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Unformatted text preview: 16 Problems and Solutions (Chapter 3) 1. A wireless receiver with an eﬁ'ective diameter of 250 cm is receiving signals at 20 GHz from a transmitter that transmits at a. power of 30 mW and a
gain of 30 dB. (5) What is the gain of the receiver antenna? (1)) What is the received power if the receiver is 5 km away from the
transmitter? [Solution] Given .1, = Eﬂ'ective diameter = 250 cm. in = Carrier frequency = 20 'GHz. 1": = ’Iransmitter power = 30 mW G: = Transmitter gain = 30 dB = 1000.
d = Distance of receiver = 5 km. A. m Eﬂ’ective area = (2%) = 4.91 m2 A = Wavelength = = 0.015 m (e) G, = Receiver entenne gain = lees = 2.74»: 105 = 54.38 dB. (b) P, = Received power at distance of 5 km = 43%;; = 4.094: W7
Watts. . Consider an antenna transmitting a. power of 5 W at 900 MHz. Calculate the received power at a. distance of 2 km if propagation is taking place in
free space. [Solution] Transmitted power = H = 5 W
Cmierﬁequency=900MHzr=>A=f =0.33m d = 2 km Assuming unit gain in free space model, i.e., G; = G, = 1
Received power can be calculated by the formula P _ GtGr‘Pt
r— 4nd)§ A
= 8.8 x 1010 W 17 3. In a cellular system, diffraction. reﬂection, and direct path take different amount of time for the signal to reach a MS. How do you diﬁ’erentiate and
use these signals? Explain clearly.Compute the level arming rate with
respect to the rms level for a vertical monopole antenna. assuming the
Rayleigh faded isotropic scattering case. The receiver speed is 20 km/hr
and the transmission occurs at 800 MHz. [Solution] Dilfraction results in smallscale fading and reﬂection results in large scale
fading. So on receiving the different signals, we can study the spatial
and temporal variations of the signal strengths with respect to distance
traveled to understand about the diﬁ'racted and reﬂected signals. If the
variations change fast with respect to time and distance, then We conclude
it to be diﬁraction and reﬂection otherwise. . . The transmission power is 40 W, under a free space propagation model, (a) “That is the transmission power in unit of dBm? (b) The receiver is in a distance of 1000 In, what is the received power,
assuming that the carrier frequency fc = 900 MHz and G; 2 GP = 1
dB? ' (c) Express the free pace path loss in dB.
[Solution] (8.) 10 x log (40 x 1000) = 46 dBm.
(b) _aee ‘ es)“ __ 40 x 1 x 1 x — (4 x 1r x 1000)2
= 2.82 x 10_a w. P. (c) Pr (dB) = 10 x log (2.82 x 103) = —75 dB. . A receiver is tuned to 1 GHz transmission and receives signals with Doppler
frequencies ranging from 10 Hz to 50 Hz when moving at a. speed of 80
km/hr. What is the fading rate? [Solution]
Moving speed 0 = 80 km/hr. = 22.22 In/sec ‘
v = 1 GHz
Thus, A = c/u,
, where c is free space propagation speed and equals to 3 x 103 rn/sec. 18 A = 0.1 m
Therefore, the fading rate
N(rm) = %
__ 2 x 22.22
_ 0.1
= 444.4 Hz. Here we are not considering the Doppler shift because the 9 is not given. . What does a. small delay spread indicate about the characteristics of a. fading channel? If the delay spread is 1 microsecond, will two different
frehuencies that are 5 MHz apart, experience correlated fading? [Solution] A all delay spread indicates that smearing or spreading out eEect is
less. The delay spread determines to what content the channel fading at
two diﬁ'erent frequencies f; and f2 are correlated. Small delay spread
indicates larger coherence bandwidth and hence correlated fading. 1 microsecond delay spread indicates
Bc .—. Coherence bandwidth = W = 159.15 kHz.
Since 1 f1  f2 > Be, correlated fading will not be experienced. . Consider an antenna transmitting at 900 MHz. The receiver is traveling at a speed of 40 Ian/h. Calculate its Doppler shift.
{Solution} 33.67 Hz 8. Repeat Problem 3.6. Calculate the average fading duration if p = 0.1. [Solution] where f," is the Doppler frequency. 19 9. Describe the consequence of the Doppler effect on the receiver in an
isotropic scattering environment. Based on your description, speculate
on the meaning of the term “Doppler Spread”. (a) Is the term “Doppler Spread” more appropriate in describing the
channel than “Doppler Shift” in a scattering environment? Why? (13) ObServe the inverse relationship that exists between “Coherence Band
width” and “Delay Spread” in a wireless channel. Attempt to simi
larly deﬁne a term “Coherence Time” that has an inverse relationship
with the “Doppler Spread”. What information does this term give
about the channel? I [Solution] The multiple reﬂections of signals cause a dynamic change in propagation
time. Each reﬂection received is liable to be suffering a different Doppler
shift. This means that a carrier transmitted on a single frequency will
receive multiple reﬂections and each reﬂection will arrive at the receiver
shifted in frequency by a different amount. The diﬂ'erence between the
highest shift and the lowest shift gives the Doppler spread. (a) Since the term Doppler spread determines the difference between
highest and lowest Doppler shifts, it doesn’t inform about the max
imum Doppler shift experienced. So Doppler spread is more appro
priate in describing the channel than “Doppler shift” in a scattering
environment. (1')) Delay spread and coherence bandwidth are parameters which de
seribe the time dispersive nature of the channel in a local area. How
ever, information about the time varying nature of the channel caused
by either relative motion between the mobile and base station, or by
movement of objects in the channel is not given. Coherence time is
the time domain dual of Doppler spread and is used to characterize
the time varying nature of the frequency dispersiveness of the channel
in the time domain. Coherence time is the time duration aver which two received signals
have a strong potential for amplitude correlation. If the reciprocal
bandwidth of the baseband signal is greater than the coherence time of the channel, then the channel will change during the transmission
of the baseband message, thus causing distortion at the receiver. 10. How can you compensate for the impact of Doppler effect in a cellular
system? Explain. [Solution] We can compensate the impact of Doppler eﬁ'ect at receiver by adaptive
detection to track the channel variations. 20 11. 12. 13. 14. How is radio propagation on land different from that in free space?
[Solution] Propagation in free space does not have any obstacles and hence it char
acterizes the most ideal situation for propagation. Whereas, radio prop
agation on land may take place close to obstacles which cause reﬂection,
diﬂ'raction, scattering. ' What is the diﬁ'erences between fast fading and slow fading?
[Solution] Slow fading is caused by movement over distances large enough to produce
gross variations in overall path length between base station and mobile
station. In other words, the long term variation in the mean level is
known as slow fading.
Rapid ﬂuctuations caused by local multipath are known as fast fad
ing. It is shortterm fading. Path loss, fading, and delay spread are the three important radio propaga
tion issues. Explain why those issues are important in a. cellular system? [Solution] The wave propagation in multipath channel depends on the actual envi
ronment including factors like antenna height, proﬁle of the buildings, road
and the terrain. Therefore we need to describe the behavior of mobile ra
dio channels using a good and relevant statistical mechanism. Thus this is
provided by studying the path loss and fading phenomenon to understand
about the characteristics of the land and terrain proﬁle. The delay spread
is also necessary to understand about the propagation channel. whether
the signal is propagating through the frequency selective or the ﬂat fading channels. A BS has a 900 MHz transmitter and a. vehicle is moving at the speed of
50 mph. Compute the received carrier frequency if the vehicle is moving (a) Directly toward the BS.
(b) Directly away from the BS.
(c) In a. direction which is 60 degree to the direction of arrival of the
transmitted signal.
[Solution]
Given
fc = Carrier frequency = 900 MHZ
A: Wavelength=ﬁ =0.3333m
v = Velocity = 50 mph = 22.22 m/s.
Let Doppler shift frequency be denoted by f4. 15. 16. 17. 21 (a) 9 = 180, (direction is towards BS.) fd = i 0039 = —67.06 Received carrier frequency: 7 f.» = f: — fd = 900 * 10° + 67.06 = 900.000067 1 10s me.
(b) 9 = 0, (direction amy from BS.) f4 = f 0059 = 67.06 Received carrier frequency: fr = for  fa = 900 1: 10° — 6?.06 = 89939993: 106 MHz.
(c) 9 a 60, (direction is towards BS.) f4 = f case = —33.53 Received carrier frequency: fr = f:  fs = 900 * 103 +3353 = 900.000033 4: 1010 MHz. What is the diversity reception? How can it be used to combat multipath?
[Solution] Diversity reception: Two frequencies, that are larger than the coherence
bandwidth, fade independently. This concept is used in diversity recep
tion. Therefore, it is a radio reception in which a resultant signal is ob
tained by combining or selecting signals, from two or more independent
sources. They are modulated with identical informationbearing signals.
(i.e., multiple copies of same message are sent using different frequencies.) Combating with multipath: The signals are chosen such that the frequen
cies are larger than coherence bandwidth, therefore their fading charac
teristics are independent at any given instant. This dfects of
correlated fading. ' 7 What is the role or usefulness of reﬂected and diffracted radio signals in
a cellular system? Explain with suitable examples. [Solution] Role of reﬂected and diﬂracted waves help in signal propagation even in
worst cases when there are no LOS component between T3: and Rx. Ex.
Tx and Rx separated by the hilly terrains, communication in urban areas with high rise buildings. What is the intersymbol interference (181)? Does it affect the transmission
rate of a digital channel? Explain clearly. [Solution] Intersymbol interference (181) is caused by time delayed multipath signals,
where in the second multipath signal is delayed such that it is received
during the next symbol. Therefore, the received signals tend to get elon
gated and smeared into each other. Thus, 181 is caused by time delayed
multipath swells. 22 18. 19. The time for which second multipath signal is delayed is delay spread Ta.
In timedispersive medium, transmission rate R for digital transmission
is limited by this delay spread Td. Therefore, for a lowerror~rate perfor mance, 1
R<2—I;; 151 also affects burst error rate of channel. A MS is not in the direct line of sight of a BS transmitting station. How
is the signal received? Explain. [Solution] With the help of miﬂtipath fading signals can reach the places inside build.
ings and places which are hidden by tall buildings and trees. With multi
path fading signals are reﬂected, diifracted and scattered which results in
multiple low power signals traveling in different directions. Consider two random variables X and Y that are independent and Gaussian
with identical variances. One is of zero mean and the other is of mean p.
Prove that the density function of Z = VX + Y is Rician distributed. {Solution} Suppose T = X 2 + 1"”. Since X and Y are statistically independent
Gaussian random variables with mean 0 and n, and common variance 0'2,
T has a noncentral chisquare distribution with oncentrality parameter
52 =0+,u2 =p2. The deofTis given by 1 _ 3 t
M'(t)=2—OEE t>0 Nowwe deﬁneenew random variableR=Ji ThepdfofRisgiven by
t 0” ’ tp
Ps(t)=;§e 73—10(35), 1320. This is the pdf of a Ricean—distributed random variable. What causes the intersymbol interference and how can you reduce the
intersymhol interference in the wireless communication system? [Solution] Intersymbol interference is caused by timedelayed multipath signals. The
most common methods to reduce intersymbol interference are: guard time,
pulse shaping, signal encoding and equalization. ...
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This note was uploaded on 12/10/2010 for the course DCIS 32 taught by Professor Go during the Spring '10 term at College of E&ME, NUST.
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