Solid lines indicate an input equal to 0, whereas dotted lines correspond to an input equal to 1... Problem 9.47Using the diagram of Figure 9.28, we see that there are only two ways to g
Trang 1
Y
j.
-
@
@
@
@@ R
1/001 0/110
0/011
1/010 1/100
0/101
1/111 0/000
01
11
10 00
3) In the next figure we draw two frames of the trellis associated with the code Solid lines indicate
an input equal to 0, whereas dotted lines correspond to an input equal to 1
v v v
v v
v v
v v
v
. . .
. . .
. . .
. . .
11 10 01 00
001 110 010 101 100 011 111 000
4) The diagram used to find the transfer function is shown in the next figure.
n
S
?
@
@
@
-
DN J
D2J
DN J
D3N J
D2J
X d
X a
Using the flow graph results, we obtain the system
X c = D3N J X a + DN J X b
X b = D2J X c + D2J X d
X d = DN J X c + DN J X d
X a = D2J X b Eliminating X b , X c and X dresults in
T (D, N, J ) = X a
X a = D
7N J3
1− DNJ − D3N J2
To find the free distance of the code we set N = J = 1 in the transfer function, so that
T1(D) = T (D, N, J ) | N =J =1= D
7
1− D − D3 = D7+ D8+ D9+· · · Hence, dfree = 7
5) Since there is no self loop corresponding to an input equal to 1 such that the output is the all
zero sequence, the code is not catastrophic
Trang 2Problem 9.47
Using the diagram of Figure 9.28, we see that there are only two ways to go from state X a to
state X a with a total number of ones (sum of the exponents of D) equal to 6 The corresponding
transitions are:
Path 1: X a D
2
→ X c D
→ X d D
→ X b
D2
→ X a
Path 2: X a D
2
→ X c → X D b → X c → X D b D
2
→ X a
These two paths correspond to the codewords
c1 = 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0,
c2 = 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0,
Problem 9.48
1) The state transition diagram and the flow diagram used to find the transfer function for this
code are depicted in the next figure
Y
j.
-
@
@
@
@@ R
1/10 0/11
0/10
1/00 1/11
0/01
1/01 0/00
01
11
10
00
n
S
?
@
@
@
-
D2N J DJ
N J
DN J
DN J
D2J
X d
X a
Thus,
X c = DN J X a + D2N J X b
X b = DJ X c + D2J X d
X d = N J X c + DN J X d
X a = DJ X b and by eliminating X b , X c and X d, we obtain
T (D, N, J ) = X a
X a = D
3N J3
1− DNJ − D3N J2
To find the transfer function of the code in the form T (D, N ), we set J = 1 in T (D, N, J ) Hence,
T (D, N ) = D
3N
1− DN − D3N
2) To find the free distance of the code we set N = 1 in the transfer function T (D, N ), so that
T1(D) = T (D, N ) | N =1= D
3
1− D − D3 = D3+ D4+ D5+ 2D6+· · · Hence, dfree = 3
Trang 33) An upper bound on the bit error probability, when hard decision decoding is used, is given by
¯
P b ≤ 1 k
ϑT (D, N ) ϑN
N =1,D= √
4p(1 −p)
Since
ϑT (D, N ) ϑN
N =1
= ϑ
ϑN
D3N
1− (D + D3)N
N =1
3
(1− (D + D3))2
with k = 1, p = 10 −6 we obtain
¯
P b ≤ D3
(1− (D + D3))2
4p(1 −p) = 8.0321 × 10 −9
Problem 9.49
1) Let the decoding rule be that the first codeword is decoded when yi is received if
p(y i |x1) > p(y i |x2)
The set of yi that decode into x1 is
Y1 ={y i : p(y i |x1) > p(y i |x2)} The characteristic function of this set χ1(yi) is by definition equal to 0 if yi ∈ Y1 and equal to 1 if
yi ∈ Y1 The characteristic function can be bounded as (see Problem 9.40)
1− χ1(yi)≤ p(y i |x2)
p(y i |x1)
1
Given that the first codeword is sent, then the probability of error is
P (error|x1) =
yi∈Y −Y1
p(y i |x1) =
yi∈Y
p(y i |x1)[1− χ1(yi)]
yi∈Y
p(y i |x1) p(y i |x2)
p(y i |x1)
1 2
yi∈Y
%
p(y i |x1)p(y i |x2)
=
2n
i=1
%
p(y i |x1)p(y i |x2)
where Y denotes the set of all possible sequences y i Since, each element of the vector yi can take
two values, the cardinality of the set Y is 2 n
2) Using the results of the previous part we have
P (error) ≤ 2
n
i=1
%
p(y i |x1)p(y i |x2) =
2n
i=1
p(y i)
1
p(y i |x1)
p(y i)
1
p(y i |x2)
p(y i)
=
2n
i=1
p(y i)
1
p(x1|y i)
p(x1)
1
p(x2|y i)
p(x2) =
2n
i=1
2p(y i)
%
p(x1|y i )p(x2|y i)
However, given the vector yi, the probability of error depends only on those values that x1 and x2
are different In other words, if x 1,k = x 2,k , then no matter what value is the kth element of yi, it
will not produce an error Thus, if by d we denote the Hamming distance between x1 and x2, then
p(x1|y i )p(x2|y i ) = p d(1− p) d
Trang 4and since p(y i) =21n, we obtain
P (error) = P (d) = 2p d2(1− p) d
2 = [4p(1 − p)] d
2
Problem 9.50
1)
Q(x) = √1
2π
∞
x
e − v2
2 dv
v= √
2t
= √1
π
∞
x
√
2
e −t2
dt
2
2
π
∞
x
√
2
e −t2
dt
2erfc
x
√
2
2) The average bit error probability can be bounded as (see (9.7.16))
¯
P b ≤ 1
k
∞
d=dfree
a d f (d)Q
1
2R c d E b
N0
= 1
k
∞
d=dfree
a d f (d)Q&
2R c dγ b
2k
∞
d=dfree
a d f (d)erfc(&
R c dγ b)
2k
∞
d=1
a d+dfreef (d + dfree)erfc(
%
R c (d + dfree)γ b)
≤ 1 2kerfc(
&
R c dfreeγ b)
∞
d=1
a d+dfreef (d + dfree)e −Rc dγ b
But,
T (D, N ) =
∞
d=dfree
a d D d N f (d)=
∞
d=1
a d+dfreeD d+dfreeN f (d+dfree )
and therefore,
ϑT (D, N ) ϑN
N =1
=
∞
d=1
a d+dfreeD d+dfreef (d + dfree)
= D dfree
∞
d=1
a d+dfreeD d f (d + dfree)
Setting D = e −Rc γ b in the previous and substituting in the expression for the average bit error probability, we obtain
¯
P b ≤ 1 2kerfc(
&
R c dfreeγ b )e R c dfreeγ b ϑT (D, N )
ϑN
N =1,D=e−Rcγb
Problem 9.51
The partition of the 8-PAM constellation in four subsets is depicted in the figure below
Trang 5u u u
u
u u u u
u u u u
Q Q QQ
c c c
#
#
#
P
!
!
!
!
!
!
1 0
1
1 0
0
3 -5
7 -1
5 -3
-3
1 3 5 7 -1
-3 -5 -7
2) The next figure shows one frame of the trellis used to decode the received sequence Each branch
consists of two transitions which correspond to elements in the same coset in the final partition level
v v v v
v v v v
@
@
@
@
@
HH
HH H
HH
HH H
7,1 -5,3
-3,5 -1,7 -1,7 -3,5
-5,3 -7,1
The operation of the Viterbi algorithm for the decoding of the sequence{−.2, 1.1, 6, 4, −3, −4.8, 3.3}
is shown schematically in the next figure It has been assumed that we start at the all zero state and that a sequence of zeros terminates the input bit stream in order to clear the encoder The numbers
at the nodes indicate the minimum Euclidean distance, and the branches have been marked with the decoded transmitted symbol The paths that have been purged are marked with an X
v v v
v v v
v v
v v
v v
v v
v v
v v v
Q Q
Q Q S S S S SS
Q Q
Q Q
Q Q
Q Q Q Q Q Q
Q Q
Q Q
S S S S SS
S S S S SS
Q Q
Q Q
Q Q
Q Q
Q Q
Q Q
Q Q
Q Q
S S S S SS
Q Q
Q Q Q
Transmitted sequence:
X
X X
X X X
X
X
X
X
X 3 14.38
-3 14.29
-5 11.09
-3 7.05 -1 11.05 -511.05 -5 15.45
5 7.05
511.45 1 15.05 3 7.05
7 6.05
5 6.05
3 10.45
1 26.45
-1 14.65
5 25.45
3 5.05
1 1.45 1.1
3 10.24
1.44 1
3.3 -4.8
-3 4
6 -.2
Trang 6Chapter 10
Problem 10.1
1) The wavelength λ is
λ = 3× 108
109 m = 3
10 m Hence, the Doppler frequency shift is
f D =± u
λ=±100 Km/hr3
10 m =±100× 103× 10
3× 3600 Hz =±92.5926 Hz
The plus sign holds when the vehicle travels towards the transmitter whereas the minus sign holds when the vehicle moves away from the transmitter
2) The maximum difference in the Doppler frequency shift, when the vehicle travels at speed 100
km/hr and f = 1 GHz, is
∆f D max = 2f D = 185.1852 Hz
This should be the bandwidth of the Doppler frequency tracking loop
3) The maximum Doppler frequency shift is obtain when f = 1 GHz + 1 MHz and the vehicle
moves towards the transmitter In this case
λmin = 3× 108
109+ 106 m = 0.2997 m
and therefore
f D max= 100× 103
0.2997 × 3600 = 92.6853 Hz Thus, the Doppler frequency spread is B d = 2f D max = 185.3706 Hz.
Problem 10.2
1) Since T m = 1 second, the coherence bandwidth
B cb = 1
2T m = 0.5 Hz and with B d = 0.01 Hz, the coherence time is
T ct = 1
2B d = 100/2 = 50 seconds (2) Since the channel bandwidth W b cb, the channel is frequency selective
(3) Since the signal duration T ct, the channel is slowly fading
Trang 7(4) The ratio W/B cb = 10 Hence, in principle up to tenth order diversity is available by subdividing the channel bandwidth into 10 subchannels, each of width 0.5 Hz If we employ binary PSK
with symbol duration T = 10 seconds, then the channel bandwidth can be subdivided into 25
subchannels, each of bandwidth T2 = 0.2 Hz We may choose to have 5th order frequency diversity and for each transmission, thus, have 5 parallel transmissions Thus, we would have a data rate
of 5 bits per signal interval, i.e., a bit rate of 1/2 bps By reducing the order of diversity, we may
increase the data rate, for example, with no diversity, the data rate becomes 2.5 bps
(5) To answer the question we may use the approximate relation for the error probability given by (10.1.37), or we may use the results in the graph shown in Figure 10.1.10 For example, for binary
PSK with D = 4, the SNR per bit required to achieve an error probability of 10 −6 is 18 dB This
the total SNR per bit for the four channels (with maximal ration combining) Hence, the SNR per bit per channel is reduced to 12 dB (a factor of four smaller)
Problem 10.3
The Rayleigh distribution is
p(α) =
α
σ2α e −α2/2σ2
α , α > 0
Hence, the probability of error for the binary FSK and DPSK with noncoherent detection averaged
over all possible values of α is
P2 =
∞
0
1
2e
−c α2Eb N0 α
σ2
α
e −α2/2σ2
α dα
2σ2
α
∞
0
αe −α2
cEb N0+2σ2α1
dα
0
x 2n+1 e −ax2
dx = n!
2a n+1 , (a > 0)
so that with n = 0 we obtain
P2 = 1
2σ2
α
∞
0
αe −α2
cEb N0+2σ2α1
dα = 1 2σ2
α
1 2
c Eb
N0 +2σ12
α
2
c Eb 2σ2
α
2 [c ¯ ρ b+ 1]
where ¯ρ b = Eb 2σ2
α
N0 With c = 1 (DPSK) and c = 12 (FSK) we have
P2=
2(1+ ¯ρ b), DPSK
1 2+ ¯ρ b , FSK
Trang 8Problem 10.4
(a)
- Matched Filter 2 ( )2
- Matched Filter 2 ( )2 n
×
-n
×
-6
?
i +? 6
cos 2πf2t
sin 2πf2t
- Matched Filter 1 ( )2
- Matched Filter 1 ( )2
n
×
-n
×
-6
?
i +? 6
cos 2πf1t
sin 2πf1t
- Matched Filter 2 ( )2
- Matched Filter 2 ( )2 n
×
-n
×
-6
?
i +? 6
cos 2πf2t
sin 2πf2t
- Matched Filter 1 ( )2
- Matched Filter 1 ( )2 n
×
-n
×
-6
?
i +? 6
cos 2πf1t
sin 2πf1t
-r1(t)
r2(t)
sample at t = kT
6
?
n +? 6
n + 6
?
-Detector select the larger
-output
(b) The probability of error for binary FSK with square-law combining for D = 2 is given in Figure 10.1.10 The probability of error for D = 1 is also given in Figure 10.1.10 Note that an increase in SNR by a factor of 10 reduces the error probability by a factor of 10 when D = 1 and by a factor
of 100 when D = 2.
Problem 10.5
(a) r is a Gaussian random variable If √
E b is the transmitted signal point, then
E(r) = E(r1) + E(r2) = (1 + k)&
E b ≡ m r
and the variance is
σ2r = σ21+ k2σ22
Trang 9The probability density function of r is
f (r) = √ 1
2πσ r e
− (r−mr)2 2σ2r
and the probability of error is
P2 =
0
−∞ f (r) dr
= √1 2π
− mr σr
− x2
2 dx
1
m2
σ2
where
m2r
σ2 = (1 + k)
2E b
σ21+ k2σ22 The value of k that maximizes this ratio is obtained by differentiating this expression and solving for the value of k that forces the derivative to zero Thus, we obtain
k = σ
2 1
σ2 2
Note that if σ1 > σ2, then k > 1 and r2 is given greater weight than r1 On the other hand, if
σ2 > σ1, then k < 1 and r1 is given greater weight than r2 When σ1 = σ2, k = 1 In this case
m2r
σ2 = 2E b
σ12 (b) When σ22 = 3σ12, k = 13, and
m2r
σ2 = (1 +
1
3)2E b
σ21+19(3σ12) =
4 3
E b
σ21
On the other hand, if k is set to unity we have
m2r
σ2 = 4E b
σ2
1 + 3σ2 1
= E b
σ2 1
Therefore, the optimum weighting provides a gain of
10 log4
3 = 1.25 dB
Problem 10.6
1) The probability of error for a fixed value of a is
P e (a) = Q
1
2a2E
N0
since the given a takes two possible values, namely a = 0 and a = 2 with probabilities 0.1 and 0.9,
respectively, the average probability of error is
P e= 0.1
2 + Q
1
8E
N0
= 0.05 + Q
1
8E
N0
Trang 10
(2) As E
N0 → ∞, P e → 0.05
(3) The probability of error for fixed values of a1 and a2 is
P e (a1, a2) = Q
1
2(a2
1+ a2
2)E
N0
In this case we have four possible values for the pair (a1, a2), namely, (0, 0), (0, 2), (2, 0), and (2, 2),
with corresponding probabilities ).01, 0.09, 0.09 and 0.81 Hence, the average probability of error is
P e = 0.01
2 + 0.18Q
1
8E
N0
+ 0.81Q
1
16E
N0
(4) As E
N0 → ∞, P e → 0.005, which is a factor of 10 smaller than in (2).
Problem 10.7
We assume that the input bits 0, 1 are mapped to the symbols -1 and 1 respectively The terminal
phase of an MSK signal at time instant n is given by
θ(n; a) = π
2
k
k=0
a k + θ0
where θ0 is the initial phase and a k is ±1 depending on the input bit at the time instant k The
following table shows θ(n; a) for two different values of θ0 (0, π), and the four input pairs of data: {00, 01, 10, 11}.
θ0 b0 b1 a0 a1 θ(n; a)
Problem 10.8
1) The envelope of the signal is
|s(t)| = %|s c (t) |2+|s s (t) |2
=
1
2E b
T b cos
2T b
+2E b
T b sin
2 πt 2T b
=
1
2E b
T b
Thus, the signal has constant amplitude
Trang 112) The signal s(t) has the form of the four-phase PSK signal with
g T (t) = cos πt
2T b
, 0≤ t ≤ 2T b
Hence, it is an MSK signal A block diagram of the modulator for synthesizing the signal is given
in the next figure
k k
l
l l
l
l
?
6
? 6
?
?
-6
?
2T b)
×
×
7 7
− π
2
×
− π
2
×
a 2n+1
a 2n
Demux Parallel Serial /
data a n
Serial
3) A sketch of the demodulator is shown in the next figure.
l
l
k
l
l k
@
@
-?
?
6
?
?
6
-?
6
-t = 2T b
t = 2T b
− π
2
7
×
×
cos(2T πt
b)
cos(2πf c t))
×
− π
2
×
7
r(t)
2T b
0 (·)dt
2T b
0 (·)dt
Threshold
Threshold
Parallel to Serial
Problem 10.9
Since p = 2, m is odd (m = 1) and M = 2, there are
N s = 2pM = 8 phase states, which we denote as S n = (θ n , a n −1 ) The 2p = 4 phase states corresponding to θ nare
Θs =
*
0, π
2, π,
3π
2 8
and therefore, the 8 states S n are
*
(0, 1), (0, −1), π
2, 1
, π
2, −1
, (π, 1), (π, −1), 3π
2 , 1
, 3π
2 , −1
8
Having at our disposal the state (θ n , a n −1 ) and the transmitted symbol a n, we can find the new phase state as
(θ n , a n −1)−→ (θ a n n+π
2a n −1 , a n ) = (θ n+1 , a n) The following figure shows one frame of the phase-trellis of the partial response CPM signal
Trang 12u u
u u u u
u
u
u u
u
u u
u
u
. . . .
. . .
.. . . . .
. . . .
. .
.
. . . .
. .
.
.
Q Q
Q Q
Q Q Q
Q Q
Q Q
Q Q Q
T T T T T T T T T T T T
Q Q
Q Q
Q Q Q
(0, 1) (0, −1)
(π2, 1)
(π2, −1) (π, 1) (π, −1)
(3π2 , 1)
(3π2 , −1)
(3π2 , −1)
(3π2 , 1) (π, −1) (π, 1)
(π2, −1)
(π2, 1) (0, −1) (0, 1)
(θ n+1 , a n)
(θ n , a n −1)
The following is a sketch of the state diagram of the partial response CPM signal
A A A A A A A A A A A AK
A A AAU
P P P P
A A A A A A A A A A AAU
1
A A A A A A A A A A A AU
A A A
A A A A A A A A A A
A A A AAK
A A A
PPP PP q
1
-1
1
-1
-1 1
-1
-1
1
-1
1
1 1
1
(π2, −1)
(π, −1)
(3π
2, −1)
(3π
2 ,1)
(π,1)
(π2,1)
(0, −1)
(0,1)
Problem 10.10
1) For a full response CPFSK signal, L is equal to 1 If h = 23, then since m is even, there are p terminal phase states If h = 34, the number of states is N s = 2p.
2) With L = 3 and h = 23, the number of states is N s = p22 = 12 When L = 3 and h = 34, the
number of states is N s = 2p22 = 32
Problem 10.11
(a) The coding gain is
R c d Hmin = 1
2 × 10 = 5 (7dB) (b) The processing gain is W/R, where W = 107Hz and R = 2000bps Hence,
W
R =
107
2× 103 = 5× 103(37dB)
...!
1
1
1
0
3 -5
7 -1
5 -3
-3
1 -1
-3 -5 -7
2) The next figure shows one frame of the trellis used...
HH H
7,1 -5 ,3
-3 ,5 -1 ,7 -1 ,7 -3 ,5
-5 ,3 -7 ,1
The operation of the Viterbi algorithm... 14.38
-3 14.29
-5 11.09
-3 7.05 -1 11.05 -5 11.05 -5 15. 45
5