Since,
R = 8 103
samples
bits
bits
sec
= 64
10 sample
sec
8
we
obtain
3
?
16 KHz
M =
4
W = 10.667 KHz M
?
=8
8 KHz
M = 16
?
Problem 5.37 :
(a) The inner product of si (t) and sj (t) is
&
n
s (t)s (t)dt =
c
i
ik
p(t
j
=
n n
k=1
&c
k=1 l=1
n
n
&
n
ik jl

Problem 5.24 :
(a) Since m2 (t) = m3 (t) the dimensionality of the signal space is two.
(b) As a basis of the signal space we consider the functions:
?
1
f1 (t) =
T
0
?
1
?T
0 t T
otherwise
0 t
f2 (t) = T1
?
The vector representation of the signals
is:
m

O
011
0
O
0111
O
010
1
1101
O
1111 1110 1100
O
O
O
0100
0000 0001 0011 0010
O
O
O
O
O
O
100
0
O1
001
O
1011
O
1010
Problem 5.23 :
The transmitted signal energy
is
A2
Eb = T
2
where T is the bit interval and A is the signal amplitude. Since both carriers a

(b) Thevariance ofthenoiseis:
n
=
|x2 dx
e
x|
2
=2
=
Hence, theSNRis:
2!
x 2
e
2
=
x dx =
3
0
2
2
SNR=
andtheprobability oferrorisgiven by:
A2
P (e)= 1 2SN R
2e
ForP (e)=105 we obtain:
ln(2105 )= 2SN R =SN R =58.534=17.6741dB
IfthenoisewasGaussian,thenth

Problem
5.21 :
The optimum decision boundary of a point is determined by the perpedicular bisectors of
each line segment connecting the point with its neighbors. The decision regions for this
QAM con- stellation are depicted in the next gure:
O
O

Problem 5.18 :
For binary phase modulation, the error probability
is
2E b
P =Q
2
A2 T
= Q
0
N0
With P2 = 106 we nd from tables
that
A2 T = 4.74 =A2 T = 44.9352 1010
0
If the data rate is 10 Kbps, then the bit interval is T = 104 and therefore, the signal

The variance of the real-valued noise component can be obtained using the relationship Re[N
]=
1
1
2
1
2
2
(N + ) to obtain : = E | vN (T = N0 T (1 e )
2
Nr
2
2
)|
N
(e) The SNR is dened as :
|vmax |
2
A Te 1
=E
|vN2(T = N e + 1
)|
0
(the same result i

Similarly we nd that R2 is the set of points (r1 , r2 ) that satisfy r2 > 0, r2 > r1 and R3 is
the region such that r2 < 0 and r2 < r1 . The regions R1 , R2 and R3 are shown in the next
gure.
R
0
2
R1
R3
(e) If the signals are equiprobable then:
P (e|m1

and therefore, the probability of error P (e|m1 ) is larger than P (e|m2 ) and P (e|m3 ). Hence,
the message m1 is more vulnerable to errors. The reason for that is that it has both
threshold lines close to it, while the other two signals have one of the

(ii) Based on (5-3-7), ( obtain the phase states :
we
15
3 3 , 9
s =
4
7 , 18 , 21
0, ,
4,
2 4
, 4
4 4
2 4
)
5
4
+
(b)
take the values 1.
(i) The combined states are S = ( , I , I ) , where *, 1/n
Hence
n
n n1
n2
I
n
there are 3 2 2 = 12 combined states

The signal y(t) has duration T = nTc and its matched lter is :
n
g(t) = y(T t) = y(nT t) = & c (nT kT t)
c
k
k
=
1
n
c
c
n
= & cni+1 (i 1)Tc t) = & cni+1 (t (i 1)Tc )
i=1
i=1
that is, a sequence of impulses starting at t = 0 and weighted by the mirror ima

Problem 5.30 :
(a) The envelope of the signal is
=
sc (t)|2 + |ss (t)|2
2Eb
t
2
2
2E
T cos 2T + b sin
=
2Eb
T
|s(t)| =
b
b
t
2Tb
Tb
b
Thus, the signal has constant amplitude.
(b) The signal s(t) is equivalent to an MSK signal. A block diagram of the modul

Problem 5.31 :
1
h = 2, L =
2
Based on (5-3-7), we obtain the 4 phase states :
s = cfw_0, /2, ,
3/2
The states in the trellis are the combination of the phase state and the correlative state,
which take the values In1 = cfw_1 . The transition from state t

(3
/2
,1)
(3/2,1)
(, 1)
(0, +1)
(, 1)
(0, 1)
(/2, 1)
(
/2,
1)
The treatment in Probl. 4.27 involved the terminal phase states only, which were
deter- mined to be cfw_/4, 3/4, 5/4, 7/4. We can easily verify that each two of the
combined states, which were

Problem 5.26 :
(a) The number of bits per symbol is
k=
4800
R
=
4800
2400
=2
Thus, a 4-QAM constellation is used for transmission. The probability of error for an Mary
k
QAM system with M = 2 , is
PM = 1
With PM = 105
1 2 1 1 Q
M
and k = 2 we
obtain
Q
Eb

Problem 5.28 :
For4phase PSK (M =4)we havethefollowingrealtionshipbetweenthesymbol rate1/T ,the
requiredbandwith W andthebitrateR =k 1/T =log2 M (see5284):
R R =W log2 M =2W =200kbits/sec
W = M
log
2
ForbinaryFSK (M =2)therequiredfrequencyseparation is1/2

(d) The following table gives the SNR per bit and the corresponding number of bits per
symbol for the constellations used in parts a)-c).
k
2
4
8
SNR (db) 9.89 14.04
28.19
As it is observed there is an increase in transmitted power of approximately 3 dB p

This is the exact symbol error probability for the 4-PSK signal, which is expected since
the vector space representations of the 4-biorthogonal and 4-PSK signals are identical.
Problem 5.15 :
(a) The output of the matched lter can be expressed as :
y(t) =

Problem 5.14 :
The following graph shows the decision regions for the four signals :
U2
A = U 1 > +|U 2|
B = U 1 < |U 2|
C
W2
C = U 2 > +|U 1|
C
D = U 2 < |U 1|
U1
W1
A

CHAPTER
5
Problem 5.1 :
(a) Taking the inverse Fourier transform of H (f ), we obtain :
h(t) = F 1 [H (f )] = F
j
1
F
1
e
j2f 1
2f T
j2
f
= sgn(t) sgn(t T ) =
tT
2
T
where sgn(x) is the signum signal (1 if x > 0, -1 if x < 0, and 0 if x = 0) and (x)
is a

Cl-lAPTl:'R
:>: Ol'TI"'~'
lffC'fl\it:RS
FOR rHE Al)l>ITl\lo wurrr GACSSIAN NOISE CfiANNEl.
329
5-Jl Determine the number of terminal phase states in the state trellis diagram for (a) a full
response binary C'PFSK with either 11 = ior !and (b) a partial-re

3JO
DIOITAL COMMUNICATIONS
elements that are -1, where w is some positive integer. The noise sequence cfw_n. is
white gaussian with variance tr".
a What is the optimum maximum likelihood detector for the two possible
transmitted signals?
b Determine the p

CHAPTER 5:
OPTIMU\.1
RECEIVERS
FOR lHE ADDITIVE WHITE GAllSSIAf'
327
NOISE CHANNEL
5
3
-5
-3
-I
-1
-3
I
5
-5
FIGURE PS-21
S-22 Specify a Gray code for the 16-QAM signal constellation shown in Fig. PS-21.
carriers
cos 2rrJ;.t and sin 21Cf.t are used to tra

328
S.26
S-27
S-28
S-29
DIGITAL
COMMUNICATIONS
b Determine the frequency response characteristic of the filter matched to s(t ).
e Consider the prefilter and the matched filler as a single "generalized matched
filter." What is the frequency response chara

332
DIGITAL
COMMUNICATIONS
S-48 A radio communication system transmits at a power level of 0.1 W at 1 GHz. The
transmitting and receiving antennas are parabolic, each having a diameter of 1 m. The
receiver is located 30 km from the transmitter.
a Determin

CH.-\PTEK
~
OPTIMl.'M
RHFVERS FOR THE ADDfrlVE '.'IHITF CiAL'SSIAl'
c Define the random variables W1 = U~
~OTSF CHAl'~EI.
331
+ V3 and W~ = U~ + U~. Then
1
D=W,-W~O 1r
Determine the probability density functions for W, and W~.
d Determine the probability

326
FIGURE
DIGITAL
C:O~MUNICATIONS
n
PS-19
a The nearest-neighbor signal points in the 8-QAM signal constellation are
separated in distance by A units. Determine the radii a and b of the inner and
outer circles.
b The adjacent signal points in the 8-PSK a

(c) The output of the correlator at t = T is :
q(T ) =
s2 ( )d
0
2
2
T
2
cos (2fc )d
A
= T2
0
However, this is the same expression with the case of the output of the matched lter
sampled at t = T . Thus, the correlator can substitute the matched lter in