978-0072957167 Chapter 12

Solutions Manual

for

Digital Communications, 5th Edition

(Chapter 12) 1

Prepared by

Kostas Stamatiou

January 29, 2008

2

Problem 12.1

g(t) = r16Ec

3Tc

cos 2π

Tct−Tc

2,0≤t≤Tc

G(f) = R∞

−∞ g(t)e−j2πf tdt

and

σ2

m= 2wmJav |G(0)|2=8

3EcTcwmJav,Ec=RcEb

m=2

JavTc

Problem 12.2

The PN spread spectrum signal has a bandwidth W and the interference has a bandwidth W1,

where W >> W1.Upon multiplication of the received signal r(t) with the PN reference at the

receiver, we have the following (approximate) spectral characteristics

6

J0W1/W

Spectrum of

Interference

W S0

1/Tb=W S0Tb

3

After multiplication with the PN reference, the interference power in the bandwidth 1/Tboccupied

by the signal is

J0W1

W1

Tb=J0W1

W Tb

Problem 12.3

The concatenation of a Reed-Solomon (31,3) code with the Hadamard (16,5) code results in an

equivalent binary code of black length n=n1n2= 31 ×16 = 496 bits. There are 15 information

Reed −Solomon code : Dmin = 31 −3 + 1 = 29

Hence, the minimum distance of the overall code is dmin = 28 ×8 = 232.A union bound on the

probability of error based on the minimum distance of the code is

JavTc

where M= 215 = 32768.Also, Eb=PavTb.Thus,

k

But kTb=nTcand dmn = 232.Hence,

Due to the large number of codewords, this union bound is very loose. A much tighter bound is

M

X

m=2

but the evaluation of the bound requires the use of the weight distribution of the concatenated

code.

Problem 12.4

For hard-decision decoding we have

PM≤(M−1) [4p(1 −p)]dmin/2= 2m+1 [4p(1 −p)]2m−2

where p=Qq4W/R

Jav/Pav Rc=Qq4Pav

Jav .Note that in the presence of a strong jammer, the

probability pis large, i.e close to 1/2. For soft-decision decoding, the error probability bound is

Jav/Pav

We select W

R=n

k=1

Rcand hence:

PM≤2m+1Qq 2m+1

Jav /Pav <2mexp −2m

Jav/Pav 

Problem 12.5

(b) The processing gain is W/R, where W= 107Hz and R= 1000bps. Hence,

(c) The jamming margin is

Jav

Pav =(2W

R)·(Rcdmin)

“Eb

Problem 12.6

We assume that the interference is characterized as a zero-mean AWGN process with power spectral

density J0. To achieve an error probability of 10−5, the required Eb/J0= 10 . Then, by using the

relation in (12-2-46) and (12-2-37), we have

2W/R

Jav/Pav =2W/R

Nu−1=Eb

J0

where R= 104bps, Nu= 30 and Eb/J0= 10.Therefore,

Problem 12.7

To achieve an error probability of 10−6,we require

Eb

J0dB

= 10.5dB

Then, the number of users of the CDMA system is

Nu=2W/R

Eb/J0+ 1

Problem 12.8

(a) We are given a system where (Jav /Pav)dB = 20 dB, R = 1000 bps and (Eb/J0)dB = 10 dB.

Hence, using the relation in (12-2-37) we obtain

2W

RdB =Jav

Pav dB +Eb

J0dB = 30 dB

(b) The duty cycle of a pulse jammer for worst-case jamming is

The corresponding probability of error for this worst-case jamming is

Problem 12.9

(a) We have Nu= 15 users transmitting at a rate of 10,000 bps each, in a bandwidth of W=

1MHz. The Eb/J0is

(c) With Nu= 30 and Eb/J0= 14.28,the processing gain should be increased to

Problem 12.10

The processing gain is given as

W

R= 500 (27 dB)

Problem 12.11

If the jammer is a pulse jammer with a duty cycle α= 0.01,the probability of error for binary

PSK is given as

P2=αQ s4W/R

Jav/Pav !

Pav

Problem 12.12

c(t) =

∞

X

n=−∞

cnp(t−nTc)

8

The power spectral density of c(t) is given by (3.4.15) as

Sc(f) = 1

Tc

Sc(f)|P(f)|2

then, Rc(m) can be represented in a discrete Fourier series as

φc(m) = 1

N

N−1

X

k=0

rC(k)ej2πmk/N , m = 0,1,…,N −1

=





1, k = 0,±N, ±2N,…

N+ 1,otherwise

Problem 12.13

Without loss of generality, let us assume that N1< N2.Then, the period of the sequence obtained

by forming the modulo-2 sum of the two periodic sequences is

Problem 12.14

(a) The period of the maximum length shift register sequence is

N= 210 −1 = 1023

Since Tb=NTc,then the processing gain is

Tc

(b) According to (12-2-37), the jamming margin is

Jav

Pav dB =2W

RbdB −Eb

J0dB

Problem 12.15

(a) The length of the shift-register sequence is

L= 2m−1 = 215 −1

10

(b) The processing gain is W/R. We have,

(c) If the noise is AWG with power spectral density N0,the probability of error expression is

Problem 12.16

(a) If the hopping rate is 2 hops/bit and the bit rate is 100 bits/sec, then, the hop rate is 200

hops/sec. The minimum frequency separation for orthogonality 2/T = 400Hz. Since there are

(b) The processing gain is W/R, where W= 13.1068 MHz and R= 100bps. Hence

Problem 12.17

(a) The total SNR for three hops is 20 ∼13 dB.Therefore the SNR per hop is 20/3. The probability

of a chip error with noncoherent detection is

2N0

11

where Ec/N0= 20/3. The probability of a bit error is

Pb= 1 −(1 −p)2

= 1 −(1 −2p+p2)

b) In the case of one hop per bit, the SNR per bit is 20, Hence,

Pb=1

2N0

Problem 12.18

(a) We are given a hopping bandwidth of 2 GHz and a bit rate of 10 kbs. Hence,

(b) The bandwidth of the worst partial-band jammer is α∗W, where

α∗= 2/(Eb/J0) = 0.2

(c) The probability of error with worst-case partial-band jamming is

P2=e−1

(Eb/J0)=e−1

10

12

Problem 12.19

The error probability for the binary convolutional code is upper-bounded as

Pb≤

∞

X

d=df ree

βdP2(d)

n=0 “1

r=0 2d−1

Problem 12.20

For hard-decision Viterbi decoding of the convolutional code, the error probability is

Pb≤

∞

X

d=df ree

βdP2(d)

Problem 12.21

For fast frequency hopping at a rate of Lhopes/bit and for soft-decision decoding, the performance

of the binary convolutional code is upper bounded as

Pb≤

∞

X

d=df ree

βdP2(Ld)

Problem 12.22

For fast frequency hopping at a rate of Lhopes/bit and for hard-decision Viterbi decoding, the

performance of the binary convolutional code is upper bounded as

Pb≤

∞

X

d=df ree

βdP2(Ld)

n=0 “γbRc

r=0 2L−1

On the other hand, if each of the L chips is detected independently, then :

∞

X

Problem 12.23

There are 64 decision variables corresponding to the 64 possible codewords in the (7,2) Reed-

Solomon code. With dmin = 6,we know that the performance of the code is no worse than the

performance of an M= 64 orthogonal waveform set, where the SNR per bit is degraded by the

factor 6/7. Thus, an upper bound on the code word error probability is

P64 ≤(M−1)P2= 63P2

r=0 11

Problem 12.24

In the worst-case partial-band interference channel, the (7,2) Reed-Solomon code provides an ef-

fective order of diversity L= 6.Hence

b

Problem 12.25

P2(a) = aQ r2aEb

J0!=a1

√2πZ+∞

r2aEb

J0

e−t2/2dt

Hence, the maximum occurs when

dP2(a)

da =1

√2πZ+∞

r2aEb

J0

e−t2/2dt −1

2raEb

πJ0

e−aEb

J0= 0 ⇒

Problem 12.26

The problem is to determine

E“exp −v

L

X

k=1

βk|2Ec+N1k|2−v

L

X

k=1

βk|N2k|2!#

where thee {βk}are fixed and the {N1k},{N2k}are complex-valued Gaussian random variables

Since the {N1k},and {N2k}are all statistically independent

E“exp −v

L

X

k=1

βk|2Ec+N1k|2!#=

L

Y

k=1

Ehexp −vβk|2Ec+N1k|2i

ψY(ju) = EejuY =1

1−juσ2exp ju Pm

i=1 m2

i

1−juσ2

k=1

Problem 12.27

The function

f(v, a) = a

1−4v2exp −2avE2

c

J0(1 + 2v)L

16

is to be minimized with respect to vand maximized with respect to a. Thus,

∂v f(v, a) = 0 ⇒8v(1 + 2v)−2aEc

J0

and ∂

∂af(v, a) = 0 ⇒1−2avEc

J0(1 + 2v)= 0

The simultaneous solution of these two equations yields the result v= 1/4 and a=3J0

Ec≤1.For

these values, the function f(v, a) becomes

Problem 12.28

The Gold code sequences of length n= 7 may be generated by the use of two feedback shift registers

of length 4 as shown below







?

–



6



–

These sequences have the following cross-correlation values

Values of correlation Frequency of this value

-5 1

Problem 12.29

The method of Omura and Levitt (1982) based on the cut-off rate R0,which is described in

Problem 12.30

(a) For the coded and interleaved DS binary PSK modulation with pulse jamming and soft-decision

decoding, the cutoff rate is

R0= 1 −log 2h1 + ae−aEc/N0i

(b)

d

da Eb

N0=−1

a2Rln a

21−R0−1+1

a2R= 0

Hence, the worst-case ais

a∗=21−R0−1e

(c) The plot of 10 log Eb

rN0versus R0,is given in the following figure:

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

2

4

6

8

10

12

14

Cut−off rate R0

10log(Eb/rN)

AWGN

Worst−case

pulse

jamming

Clearly, there is no penalty in SNR due to worst case pulse jamming for rates below 0.548. Even

for R0= 0.8 the SNR loss is relatively small. As R0→1,the relative difference between pulse

jamming and AWGN becomes large.

Problem 12.31

(a)

R0= log 2q−log 2[1 + (q−1)aexp(−aEc/2N0)] ⇒

(b)

d

da Eb

N0=−2

a2Rln (q−1)a

q2−R0−1+2

a2R= 0 ⇒

(c) The plots are given in the following figure

0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

5

10

15

20

25

R0/log2(q)

10log(Eb/rN)

q=2

q=4

q=32

q=8

For low rates, the loss due to partial band jamming is negligible if coding is used. Increasing q

reduces the SNR/bit at low rates. At very high rates, a large qimplies a large SNR loss. For q= 2,

there is a 3dB loss relative to binary PSK. As q→ ∞,the orthogonal FSK approaches -1.6dB as

R0→0.