# fileChapter 3 Modulation and Frequency-division Multiple Access Problem 3.1 Show that amplitude...

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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 tftfAA

tftmAts

mcmcmc

cmmc

cc

−++=

= =

ππ

ππ π

2cos2cos 2 1

)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 t kf

k ktm π π

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

10 2cos

10 2cos

)2/( )2/sin(

2 1

2cos 10

2cos )2/(

)2/sin()(

k

c c

c c

c k

c

t kf

ft kf

f k

k

tft kf

k ktm

ππ π

π

ππ π

π

which consists of side frequencies

±

10 c

c kff with decreasing amplitude

)2/( 2/sin(

π π

k k 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 16 Wt

p t Wt W t

πρ ρ

= −

(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),

( )( )

−≥

−

otherwise0

0 2 1

)( WfWfP ≤≤=

Hence,

( )

( )WtW Wt

WtW

dfe W

tp W

W

ftj

2sinc2 2

2sin2

2 1)( 2

=

=

= ∫ −

π π

π

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

( )

otherwise 0

20for 4

cos 2 1

=

≤≤

= Wff

WW fP π

Hence,

( )

( ) ( )

( )( )

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

( )( ) ( )WtWt W

Wt WtWtWtWtW

Wt

WtWtWtWt W

t W

t W

W

t W

t W

W

W

dft W

ft W

f W

dfftf WW

ef WW

tp

W

W

W

W

W

ftj

π π

π ππ

π

ππππ

π

π

π

π

ππ

ππ

π π

4cos 81 24

81 4cos814cos8122

81

4 2

sin814 2

sin81 22

2 4 1

2 4 12sin

2 4 1

2 4 12sin

2 1

2 4 1cos2

4 1cos

2 1

2cos 4

cos 2 1

4 cos

2 1)(

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

( )

+−=

π ρρρ 4120 W

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

tW π

ρπ 2

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

t ρ8 1= . To do so, we apply LHopitals rule,

obtaining

( )( ) ( )( ){ }

( )( ){ }

( ) ( )

−+

+=

+++

+−

−−−=

=

−

++− =

ρ π

πρ π

π ρ

π ρ

πρπ ρ

πρπ ρ

πρ

ρπρ

ρπρρπ

ρ

4 cos21

4 sin21

44 sin1

44 cos4

44 cos1

2

8 1 ,

281

12cos812sin 2

8 1

2

W

W

W t

WtWt dt d

tWWttW dt d

W W

p

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

58.021 22

1 4 1

2 1 ≈

+=

πW p

W

For 1=ρ , we have

1 8 1

2 1 =

W p

W

A cautionary note is in order. It is tempting to apply LHopitals 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 LHopitals 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 Parsevals 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 W

T 21 = . Accordingly, the frequency

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