000_SOLN_Chapte1.pdf

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Wireless Communication and Networks Upena Dalal

Oxford University Press 2014. All rights reserved.

Chapter 1

Problem 1.1 If the bit rate is to be maintained to 10 Mbps, what modifications should be

made in the system to cope up with SNR variations between 10 dB and 20 dB.

Solution:

SNR1=10 dB

SNR2=20 dB

SNRdB=10log10SNR

SNR1=antilog(10/10)=1

SNR2=antilog(20/10)=100

Considering simple binary transmission

BW=1/2bitrate W1=5 Mbps.

Now SNR2 SNR1W1/W2

Taking log on both the sides

logSNR2 = (W1/W2)logSNR1

(20/10) = (W1/W2)

(W1/W2) = 2

W2 = W1/2 reduce the BW by 2.

We need to change the modulation scheme to maintained the bit rate so instead of BPSK,

QPSK scheme to be implemented.


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Problem 1.2 If square pulses, each of the duration 0.05 s are to be transmitted at a carrier

frequency of 100 MHz, what will be the shape of the spectrum? According to this spectrum

find out the (a) null to null (significant energy) bandwidth, (b) fractional power containment

bandwidth, (c) bounded PSD, (d) absolute bandwidth.

Solution:

Hint is already given as follows

Fractional power containment bandwidth: According to FCC rules, the occupied bandwidth is

the band that levels exactly 0.5% of the signal power above the upper band limit and exactly

0.5% of the signal power below the lower band limit. Thus 99% of the signal power is inside

the occupied band.

Bounded power spectral density: Typical attenuation level might be 35 or 50 dB.

Absolute bandwidth: It is the interval between frequencies outside of which the spectrum is

zero. However, for all realizable waveforms absolute bandwidth is infinite.

Considering the case of BPSK

The shape of the spectrum is sinc type from the equation of power spectral density

(a)

0 to first null = (1/0.05 s)= 20 MHz

Same happens at 100 MHz carrier

Null to null BW 120-80

40 MHz


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Non-regulatory definition of null to null bandwidth is the width of the main spectral lobe of

the PSD bounded by spectra nulls if they exist

(b) The fractional power containment bandwidth is the most appropriate measure of

necessary bandwidth because it is a measure of the integrated power spectrum density and

can be related to system performance. It is 99% power containment bandwidth and defined as

dffS

)( Where, is the decimal equivalent of percentage power (Here 0.99 for 99%)

Approximately 5 nulls contain the most of the energy and with that assumption 99 MHz on

both the ends means 198 MHz


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(c) Half power bandwidth is defined as 0.88 R means considering both the sides it is

approximately 35.2 MHz, means bounded power spectral density is slightly higher than this.

(d) absolute BW =


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Problem 1.3 Energies of signalsg1(t) andg2(t) areEg1andEg2 respectively.

(a) Show that in general the energy of the signalg1(t)+g2(t) is notEg1+Eg2

(b) Under what condition is the energy ofg1(t)+g2(t) equal toEg1+Eg2

(c) Can the energy of the signalg1(t)+g2(t) be zero? If so, under what conditions?

Solution:

(a) The total energy of [g1(t)+g2(t)]=Eg1+g2

= (

g1(t)+g2(t))

2 dt

= (

g12

(t)dt+ (

g22

(t)dt+2 (

g1(t)g2(t)) dt

=Eg1+Eg2 +2 (

g1(t)g2(t)) dt

Eg1+Eg2

(b) From the result in part (a) if the third term of the equation becomes zero i.e.

2 (

g1(t)g2(t)) dt= 0 then and then only total energy of (g1(t)+g2(t)) equals toEg1+

Eg2.

(c) Eg1+g2 = 0 if and only ifEg1=Eg2=0. This is possible only with the condition

g1(t)=g2(t)=0.


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Problem 1.4 Determine the energy spectral density of the square pulses(t)=rect(t/T), where

rect(t/T) equals 1, forT/2 tT/2 and equals zero elsewhere. Calculate the normalizedenergyEsin the pulse.

Solution:

s(t)=rect(t/T)

rect(t/T) = 1, forT/2 tT/2

=0, elsewhere

Es= 1 /

/

= T/2

(-T/2)

Es=T

ESD=Es=

2(t)dt

= 1 /

/

= T


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Problem 1.5 The input x and output y of a certain nonlinear channel are related as

y=x+0.22x3. The input signalx(t) is a sum oftwo modulated signals

x(t)=x1(t)cos1t+ x2(t)cos2t

Where, theX1() andX2()are shown in Fig

1=2(100 x 103)and 2=2(100 x 10

3),

(a)Sketch the spectra of the input signalx(t)and output signaly(t).(b)Can signalsx1(t)andx2(t)be recovered without distortion and interference from

the outputy(t)?

Solution:y=x+0.22x3

= x1(t)cos1t+ x2(t)cos2t +0.22(x1(t)cos1t+ x2(t)cos2t)3

{Writing x1for x1(t) and x2for x2(t) }

= (x1+ 0.165x23 + 1.5x1x22 )cos 1t +(x2+ 0.165x22 + 1.5x22 )cos 2t


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+0.75 x2x12cos(21+2)t +0.75 x1x2

2cos(22+1)t

+0.75 x2x12cos(21-2)t +0.75 x1x2

2cos(22-1)t

+ 0.25x13cos 31t+ 0.25x2

3cos 32t

(a)

Thus, the input spectra has two frequencies 1 and 2

the output spectra are centered at frequencies

1, 2, 21+2,21-2, 22+1, 22-1, 31, 32

(b)x1(t)andx2(t) each has bandwidth 5 kHz , so x1x22, x2x1

2, x13, x2

3each has bandwidth

=15 kHz . Spectra centered at 1, 2,21-2 and 22-1, overlap. Similarly, spectra at

21+2, 22+1, 31 and 32also overlap. Hence, it is not possible to recoverx1(t)andx2(t)

from y(t).


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Problem 1.6Show that an arbitrary functions(n) can be represented by a sum of even

functionse(n) and odd functionso(n), i.e.

()= () ()

Solution:

()=()+()

. (1)

()=()()

. (2)

By adding (1) and (2) we will get

()= () ()


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Problem 1.8 A channel bandwidth is 2 MHz and SNR= 25 dB. Using Shannons formula,

find the channel capacity. If we assume that we can achieve this limit based on Nyquists

formula, find the number of signalling levels required

Solution:

Shannons formula

W=2 MHz

SNR= 25db

C= W log2(1 + SNR)

C=16.62M bits/sec

= 2 log (Nyquist formula)

M=17.81

M=18 levels approximately


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Problem 1.9 A system with digital signalling is operated at 4800 bits per second. If the signal

element encodes a 4-bit word, what is the minimum required bandwidth?

Solution:

4800/4 = 1200 words /sec

BW= 4.8 kHz


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Problem 1.10For the signal shown in Fig. 1.14 , find the type of signal and suitable measure

to analyze it.

Solution:

The signal is a periodic ramp or sawtooth wave and a power signal and can be analysed with

the help of Fourier series. It can also be considered the extreme case of an

asymmetric triangular wave. While a square wave is constructed from only odd harmonics, a

sawtooth wave spectrum contains both even and odd harmonics of the fundamental

frequency. It contains all the integer harmonics.Detail Fourier series can be found from any

book on signal processing.

000_SOLN_Chapte1.pdf

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