t)dt
=

2
0
ψ
1
(t)ψ
2
(t)dt +

4
2
ψ
1
(t)ψ

2
(t)dt
=
1
4

2
0
dt −
1
4

4
2
dt =
1
4
× 2 −
1
4
× (4 − 2)
=0
Similarly,
c
13
=

∞
−∞
ψ

1
(t)ψ
3
(t)dt =

4
0
ψ
1
(t)ψ
3
(t)dt
=
1
4

1
0
dt −
1
4

2
1
dt −
1
4

3
2

dt +
1
4

4
3
dt
=0
and
c
23
=

∞
−∞
ψ
2
(t)ψ
3
(t)dt =

4
0
ψ
2
(t)ψ
3
(t)dt
=
1

4

1
0
dt −
1
4

2
1
dt +
1
4

3
2
dt −
1
4

4
3
dt
=0
Thus, the signals ψ
n
(t) are orthogonal.
b) We first determine the weighting coefficients
x
n

=

∞
−∞
x(t)ψ
n
(t)dt, n =1, 2, 3
x
1
=

4
0
x(t)ψ
1
(t)dt = −
1
2

1
0
dt +
1
2

2
1
dt −
1
2


3
2
dt +
1
2

4
3
dt =0
x
2
=

4
0
x(t)ψ
2
(t)dt =
1
2

4
0
x(t)dt =0
x
3
=

4

0
x(t)ψ
3
(t)dt = −
1
2

1
0
dt −
1
2

2
1
dt +
1
2

3
2
dt +
1
2

4
3
dt =0
As it is observed, x(t) is orthogonal to the signal waveforms ψ
n

(t), n =1, 2, 3 and thus it can not
represented as a linear combination of these functions.
Problem 7.4
a) The expansion coefficients {c
n
}, that minimize the mean square error, satisfy
c
n
=

∞
−∞
x(t)ψ
n
(t)dt =

4
0
sin
πt
4
ψ
n
(t)dt
162
Hence,
c
1
=


4
0
sin
πt
4
ψ
1
(t)dt =
1
2

2
0
sin
πt
4
dt −
1
2

4
2
sin
πt
4
dt
= −
2
π
cos

πt
4




2
0
+
2
π
cos
πt
4




4
2
= −
2
π
(0 − 1) +
2
π
(−1 − 0)=0
Similarly,
c
2

=

4
0
sin
πt
4
ψ
2
(t)dt =
1
2

4
0
sin
πt
4
dt
= −
2
π
cos
πt
4




4

0
= −
2
π
(−1 − 1) =
4
π
and
c
3
=

4
0
sin
πt
4
ψ
3
(t)dt
=
1
2

1
0
sin
πt
4
dt −

1
2

2
1
sin
πt
4
dt +
1
2

3
2
sin
πt
4
dt −
1
2

4
3
sin
πt
4
dt
=0
Note that c
1

, c
2
can be found by inspection since sin
πt
4
is even with respect to the x = 2 axis and
ψ
1
(t), ψ
3
(t) are odd with respect to the same axis.
b) The residual mean square error E
min
can be found from
E
min
=

∞
−∞
|x(t)|
2
dt −
3

i=1
|c
i
|
2

Thus,
E
min
=

4
0

sin
πt
4

2
dt −

4
π

2
=
1
2

4
0

1 − cos
πt
2


dt −
16
π
2
=2−
1
π
sin
πt
2




4
0
−
16
π
2
=2−
16
π
2
Problem 7.5
a) As an orthonormal set of basis functions we consider the set
ψ
1
(t)=


10≤ t<1
0 o.w
ψ
2
(t)=

11≤ t<2
0 o.w
ψ
3
(t)=

12≤ t<3
0 o.w
ψ
4
(t)=

13≤ t<4
0 o.w
In matrix notation, the four waveforms can be represented as





s
1
(t)
s

2
(t)
s
3
(t)
s
4
(t)





=





2 −1 −1 −1
−2110
1 −11−1
1 −2 −22











ψ
1
(t)
ψ
2
(t)
ψ
3
(t)
ψ
4
(t)





163
Note that the rank of the transformation matrix is 4 and therefore, the dimensionality of the
waveforms is 4
b) The representation vectors are
s
1
=

2 −1 −1 −1


s
2
=

−2110

s
3
=

1 −11−1

s
4
=

1 −2 −22

c) The distance between the first and the second vector is
d
1,2
=

|s
1
− s
2
|
2
=






4 −2 −2 −1




2
=
√
25
Similarly we find that
d
1,3
=

|s
1
− s
3
|
2
=






10−20




2
=
√
5
d
1,4
=

|s
1
− s
4
|
2
=





111−3





2
=
√
12
d
2,3
=

|s
2
− s
3
|
2
=





−3201




2
=
√
14

d
2,4
=

|s
2
− s
4
|
2
=





−333−2




2
=
√
31
d
3,4
=

|s

3
− s
4
|
2
=





013−3




2
=
√
19
Thus, the minimum distance between any pair of vectors is d
min
=
√
5.
Problem 7.6
As a set of orthonormal functions we consider the waveforms
ψ
1
(t)=


10≤ t<1
0 o.w
ψ
2
(t)=

11≤ t<2
0 o.w
ψ
3
(t)=

12≤ t<3
0 o.w
The vector representation of the signals is
s
1
=

222

s
2
=

200

s
3

=

0 −2 −2

s
4
=

220

Note that s
3
(t)=s
2
(t) − s
1
(t) and that the dimensionality of the waveforms is 3.
164
Problem 7.7
The energy of the signal waveform s

m
(t)is
E

=

∞
−∞



s

m
(t)


2
dt =

∞
−∞





s
m
(t) −
1
M
M

k=1
s
k
(t)






2
dt
=

∞
−∞
s
2
m
(t)dt +
1
M
2
M

k=1
M

l=1

∞
−∞
s
k
(t)s
l
(t)dt

−
1
M
M

k=1

∞
−∞
s
m
(t)s
k
(t)dt −
1
M
M

l=1

∞
−∞
s
m
(t)s
l
(t)dt
= E +
1
M

2
M

k=1
M

l=1
Eδ
kl
−
2
M
E
= E +
1
M
E−
2
M
E =

M − 1
M

E
The correlation coefficient is given by
γ
mn
=


∞
−∞
s

m
(t)s

n
(t)dt


∞
−∞
|s

m
(t)|
2
dt

1
2


∞
−∞
|s

n
(t)|

2
dt

1
2
=
1
E


∞
−∞

s
m
(t) −
1
M
M

k=1
s
k
(t)

s
n
(t) −
1
M

M

l=1
s
l
(t)

dt
=
1
E



∞
−∞
s
m
(t)s
n
(t)dt +
1
M
2
M

k=1
M

l=1


∞
−∞
s
k
(t)s
l
(t)dt

−
1
E


1
M
M

k=1

∞
−∞
s
n
(t)s
k
(t)dt +
1
M
M


l=1

∞
−∞
s
m
(t)s
l
(t)dt

=
1
M
2
ME−
1
M
E−
1
M
E
M−1
M
E
= −
1
M − 1
Problem 7.8
Assuming that E[n

2
(t)] = σ
2
n
, we obtain
E[n
1
n
2
]=E


T
0
s
1
(t)n(t)dt


T
0
s
2
(v)n(v)dv

=

T
0


T
0
s
1
(t)s
2
(v)E[n(t)n(v)]dtdv
= σ
2
n

T
0
s
1
(t)s
2
(t)dt
=0
where the last equality follows from the orthogonality of the signal waveforms s
1
(t) and s
2
(t).
Problem 7.9
a) The received signal may be expressed as
r(t)=

n(t)ifs
0

(t) was transmitted
A + n(t)ifs
1
(t) was transmitted
165
Assuming that s(t) has unit energy, then the sampled outputs of the crosscorrelators are
r = s
m
+ n, m =0, 1
where s
0
=0,s
1
= A
√
T and the noise term n is a zero-mean Gaussian random variable with
variance
σ
2
n
= E

1
√
T

T
0
n(t)dt
1

√
T

T
0
n(τ)dτ

=
1
T

T
0

T
0
E [n(t)n(τ)] dtdτ
=
N
0
2T

T
0

T
0
δ(t −τ)dtdτ =
N
0

2
The probability density function for the sampled output is
f(r|s
0
)=
1
√
πN
0
e
−
r
2
N
0
f(r|s
1
)=
1
√
πN
0
e
−
(r−A
√
T )
2
N
0

Since the signals are equally probable, the optimal detector decides in favor of s
0
if
PM(r, s
0
)=f(r|s
0
) >f(r|s
1
) = PM(r, s
1
)
otherwise it decides in favor of s
1
. The decision rule may be expressed as
PM(r, s
0
)
PM(r, s
1
)
= e
(r−A
√
T )
2
−r
2
N
0

= e
−
(2r−A
√
T )A
√
T
N
0
s
0
>
<
s
1
1
or equivalently
r
s
1
>
<
s
0
1
2
A
√
T
The optimum threshold is

1
2
A
√
T .
b) The average probability of error is
P (e)=
1
2
P (e|s
0
)+
1
2
P (e|s
1
)
=
1
2

∞
1
2
A
√
T
f(r|s
0
)dr +

1
2

1
2
A
√
T
−∞
f(r|s
1
)dr
=
1
2

∞
1
2
A
√
T
1
√
πN
0
e
−
r
2

N
0
dr +
1
2

1
2
A
√
T
−∞
1
√
πN
0
e
−
(r−A
√
T )
2
N
0
dr
=
1
2

∞

1
2

2
N
0
A
√
T
1
√
2π
e
−
x
2
2
dx +
1
2

−
1
2

2
N
0
A
√

T
−∞
1
√
2π
e
−
x
2
2
dx
= Q

1
2

2
N
0
A
√
T

= Q

√
SNR

166
where

SNR =
1
2
A
2
T
N
0
Thus, the on-off signaling requires a factor of two more energy to achieve the same probability of
error as the antipodal signaling.
Problem 7.10
Since the rate of transmission is R =10
5
bits/sec, the bit interval T
b
is 10
−5
sec. The probability
of error in a binary PAM system is
P (e)=Q


2E
b
N
0

where the bit energy is E
b
= A

2
T
b
. With P (e)=P
2
=10
−6
, we obtain

2E
b
N
0
=4.75 =⇒E
b
=
4.75
2
N
0
2
=0.112813
Thus
A
2
T
b
=0.112813 =⇒ A =

0.112813 × 10

5
= 106.21
Problem 7.11
a) For a binary PAM system for which the two signals have unequal probability, the optimum
detector is
r
s
1
>
<
s
2
N
0
4
√
E
b
ln
1 − p
p
= η
The average probability of error is
P (e)=P (e|s
1
)P (s
1
)+P (e|s
2
)P (s

2
)
= pP(e|s
1
)+(1− p)P (e|s
2
)
= p

η
−∞
f(r|s
1
)dr +(1− p)

∞
η
f(r|s
1
)dr
= p

η
−∞
1
√
πN
0
e
−

(r−
√
E
b
)
2
N
0
dr +(1− p)

∞
η
1
√
πN
0
e
−
(r+
√
E
b
)
2
N
0
dr
= p
1
√

2π

η
1
−∞
e
−
x
2
2
dx +(1− p)
1
√
2π

∞
η
2
e
−
x
2
2
dx
where
η
1
= −

2E

b
N
0
+ η

2
N
0
η
2
=

2E
b
N
0
+ η

2
N
0
Thus,
P (e)=pQ


2E
b
N
0
− η


2
N
0

+(1− p)Q


2E
b
N
0
+ η

2
N
0

b) If p =0.3 and
E
b
N
0
= 10, then
P (e)=0.3Q[4.3774] + 0.7Q[4.5668] = 0.3 ×6.01 ×10
−6
+0.7 × 2.48 × 10
−6
=3.539 × 10
−6

167
If the symbols are equiprobable, then
P (e)=Q[

2E
b
N
0
]=Q[
√
20]=3.88 × 10
−6
Problem 7.12
a) The optimum threshold is given by
η =
N
0
4
√
E
b
ln
1 − p
p
=
N
0
4
√
E

b
ln 2
b) The average probability of error is (η =
N
0
4
√
E
b
ln 2)
P (e)=p(a
m
= −1)

∞
η
1
√
πN
0
e
−(r+
√
E
b
)
2
/N
0
dr

+p(a
m
=1)

η
−∞
1
√
πN
0
e
−(r−
√
E
b
)
2
/N
0
dr
=
2
3
Q

η +
√
E
b


N
0
/2

+
1
3
Q

√
E
b
− η

N
0
/2

=
2
3
Q


2N
0
/E
b
ln 2
4

+

2E
b
N
0

+
1
3
Q


2E
b
N
0
−

2N
0
/E
b
ln 2
4

Problem 7.13
a) The maximum likelihood criterion selects the maximum of f (r|s
m
) over the M possible trans-

mitted signals. When M = 2, the ML criterion takes the form
f(r|s
1
)
f(r|s
2
)
s
1
>
<
s
2
1
or, since
f(r|s
1
)=
1
√
πN
0
e
−(r−
√
E
b
)
2
/N

0
f(r|s
2
)=
1
√
πN
0
e
−(r+
√
E
b
)
2
/N
0
the optimum maximum-likelihood decision rule is
r
s
1
>
<
s
2
0
b) The average probability of error is given by
P (e)=p

∞

0
1
√
πN
0
e
−(r+
√
E
b
)
2
/N
0
dr +(1− p)

0
−∞
1
√
πN
0
e
−(r−
√
E
b
)
2
/N

0
dr
168
= p

∞
√
2E
b
/N
0
1
√
2π
e
−
x
2
2
dx +(1− p)

−
√
2E
b
/N
0
−∞
1
√

2π
e
−
x
2
2
dx
= pQ


2E
b
N
0

+(1− p)Q


2E
b
N
0

= Q


2E
b
N
0


Problem 7.14
a) The impulse response of the filter matched to s(t)is
h(t)=s(T −t)=s(3 − t)=s(t)
where we have used the fact that s(t) is even with respect to the t =
T
2
=
3
2
axis.
b) The output of the matched filter is
y(t)=s(t) s(t)=

t
0
s(τ)s(t −τ)dτ
=




























0 t<0
A
2
t 0 ≤ t<1
A
2
(2 − t)1≤ t<2
2A
2
(t − 2) 2 ≤ t<3
2A
2
(4 − t)3≤ t<4
A
2

(t − 4) 4 ≤ t<5
A
2
(6 − t)5≤ t<6
06≤ t
A scetch of y(t) is depicted in the next figure




❅
❅
❅✁
✁
✁
✁
✁
✁❆
❆
❆
❆
❆
❆

❅
❅
❅
2A
2
A

2
135642
c) At the output of the matched filter and for t = T = 3 the noise is
n
T
=

T
0
n(τ)h(T − τ )dτ
=

T
0
n(τ)s(T − (T − τ))dτ =

T
0
n(τ)s(τ)dτ
The variance of the noise is
σ
2
n
T
= E


T
0


T
0
n(τ)n(v)s(τ)s(v)dτdv

=

T
0

T
0
s(τ)s(v)E[n(τ)n(v)]dτdv
169
=
N
0
2

T
0

T
0
s(τ)s(v)δ(τ −v)dτ dv
=
N
0
2

T

0
s
2
(τ)dτ = N
0
A
2
d) For antipodal equiprobable signals the probability of error is
P (e)=Q



S
N

o

where

S
N

o
is the output SNR from the matched filter. Since

S
N

o
=

y
2
(T )
E[n
2
T
]
=
4A
4
N
0
A
2
we obtain
P (e)=Q



4A
2
N
0


Problem 7.15
a) Taking the inverse Fourier transform of H(f ), we obtain
h(t)=F
−1
[H(f)] = F

−1

1
j2πf

−F
−1

e
−j2πfT
j2πf

= sgn(t) − sgn(t − T )=2Π

t −
T
2
T

b) The signal waveform, to which h(t) is matched, is
s(t)=h(T −t)=2Π

T −t −
T
2
T

=2Π

T

2
− t
T

= h(t)
where we have used the symmetry of Π

t−
T
2
T

with respect to the t =
T
2
axis.
Problem 7.16
If g
T
(t) = sinc(t), then its matched waveform is h(t) = sinc(−t) = sinc(t). Since, (see Problem
2.17)
sinc(t)  sinc(t) = sinc(t)
the output of the matched filter is the same sinc pulse. If
g
T
(t) = sinc(
2
T
(t −
T

2
))
then the matched waveform is
h(t)=g
T
(T −t) = sinc(
2
T
(
T
2
− t)) = g
T
(t)
170
where the last equality follows from the fact that g
T
(t) is even with respect to the t =
T
2
axis. The
output of the matched filter is
y(t)=F
−1
[g
T
(t) g
T
(t)]
= F

−1

T
2
4
Π(
T
2
f)e
−j2πfT

=
T
2
sinc(
2
T
(t − T )) =
T
2
g
T
(t −
T
2
)
Thus the output of the matched filter is the same sinc function, scaled by
T
2
and centered at t = T.

Problem 7.17
1) The output of the integrator is
y(t)=

t
0
r(τ)dτ =

t
0
[s
i
(τ)+n(τ)]dτ
=

t
0
s
i
(τ)dτ +

t
0
n(τ)dτ
At time t = T we have
y(T)=

T
0
s

i
(τ)dτ +

T
0
n(τ)dτ = ±

E
b
T
T +

T
0
n(τ)dτ
The signal energy at the output of the integrator at t = T is
E
s
=


±

E
b
T
T


2

= E
b
T
whereas the noise power
P
n
= E


T
0

T
0
n(τ)n(v)dτdv

=

T
0

T
0
E[n(τ)n(v)]dτdv
=
N
0
2

T

0

T
0
δ(τ −v)dτ dv =
N
0
2
T
Hence, the output SNR is
SNR =
E
s
P
n
=
2E
b
N
0
2) The transfer function of the RC filter is
H(f)=
1
1+j2πRCf
Thus, the impulse response of the filter is
h(t)=
1
RC
e
−

t
RC
u
−1
(t)
171
and the output signal is given by
y(t)=
1
RC

t
−∞
r(τ)e
−
t−τ
RC
dτ
=
1
RC

t
−∞
(s
i
(τ)+n(τ))e
−
t−τ
RC

dτ
=
1
RC
e
−
t
RC

t
0
s
i
(τ)e
τ
RC
dτ +
1
RC
e
−
t
RC

t
−∞
n(τ)e
τ
RC
dτ

At time t = T we obtain
y(T)=
1
RC
e
−
T
RC

T
0
s
i
(τ)e
τ
RC
dτ +
1
RC
e
−
T
RC

T
−∞
n(τ)e
τ
RC
dτ

The signal energy at the output of the filter is
E
s
=
1
(RC)
2
e
−
2T
RC

T
0

T
0
s
i
(τ)s
i
(v)e
τ
RC
e
v
RC
dτdv
=
1

(RC)
2
e
−
2T
RC
E
b
T


T
0
e
τ
RC
dτ

2
= e
−
2T
RC
E
b
T

e
T
RC

− 1

2
=
E
b
T

1 − e
−
T
RC

2
The noise power at the output of the filter is
P
n
=
1
(RC)
2
e
−
2T
RC

T
−∞

T

−∞
E[n(τ)n(v)]dτdv
=
1
(RC)
2
e
−
2T
RC

T
−∞

T
−∞
N
0
2
δ(τ −v)e
τ+v
RC
dτdv
=
1
(RC)
2
e
−
2T

RC

T
−∞
N
0
2
e
2τ
RC
dτ
=
1
2RC
e
−
2T
RC
N
0
2
e
2T
RC
=
1
2RC
N
0
2

Hence,
SNR =
E
s
P
n
=
4E
b
RC
TN
0

1 − e
−
T
RC

2
3) The value of RC that maximizes SNR, can be found by setting the partial derivative of SNR
with respect to RC equal to zero. Thus, if a = RC, then
ϑSNR
ϑa
=0=(1−e
−
T
a
) −
T
a

e
−
T
a
= −e
−
T
a

1+
T
a

+1
Solving this transcendental equation numerically for a, we obtain
T
a
=1.26 =⇒ RC = a =
T
1.26
Problem 7.18
1) The matched filter is
h
1
(t)=s
1
(T −t)=

−
1

T
t +1, 0 ≤ t<T
0 otherwise
172
The output of the matched filter is
y
1
(t)=

∞
−∞
s
1
(τ)h
1
(t − τ)dτ
If t ≤ 0, then y
1
(t)=0,If0<t≤ T , then
y
1
(t)=

∞
0
τ
T

−
1

T
(t − τ)+1

dτ
=

t
0
τ

1
T
−
t
T
2

dτ +
1
T
2

t
0
τ
2
dτ
= −
t
3

6T
2
+
t
2
2T
If T ≤ t ≤ 2T , then
y
1
(t)=

T
t−τ
τ
T

−
1
T
(t − τ)+1

dτ
=

T
t−τ
τ

1
T

−
t
T
2

dτ +
1
T
2

T
t−τ
τ
2
dτ
=
(t − T )
3
6T
2
−
t − T
2
+
T
3
For 2T<0, we obtain y
1
(t) = 0. In summary
y

1
(t)=









0 t ≤ 0
−
t
3
6T
2
+
t
2
2T
0 <t≤ T
(t−T )
3
6T
2
−
t−T
2
+

T
3
T<t≤ 2T
02T<t
Asketchofy
1
(t) is given in the next figure. As it is observed the maximum of y
1
(t), which is
T
3
,
is achieved for t = T .
T/3
T
2T
2) The signal waveform matched to s
2
(t)is
h
2
(t)=

−1, 0 ≤ t ≤
T
2
2,
T
2
<t≤ T

The output of the matched filter is
y
2
(t)=

∞
−∞
s
2
(τ)h
2
(t − τ)dτ
If t ≤ 0ort ≥ 2T , then y
2
(t)=0. If0<t≤
T
2
, then y
2
(t)=

t
0
(−2)dτ = −2t.If
T
2
<t≤ T , then
y
2
(t)=


t−
T
2
0
4dτ +

T
2
t−
T
2
(−2)dτ +

t
−
T
2
dτ =7t −
9
2
T
173
If T<t≤
3T
2
, then
y
2
(t)=


T
2
t−T
4dτ +

t−
T
2
T
2
(−2)dτ +

T
t−
T
2
dτ =
19T
2
− 7t
For,
3T
2
<t≤ 2T , we obtain
y
2
(t)=

T

t−T
(−2)dτ =2t − 4T
In summary
y
2
(t)=

















0 t ≤ 0
−2t 0 <t≤
T
2
7t −
9
2

T
T
2
<t≤ T
19T
2
− 7tT<t≤
3T
2
2t − 4T
3T
2
<t≤ 2T
02T<t
A plot of y
2
(t) is shown in the next figure

◗
◗
◗
◗
☞
☞
☞
☞
☞
☞
☞
☞

☞
▲
▲
▲
▲
▲
▲
▲
▲
▲
✑
✑
✑
✑
T
5T
2
−T
2T
3) The signal waveform matched to s
3
(t)is
h
3
(t)=

20≤ t ≤
T
2
0

T
2
<t≤ T
The output of the matched filter is
y
3
(t)=h
3
(t) s
3
(t)=

4t − 2T
T
2
≤ t<T
−4t +6TT≤ t ≤
3T
2
In the next figure we have plotted y
3
(t).

✁
✁
✁
✁
✁
✁❆
❆

❆
❆
❆
❆
3T
2
T
T
2
2T
Problem 7.19
1) Since m
2
(t)=−m
3
(t) the dimensionality of the signal space is two.
2) As a basis of the signal space we consider the functions
ψ
1
(t)=

1
√
T
0 ≤ t ≤ T
0 otherwise
ψ
2
(t)=






1
√
T
0 ≤ t ≤
T
2
−
1
√
T
T
2
<t≤ T
0 otherwise
174
The vector representation of the signals is
m
1
=[
√
T, 0]
m
2
=[0,
√
T ]

m
3
=[0, −
√
T ]
3) The signal constellation is depicted in the next figure
✈
✈
✈
(0, −
√
T )
(0,
√
T )
(
√
T,0)
4) The three possible outputs of the matched filters, corresponding to the three possible transmitted
signals are (r
1
,r
2
)=(
√
T + n
1
,n
2
), (n

1
,
√
T + n
2
) and (n
1
, −
√
T + n
2
), where n
1
, n
2
are zero-
mean Gaussian random variables with variance
N
0
2
. If all the signals are equiprobable the optimum
decision rule selects the signal that maximizes the metric
C(r · m
i
)=2r · m
i
−|m
i
|
2

or since |m
i
|
2
is the same for all i,
C

(r · m
i
)=r · m
i
Thus the optimal decision region R
1
for m
1
is the set of points (r
1
,r
2
), such that (r
1
,r
2
) · m
1
>
(r
1
,r
2

) · m
2
and (r
1
,r
2
) · m
1
> (r
1
,r
2
) · m
3
. Since (r
1
,r
2
) · m
1
=
√
Tr
1
,(r
1
,r
2
) · m
2

=
√
Tr
2
and
(r
1
,r
2
) · m
3
= −
√
Tr
2
, the previous conditions are written as
r
1
>r
2
and r
1
> −r
2
Similarly we find that R
2
is the set of points (r
1
,r
2

) that satisfy r
2
> 0, r
2
>r
1
and R
3
is the
region such that r
2
< 0 and r
2
< −r
1
. The regions R
1
, R
2
and R
3
are shown in the next figure.






❅
❅

❅
❅
❅
❅
0
R
3
R
2
R
1
5) If the signals are equiprobable then,
P (e|m
1
)=P (|r −m
1
|
2
> |r − m
2
|
2
|m
1
)+P (|r −m
1
|
2
> |r − m
3

|
2
|m
1
)
When m
1
is transmitted then r =[
√
T + n
1
,n
2
] and therefore, P (e|m
1
) is written as
P (e|m
1
)=P (n
2
− n
1
>
√
T )+P (n
1
+ n
2
< −
√

T )
175
Since, n
1
, n
2
are zero-mean statistically independent Gaussian random variables, each with variance
N
0
2
, the random variables x = n
1
− n
2
and y = n
1
+ n
2
are zero-mean Gaussian with variance N
0
.
Hence,
P (e|m
1
)=
1
√
2πN
0


∞
√
T
e
−
x
2
2N
0
dx +
1
√
2πN
0

−
√
T
−∞
e
−
y
2
2N
0
dy
= Q


T

N
0

+ Q


T
N
0

=2Q


T
N
0

When m
2
is transmitted then r =[n
1
,n
2
+
√
T ] and therefore,
P (e|m
2
)=P (n
1

− n
2
>
√
T )+P (n
2
< −
√
T )
= Q


T
N
0

+ Q


2T
N
0

Similarly from the symmetry of the problem, we obtain
P (e|m
2
)=P (e|m
3
)=Q



T
N
0

+ Q


2T
N
0

Since Q[·] is momononically decreasing, we obtain
Q


2T
N
0

<Q


T
N
0

and therefore, the probability of error P(e|m
1
) is larger than P(e|m

2
) and P (e|m
3
). Hence, the
message m
1
is more vulnerable to errors.
Problem 7.20
The optimal receiver bases its decisions on the metrics
PM(r, s
m
)=f(r|s
m
)P (s
m
)
For an additive noise channel r = s
m
+ n,so
PM(r, s
m
)=f(n)P (s
m
)
where f(n)istheN-dimensional PDF for the noise channel vector. If the noise is AWG, then
fn)=
1
(πN
0
)

N
2
e
−
|r−s
m
|
2
N
0
Maximizing f(r|s
m
)P (s
m
) is the same as minimizing the reciprocal e
|r−s
m
|
2
N
0
/P (s
m
), or by taking
the natural logarithm, minimizing the cost
D(r, s
m
)=|r − s
m
|

2
− N
0
P (s
m
)
This is equivalent to the maximization of the quantity
C(r, s
m
)=r · s
m
−
1
2
|s
m
|
2
+
N
0
2
ln P (s
m
)
176
If the vectors r, s
m
correspond to the waveforms r(t) and s
m

(t), where
r(t)=
N

i=1
r
i
ψ
i
(t)
s
m
(t)=
N

i=1
s
m,i
ψ
i
(t)
then,

∞
−∞
r(t)s
m
(t)dt =

∞

−∞
N

i=1
r
i
ψ
i
(t)
N

j=1
s
m,j
ψ
j
(t)dt
=
N

i=1
N

j=1
r
i
s
m,j

∞

−∞
ψ
i
(t)ψ
j
(t)dt
=
N

i=1
N

j=1
r
i
s
m,j
δ
i,j
=
N

i=1
r
i
s
m,i
= r ·s
m
Similarly we obtain


∞
−∞
|s
m
(t)|
2
dt = |s
m
|
2
= E
s
m
Therefore, the optimal receiver can use the costs
C(r, s
m
)=

∞
−∞
r(t)s
m
(t)dt −
1
2

∞
−∞
|s

m
(t)|
2
dt +
N
0
2
ln P (s
m
)
=

∞
−∞
r(t)s
m
(t)dt + c
m
to base its decisions. This receiver can be implemented using M correlators to evaluate

∞
−∞
r(t)s
m
(t)dt.
The bias constants c
m
can be precomputed and added to the output of the correlators. The struc-
ture of the receiver is shown in the next figure.
❦

❦
❦❦
❦
❦
❄
❄
❄
✲
✲
✲
✲
✲
✲
✲
✲
✲
✲
✲
✲
✲
✲
✻
✻
✻
.
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largest
the
Select
c
M
c
2
c
1
r · s
1
r · s
M
r · s
2

(·)dt

(·)dt

(·)dt
s
M
(t)

s
2
(t)
s
1
(t)
r(t)
Parallel to the development of the optimal receiver using N filters matched to the orthonormal
functions ψ
i
(t), i =1, ,N, the M correlators can be replaced by M equivalent filters matched to
the signal waveforms s
m
(t). The output of the m
th
matched filter h
m
(t), at the time instant T is

T
0
r(τ)h
m
(T −τ)dτ =

T
0
r(τ)s
m
(T −(T −τ))dτ

=

T
0
r(τ)s
m
(τ)dτ
= r ·s
m
177