Chapter 3 Modulation and Frequency-division Multiple Access

Problem 3.1 Show that amplitude modulation is a nonlinear process, as it violates the principle of superposition. Solution By definition

( ) ( )tftmkAts cac π2cos)(1)( += (1) where m(t) is the message signal. Suppose m(t) is defined by the linear combination

)()()( 2211 tmatmatm += (2) Substituting Eq. (2) into (1):

( ) )2cos()()(1)( 2211 tftmaktmakAts caac π++= (3) For s(t) to satisfy the principle of superposition, we require s(t) to be a linear combination of )(1 tm and

)(2 tm . From Eq. (3), we clearly see that the principle is violated because of the presence of an unmodulated carrier )2cos( tfA cc π . Problem 3.2 Consider the sinusoidal modulating signal

( )( ) cos 2m mm t A f tπ= Show that the use of DSB-SC modulation produces a pair of side frequencies, one at c mf f+ and the other at c mf f− , where cf is the carrier frequency. What is the condition that the modulator has to satisfy in order to make sure that the two side-frequencies do not overlap? Solution By definition

( )( ) ( )( )[ ]tfftffAA

tftfAAtftmAts

mcmcmc

cmmc

cc

−++=

==

ππ

πππ

2cos2cos21

)2cos()2cos()2cos()()(

(1)

Eq. (1) shows that the DSB-SC modulated signal consists of two side frequencies, one at mc ff + and the other at mc ff − . For the two side frequencies not to overlap, we require that the carrier frequency cf satisfies the condition

mcmc ffff −>+ or, equivalently, mc ff > . Problem 3.3 Consider a binary data stream m(t) in the form of a square wave with amplitudes ±1, centered on the origin. Determine the spectrum of the BPSK signal obtained by multiplying m(t) by a sinusoidal carrier whose frequency is ten times that of the fundamental frequency of the square wave. Solution The Fourier series expansion of the square wave )(tm is defined by

∑=

=,...5,3,1

)10

2cos()2/(

)2/sin()(k

c tkf

kktm ππ

π (1)

which represents a periodic waveform symmetric about the origin. The BPSK signal is defined by

)2cos()()( tfAtmts cc π= (2) where the carrier frequency cf is 10 times the fundamental frequency of )(tm ; that is k in Eq. (1). Setting

cA = 1 for convenience of presentation, Eq. (1) and (2) yield

( )

∑

∑

=

=

−+

+=

=

,...5,3,1

,...5,3,1

102cos

102cos

)2/()2/sin(

21

2cos10

2cos)2/(

)2/sin()(

k

cc

cc

ck

c

tkf

ftkf

fk

k

tftkf

kktm

πππ

π

πππ

π

which consists of side frequencies

±

10c

ckff with decreasing amplitude

)2/(2/sin(

ππ

kk for ,...5,3,1=k

Problem 3.4 (a) Starting with the RC spectrum P(f) of Eq. (3.17), evaluate the inverse Fourier transform of P(f) and thus show that

( ) ( )2 2 2

cos 2( ) sinc 2

1 16Wt

p t WtW t

πρρ

= −

(b) Determine P(f) and p(t) for the special case of ρ = 1, which is known as the full-cosine roll-off pulse.

Solution From Eq. (3.17),

( )( )

−≥

−<≤

−−+

<≤

=

1

11

1

20

212

cos141

021

)(

fWf

fWffWfWW

ffW

fP ρρ

π

Assuming zero phase, the inverse Fourier transform of )( fP is defined by

∫∞

∞−

= dfefPtP ftj π2)()( (1)

Since )( fP is an even function of f, we have

( )

( )( ) ( )

( ) ( )( ) ( )∫ ∫

∫ ∫

∫

∫

− −

−

∞

∞

∞−

−−+=

−−++=

=

=

1 1

1

1 1

1

2

0

2

0

20

2cos12

cos212cos5.1

2cos12

cos1412)2cos(

212

)2cos()(2

2cos)()(

fW fW

f

f fW

f

dfftWfWW

dfftW

dfftWfWW

dfftW

dfftfP

ftfPtp

πρρ

ππ

πρρ

ππ

π

π

Problem 3.5 (a) Starting with Eq. (3.20), derive the root RC pulse shape p(t) of Eq. (3.21).

(b) Evaluate p(t) at (i) t = 0, and (ii) 1

8t

Wρ= ± .

(c) Show that the p(t) derived in part (a) satisfies the orthogonality constraint described in Eq. (3.22)

Solution (a) The derivation of the root raised cosine pulse for 10 ≤≤ ρ follows a similar procedure described for

the ordinary raised cosine pulse. As checks on the formula of Eq. (3.21), we may consider the following two limiting cases:

(i) 0=ρ , for which Eq. (3.20) reduces to

otherwise0

021

)( WfWfP ≤≤=

Hence,

( )

( )WtWWt

WtW

dfeW

tpW

W

ftj

2sinc22

2sin2

21)( 2

=

=

= ∫−

ππ

π

which checks the formula of Eq. (3.21) for 0=ρ . (ii) For 1=ρ , we have

( )

otherwise 0

20for 4

cos21

=

≤≤

= Wff

WWfP π

Hence,

( )

( ) ( )

( )( )

( ) ( ) ( ) ( )( )( )

( )( ) ( )WtWtW

WtWtWtWtWtW

Wt

WtWtWtWtW

tW

tW

W

tW

tW

W

W

dftW

ftW

fW

dfftfWW

efWW

tp

W

W

W

W

W

ftj

ππ

πππ

π

ππππ

π

π

π

π

ππ

ππ

π π

4cos8124

814cos814cos8122

81

42

sin8142

sin8122

241

2412sin

241

2412sin

21

241cos2

41cos

21

2cos4

cos21

4cos

21)(

2

2

2

2

0

2

2

2

2

2

−=

−++−=

−

−++

+−

=

−

−

+

+

+

=

−+

+=

=

=

∫

∫

∫

−

−

which also checks with Eq. (3.21) for 1=ρ . (b) Evaluate p(t) at (i) t = 0, and (ii) t = 1/(8ρW). (i) Evaluating the root raised cosine pulse of Eq. (3.21) at t=0, we get

( )

+−=

πρρρ 4120 W

where the term ρ−1 follows from the sinc function ( )( )Wt

tWπ

ρπ2

12sin − evaluated at t=0.

(ii) We first note that the root raised cosine pulse is an even function of time t, which means that

)()( tptp =− .

Hence, we only need to evaluate p(t) at W

tρ81= . To do so, we apply L�Hopital�s rule,

obtaining

( )( ) ( )( ){ }

( )( ){ }

( ) ( )

−+

+=

+++

+−

−−−=

=

−

++−=

ρπ

πρπ

πρ

πρ

πρπρ

πρπρ

πρ

ρπρ

ρπρρπ

ρ

4cos21

4sin21

44sin1

44cos4

44cos1

2

81 ,

281

12cos812sin2

81

2

W

W

Wt

WtWtdtd

tWWttWdtd

WW

p

As a check on this result, we do the following evaluations: For 5.0=ρ , we have

58.02122

141

21 ≈

+=

πWp

W

For 1=ρ , we have

181

21 =

Wp

W

A cautionary note is in order. It is tempting to apply L�Hopital�s rule individually to the

additive terms that constitute the formula of Eq. (3.21). This is wrong. The proper way to proceed is, first of all, to put the two terms on a common denominator, and then apply L�Hopital�s rule to the numerator and denominator of the resulting composite expression. (Note: In the first printing of the book, there was a minor error in the answer. Specifically, the scaling factor should be W and not W2 .)

(c) To prove the orthogonality property of the root raised cosine pulse, it is best to do the proof in the frequency domain by invoking Parseval�s theorem in Fourier theory. Examining the frequency plots presented in Fig. 3.10 (a), we see that ( )fP is zero outside a frequency band the width of which depends on the roll-off factor ρ . Specifically,

(i) For 0=ρ , we have the regular raised cosine pulse, with the frequency band occupying the interval ( )WW ,− . The frequency shifting theorem teaches us that the Fourier transform of

( )nTtp − for integer n is the same as that of ( )tp except for a shift by an integer multiple of

WT

21 = . Accordingly, the frequency bands occupied by )(tp and )( nTtp − have no

overlaps, which means that )(tp and )( nTtp − are indeed orthogonal for 0=ρ .

(ii) The property also holds for 5.0=ρ , in which case the frequency band occupied by )(tp occupies the interval ( )WW 5.1,5.1− . Here again, we find that the frequency bands occupied by )(tp and )( nTtp − for integer n do not overlap.

(iii) A similar result holds for .1=ρ Based on these results, we may state that the orthogonality

property of ( )tp holds for all values of ρ . Problem 3.6 Construct a table summarizing the signal-space characterization of the QPSK signal constellation described in Fig. 3.14(c). Solution Examining Fig. 3.14(c): (i) For dibit 00, we see bE21 =φ , and 02 =φ .

(ii) For dibit 01, we see 01 =φ , and bE22 =φ .

(iii) For dibit 11, we see bE21 −=φ , and 02 =φ .

(iv) For dibit 10, we see 01 =φ , and bE22 −=φ . Problem 3.7 What conclusion can you draw from the signal-space diagram of Fig. 3.16 for BFSK, compared with that of Fig. 3.14(a) for BPSK? Solution The signal space diagram for BFSK is two-dimensional, as shown by

b

b

E

E

=

=

2

1

φ

φ

On the other hand, the signal-space diagram for BPSK is one dimensional, as shown by bE±=1φ Problem 3.8 How is BFSK extended to M-ary FSK? Solution The number of transmitted symbols in BFSK is two. Correspondingly, the signal-space diagram is two-dimensional. Generalizing this result to an arbitrary number of transmitted symbols, M, we find that the signal-space diagram of M-ary FSK is M-dimensional. Problem 3.9 Assume that the frequencies 1f and 2f are both odd integer multiplies of 1/4T. Then show that the MSK signal may be expressed in terms of the orthonormal pair of coordinates

( ) ( )12 cos cos 2 0

2 ct t f t t TT T

πφ π = ≤ ≤

and

( ) ( )22 sin sin 2 0

2 ct t f t t TT T

πφ π = ≤ ≤

In particular, show that

1 1 2 2( ) ( ) ( )s t s t s tφ φ= + where

( )( )

2

1 10

( ) ( )

cos 0

T

s s t t dt T t T

E

φ

θ

= − ≤ ≤

=

∫ (1)

and

( )( )2 2( ) ( ) 0 2

sin

T

T

s s t t dt t T

E T

φ

θ−

= ≤ ≤

=

∫ (2)

Hence, using the formulas of Eqs. (1) and (2), verify the entries for s1 and s2 in Table 3.3. Solution The MSK signal is defined by

( )( )

( )( )

+

+=

2 symbolfor 02cos2

1 symbolfor 02cos2

)(

2

1

θπ

θπ

tfTE

tfTE

tsb

b

Let T

nf4

11 = for symbol 1, and

Tnf4

22 = for symbol 0, where both 1n and 2n are odd integers, and 21 nn ≠ .

We may also express the MSK signal as

( )( ) ( ) ( )( ) ( )tftTE

tftTE

ts cb

cb πθπθ 2sinsin

22coscos

2)( −=

The in-phase component is

( )( )

( )( )

=

=

tTT

E

tTE

ts

b

bI

2cos0cos

2

cos2

)(

πθ

θ

For TtT ≤≤− , ( )0θ assumes the value 0 or π ; hence,

TtTtTT

Ets bI ≤≤−

±=

2cos

2)( π

Similarly, we may express the quadrature component as

( )( )

( )( ) ( )( )

TttTT

E

tT

TTTE

tTEts

b

b

bQ

20 2

sin2

2sinsinsin

2

sin2

)(

≤≤

±=

=

=

π

πθθ

θ

where ( )2πθ ±=T in the interval Tt 20 ≤≤ .

Therefore,

( )( ) TtTE

dtttsS

b

T

T

≤≤−=

= ∫−

0cos

)()( 11

θ

φ (3)

and

( )( ) TtTE

dtttsS

b

T

20 sin

)()(2

022

≤≤−=

= ∫θ

φ (4)

With 1S and 2S defined in Eq. (3) and (4), the results presented in Table 3.3 immediately follow. Problem 3.10 Using the results of Problem 3.9, construct the waveform of the MSK signal for the binary sequence 1101000, assuming that f1 = 5/(4T), and f2 = 3/(4T) by plotting the components s1φ1(t) and s2φ2(t).

Solution See Figure 3.19 on page 138 of the text book. Problem 3.11 Find the frequencies at which the baseband power spectrum of the MSK signal attains its minimum value of zero. Solution The baseband power spectrum of the MSK signal is defined by

( ) 2

222 1162cos32

)(

−

=fT

TfEtS b

Bπ

π

For T

T43= , we have

( )

( )1671

16916116

02

3cos2cos

22 −=

−=−

=

=

fT

Tf ππ

Hence, ( ) 0=fSB for T

T43= .

The baseband power spectrum ( )fSB again assumes a value of zero at T

f45= , in which case

( ) 025cos2cos =

= ππTf

Indeed, we may go on to say that ( ) 0=fSB at Tnf

4= for ,...7,5,3=n

Problem 3.12 Alternatively, we can derive the Wiener�Hopf equation (3.89) by differentiating the cost function J with respect to the weighting parameter vector a, setting the result equal to zero, and then solving for a. Show that this procedure also leads to Eq. (3.89). Solution From Eq.(3.87),

arraRaa ��� −−+= PJ (1)

where the parameter vector a is complex valued. Differentiation with respect to a complex value parameter requires special care. To simplify matters, we first cinsider the scalar case,

βα ja +=

in which case the correlation matrix R assumes a real value ρ , and the cross-correlation vector γ is a complex valued scalar. Correspondingly, the cost function simplifies to

aaaaPJ *** γγ −−+= ρ which is rewritten in the expanded form

( ) ( ) ( )γγγγ −−+−++= **22 βαβαρ jPJ Differentiating J with respect to α :

( ) ( )( )

* *2

2 2Re

J jρα βα

ρα

∂ = − + − −∂

= −

γ γ γ γ

γ

Therefore setting 0Jα

∂ =∂

and solving for α :

( )Re

optαρ

=γ

(2)

Next, differentiating J with respect to β :

( )( )

*2

2 2 Im

J jρββ

ρβ

∂ = − −∂

= −

γ γ

γ

Setting 0Jβ

∂ =∂

and solving for β :

( )Im

optβρ

=γ

(3)

Therefore, combining Eqs. (2) and (3):

( ) ( )( )1 Re Im

opt opt opta j

j

α β

ρ

ρ

= +

= +

=

γ γ

γ

which is the scalar version of Eq. (3.89) To generalize the solution for a matrix-valued scenario, let the thk element of parameter vector a be expressed as

k k ka jα β= + Correspondingly, define the derivative of cost function J with respect to ka as

12k k k

J j Ja α β

∂ ∂ ∂= − ∂ ∂

where the scaling factor ½ is introduced to satisfy the requirements

1kk

aa∂ =

∂

and

** 0k kk k

a aa a∂ ∂= =

∂ ∂

We may therefore go on to define

1 112

m m

j

j

α β

α β

∂ ∂ − ∂ ∂ ∂ =∂ ∂ ∂ − ∂ ∂

aM

and

1 1

*

12

m m

j

a

α β

α β

∂ ∂ + ∂ ∂ ∂ =∂ ∂ ∂ + ∂ ∂

M (4)

where m is the dimension of parameter vector a . In what follows, we adopt the definition of Eq. (4). On this basis, it is a straightforward matter to show that

( )

( )

( )

�*

�*

�*

∂ =∂∂ =

∂∂ =

∂

a γ 0a

γ a γa

a Ra Raa

Accordingly, applying the definition of Eq. (4) to the cost function of Eq. (1), we get

*

J∂ = −∂

Ra γa

Setting this result equal to zero and then solving for the optimum parameter vector:

1opt

−=a R γ Problem 3.13 If the transmission frequency is 1.9 GHz and the receiver uses a simple quartz crystal for regenerating the channel frequency, what is the range of residual frequency error on the baseband signal? A quartz crystal typically has an accuracy of 10 parts per million (ppm). If the data rate is 9.6 kilobits/s, how much phase rotation will occur during one symbol period with this error? Solution1 A relative error of 10 ppm on a 1.9 GHz carrier implies a frequency error of

( )( )kHz 19

109.11010 96

±=××±=∆ −f

With a symbol rate of 9.6 kilobits/s, the phase rotation over one symbol period is bounded by

1 The answer given in the first printing of the text is in error.

radians 4cycles 2

kilobits/s 6.9

19

π

φ

±=±=

±=∆

Problem 3.14 Show that frequency shifting and phase rotation do not affect the statistical properties of zero-mean white Gaussian noise. Solution Let { }nZ be a sequence of white complex Gaussian noise samples. Let nnn Zαψ = where nα is a complex scalar. Rewriting this in terms of real and imaginary

( )( )( ) ( )n

rnn

inn

inn

rn

nnin

rnn

yxjyxjyxj

ααααααψ

++−=

++=

Since the product of a scalar and a Gaussian random variable is another random variable and the sum of the two Gaussian random variables is Gaussian, the variable nψ is a complex Gaussian random variable. In phasor notation a frequency shift can be represented as

nnTj

n Ze ωψ = Consequently, nψ is complex Gaussian

[ ] [ ]0=

= nnTj

n Ze EE ωψ

and

[ ] [ ][ ]

)()(

0

0

)(*

kNkNeZZe

ZeZe

jwkTknn

jwkTkn

Tknjn

jwnTknn

δδ

ψψ ω

==

=

=

−

+−

++−

+

EEE

So nψ is also white with the same mean and variance as nZ . It can similarly be shown that for a phase rotation φje , the statistical properties of nψ are the same as nZ . Problem 3.15 What is the round-trip path delay for a 60-m path with CT-2? Does this amount of delay explain the amount of allocated guard intervals? Solution

For a 60-m path the round trip propagation delay is

dsmicrosecon 4.0103

)60)(2(8

=×

=

=− cdT tripround

The guard interval is 5.5 or 6.5 bits. The larger corresponds to a period of

rTguard

bits Guard=

where r is the transmission rate of 72 kHz. Therefore, the guard time is dsmicrosecon 83=guardT The extra guard time is for processing time and other timing errors with the terminals. Problem 3.16 The maximum frequency error allowed in the CT-2 standard is 10 kHz or less. What is the relative frequency error, expressed in parts per million (ppm)? Solution A 10 kHz error with a minimum transmission frequency of 864.15 MHz implies a relative frequency error of

ppm 6.11

101015.864

1010

10

66

3

6

=

××

×=

×∆=∆T

r fff

Problem 3.17Figure 3.37(a) presents the block diagram of a passband digital phase modulator, which lends itself to VLSIimplementation (Steele and Hanzo, 1999). The premodulation filter of impulse response h(t) is designed to produce adata-dependent phase signal θ(t), which addresses two read-only-memory (ROM) units to yield values of thetrigonometric terms cos(θ(t)) and sin(θ(t)). The resulting digital signals are converted into analog form.(a) In effect, the baseband model of Fig. 3.10(a) is being extended to deal with nonlinear phase modulation. This

extension is however subject to the assumption that the essentially highest frequency component of both cos(θ(t))and sin(θ(t)) is less than the carrier frequency fc. Justify this assumption.

(b) Under this assumption, show that the radiated output of Fig. 3.37(a) can be formulated as the phase-modulatedsignal

where A is a constant amplitude.

(c) The premodulation filter can be implemented in the form of a tapped-delay-line filter, as in Fig. 3.37(b). Justifythis method of implementation.

Solution(a) The signal θ(t) at the premodulation filter output is the convolution of the input data stream

{bk(t)} and the filter’s impulse response h(t):

(1)

In general, θ(t) may occupy an infinitely-wide frequency range. For s(t) to be uniquely definedwith respect to θ(t), the product terms cos(θ(t))cos(2πfct) and sin(θ(t))sin(2πfct) must not

s t( ) A 2π f ct θ t( )+( )cos=

Figure 3.37: Block diagrams for Problem 3.17

.

COS

SIN DAC

DAC

Σ

Inputbinary datastream{bk} Premodulation

filterh(t)

θ(t)Phase-modulatedsignal s(t)

Acos(2πfct)

Asin(2πfct)

Read-onlymemories

Digital-to- analogconverters

(a)

Σ Σ Σ Σ

. . .. . . .

. . .

T T T{bk}

h0 h1 h2 hK-2 hK-1

θ(t)

(b)

θ t( ) bk τ( )h t τ–( ) τd∞–

∞∫=

exhibit spectral overlaps. In other words, the highest significant frequency component of bothcos(θ(t)) and sin(θ(t)) must be less than the carrier frequency fc. Clearly, the meaning of“significant” frequency is a matter of how to define the bandwidth of cos(θ(t)) and sin(θ(t));the larger we demand the bandwidth to be, the higher will the significant frequencycomponent be.

(b) Summing two channel outputs of Figure 3.37(a) with a plus sign used for the lower channeland a minus sign for the upper channel, we may write

(c) According to Figure 3.37(b):

(2)

Assume that h(t) is causal, which is a necessary requirement for real-time operation, that is,h(t) = 0 for t < 0. Then Eq. (1) may be written as

(Convolution is commutative)

Let τ = kT, where T is the sampling interval and k is an integer. Assuming that T is sufficientlysmall to satisfy the sampling theorem applied to h(τ) and setting dτ = T, we have

(3)

Define,

If we further assume that hk is effectively zero for t > KT, that is, k > K-1, Eq. (3) reduces to(2).

s t( ) A θ t( )( ) 2π f ct( ) A θ t( )( ) 2π f ct( )sinsin–coscos=

A 2π f ct θ t( )+( )cos=

θ t( ) hkbk t kT–( )k=0

K -1

∑=

θ t( ) h τ( )bk t τ–( ) τd0

∞∫=

h τ( )bk t τ–( ) τd0

∞∫=

θ t( ) h kT( )bk t kT–( )Tk=0

∞

∑=

hk Th kT( )= k 0 1 … ∞, , ,=

Problem 3.18(a) Construct and label the constellations of M-ary PSK for (i) M = 8, and (ii) M = 16.(b) Discuss the differences that distinguish 16-PSK considered in part (a) and 16-QAM described in Fig. 3.13, doing

so in the context of information transmission over a wireless channel.

Solution(a) For M = 8, the 8-PSK constellation is

For M = 16, the 16-PSK constellation is

(b) For 16-QAM, the constellation is

16-QAM is a hybrid form of amplitude and phase modulations. Hence, for it to work properlyon a wireless channel, the channel has to be linear; otherwise, the received signal will bedistorted, which is undesirable. On the other hand, 16-PSK is a special case of phase

..

.. .

.

.0

45o

Real

Imaginary

45o

.

Real22.5

o

22.5o

. Imaginary

22.5

o

...... . . . . .

.

....

22.5o

Imaginary

Real.. ..

.....

. .....

.0

modulation. Therefore, so long as the wireless channel is nonlinear with no AM to PMconversion, the 16-PSK, in theory, is impervious to nonlinear distortion.

Problem 3.19Quadriphase-shift keying and minimum-shift keying provide two spectrally efficient methods for the digitaltransmission of binary data over a wireless channel. List the advantages and disadvantages of these two methods ofdigital modulation.

Solution

Problem 3.20The π/4-shifted DQPSK is characterized by two combined features:• the use of 8 carrier-phase states, and• the transmission of an information-bearing signal in the differential carrier phase.Specifically, the differential carrier phase ∆θk is governed by the mapping

where {bk} denotes the incoming data stream.

QPSK MSK

Computationalcomplexity

QPSK is an example of linearmodulation; hence, it offers thepractical advantage of simplicity inimplementation in both the trans-mitter and receiver

MSK is a continuous-phasefrequency modulation, whichmakes it to be a special form ofnonlinear modulation. Accord-ingly, it is more complex thanQPSK in the implementation ofboth the transmitter and receiver.

Spectral occupancy For f >> 1/Tb, where Tb is the bitduration, the baseband powerspectral density of QPSK falls offas the inverse square of frequency

For f >> 1/Tb, the baseband powerspectral density of MSK falls offas the inverse fourth power offrequency. Accordingly, MSKdoes not produce as much interfer-ence outside the signal band ofinterest as QPSK. This is a highlydesirable characteristic of MSK,especially when the requirement isto operate with a bandwidth limita-tion. This desirable characteristicis furether enhanced through theuse of a Gaussian prefilter, result-ing in the so-called GMSK.

∆θk

3π 4⁄– for bk=-3

π 4⁄– for bk=-1

+π 4⁄ for bk=+1

+3π 4⁄ for bk=+3

=

Formulate the expression or the complex envelope of the π/4-shifted DQPSK signal.

Solution

The DQPSK signal is defined by

where E is the signal energy per symbol, and T is the symbol period. Hence, the complex envelopeof s(t) is

Problem 3.21Continuing with the π/4-shifted DQPSK described in Problem 3.20, suppose that the incoming pulse amplitude isshaped in accordance with square root raised cosine spectrum discussed in Section 3.4.1.(a) Using simulation, compute the phase trajectory of the π/4-shifted DQPSK signal by plotting its quadrature

component versus the in-phase component for an incoming random quaternary sequence.(b) Demonstrate that the phase trajectory does not pass through the origin. What are the practical implications of this

property in the context of information transmission over a wireless channel?

π4----shifted

s t( )

2ET

------- 2π f ct 3π4

------– cos for bk 3–=

2ET

------- 2π f ct π4---–

cos for bk 1–=

2ET

------- 2π f ct π4---+

cos for bk +1=

2ET

------- 2π f ct 3π4

------+ cos for bk +3=

=

s̃ t( )

2ET

------- j3π4

---------– exp for bk 3–=

2ET

------- jπ4

------– exp for bk 1–=

2ET

-------jπ4

------ exp for bk +1=

2ET

-------j3π4

--------- exp for bk +3=

=

Solution

Problem 3.22In this problem, we explore the effect of a multipath channel on the waveform of three digitally modulated signals.The multipath channel is represented by the simple tapped-delay model of Fig. 3.38, where T denotes the symbolduration. Values of the tap-weights are

The incoming binary sequence is ...0000110101110101...

(a) Plot the waveforms of the modulated signals at the model input, using the following methods:(i) QPSK(ii) OPSK(iii) π/4-shifted QPSK

(b) For each of these methods, plot the waveform produced at the model output.(c) What conclusions can you draw from the results of parts (a) and (b)?

−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

1Phase Trajectory

w0 0.25=

w1 0.25=

w2 0.25=

Figure 3.38: Tapped-delay-line model of multipath channel

Σ Σ

T T. .Input

w0 w1 w2

Output

Solution

0 T 2T 3T 4T−1.5

−1

−0.5

0

0.5

1

1.5Undistorted QPSK Modulation Sequence: −1 1 1 −1 1 −1 −1 −1

0 T 2T 3T 4T 5T 6T−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8Distorted QPSK Modulation Sequence: −1 1 1 −1 1 −1 −1 −1

0 T 2T 3T 4T−1.5

−1

−0.5

0

0.5

1

1.5Undistorted OQPSK Modulation Sequence: −1 1 1 −1 1 −1 −1 −1

0 T 2T 3T 4T 5T 6T−2

−1.5

−1

−0.5

0

0.5

1

1.5

2Distorted OQPSK Modulation Sequence: −1 1 1 −1 1 −1 −1 −1

Problem 3.23Repeat Problem 3.22, this time using the more difficult set of tap-weights:

Why is this set of tap-weights more difficult to deal with than those of Problem 3.22?

0 T 2T 3T 4T−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1pi/4 QPSK Modulation Sequence: −1 1 1 −1 1 −1 −1 −1

0 T 2T 3T 4T 5T 6T−1.5

−1

−0.5

0

0.5

1

1.5Distorted pi/4 QPSK Modulation Sequence: −1 1 1 −1 1 −1 −1 −1

w0 0.5=

w1 0.5=

w2 0.5=

Solution

0 T 2T 3T 4T−1.5

−1

−0.5

0

0.5

1

1.5Undistorted QPSK Modulation Sequence: −1 1 1 −1 1 −1 −1 −1

0 T 2T 3T 4T 5T 6T−2

−1.5

−1

−0.5

0

0.5

1

1.5

2Distorted QPSK Modulation Sequence: −1 1 1 −1 1 −1 −1 −1

0 T 2T 3T 4T−1.5

−1

−0.5

0

0.5

1

1.5Undistorted OQPSK Modulation Sequence: −1 1 1 −1 1 −1 −1 −1

0 T 2T 3T 4T 5T 6T−2

−1.5

−1

−0.5

0

0.5

1

1.5

2Distorted OQPSK Modulation Sequence: −1 1 1 −1 1 −1 −1 −1

Comparing the waveforms plotted in Figure 2 this solution with the corresponding figure in thesolution to Problem 3.23, we see that the second channel model has a deeper null and therefore amore deleterious effect on the output waveform than the first channel model.

0 T 2T 3T 4T−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1pi/4 QPSK Modulation Sequence: −1 1 1 −1 1 −1 −1 −1

0 T 2T 3T 4T 5T 6T−1.5

−1

−0.5

0

0.5

1

1.5Distorted pi/4 QPSK Modulation Sequence: −1 1 1 −1 1 −1 −1 −1

Problem 3.24Equations (3.58) and (3.59) define the two coordinates for signal-space analysis of MSK. Given these twoorthonormal coordinates, depict the signal-space representation of the MSK signal.

Solution

Problem 3.25Equation (3.67) defines the impulse response of the pulse-shaping filter h(t) used to generate a GMSK signal.(a) Show that the time function h(t) satisfies all the properties of a probability density function.(b) Expanding on the interpretation of h(t) as a probability density function, show that the variance of the

distribution is proportional to (T/W)2. What is the significance of this interpretation?

Solution(a) From Eq. (3.67):

, T = symbol duration

Let

that is,

RegionZ2

RegionZ4

RegionZ1

RegionZ3

Message point m2: Symbol 1[θ(0) = π,θ(Tb) = -π/2]

Message point m1: Symbol 0[θ(0) = 0,θ(Tb) = -π/2]

Message point m3: Symbol 0[θ(0) = π,θ(Tb) = π/21]

Message point m2: Symbol 1[θ(0) = π,θ(Tb) = π/21]

Decisionboundary

Decisionboundary

φ2

φ1

Eb

Eb

Eb

Eb-

-

. .

..

Signal-space diagram for MSK system

h t( ) 2π2elog

-------------= W2π2

2elog-------------W

2t2

–

exp

2π2elog

-------------WT( )T

-------------- 2π2

2elog------------- WT( )2 t

T---

2–

exp=

2σ2 2elog

2π2WT( )2

--------------------------= 2πσ2elog

2π WT( )-------------------------=

and Th(Tx) = f(x)

We may then rewrite h(t) as the new function

which represents the probability density function of a Gaussian distributed random variablewith zero mean and variance σ2. As such, the total area under h(t) or that under f(x) is unity.That is

(b) The variance of f(x) and therefore that of h(t) is inversely proportional to (WT)2. Thus byreducing the time-bandwidth product WT, the variance is increased, that is, two things happen:(i) The smoothness of the filtered “frequency-pulse” is increased.(ii) The bandwidth occupancy of the GMSK signal is decreased.Point (ii) is precisely what we see in Fig. 3.21, p.142 of the book.

Problem 3.26The tamed frequency modulation (TFM), due to deJager and Dekker (1978), is designed to provide a frequency-modulated signal whose power spectrum is compact without sidelobes. This desirable spectral characteristic isachieved by careful control of the phase transitions of the frequency-modulated signal. Figure 3.39 depicts the TFMmodulator that consists of a premodulation filter feeding a voltage-controlled oscillator. The premodulation filteritself consists of the tapped-delay-line filter cascaded with a low-pass filter whose impulse response is the inverseFourier transform of the example transfer function

where T is the symbol duration.(a) Show that the overall transfer function of the corresponding premodulation filter is given by

tT--- x=

f x( ) 1

2πσ-------------- x

2

2σ2---------–

exp=

h t( ) td∞–

∞∫ f x( ) xd

∞–

∞∫ 1= =

H 0 f( ) πfT( ) πfT( )sin⁄ for 0 f≤ 1 2T⁄≤0 otherwise

=

H f( )πfT( )

πfT( )sin---------------------- πfT( )2

cos for 0 f≤ 1 2T⁄≤

0 otherwise

=

(b) Using computer simulation, plot the overall impulse response of the premodulation filter, denoted by h(t).(c) The filter defined in part (b) is noncausal. Propose the use of a delay that would make the filter causal for all

practical purposes.(d) How does the impulse response of the premodulation filter computed in part (b) compare with that of the

premodulation filter used in the GMSK modulator?(e) Discuss the practical benefit that could be gained by using TFM for a FDMA system.

Solution(a) The overall transfer function of the premodulation filter is

(1)

Using the relation cos2θ = 2cos2θ-1, we may rewrite Eq. (1) as

(b) Recognizing that, by definition, the transfer function of a linear filter is the Fourier transformof the filter’s impulse response, we may use a computer to calculate the inverse transform ofthe H(f) in part (a). The result of the computation is plotted below:

Figure 3.39: Block diagram of TFM modulator

Σ

T

-T

.

Premodulation filter: Transfer function H(f)

1/4

1/2

1/4

H0(f)Voltage-controlledoscillator

TFMsignal

Inputbinarysequence

H f( ) 14---e

j2πfT– 12---

14---e

j2πfT+ +

H 0 f( )=

12--- 2πfT( ) 1

2---+cos

H 0 f( )=

H f( ) πfT( )H 0 f( )2cos=

πfT( )πfT( )sin

---------------------- πfT( )2cos for 0 f

12T-------≤ ≤

0 otherwise

=

(c) From the plot of part (b), we may infer that a delay of 2.5T would make the premodulationfilter effectively causal.

(d) Comparing the impulse response plotted in part (b) with that of Figure 3.20 of the textbook,p.141, for the GMSK, we may make the following observations:(i) At a coarse level, the two impulse responses are similar.(ii) At a very fine level, we see that the impulse response of the pulse-shaping filter for the

GMSK is always nonnegative, whereas the impulse response of the pulse-shaping filter forthe tamed-frequency modulation assumes negative values, albeit somewhat small, for thetwo time intervals 0 < (t/T) < 0.5 and 4.5 < (t/T) < 5.

(e) At frequencies +1/T away from the carrier frequency fc, the power spectrum of a tamed-frequency modulated (TFM) signal drops by about 60 dB below at fc. Thus, by using TFM asa basis for FDMA, the practical benefit would be that of making the leakage from one user toanother insignificantly small.

Problem 3.27An FDMA system using frequency modulation accommodates a total of N = 100 mobile users assigned to a particularcell. The largest frequency component of the speech signal is W = 3.4 kHz. Using Carson’s rule, determine thebandwidth of the uplink and downlink of the system for each of the following frequency deviations:(a) D = 1(b) D = 2

0 0.1 0.2 0.3 0.4 0.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Frequency

Am

plitu

deSpectrum

H(f) = (πfT)/sin(πfT) cos(πfT)2

0 5 10 150

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Time

Am

plitu

de

Impulse Response

IFFT( H(f) )

(c) D = 3

Solution(a) According to Carson’s rule, the bandwidth of the FM signal is

Also, by definition,

Therefore with W = 3.4 kHz and D = 1, we have

and

The bandwidth of the uplink and downlink isN = number of users

(b) For D = 2, we have

(c) For D = 3, we have

Problem 3.28The use of frequency hopping makes it possible for the carrier frequency to hop randomly from one frequency toanother. Discuss how the use of frequency hopping can improve the performance of an FDMA system operating in awireless communication environment.

SolutionThe primary motivation of frequency hopping is security. More specifically, two things are done,one at the transmitter and the other at the receiver:

• hopping the carrier frequency of a frequency-modulated signal in accordance with a pseudo-noise generator in the transmitter, and

• using an identical version of the pseudo-noise generator in the receiver to undo the frequency-hopping.

BT 2∆f 1 1D----+

=

D ∆fW------=

∆f DW 3.4 kHz= =

BT 2 3.4 1 1+( )× 13.6 kHz= =

Btotal N BT×=

100 13.6 kHz = 1.36 MHz×=

∆f 2 3.4× 6.8 kHz= =

BT 2 6.8 1 12---+

× 20.4 kHz= =

Btotal 100 20.4 kHz× 2.04 MHz= =

∆f 3 3.4× 10.2 kHz= =

BT 2 10.2 1 13---+

× 27.2 kHz= =

Btotal 100 27.2 kHz× 2.72 MHz= =

Accordingly, a signal produced by a jammer lying in the frequency band of the modulated signalwill undergo frequency hopping of its own in the receiver. The net effect is a securecommunication system with respect to jammers trying to interfere with the operation of thecommunication system.

Problem 3.29As a measure of the adjacent channel interference problem illustrated in Fig. 3.24, consider the index of performance

where PSL is the power spilling over into a channel of interest due to the sidelobes of an adjacent subchannel, andPML is the power produced in that channel due to its own main lobe.(a) Using Eq. (3.67), derive a formula for ACI.(b) Calculate the index of performance for the example illustrated in Fig. 3.24.

Solution(a) We are given the energy spectral density

The main lobe occupies the band -(1/T) < f < (1/T). The adjacent band on the left lies insidethe sidelobe -(3/T) < f < -(1/T). The adjacent band on the right lies inside the sidelobe (1/T) < f< (3/T). The index of performance is

(b) For the example illustrated in Figure 3.24, T = 1; hence

Numerical integration of the definition intergrals in the numerator and denominator yields

PSL (Spillage power) = 0.0318PML (Main lobe power) = 0.9028ACI = PSL/PML = 0.0352

Problem 3.30A general formula for assessing the adjacent-channel interference problem is

ACIPSL

PML-----------=

P f( ) 2 c2fT( )sin=

ACIc fT( )2

sin fd3 T⁄–

1 T⁄–

∫fT( )2

sin fd1 T⁄–

1 T⁄∫

----------------------------------------------fT( )2

sin fd1 T⁄

3 T⁄∫

fT( )2sin fd

1 T⁄–

1 T⁄∫-------------------------------------------= =

ACIf( )2

sin fd1

3

∫c f( )2

sin fd1–

1

∫-------------------------------------=

ACI δf( )G f( ) H f δf–( ) 2

fd∞–

∞∫

G f( ) H f( ) 2fd

∞–

∞∫

--------------------------------------------------------------=

where G(f) is the power spectral density of the input signal, H(f) is the frequency response of the band-pass filter usedto separate adjacent channels, and δf is the frequency separation between the two channels. Justify the validity of thisformula.

SolutionFirst of all, given a filter with transfer function H(f) and an input signal with power spectraldensity G(f), the power spectral density of the output signal, denoted by Gout(f), is defined by

With the filter assumed to be of the band-pass variety, it is logical to center the passband of thefilter on the carrier frequency pertaining to the channel of interest. Typically, the filter is selectiveenough for the mainlobe of G(f) to lie inside the pass band of H(f). Hence, the integral

defines the average power output of the filter due to the mainlobe of G(f).Next, let δf denote the separation between two adjacent channels. Hence, the frequency-

translated transfer function H(f - δf) would be centered on the adjacent channel to the right of thechannel of interest if δ(f) is positive and to the left of the channel of interest if δ(f) is negative. Ineither case, assuming that |H(f)| is symmetrical about the midband frequency, we find that

defines the average power output of the filter due to either of the two sidelobes of G(f) adjacent tothe mainlobe.

Accordingly, the index of performance

defines the adjacent channel interference as stated in the problem.

Problem 3.31One way of linearizing a nonlinear power amplifier is to predistort the input signal. In effect, the cascade connectionof two nonlinear components behaves like a linear memoryless system. Discuss the rationale of how such a schemecan be implemented.

SolutionConsider the cascaded system:

Based on this figure, we may formulate the following relationships:

Gout f( ) H f( ) 2G f( )=

H f( ) 2G f( ) fd

∞–

∞∫

H f δf–( ) 2G f( ) fd

∞–

∞∫

ACI δf( )H f δf–( ) 2

G f( ) fd∞–

∞∫

H f( ) 2G f( ) fd

∞–

∞∫

-------------------------------------------------------------=

Modulatedsignals(t)

Outputsignaly(t)

Linearizingdevice

Nonlinearpoweramplifier

A( ). L( ).

x t( ) A s t( )( )=

y t( ) L x t( )( )=

(1)Ideally, we would like to design the linearizing device to produce an output that is a scaled versionof the original modulation signal s(t), as shown by

(2)where k is a constant. From Eqs. (1) and (2):

(3)To satisfy this relationship, we require the condition

(4)

where the input-output mapping is the inverse of the power amplifier’s input-outputmapping . In other words,

which follows from the identity

The conclusion to be drawn from Eq. (4) is summed up as follows:(i) The linearizing device is itself nonlinear.(ii) The input-output mapping of the linear device is the inverse of the input-output mapping of

the power amplifier, except for a scaling factor.

To illustrate what we mean by this characterization of the linearizing device, suppose the input-output characteristic of the power amplifier is defined by the quadratic equation:

(5)

where a1 and a2 are constants. Correspondingly, let the input-output characteristic of thelinearizing device be defined by

(6)

Substituting Eq. (6) into (5):

From this equation, it is obvious that with the characteristics assumed in Eqs. (5) and (6), perfectlinearization is not feasible.

The best we can do is some form of approximation, as shown by:(7a)

(7b)

(7c)

(7d)

How do we satisfy these relationships? Here is one possibility:

L A s t( )( )( )=

y t( ) ks t( )=

ks t( ) L A s t( )( )( )=

L .( ) k A1– .( )=

A1– .( )

A .( )

L A s t( )( )( ) k A1–

A s t( )( )( )=

ks t( )=

A1–

A .( )( ) 1=

x t( ) a1s t( ) a2s2

t( )+=

y t( ) b1x t( ) b2x2

t( )+=

y t( ) b1 a1s t( ) a2s2

t( )+( ) b2 a1s t( ) a2s2

t( )+( )2

+=

b1a1s t( ) b1a2 b2a12

+( )s2

t( ) 2b2a2s3

t( ) b2a22s

4t( )+ + +=

b1a1 k=

b1a2 b2a12

+ 0=

2b2a2 0≈

b2a22 0≈

1. The magnitude of constant a2 is small (i.e., the second-order nonlinearity characterizing thepower amplifier is of a mild sort).

2. The magnitude of constant b2 is small too (i.e., the second-order nonlinearity characterisingthe linearizing device is also of a mild sort).

3. With both a2 and b2 having small magnitudes, the product terms 2b2a2 and can assume

very small values, both being close enough to zero to satisfy Eqs. (7a) and (7b).4. To satisfy Eq. (7c):

which is satisfied by choosing

Problem 3.32Plot the loss in dB as a function of residual phase error using BPSK modulation. For typical modem implementations,there is an allocated implementation margin that may range from 0.5dB to 2dB, depending upon the application. Theimplementation margin includes all losses due to nonideal implementation of the modem. If the portion of theimplementation margin allocated for phase errors is 0.25 dB, what is the maximum phase error allowed if the target

BER is 10-5?

SolutionWith BPSK modulation the error rate in additive white Gaussian noise from Table 3.4 is

Considering Fig. 3.14 if there was a residual phase error θ in the recovered amplitude theprojection onto the real axis would be

instead of for data bits bk = +1. Consequently the degradation in performance is

This is illustrated below. From the above analysis it is clear that the degradation does not dependon error rate. A degradation of less than 0.25 dB implies a phase error of less than 13 degrees.

b2a22

ka1-----

a2 b2a12

+ 0=

b2

ka2

a13

--------=

Pe12---erfc

Eb

N 0-------

=

Eb θcos±

Eb±

10Eb θ2

cos

Eb---------------------

10log 20 θcos10log=

Problem 3.33The statement “The residual frequency error is small compared with the signal bandwidth” implies that there is

minimal phase rotation over between two successive symbols. If the maximum phase rotation permitted is 10o, whatis the maximum frequency error that would be permitted as a fraction of the symbol rate. What does this imply aboutthe accuracy of the local oscillator for down-converting the received signal?

SolutionLet fs be the symbol rate. A phase rotation of 10˚ over one symbol period corresponds to 10/360cycles of frequency error over a period 1/fs. Consequently the maximum frequency error is

or 3% of the symbol rate at most. The local oscillator must introduce an error that is no more than3% of the symbol rate. For example, if the transmission frequency is 800 MHz and the symbolrate is 10 kHz, then the maximum frequency error is 300 Hz. This is an error of less than 0.5 ppmof the transmission frequency. Fortunately, there are receiver techniques for alleviating thisrequirement on the frequency accuracy.

Problem 3.34Show that pilot symbols can be used to track both fading and small residual frequency errors in the receiver. What arethe constraints on this residual frequency error?

SolutionTo model both fading and frequency errors in the received signal, we combine Eq. (3.74) and(3.78) to obtain

f e10 360⁄

1 f s⁄-------------------≤ 0.03 f s=

x̃ t( ) α̃ t( )ej 2π∆ft+θ( )

m t( ) w̃ t( )+=

If the change due to fading and frequency is small over a symbol period, T, then, analogous to Eq.(3.79), we obtain at the output of the matched filter

and, for known pilot symbols, Eq. (3.80) becomes

Thus hKi is a noisy estimate of the fading and frequency error. This estimate can be improved in asimilar manner to that described following Eq. (3.80).

With this method of tracking the phase, the frequency error must be such that the phase error

between pilot symbols, given by plus the rotation due to fading is small; at least lessthan π radians to prevent any ambiguity.

Problem 3.35The last column of Table 3.4 lists the exact formulas for the bit error rates of different digital modulation schemesoperating over a slow Rayleigh fading channel. The parameter γ0 denotes the mean value of the received signal-energy-to-noise spectral density ratio.(a) Derive these exact formulas.(b) Assuming that γ0 is large compared with unity, find the approximate forms of these formulas.

Solution(a) With slow fading, at any particular instant in time, the channel appears to be a white Gaussian

noise channel with instantaneous signal-to-noise ratio

(1)

where α is the fading gain. The corresponding instantaneous error rate is , whereis the error rate in additive white Gaussian noise with no fading. For example, for coherentdetection of BPSK

(2)

Since α has a Rayleigh distribution, α2 has a Chi-squared distribution with 2 degrees offreedom (See Appendix C.5). Since Eb/N0 is constant, also has a Chi-square distributiongiven by

(3)

where is the average value of Eq.(1) (see Eq.(3.91)). If we average Eq.(2) over all values of, then the error rate in slow Rayleigh fading is

(4)

If we evaluate Eq.(4) for the case of coherent BPSK, using integration by parts we obtain

(5)

yk αkej 2π∆fT +θ( )

bk wk+=

hKi αKiej 2π∆KiT +θ( )

wKi+=

ej 2π∆fKT( )

γEb

N 0-------α2

=

Pe γ( ) Pe γ( )

Pe γ( ) erfc γ( )=

γ

p γ( ) 1γ0-----e

γ γ0⁄–= γ 0≥

γ0

γ

P γ0( ) Pe γ( ) p γ( ) γd0

∞∫=

P γ0( ) eγ γ0⁄– erfc γ( )

2---------------------

0

∞–

1

π------- e

γ γ0⁄–

0

∞∫–=

eγ–

γ-------dγ

where the formula for erfc(x) may be found in Appendix E. Evaluating the first term of Eq.(4)

and substituting in the second term, we obtain

(6)

If we recognize the second term as the Gaussian integral

(7)

and identifying 2σ2 = 1/(1+1/γ0), we have

(8)

(b) If γ0 >> 1, then

(9)

where the third line comes from the approximation for x<<1, and the fourth

line uses the approximation for large γ0. The results for other modulations and

detection strategies follow similarly.

Problem 3.36A digital communication system uses MSK for information transmission. The requirement is to do the transmission

with a bit error rate that must not exceed 10-4. Calculate the minimum signal-to-noise ratio needed to meet thisrequirement for the following two scenarios:(a) Additive white Gaussian noise channel.(b) Rayleigh fading channel.

Solution(a) From Fig 3.32, we obtain the error rate with MSK is 10-4 in additive white Gaussian noise

when the SNR is approximately 8.3 dB.

s2 γ=

P γ0( ) 12---

1

π------- e

s21+1 γ0⁄– eγ–

γ-------dγ

0

∞∫–=

1

2πσ-------------- e

s2 2σ2⁄–sd

0

∞∫ 1

2---=

P γ0( ) 12---

1

π------- 1

2--- 2π

γ0

2 1 γ0+( )----------------------

–=

12--- 1

γ0

1 γ0+--------------–

=

P γ0( ) 12--- 1

γ0

1 γ0+--------------–

=

12--- 1 1 1

1 γ0+--------------––

=

12--- 1 1 1

2 1 γ0+( )----------------------––

≈

14γ0--------≈

1 x– 1 x 2⁄–≈1 γ0+( ) γ0≈

(b) From Table 3.4, we obtain the error rate with MSK is 10-4 in Rayleigh fading when

solving for γ0 we obtain that an SNR of 34 dB is required.

Problem 3.37In this problem, we address the issue of evaluating the power spectrum of an OFDM signal. From the discussionpresented in Section 3.13, we may treat the OFDM signal as a modulated set of orthogonal subcarriers whosefrequencies are separated by the reciprocal of the symbol duration τ. Consider an incoming signal constellationcharacterized by two features:• zero mean, and• amplitude-shaping pulse p(f).(a) Derive the expression for the power spectrum of the complex envelope of the OFDM signal.(b) Plot the power spectrum derived in part (a) for the following specifications:

(i) Number of subcarriers, N = 16(ii) Pulse-amplitude shaping pulse in the form of a rectangular function of time t.

(c) Repeat the power spectrum computation of part (b) for N = 48.

Solution(a) From Eq.(3.95) and (3.96), the time domain representation of the OFDM signal in the kth

symbol period is

(1)

where we have relabeled the pulse shape p(t) instead of g(t). The Fourier Transform ofindividual OFDM symbol is therefore

(2)

The spectrum of an individual OFDM symbol is the magnitude squared of Eq.(2). For a longertime interval, if the data are + and independently distributed, then the spectrum can beapproximated by

(3)

(b) For a rectangular pulse of duration T, P(f) = Tsinc(fT). Substituting this in Eq. (3), spectra for

N=16 carriers is given by where fi = i/T. This is plotted below.

10 4– 12--- 1

γ0

1 γ0+--------------–

=

s̃ t( ) bk i, p t kT–( ) j2π f it( )expi=- N 2+1⁄–

N

∑=

S f( ) bk i, P f f i–( )i=- N 2+1⁄–

N 2⁄

∑=

S f( ) 2P f f i–( ) 2

i=- N 2+1⁄–

N 2⁄

∑=

S f( ) 2T c2

fT 1–( )sini=-7

8

∑=

(c) The power spectrum in this case is shown in the following figure.