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Examen

Wireless Communications – Solutions Manual for Exercises – Theodore S. Rappaport, 2nd Edition – Complete Answer Guide

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wireless communication solutions Rappaport answer key signal propagation mobile radio channels modulation techniques multipath fading cellular concept CDMA FDMA TDMA antenna theory telecommunication engineering

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SOLUTIONS MANUAL

WIRELESS COMMUNICATIONS AND
NETWORKS
SECOND EDITION




WILLIAM STALLINGS




@Computer_IT_Engineering

, TABLE OF CONTENTS




Chapter 2: Transmission Fundamentals ....................................................................... 5
Chapter 3: Communication Networks ......................................................................... 8
Chapter 4: Protocols and the TCP/IP Suite ............................................................... 14
Chapter 5: Antennas and Propagation ....................................................................... 17
Chapter 6: Signal Encoding Techniques ..................................................................... 22
Chapter 7: Spread Spectrum ........................................................................................ 28
Chapter 8: Coding and Error Control ......................................................................... 34
Chapter 9: Satellite Communications.......................................................................... 44
Chapter 10: Cellular Wireless Networks ...................................................................... 48
Chapter 11: Cordless Systems and Wireless Local Loop ........................................... 54
Chapter 12: Mobile IP and Wireless Access Protocol ................................................. 56
Chapter 13: Wireless LAN Technology ........................................................................ 59
Chapter 14: Wi-Fi and the IEEE 802.11 Wireless LAN Standard .............................. 61
Chapter 15: Bluetooth and IEEE 802.15 ........................................................................ 65




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@Computer_IT_Engineering

, CHAPTER 2
TRANSMISSION FUNDAMENTALS

A NSWERS TO Q UESTIONS
2.1 A continuous or analog signal is one in which the signal intensity varies in a
smooth fashion over time while a discrete or digital signal is one in which the signal
intensity maintains one of a finite number of constant levels for some period of time
and then changes to another constant level.

2.2 Amplitude, frequency, and phase are three important characteristics of a periodic
signal.

2.3 2π radians.

2.4 The relationship is f = v, where  is the wavelength, f is the frequency, and v is the
speed at which the signal is traveling.

2.5 The spectrum of a signal consists of the frequencies it contains; the bandwidth of a
signal is the width of the spectrum.

2.6 Attenuation is the gradual weakening of a signal over distance.

2.7 The rate at which data can be transmitted over a given communication path, or
channel, under given conditions, is referred to as the channel capacity.

2.8 Bandwidth, noise, and error rate affect channel capacity.

2.9 With guided media, the electromagnetic waves are guided along an enclosed
physical path, whereas unguided media provide a means for transmitting
electromagnetic waves through space, air, or water, but do not guide them.

2.10 Point-to-point microwave transmission has a high data rate and less attenuation
than twisted pair or coaxial cable. It is affected by rainfall, however, especially
above 10 GHz. It is also requires line of sight and is subject to interference from
other microwave transmission, which can be intense in some places.

2.11 Direct broadcast transmission is a technique in which satellite video signals are
transmitted directly to the home for continuous operation.

2.12 A satellite must use different uplink and downlink frequencies for continuous
operation in order to avoid interference.

2.13 Broadcast is omnidirectional, does not require dish shaped antennas, and the
antennas do not have to be rigidly mounted in precise alignment.

2.14 Multiplexing is cost-effective because the higher the data rate, the more
cost-effective the transmission facility.
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@Computer_IT_Engineering

, 2.15 Interference is avoided under frequency division multiplexing by the use of guard
bands, which are unused portions of the frequency spectrum between
subchannels.

2.16 A synchronous time division multiplexer interleaves bits from each signal and
takes turns transmitting bits from each of the signals in a round-robin fashion.


A NSWERS TO PROBLEMS
2.1 Period = 1/1000 = 0.001 s = 1 ms.

2.2 a. sin (2ft – ) + sin (2ft + ) = 2 sin (2ft + ) or 2 sin (2ft – ) or - 2 sin (2ft)
b. sin (2ft) + sin (2ft – ) = 0.

2.3
N C D E F G A B C
F 264 297 330 352 396 440 495 528
D 33 33 22 44 44 55 33
W 1.25 1.11 1 0.93 0.83 0.75 0.67 0.63

N = note; F = frequency (Hz); D = frequency difference; W = wavelength (m)

2.4 2 sin(4t + ); A = 2, f = 2,  = 

2.5 (1 + 0.1 cos 5t) cos 100t = cos 100t + 0.1 cos 5t cos 100t. From the trigonometric
identity cos a cos b = (1/2)(cos(a + b) + cos(a – b)), this equation can be rewritten as
the linear combination of three sinusoids:
cos 100t + 0.05 cos 105t + 0.05 cos 95t

2.6 We have cos2x = cos x cos x = (1/2)(cos(2x) + cos(0)) = (1/2)(cos(2x) + 1). Then:
f(t) = (10 cos t)2 = 100 cos2t = 50 + 50 cos(2t). The period of cos(2t) is  and therefore
the period of f(t) is .

2.7 If f1(t) is periodic with period X, then f1(t) = f1(t +X) = f1(t +nX) where n is an integer
and X is the smallest value such that f1(t) = f1(t +X). Similarly, f2(t) = f2(t +Y) = f2(t +
mY). We have f(t) = f1(t) + f2(t). If f(t) is periodic with period Z, then f(t) = f(t + Z).
Therefore f1(t) + f2(t) = f1(t + Z) + f2(t + Z). This last equation is satisfied if f1(t) = f1(t
+ Z) and f2(t) = f2(t + Z). This leads to the condition Z = nX = mY for some integers
n and m. We can rewrite this last as (n/m) = (Y/X). We can therefore conclude that
if the ratio (Y/X) is a rational number, then f(t) is periodic.

2.8 The signal would be a low-amplitude, rapidly changing waveform.

2.9 Using Shannon's equation: C = B log2 (1 + SNR)
We have W = 300 Hz (SNR)dB = 3
Therefore, SNR = 100.3

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@Computer_IT_Engineering

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Subido en
19 de julio de 2025
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