Digital Communications 5th Ed SolutionsChap3

Solutions Manual

for

Digital Communications, 5th Edition

(Chapter 3) 1

Prepared by

Kostas Stamatiou

January 11, 2008

1PROPRIETARY MATERIAL. cThe McGraw-Hill Companies, Inc. All rights reserved. No part of thisManual may be displayed, reproduced or distributed in any form or by any means, without the prior written

permission of the publisher, or used beyond the limited distribution to teachers and educators permitted byMcGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using

it without permission.

2Problem 3.1

Assuming M is even we have

12 + 32 + 52 + . . .(M 1)2 =Mi=1

i2 M/2k=1

(2k)2

=M(M + 1)(2M + 1)

6 4

M/2k=1

k2

=M(M + 1)(2M + 1)

6 4M/2(M/2 + 1)(M + 1)

6

=M(M2 1)

6

Problem 3.2

s1 =(E , 0)

s2 =(E , 0

)s3 =

(0,E

)s4 =

(0,E

)

6

-

f2

f1

s1s2

s4

s3

o o

o

o

As we see, this signal set is indeed equivalent to a 4-phase PSK signal.

PROPRIETARY MATERIAL. cThe McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual, you are using it without permission.

3Problem 3.3

1.2. The signal space diagram, together with the Gray encoding of each signal point is given

in the following figure :

00

01

11

10

The signal points that may be transmitted at times t = 2nT n = 0, 1, ... are given with blank

circles, while the ones that may be transmitted at times t = 2nT + 1, n = 0, 1, ... are given with

filled circles.

Problem 3.4

1. Consider the QAM constellation of Fig. P3-4. Using the Pythagorean theorem we can find the

radius of the inner circle as:

a2 + a2 = A2 = a = 12A

The radius of the outer circle can be found using the cosine rule. Since b is the third side of a

triangle with a and A the two other sides and angle between then equal to = 75o, we obtain:

b2 = a2 +A2 2aA cos 75o = b = 1 +3

2A

2. If we denote by r the radius of the circle, then using the cosine theorem we obtain:

A2 = r2 + r2 2r cos 45o = r = A22


43. The average transmitted power of the PSK constellation is:

PPSK = 81

8(

A22

)2= PPSK =

A2

22whereas the average transmitted power of the QAM constellation:

PQAM =1

8

(4A2

2+ 4

(1 +3)2

4A2

)= PQAM =

[2 + (1 +

3)2

8

]A2

The relative power advantage of the PSK constellation over the QAM constellation is:

gain =PPSKPQAM

=8

(2 + (1 +3)2)(2 2) = 1.5927 dB

Problem 3.5

1. Although it is possible to assign three bits to each point of the 8-PSK signal constellation so that

adjacent points differ in only one bit, (e.g. going in a clockwise direction : 000, 001, 011, 010, 110,

111, 101, 100). this is not the case for the 8-QAM constellation of Figure P3-4. This is because there

are fully connected graphs consisted of three points. To see this consider an equilateral triangle

with vertices A, B and C. If, without loss of generality, we assign the all zero sequence {0, 0, . . . , 0}to point A, then point B and C should have the form

B = {0, . . . , 0, 1, 0, . . . , 0} C = {0, . . . , 0, 1, 0, . . . , 0}where the position of the 1 in the sequences is not the same, otherwise B=C. Thus, the sequences

of B and C differ in two bits.

2. Since each symbol conveys 3 bits of information, the resulted symbol rate is :

Rs =90 106

3= 30 106 symbols/sec

Problem 3.6

The constellation of Fig. P3-6(a) has four points at a distance 2A from the origin and four points

at a distance 22A. Thus, the average transmitted power of the constellation is:

Pa =1

8

[4 (2A)2 + 4 (2

2A)2

]= 6A2


5The second constellation has four points at a distance7A from the origin, two points at a dis-

tance3A and two points at a distance A. Thus, the average transmitted power of the second

constellation is:

Pb =1

8

[4 (

7A)2 + 2 (

3A)2 + 2A2

]=

9

2A2

Since Pb < Pa the second constellation is more power efficient.

Problem 3.7

One way to label the points of the V.29 constellation using the Gray-code is depicted in the next

figure.

O

O

O

O

O

O

O

O O O O O O

O

O

O

0000 0001 0011 00101100111011111101

0100

0101

0111

0110

1000

1001

1011

1010

Problem 3.8

We assume that the input bits 0, 1 are mapped to the symbols -1 and 1 respectively. The terminal


6phase of an MSK signal at time instant n is given by

(n;a) =pi

2

kk=0

ak + 0

where 0 is the initial phase and ak is 1 depending on the input bit at the time instant k. Thefollowing table shows (n;a) for two different values of 0 (0, pi), and the four input pairs of data:

{00, 01, 10, 11}.

0 b0 b1 a0 a1 (n;a)

0 0 0 -1 -1 pi0 0 1 -1 1 0

0 1 0 1 -1 0

0 1 1 1 1 pi

pi 0 0 -1 -1 0

pi 0 1 -1 1 pi

pi 1 0 1 -1 pi

pi 1 1 1 1 2pi

Problem 3.9

1.

(i) There are no correlative states in this system, since it is a full response CPM. Based on (3-3-16),

we obtain the phase states :

s =

{0,2pi

3,4pi

3

}(ii) Based on (3-3-17), we obtain the phase states :

s =

{0,3pi

4,3pi

2,9pi

4 pi

4, pi,

15pi

4 7pi

4,18pi

4 pi

2,21pi

4 5pi

4

}

2.

(i) The combined states are Sn = (n, In1, In2) , where{, In1/n2

}take the values 1. Hence

there are 3 2 2 = 12 combined states in all.(ii) The combined states are Sn = (n, In1, In2) , where

{, In1/n2

}take the values 1. Hence

there are 82 2 = 32 combined states in all.


7Problem 3.10

The bandwidth required for transmission of an M -ary PAM signal is

W =R

2 log2MHz

Since,

R = 8 103 samplessec

8 bitssample

= 64 103 bitssec

we obtain

W =

16 KHz M = 4

10.667 KHz M = 8

8 KHz M = 16

Problem 3.11

The autocorrelation function for u(t) is :

Ruu(t) = E [u(t+ )u(t)]

=

n=

m=E (ImI

n)E [u(t+ mT )u(t nT )]

=

n=

m=Rii(m n)E [u(t+ mT )u(t nT )]

=

m=Rii(m)

n=E [u(t+ mT nT )u(t nT )]

=

m=Rii(m)

n=

T0

1T u(t+ mT nT )u(t nT )d

Let a = + nT, da = d, and a (,). Then :

Ruu(t) =

m=Rii(m)

n=

(n+1)TnT

1T u(t+ mT a)u(t a)da

=

m=Rii(m)1T

u(t+ mT a)u(t a)da

= 1T

m=Rii(m)Ruu( mT )


8Thus we have obtained the same autocorrelation function as given by (4.4.11). Consequently the

power spectral density of u(t) is the same as the one given by (4.4.12) :

Suu(f) =1

T|G(f)|2 Sii(f)

Problem 3.12

The 16-QAM signal is represented as s(t) = In cos 2pift+Qn sin 2pift, where In = {1,3} , Qn ={1,3} . A superposition of two 4-QAM (4-PSK) signals is :

s(t) = G [An cos 2pift+Bn sin 2pift] + Cn cos 2pift+ Cn sin 2pift

where An, Bn,Cn,Dn = {1} . Clearly : In = GAn + Cn, Qn = GBn +Dn. From these equationsit is easy to see that G = 2 gives the requires equivalence.

Problem 3.13

We have that Suu(f) = 1T |G(f)|2 Sii(f) But E(In) = 0, E(|In|2

)= 1, hence : Rii (m) =

1, m = 00, m 6= 0 . Therefore : Sii(f) = 1 Suu(f) = 1T |G(f)|2 .

1. For the rectangular pulse :

G(f) = ATsinpifT

pifTej2pifT/2 |G(f)|2 = A2T 2 sin

2pifT

(pifT )2

where the factor ej2pifT/2 is due to the T/2 shift of the rectangular pulse from the center t = 0.

Hence :

Suu(f) = A2T sin2pifT

(pifT )2


90

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-5 -4 -3 -2 -1 0 1 2 3 4 5fT

Sv(f)

2. For the sinusoidal pulse : G(f) = T0 sin

pitT exp(j2pift)dt. By using the trigonometric identity

sinx = exp(jx)exp(jx)2j it is easily shown that :

G(f) =2AT

pi

cos piTf

1 4T 2f2 ej2pifT/2 |G(f)|2 =

(2AT

pi

)2 cos 2piTf(1 4T 2f2)2

Hence :

Suu(f) =(2A

pi

)2T

cos 2piTf

(1 4T 2f2)2

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-5 -4 -3 -2 -1 0 1 2 3 4 5fT

Sv(f)

3. The 3-db frequency for (a) is :

sin 2pif3dbT

(pif3dbT )2 =

1

2 f3db = 0.44

T


10

(where this solution is obtained graphically), while the 3-db frequency for the sinusoidal pulse on

(b) is :cos 2piTf

(1 4T 2f2)2 =1

2 f3db = 0.59

T

The rectangular pulse spectrum has the first spectral null at f = 1/T, whereas the spectrum of the

sinusoidal pulse has the first null at f = 3/2T = 1.5/T. Clearly the spectrum for the rectangular

pulse has a narrower main lobe. However, it has higher sidelobes.

Problem 3.14

1. Bn = In + In1. Hence :

In In1 Bn

1 1 2

1 1 01 1 01 1 2

The signal space representation is given in the following figure, with P (Bn = 2) = P (Bn = 2) =1/4, P (Bn = 0) = 1/2.

-o oI

-2 2

Bn

0

2.

RBB(m) = E [Bn+mBn] = E [(In+m + In+m1) (In + In1)]

= Rii(m) +Rii(m 1) +Rii(m+ 1)Since the sequence {In} consists of independent symbols :

Rii(m) =

E [In+m]E [In] = 0 0 = 0, m 6= 0E [I2n] = 1, m = 0

Hence :

RBB(m) =

2, m = 0

1, m = 10, o.w

and

SBB(f) =

m=RBB(m) exp(j2pifmT ) = 2 + exp(j2pifT ) + exp(j2pifT )= 2 [1 + cos 2pifT ] = 4 cos 2pifT


11

A plot of the power spectral density SB(f) is given in the following figure :

0.5 0.4 0.3 0.2 0.1 0 0.1 0.2 0.3 0.4 0.50

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Normalized frequency fT

Power spectral density of B

3. The transition matrix is :

In1 In Bn In+1 Bn+1

1 1 2 1 21 1 2 1 01 1 0 1 01 1 0 1 21 1 0 1 21 1 0 1 01 1 2 1 01 1 2 1 2

The corresponding Markov chain model is illustrated in the following figure :

"!#

"!#

"!# @@R @@

@

@I

@@

-

?

2 -2

1/4 1/4

1/2

1/2

1/2

1/2

1/2

0

Problem 3.15

1. In = an an2, with the sequence {an} being uncorrelated random variables (i.e E (an+man) =


12

(m)). Hence :

Rii(m) = E [In+mIn] = E [(an+m an+m2) (an an2)]= 2(m) (m 2) (m+ 2)

=

2, m = 0

1, m = 20, o.w.

2. Suu(f) = 1T |G(f)|2 Sii(f) where :Sii(f) =

m=Rii(m) exp(j2pifmT ) = 2 exp(j4pifT ) exp(j4pifT )

= 2 [1 cos 4pifT ] = 4 sin 22pifTand

|G(f)|2 = (AT )2(sinpifT

pifT

)2Therefore :

Suu(f) = 4A2T(sinpifT

pifT

)2sin 22pifT

3. If {an} takes the values (0,1) with equal probability then E(an) = 1/2 and E(an+man) = 1/4, m 6= 01/2, m = 0

= [1 + (m)] /4. Then :

Rii(m) = E [In+mIn] = 2Raa(0) Raa(2)Raa(2)= 14 [2(m) (m 2) (m+ 2)]

and

Sii(f) =

m=Rii(m) exp(j2pifmT ) = sin 22pifTSuu(f) = A2T

(sinpifTpifT

)2sin 22pifT

Thus, we obtain the same result as in (b) , but the magnitude of the various quantities is reduced

by a factor of 4 .

Problem 3.16

We may use the result in (3.4.27), where we set K = 2, p1 = p2 = 1/2 :

S(f) = 1T 2

l=

2

i=1

1

2Si

(l

T

)2

(f l

T

)+

1

T

2i=1

1

4|Si(f)|2 2

T

1

4Re [S1(f)S

2(f)]


13

To simplify the computations we may define the signals over the symmetric interval T/2 t T/2. Then :

Si(f) =T

2j

[sinpi(f fi)Tpi(f fi)T

sinpi(f + fi)T

pi(f + fi)T

](the well-known rectangular pulse spectrum, modulated by sin 2pifit) and :

|Si(f)|2 =(T

2

)2 [(sinpi(f fi)Tpi(f fi)T

)2+

(sinpi(f + fi)T

pi(f + fi)T

)2]

where the cross-term involving the product sinpi(ffi)Tpi(ffi)T sinpi(f+fi)Tpi(f+fi)T

is negligible when fi >> 0. Also

:

S1(lT

)= T2j

[sinpi( l

T n2T

)T

pi( lT n2T

)T sinpi(

lT+ n2T

)T

pi( lT+ n2T

)T

]

= T2j

[sin(pilpin

2)

(pilpin2)

sin(pil+pin2)

(pil+pin2)

]

= T2j 2l(1)l+1(sin pin2

)/pi(l2 n2/4)

= lTj (1)l+1sin pin

2

pi(l2n2/4)

and similarly for S2(lT ) (with m instead of n). Note that if n(m) is even then S1(2)(

lT ) = 0 for all l

except at l = n(m)/2, where S1(2)(n(m)2T ) = T2j . For this case

1

T 2

l=

2

i=1

1

2Si

(l

T

)2

(f l

T

)=

1

16

[(f n

2T

)+

(f +

n

2T

)+

(f m

2T

)+

(f m

2T

)]

The third term in (3.4.27) involves the product of S1(f) and S2(f) which is negligible since they

have little spectral overlap. Hence :

S(f) = 116

[(f n

2T

)+

(f +

n

2T

)+

(f m

2T

)+

(f m

2T

)]+

1

4T

[|S1(f)|2 + |S2(f)|2

]In comparison with the spectrum of the MSK signal, we note that this signal has impulses in the

spectrum.

Problem 3.17

MFSK signal with waveforms : si(t) = sin2piitT , i = 1, 2, ...,M 0 t T

The expression for the power density spectrum is given by (3.4.27) with K =M and pi = 1/M.

From Problem 4.23 we have that :

Si(f) =T

2j

[sinpi(f fi)Tpi(f fi)T

sinpi(f + fi)T

pi(f + fi)T

]


14

for a signal si(t) shifted to the left by T/2 (which does not affect the power spectrum). We also

have that :

Si

(nT

)=

T/2j, n = i0, o.w.

Hence from (3.4.27) we obtain :

S(f) = 1T 2

(1M

)2 (T 24

)Mi=1 [(f fi) + (f + fi)]

+ 1T(1M

)2Mi=1 |Si(f)|2

2TM

i=1

Mj=i+1

(1M

)2Re[Si(f)S

j (f)

]

=(

12M

)2Mi=1 [(f fi) + (f + fi)] + 1TM2

Mi=1 |Si(f)|2

2TM2

Mi=1

Mj=i+1Re

[Si(f)S

j (f)

]

Problem 3.18

QPRS signal v(t) =

n (Bn + jCn)u(t nT ), Bn = In + In1, Cn = Jn + Jn1.

1. Similarly to Problem 3.11, the sequence Bn can take the values : P (Bn = 2) = P (Bn =

2) = 1/4, P (Bn = 0) = 1/2. The same holds for the sequence Cn; since these two sequences areindependent :

P {Bn = i, Cn = j} = P {Bn = 1}P {Cn = j}

Hence, since they are also in phase quadrature the signal space representation will be as shown in

the following figure (next to each symbol is the corresponding probability of occurrence) :


15

6

-

-

o o

o

o o

o

o o

Cn

Bn

1/16

1/161/16

1/16

1/8

o1/8

1/8

1/41/8

2

2. If we name Zn = Bn + jCn :

RZZ(m) =12E [(Bn+m + jCn+m) (Bn jCn)]

= 12 {E [Bn+mBn] + E [Cn+mCn]} = 12 (RBB(m) +RCC(m)) = RBB(m) = RCC(m)

since the sequences Bn, Cn are independent, and have the same statistics. Now, from Problem 3.11

:

RBB(m) =

2, m = 0

1, m = 10, o.w

= RCC(m) = RZZ(m)

Hence, from (4-4-11) :

Rvs() =1

T

m=

RBB(m)Ruu( mT ) = Rvc() = Rv()

Also :

Svs(f) = Svc(f) = Sv(f) = 1T|U(f)|2 SBB(f)

since the corresponding autocorrelations are the same . From Problem 3.11 : SBB(f) = 4 cos 2pifT ,so

Svs(f) = Svc(f) = Sv(f) = 4T|U(f)|2 cos 2pifT

Therefore, the composite QPRS signal has the same power density spectrum as the in-phase and

quadrature components.

3. The transition probabilities for the Bn, Cn sequences are independent, so the probability of

a transition between one state of the QPRS signal to another state, will be the product of the


16

probabilities of the respective B-transition and C-transition. Hence, the Markov chain model will

be the Cartesian product of the Markov model that was derived in Problem 3.11 for the sequence

Bn alone. For example, the transition probability from the state (Bn, Cn) = (0, 0) to the same

state will be : P (Bn+1 = 0|Bn = 0) P (Cn+1 = 0|Cn = 0) = 12 12 = 14 and so on. Below, we give apartial sketch of the Markov chain model; the rest of it can be derived easily, from the symmetries

of this model.

@@R

-

?

@@@@@@@@@R

@@

@@

@@

@@

@@I

@@I

@@

@@

I@@@@@@@@@R

-2,0

0,-2 2,-2

1/41/4

1/4

1/8

1/16

1/4

1/8

1/4

1/4 1/8

-2,-2

0,0 2,0

-2,2 0,2 2,2

Problem 3.19


17

1. Since : a = 0, 2a = 1, we have : Sss(f) = 1T |G(f)|2 . But :

G(f) = T2sinpifT/2pifT/2 e

j2pifT/4 T2 sinpifT/2pifT/2 ej2pif3T/4

= T2sinpifT/2pifT/2 e

jpifT (2j sinpifT/2)

= jT sin2pifT/2pifT/2 e

jpifT

|G(f)|2 = T 2(sin 2pifT/2pifT/2

)2

Sss(f) = T(sin 2pifT/2pifT/2

)2

2. For non-independent information sequence the power spectrum of s(t) is given by : Sss(f) =1T |G(f)|2 Sbb(f). But :

Rbb(m) = E [bn+mbn]

= E [an+man] + kE [an+m1an] + kE [an+man1] + k2E [an+m1an1]

=

1 + k2, m = 0

k, m = 10, o.w.

Hence :

Sbb(f) =

m=

Rbb(m)ej2pifmT = 1 + k2 + 2k cos 2pifT

We want :

Sss(1/T ) = 0 Sbb(1/T ) = 0 1 + k2 + 2k = 0 k = 1

and the resulting power spectrum is :

Sss(f) = 4T(sin 2pifT/2

pifT/2

)2sin 2pifT

3. The requirement for zeros at f = l/4T, l = 1,2, ... means : Sbb(l/4T ) = 0 1 + k2 +2k cos pil/2 = 0, which cannot be satisfied for all l. We can avoid that by using precoding in the


18

form :bn = an + kan4. Then :

Rbb(m) =

1 + k2, m = 0

k, m = 40, o.w.

Sbb(f) = 1 + k

2 + 2k cos 2pif4T

and , similarly to (b), a value of k = 1, will zero this spectrum in all multiples of 1/4T.

Problem 3.20

1. The power spectral density of the FSK signal may be evaluated by using equation (3-4-27) with

K = 2 (binary) signals and probabilities p0 = p1 =12 . Thus, when the condition that the carrier

phase 0 and and 1 are fixed, we obtain

S(f) = 14T 2

n=

|S0(nT) + S1(

n

T)|2(f n

T) +

1

4T|S0(f) S1(f)|2

where S0(f) and S1(f) are the fourier transforms of s0(t) and s1(t). In particular :

S0(f) =

T0

s0(t)ej2pift dt

=

2EbT

T0

cos(2pif0t+ 0)ej2pift dt, f0 = fc f

2

=

TEb2

[sinpiT (f f0)pi(f f0) +

sinpiT (f + f0)

pi(f + f0)

]ejpifT ej0

Similarly :

S1(f) =

T0

s1(t)ej2pift dt

=

TEb2

[sinpiT (f f1)pi(f f1) +

sinpiT (f + f1)

pi(f + f1)

]ejpifT ej1

where f1 = fc +f2 . By expressing S(f) as :

S(f) = 14T 2

n=

[|S0(n

T)|2 + |S1(n

T)|2 + 2Re[S0(n

T)S1(

n

T)]](f n

T)

+1

4T

[|S0(f)|2 + |S1(f)|2 2Re[S0(f)S1(f)]]PROPRIETARY MATERIAL. cThe McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual, you are using it without permission.

19

we note that the carrier phases 0 and 1 affect only the terms Re(S0S1). If we average over the

random phases, these terms drop out. Hence, we have :

S(f) = 14T 2

n=

[|S0(n

T)|2 + |S1(n

T)|2](f n

T)

+1

4T

[|S0(f)|2 + |S1(f)|2]where :

|Sk(f)|2 = TEb2

sinpiT (f fk)pi(f fk) +sinpiT (f + fk)

pi(f + fk)

2

, k = 0, 1

Note that the first term in S(f) consists of a sequence of samples and the second term constitutesthe continuous spectrum.

2. Note that :

|Sk(f)|2 = TEb2

[(sinpiT (f fk)pi(f fk)

)2+

(sinpiT (f + fk)

pi(f + fk)

)2]

because the productsinpiT (f fk)pi(f fk)

sinpiT (f + fk)

pi(f + fk) 0

if fk is large enough. Hence |Sk(f)|2 decays proportionally to 1(ffk)2 1f2 for f fc. Conse-

quently, S(f) exhibits the same behaviour.

Problem 3.21

1) The power spectral density of X(t) is given by

Sx(f) = 1TSi(f)|U(f)|2

The Fourier transform of u(t) is

U(f) = F [u(t)] = AT sinpifTpifT

ejpifT

Hence,

|U(f)|2 = (AT )2sinc2(fT )and therefore,

Sx(f) = A2TSi(f)sinc2(fT ) = A2T sinc2(fT )


20

2) If u1(t) is used instead of u(t) and the symbol interval is T , then

Sx(f) = 1TSa(f)|U1(f)|2

=1

T(A2T )2sinc2(f2T ) = 4A2T sinc2(f2T )

3) If we precode the input sequence as bn = In + In1, then

Rb(m) =

1 + 2 m = 0

m = 10 otherwise

and therefore, the power spectral density Sb(f) is

Sb(f) = 1 + 2 + 2 cos(2pifT )

To obtain a null at f = 13T , the parameter should be such that

1 + 2 + 2 cos(2pifT )|f= 1

3T

= 0

and does not have a real-valued solution. Therefore the above precoding cannot result in a PAM

system with the desired spectral null.

4) The answer to this question is no. This is because Sb(f) is an analytic function and unless itis identical to zero it can have at most a countable number of zeros. This property of the analytic

functions is also referred as the theorem of isolated zeros.

Problem 3.22

1. Since X(t) = Re [

n=(an + jbn)u(t nT )ej2pifct], the lowpass equivalent signal is Xl(t) =n=(an + jbn)u(t nT ). The in-phase and quadrature components are thus given by

Xi(t) =

n= an cos 2pifct and Xq(t) =

n= bn sin 2pifct.

2. Defining In = an + jbn, from the values of (an, bb) pair, it is clear that E[In] = 0 and

Ri(m) = E[In+mIn] =

{1 m = 0

0 m 6= 0

and Si(f) = 1. Also note that U(f) = 2T sinc2(2Tf) and from 3.4-16

SXl(f) =1

TSi(f)|U(f)|2 = 4T sinc4(2Tf)


21

Using 2.9-14,

SX(f) = 14

[4T sinc4(2T (f fc)) + 4T sinc4(2T (f + fc))

]= T

[sinc4(2T (f fc)) + T sinc4(2T (f + fc))

]3. This is equivalent to a precoding of the form Jn = In + In1 where Jn = cn + jdn. Using

3.4-20 we have

SYl(f) = SXl(f)1 + ej2pifT 2

= 4T sinc4(2Tf)(1 + ||2 + 2Re[ej2pifT ]

)To have no DC components the PSD must vanish at f = 0, or 1+ ||2+2Re[] = 0 resultingin 1+2r+2r+

2i = 0. This relation is satisfied for infinitely many s, for instance = 1.

Problem 3.23

1. Since s(t) =

n= ng(t nT ) cos(2pif0t+ n), we have sl(t) =

n= nejng(t nT ).

We know that Ssl(f) = 1T Sa(f)|G(f)|2, where an = nejn s are all independent eachwith probability of 18 . Since g(t) =

(tT/2T/2

), therefore G(f) = ejpifT T2 sinc

2(Tf2

), and

|G(f)|2 = T 24 sinc4(Tf2

). We need to find Sa(f) =

m=Ra(m)e

j2pifmT , but

Ra(m) = E[am+nan]

=

{E[|an|2], m = 0|E(an)|2, m 6= 0

=

18

7i=1 |i|2, m = 0

164

8i=1 ieji2 , m 6= 0=

{18

2, m = 0164 ||2, m 6= 0

From here we have

Sa(f) = 164||2

m=

ej2pifmT +

(1

82 1

64||2

)

=1

8

(2 1

8||2

)+||264T

m=

(f m

T

)


22

and

Ssl(f) =1

TSa(f)|G(f)|2

=T

4sinc4

(Tf

2

)[1

8

(2 1

8||2

)+||264T

m=

(f m

T

)]

=T

32

[(2 1

8||2

)sinc4

(Tf

2

)+||28T

m=

sinc4(m2

)(f m

T

)]

2. In this case = 0 and 2 = a2+b2

4 , therefore, Ssl(f) = T (a2+b2)128 sinc

4(Tf2

).

3. For a = b the constellation is on a circle at angles 45 apart, therefore it is 8PSK and

Ssl(f) = Ta2

64 sinc4(Tf2

).

4. bns will still be independent and equiprobable, therefore the previous parts will not change.

Problem 3.24

1. In general

Sv(f) = 1T|G(f)|2Si(f)

where

Si(f) =

m=

Ri(m)ej2pifmT

and Ri(m) = E[InIn+m]. Since the sequence In is iid we have

Ri(m) =

{(E[In])

2 m 6= 0E[I2n]

m = 0=

{0 m 6= 014(2

2 + 22) = 2 m = 0= 2(m)

Hence Si(f) =

m=Ri(m)ej2pifmT = 2 and Sv(f) = 1T |G(f)|2Si(f) = 2T |G(f)|2, where

G(f) = F [g(t)] = T2 ejpifT/2 sinc(Tf/2) + TejpifT sinc(Tf).2. In this cases the signaling interval is 2T and Jn substitutes In. We have

Rj(m) = E[JnJn+m] = E [(In1 + In + In+1)(In+m1 + In+m + In+m+1)]

= 3Ri(m) + 2Ri(m+ 1) + 2Ri(m 1) +Ri(m+ 2) +Ri(m 2)= 6(m) + 4(m 1) + 4(m+ 1) + 2(m 2) + 2(m+ 2)

=

6 m = 0

4 m = 12 m = 20 otherwise


23

Hence Sj(f) =

m=Rj(m)ej2pimf2T = 6 + 8 cos(4pifT ) + 4 cos(8pifT ) and

Sw(f) = 12T|G(f)|2Sj(f) = |G(f)|

2

T(3 + 4 cos 4pifT + 2cos 8pifT )

Problem 3.25

1. We have

Ra(m) = E[an+man]

=

{E[a2n] m = 0

(E[an])2 m 6= 0

=

{54 m = 0116 m 6= 0

Using Sa(f) =

m=Ra(m)ej2pifmT , we have

Sa(f) = 1916

+1

16

m=

ej2pifmT

=19

16+

1

16T(f m

T

)

and since g(t) = sinc(t/T ), we have G(f) = T(Tf), hence |G(f)|2 = T 2(Tf) and

Sv(f) = 1T

(T 2(Tf)

) [1916

+1

16T(f m

T

)]

resulting in

Sv(f) = 1916T(Tf) +

1

16(f)

2. The power spectral density is multiplied by1 + ej2pifT ej4pifT 2 = 3 2 cos(4pifT ).

Therefore

Su(f) = 1916

(3 2 cos 4pifT )T(f) + 116(f)

3. In this case Sw(f) is multiplied by1 + jej2pifT 2 = (2 + 2 sin 2pifT ) and

Sw(f) = 198(1 + sin 2pifT )T(f) +

1

8(f)


24

Problem 3.26

First note that

Ra(m) = E[anan+m] =

{1, n = m

0, otherwise.

1. For QPSK we have Xl =

n=(a2n+ ja2n+1)g2T (t2nT ) =

n= Ing2T (t2nT ) whereIn = a2n + ja2n+1 and therefore

Ri(m) = E[In+mIn]

= E[(a2(n+m) + ja2n+2m+1)(a2n ja2n+1)

]= [2Ra(2m) + jRa(2m+ 1) jRa(2m 1)]

=

{2, m = 0

0, otherwise.

Therefore Si(f) =

m=Ri(m)ej4pifmT = 2. Also note that g2T (t) =

(tT2T

), and

therefore |G2T (f)|2 = 4T 2 sinc2(2Tf). Substituting into SXl(f) = 1T Si(f)|G2T (f)|2, we haveSXl(f) = 8T sinc2(2Tf).

2. Here Xl(t) =

n= a2ng2T (t 2nT ) + j

n= a2n+1g2T (t (2n+ 1)T ), and

RXl(t+ , t) = E [Xl(t+ )Xl (t)]

= E

[(

n=

a2ng2T (t+ 2nT ) + j

n=

a2n+1g2T (t+ (2n + 1)T ))

(

m=

a2mg2T (t 2mT ) j

m=

a2m+1g2T (t (2m+ 1)T ))]

=

n=

[g2T (t+ 2nT )g2T (t 2nT )

+ g2T (t+ (2n + 1)T )g2T (t (2n + 1)T )]

=

n=

g2T (t+ nT )g2T (t nT )

Same as BPSK with g(t) = g2T (t), therefore we have the same spectrum as in part 1.

3. The only difference here is that instead of the Fourier transform of the rectangular signal we

have to use the Fourier transform of the sinusoidal pulse g1(t), nothing else changes. Noting

that sin(x) = ejxejx

2j the Fourier transform of the sinusoidal pulse in the problem can be

obtained by direct application of the definition of Fourier transform

G1(f) =

2T0

sin

(pit

2T

)ej2pift dt = = 4T

pi

cos 2piTf

1 16T 2f2 ej2pifT


25

and therefore

S(f) = 32Tpi2

cos2 2piTf

(1 16T 2f2)2

4. The envelope of Xl, is in general |Xl(t)|, so we need to show that |Xl(t)| is independent of t.We have

Xl(t) =

n=

a2ng2T (t 2nT ) + j

n=

a2n+1g2T (t (2n + 1)T )

Therefore,

|Xl(t)|2 =(

n=

a2ng2T (t 2nT ))2

+

(

n=

a2n+1g2T (t (2n + 1)T ))2

=

n=

g22T (t 2nT ) +

n=

g22T (t (2n + 1)T )

=

[sin2

(pit

2T

)+ cos2

(pit

2T

)]for all t

= 1

where we have used the following facts:

(a) the duration of g2T (t) is 2T and hence g2T (t 2nT ) and g2T (t 2mT ) for m 6= n arenon-overlapping and therefore in the expansion of the squares the cross terms vanish

(same is true for g2T (t (2n + 1)T ) and g2T (t (2m+ 1)T ).)(b) For all n, a2n = 1.

Problem 3.27

Since ans are iid

Ra(m) = E[anan+m] =

{E[a2n], m = 0

E[an]E[an+m], m 6= 0=

{12 , m = 014 , m 6= 0

1. bn = an1 an, hence P (bn = 0) = P (an = an1 = 0) + P (an = an1 = 1) = 12 andP (bn = 1) =

12 . This means that bns individually have the same probabilities as ans but of

course unlike ans they are not independent. In fact bn depends on bn1 and bn+1 and since

Rb(m) = E[bnbn+m] = E[(an1 an)(an+m1 an+m)], we conclude that if m 6= 0,1 thenRb(m) = E[bn][bn+m] =

14 and Rb(0) = E[b

2n] =

12 . On the other hand, for m = 1, we have

to consider different values that an1, an, and an+1 can assume. In order for bnbn+1 to be

1, we have to have bn = 1 and bn+1 = 1. This means that an1 = an+1 6= an and this can


26

happen in two cases; an1 = an+1 = 1, an = 0, and an1 = an+1 = 0, an = 1, each with

probability 1/8. Therefore, Rb(1) = 1/4 and

Rb(m) =

{12 , m = 014 , m 6= 0

and S(f) =m=Rb(m)ej2pifmT = 14+14m= ej2pifmT . Also, |G(f)|2 = T 2 sinc2(Tf),and

Sv(f) = 1T|G(f)|2Sb(f) = T sinc2(Tf)

[1

4+1

4

m=

ej2pifmT

]

We can simplify this using the relation

n= ej2pifnT = 1T

n=

(f nT

)to obtain

Sv(f) = T sinc2(Tf)[1

4+

1

4T

m=

(f m

T

)]

=T

4sinc2(Tf) +

1

4sinc2(T

m

T)

m=

(f m

T

)

=T

4sinc2(Tf) +

1

4(f)

2. Here, Rb(m) = E[(an + an1)(an+m + an+m1)] = 2Ra(m) + Ra(m + 1) + Ra(m 1), andhence

Rb(m) =

32 , m = 054 , m = 11, otherwise

and

Sb(f) =

m=

Rb(m)ej2pifmT =

1

2+1

4

(ejpifT + ej2pifT

)+

m=

ej2pifmT

Therefore,

Sv(f) = T sinc2(Tf)[1

2+1

4

(ej2pifT + ej2pifT

)+

m=

ej2pifmT

]

= T sinc2(Tf)

[1

2+1

2cos(2pifT ) +

1

T

m=

(f m

T

)]

=T

2sinc2(Tf) [1 + cos(2pifT )] +

m=

sinc2(m)(f m

T

)

=T

2sinc2(Tf) [1 + cos(2pifT )] + (f)


27

Problem 3.28

Here we have 8 equiprobable symbols given by the eight points in the constellation.

1. Obviously E(an) = 0 and

E[anam] =

{E[an]E[a

m] = 0 n 6= m

E[|an|2] = r21+r2

2

2 n = m

Hence Sa(f) = r21+r2

2

2 . Also obviously |G(f)|2 = T 2 sinc2(Tf). Therefore

Sl(f) = T r21 + r

22

2sinc2(Tf)

2. S(f) = 14Sl(f f0) + 14Sl(f + f0).3. In this case Sl(f) = Tr2 sinc2(Tf). An example of the plot is shown below

6 4 2 0 2 4 60

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Problem 3.29

The MSK and offset QPSK signals have the following form :

v(t) =n

[anu(t 2nT ) jbnu(t 2nT T )]


28

where for the QPSK :

u(t) =

1, 0 t 2T0, o.w.

and for MSK :

u(t) =

sin

pit2T , 0 t 2T0, o.w.

The derivation is identical to that given in Sec. 3.4.2 with 2T substituted for T. Hence, the result

is:

Rvv() =12T

m=Rii(m)Ruu( m2T )

= 12T

m=

(2a +

2b

)(m)Ruu( m2T )

= 2a

T Ruu()

and :

Svv(f) = 2a

T|U(f)|2

For the rectangular pulse of QPSK, we have :

Ruu() = 2T

(1 | |

2T

), 0 | | 2T

For the MSK pulse :

Ruu() = u(t+ )u

(t)dt = 2T0 sin

pit2T sin

pi(t+)2T dt

= T(1 | |2T

)cos pi| |2T +

pi sin

pi| |2T

(check for tha 1/2 factor in the defn. Of autocorr. Function!)

Problem 3.30

a. For simplicity we assume binary CPM. Since it is partial response :

q(T ) = T0 u(t)dt = 1/4

q(2T ) = 2T0 u(t)dt = 1/2, q(t) = 1/2, t > 2T

so only the last two symbols will have an effect on the phase :

R(t; I) = 2pihn

k= Ikq(t kT ), nT t nT + T= pi2

n2k= Ik + pi (In1q(t (n 1)T ) + Inq(t nT )) ,


29

It is easy to see that, after the first symbol, the phase slope is : 0 if In,In1 have different signs,

and sgn(In)pi/(2T ) if In,In1 have the same sign. At the terminal point t = (n+1)T the phase is :

R((n + 1)T ; I) =pi

2

n1k=

Ik +pi

4In

Hence the phase tree is as shown in the following figure :

!!!!!

aaaaa@@@@@@@@@

@@@@@

@

o

o o

oo

o

+1 -1

+1

+1

+1

+1+1

-1

-1

-1

-1

o

o o

oo

o

-1 +1

-1

-1

-1

-1-1

+1

+1

+1

+10

pi/4

3pi/4

5pi/4

7pi/4

pi/4

3pi/4

5pi/4

7pi/4

t=0 t=T t=2T t=3T

b. The state trellis is obtained from the phase-tree modulo 2pi:


30

B

BBBBBBBBBBBBBBBB

@@@@@@@@@@@@@@@@@

@@@@@@@@@@

BBBBBBBBBBBBB

o o o

o o

o o

o o o o

o

o

o

+1 +1-1 -1

-1 -1+1 +1

+1

+1

-1

-1

-1+1 +1

+1-1 -1

+1 +1

+1

-1-1

-1

+1

-1

+1

t=0 t=T t=2T t=3T t=4T

0

0 = pi/4

1 = 3pi/4

2 = 5pi/4

3 = 7pi/4

(c) The state diagram is shown in the following figure (with the (In,In1) or (In1,In) that cause

the respective transitions shown in parentheses)

PPPPPP

PPPP

PPi

BBBBBB

BBBBBB

PPPPPPq

PPPP

PPi

B

BBBBB

BBBBBB

@@..........................@

@I..................

........

@@..........................@@R

..........................

(1,1)

(1,1)

(1,1)

(1,1)

(-1,-1)

(-1,-1)

(-1,-1)

(-1,-1)

(-1,,1)

(-1,1)(-1,1)

(-1,1)

0 = pi/4

1 = 3pi/42 = 5pi/4

3 = 7pi/4q

Problem 3.31


31

R(t; I) = 2pih

nk=

Ikq(t kT )

1. Full response binary CPFSK (q(T ) = 1/2):

(i) h = 2/3. At the end of each bit interval the phase is : 2pi 2312

nk= Ik =

2pi3

nk= Ik. Hence

the possible terminal phase states are {0, 2pi/3, 4pi/3} .(ii) h = 3/4. At the end of each bit interval the phase is : 2pi 34

12

nk= Ik =

3pi4

nk= Ik. Hence

the possible terminal phase states are {0, pi/4, pi/2, 3pi/4, pi, 5pi/4, 3pi/2, 7pi/4}

2. Partial response L = 3, binary CPFSK : q(T ) = 1/6, q(2T ) = 1/3, q(3T ) = 1/2. Hence, at the

end of each bit interval the phase is :

pih

n2k=

Ik + 2pih (In1/3 + In/6) = pih

n2k=

Ik +pih

3(2In1 + In)

The symbol levels in the parenthesis can take the values {3,1, 1, 3} . So :(i) h = 2/3. The possible terminal phase states are :

{0, 2pi/9, 4pi/9, 2pi/3, 8pi/9, 10pi/9, 4pi/3, 14pi/9, 16pi/9}

(ii) h = 3/4. The possible terminal phase states are : {0, pi/4, pi/2, 3pi/4, pi, 5pi/4, 3pi/2, 7pi/4}

Problem 3.32

We are given by Equation (3.3-33) that the pulses ck(t) are defined as

ck(t) = s0(t)L1n=1

s0[t+ (n+ Lak,n)], 0 t T minn[L(2 ak,n n]

Hence, the time support of the pulse ck(t) is

0 t T minn[L(2 ak,n) n]

We need to find the index n which minimizes S = L(2 ak,n) n, or equivalently maximizesS1 = Lak,n + n :

n = argmaxn

[Lak,n + n], n = 1, ..., L 1, ak,n = 0, 1It is easy to show that

n = L 1 (3.0.1)


32

if all ak,n, n = 0, 1, ..., L 1 are zero (for a specific k), andn = max {n : ak,n = 1} (3.0.2)

otherwise.

The first case (1) is shown immediately, since if all ak,n, n = 0, 1, ..., L 1 are zero, thenmaxn S1 = maxn n, n = 0, 1, ..., L 1. For the second case (2), assume that there are n1, n2such that : n1 < n2 and ak,n1 = 1, ak,n2 = 0. Then S1(n1) = L + n1 > n2(= S1(n2)), since

n2 n1 < L 1 due to the allowable range of n.So, finding the binary representation of k, k = 0, 1, ..., 2L11, we find n and the corresponding

S(n) which gives the extent of the time support of ck(t):

k = 0 ak,L1 = 0, ..., ak,2 = 0, ak,1 = 0 n = L 1 S = L+ 1k = 1 ak,L1 = 0, ..., ak,2 = 0, ak,1 = 1 n = 1 S = L 1k = 2/3 ak,L1 = 0, ..., ak,2 = 1, ak,1 = 0/1 n = 2 S = L 2

and so on, using the binary representation of the integers between 1 and 2L1 1.

Problem 3.33

sk(t) = Iks(t) Sk(f) = IkS(f), E(Ik) = i, 2i = E(I2k) 2iKk=1

pkSk(f)

2

= |S(f)|2Kk=1

pkIk

2

= 2i |S(f)|2

Therefore, the discrete frequency component becomes :

2iT 2

n=

S (nT

)2 (f nT

)

The continuous frequency component is :

1T

Kk=1 pk(1 pk) |Sk(f)|2 2T

i

33

Thus, we have obtained the result in (4.4.18)

Problem 3.34

The line spectrum in (3.4.27) consists of the term :

1

T 2

n=

Kk=1

pkSk

(nT

)2

(f n

T

)

Now, if

Kk=1pksk(t) = 0, then

Kk=1pkSk(f) = 0, f. Therefore, the condition

Kk=1pksk(t) = 0

is sufficient for eliminating the line spectrum.

Now, suppose that

Kk=1pksk(t) 6= 0 for some t [t0, t1]. For example, if sk(t) = Iks(t), then

Kk=1pksk(t) = s(t)

Kk=1pkIk, where

Kk=1pkIk i 6= 0 and s(t) is a signal pulse. Then, the

line spectrum vanishes if S(n/T ) = 0 for all n. A signal pulse that satisfies this condition is shown

below :

-

6

-T

Tt

s(t)

1

In this case, S(f) = T(sinpiTfpiTf

)sinpiTf, so that S(n/T ) = 0 for all n. Therefore, the condition

Kk=1pksk(t) = 0 is not necessary.


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