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Unformatted text preview: CHAPTER 8 Problem 8.1 2
(a) Free space loss =1010g (ﬁt—d)
10 2010g (Wde
10 3 x 10 /4 x 10 88 dB (b) The power gain of each antenna is 1010g10Gr = lOlogloGt = 1010g (“n—2)“)
10 k
= 1010g (4xnxnx20.6)
10 (3/40)
= 36.24 dB (c) Received Power: Transmitted power +G,  Free space loss = 1 + 36.24 — 88
= —50.76 dBW Problem 8.2 The antenna gain and free—space loss at 12 GHz can be calculated by simply adding 2010g10(12/4)
for the values calculated in Problem 8.1 for downlink frequency 4 GHz. Speciﬁcally, we have: (a) Free—space loss: 88 + 2010g10(3)
= 97.54 dB (b) Power gain of each antenna
= 36.24 + 2010g10(3) = 45.78 dB
(0) Received power = —50.76 dBW
The important points to note from the solutions to Problems 8.1 and 8.3 are: 1. Increasing the operating frequency produces a corresponding increase in freespace loss, and an
equal increase in the power gain of each antenna. 2. The net result is that, in theory, the received power remains unchanged. 436 Problem 8.3 The Friis free—space equation is given by A. 2
Pr = PthGr(m) (a) Using the relationship A2 i2
A, = EEG” and A: = ﬁGt , we may write
47tA 47IA 2
1).: Isl—2 2] [ g] [.97.]
A A n
P A A,
= £412 (1)
41m, A 2
(b) Pr ' P[?] Grfml
P A G
= t [2" (2)
4nd In both Eqs. (1) and (2) the dependent variable is the received signal power, but the
independent variables are different. (c) Equation (1) is the appropriate choice for calculating Pr performance when the dimensions of both the transmitting and receiving antennas are already ﬁxed. Equation (1) states that for
ﬁxed size antennas, the received power increases as the wavelength is decreased. Equation (2) is the appropriate choice when both At and G, are ﬁxed and the requirement is to determine the required value of the average transmitted power Pt in order to realize a speciﬁed
P I. Problem 8.4 The free space loss is given by 437 4n 2
Lfree space 2 (T According to the above formulation for free space loss, free space loss is frequency dependent. Path loss, as characterized in this formulation, is a deﬁnition based on the use of an isotropic
receiving antenna (G, = 1). The power density, p(d), is a function of distance and is equal to EIRP
Nd) = _2
4nd The received power of an isotropic antenna is equal to 2 A
, puma; “U
H EIRP X L2
47rd2 4“ EIRP
(:11 2
A : EIRP[Lfree—space (1) Equation (1) states the power received by an isotropic antenna is equal to the effective transmitth
power EIRP, reduced only by the path loss. However, when the receiving antenna is not isotropic,
the received power is modiﬁed by the receiving antenna gain 0,, that is, Eq. (1) is multiplied by Gr
Problem 8.5 In a satellite communication system, satellite power is limited by the permissible antenna size.
Accordingly, a sensible design strategy is to have the path loss on the downlink smaller than the pass loss on the uplink. Recognizing the inverse dependence of path loss on the wavelength A, it
follows that we should have )‘uplink < A'downlink or, equivalently, fuplink >fdownlink 438 Problem 8.6 Received power in dBW is deﬁned by Pr=EIRP + Gr—Freespace loss
For these three components, we have
(1) EIRP = 1010g10(P,G,) = 1010g10Pt + 1010g10(G,) = 1010g10(0.1)+1010g10(G,) Transmit antenna gain (in dB): lOlogloGt = 1010g10( (3/40)2 = 30.89 dB (2) Receive antenna gain: lOloglOGr = 1010g10[ (3/40)2 = 49.84 dB (3) Freespace loss: 4xan
L = 2 ——
p Olog10( A J 7
_ 4x7: x4 x 10
‘ 201°g10( 3/40 ) 196.25 dB Hence, using Eqs. (1) to (5), we ﬁnd that 4xnx0.7xn/4j 4><1t><0.55><1t><52 439 J (1) (2) (3) (4) (5) Pr = 1010g10(0.1) + 30.89+49.84 —— 196.52 ll ~ 206.52 + 8.073 —125 dBW Problem 8.7 (a) RMS value of thermal noise = «I E [v2] = A/4kTRAf volts, where k is Boltzmann’s constant equal to 1.38 x 1023, T is the absolute temperature in degrees Kelvin, and R is the resistance
in ohmns. Hence, A/4 x138 ><10'23 x 290 x 75 ><1>< 106
—17
A/4x1.38 x290><75 x10 = 1.096 x10'6 volts RMS value (b) The maximum available noise power delivered to a matched load is kTAf 1.38 x10‘23 x 290 ><106 4.0 x 10'15 watts Problem 8.8
The wavegude loss is 1 dB; that is, Gwaveguide = 078 The noise temperature at the input to the LNA due to the combined presence of antenna and
waveguide is T=G e waveguide X Tantenna + (1 — Gwaveguide)Twaveguide = 0.78 x 50 + 290(1* 0.78) = 102.8K The overall noise temperature of the system is 440 500 1000
Tsystem 2 Te + 50 + 2T0 + 2—00— 160.3K
The system noise temperature referred to the antenna terminal is
160.3/0.78 = 205.5K Problem 8.9 In this problem, we are given the noise ﬁgures (F) and the available power gains (G) of the devices. By using the following relationship, we can estimate the equivalent noise temperature of
each device: F T+ Te
‘ T
Te = T(F1) where T is room temperature (290K) and Te is the equivalent noise temperature. (a) The equivalent noise temperatures of the given four components are Waveguide
Twaveguide = 290(2 '— 1) = 290K Mixer T = 290(3 — 1) = 580K mixer Low—noise RF ampliﬁer
TRF = 290(1.71) = 203K IF ampliﬁer TIF = 290(5— 1)
= 1160K (b) The effective noise temperature at the input to the LNA due to the antenna and waveguide is 441 T=G e ><T +(1—G )xT waveguide antenna waveguide waveguide = 0.2 x 50 + 290(1— 0.2)
= 242K The effective noise temperature of the system is T T
T  T +T + ”WW—IF
system e RF GRF GRF X Gmixer
580 1160
= 2 2 2 3 —
4 + 0 + 10 +10><5
= 526.2K Problem 8.10 (a) For the uplink power budget, the ratio]?! is given by C G _ =¢—G,—BO,+——k—L
N uplink S T r
where (1)3 = Power density at saturation
G 1 = Gain of 1m2 BO 1 = Power back—off % 2 Figure of Merit k = Boltzmann constant in dBK
L, = Losses due to rain For the given satellite system, we have 9 = —81—44.5—0.0+1.9+228.6+0.0
N uplink 105.0 dBHz 442 where we have used the following gain of 1m2 antenna: G1=1010g ( 4><n><1 2)
10 (3 x108/14 x109)
= 44.5 dB Boltzmann constant k = —228.6 dB E
(b) Given the data rate in the uplink = 33.9 Mb/s and link margin of 6 dB, the required N—b is 0 E
(—3 = (£0) — (1010g10M+ IOlogloR)
N required N uplink = 105 — 6 — 1010g10(33.9 x 106)
= 105675.3
= 23.7 dB Equivalently, we have Given the use of 8—PSK, the symbol error rate is deﬁned by P6 = erfc(A/Nf—Osin(n/8)) For 8—PSK E 3E 17
_=_=234 3:702
N0 NO X Hence, Pe erfc(A/702 x sin(7t/8))
0 R 443 This result further conﬁrms the statement we made in Example 8.2 in that the satellite
communication system is essentially downlink—limited. Recognizing that we have more powerful
resources available at an earth station then at a satellite, it would seem reasonable that the BER at
the satellite can be made practically zero by transmitting enough signal power along the uplink. Problem 8.11 For the downlink, the relationship between E
(£0) and (—3 , expressed in decibels, is described by
N N req (£0) = (—E—bo) + lOlogM+1010gR (1)
N downlink N req where M is the margin and R is the bit rate in bits/second. Solving Eq. (1) for the the link margin in dB and evaluating it for the problem at hand, we get
lOlogloM = 85 — 10—1010g10(106) = 5dB For the downlink budget, the equation for (NEG) , expressed in decibels, is as follows: Col (0')
—' = EIRP + — —L — lOlog k
(N downlink T dB freespace 1 0 where k is Boltzmann’s constant. For a satisfactory reception at any situation, we consider additional losses due to rain etc. up to the
calculated link margin of 5 dB. Hence, we may write C G
(F011 1. k = EIRP + (%)dB — Lfreespace _ 1010g10k — lOlog10M(dB) (2)
own in
where
EIRP = 57 dBW
Lfreespace = freespace 1053 444 = 92.4 + 2010g10( 12.5) + 2010g10(40, 000) = 206 dB
lOloglok = 228.6 dBK 1010g10M = 5 dB
Using these values in Eq. (2) and solving for G,/T, we get G (.1) :85—57+206—228.6+5
T dB = 10.4 dB With T: 310K, we thus ﬁnd G, = 10.4+ lOlog10(310) = 35.31 dB The receiving antenna gain in is given by 4TCAT])
101 G = 101
0g“) ’ og10( A2 For a dish antenna (circular) with diameter D, the area A equals nD2/4. Thus, 1010g10Gr = 2010g10D + 2010g10f +1010g10(n)+ 20.4(dB) where D is measured in meters and f is measured in GHz. Solving for the antenna diameter for the
given system, we ﬁnally get Dmin = 0.6 meters Problem 8.12 (a) Similarities between satellite and wireless communications:  They are both bandwidthlimited.  They both rely on multipleaccess techniques for their operation. ° They both have uplink and downlink data transmissions. ' The performance of both systems is inﬂuenced by intersymbol interference
and external interference signals. 445 (b) Major differences between satellite and wireless communications: ' Multipath fading and user mobility are characteristic features of wireless
communications, which have no counterparts in satellite communications.  The carrier frequency for satellite communications is in the gigahertz range
(Ku—band), whereas in satellite communications it is in the megahertz range. ' Satellite communication systems provide broad area coverage, whereas wireless
communications provide local coverage with provision for mobility
in a cellular type of layout. Problem 8.13 In a wireless communication system, transmit power is limited at the mobile unit, whereas no
such limitation exists at the base station. A sensible design strategy is to make the path loss (i.e.,
freespace loss) on the downlink as small as possible, which, in turn, suggests that we make (Path 1055)uplink < (Path loss)downlink
Recognizing that path loss is inversely proportional to wavelength, it follows that )‘uplink > A downlink or, equivalently, fuplink <fdownlink Problem 8.14 The phase difference between the direct and reﬂected waves can be expressed as wetlrhbzhmﬁl— raw where K is the wave length. For large d, Eq. (1) may be approximated as ~ 4n(hbhm)
¢ ~ Ad radians With perfect reﬂection (i.e., reﬂected coefﬁcient of the ground is 1) and assuming small 4) (i.e.,
large d), the received power P, is deﬁned by  2 2 471: h h
P, = Poll—e’¢ :Posin [—(b—MU Rd 446 4n(hbhm) 2
z 2
P( M ) ( >
A! 2
WhCI’C P0 = PthGm(m) (3) Using Eq. (3) in (2): 4712(hbh ) 2 )V 2
Pr = P’Gme(—Adlj (ml
2 2
h h
= PthGm[—[:me] which shows that the received power is inversely proportional to the fourth power of distance d
between the two antennas. Problem 8.15 The complex (baseband) impulse response of a wireless channel may be described by N) I30) 2 ale_ 15(z‘—‘c)+aze_j¢25(t—1:) (1) where the amplitudes a1 and a2 are Rayleigh distributed, and the phase angles (1)1 and (1)2 are uniformly distributed. This model assumes (2) the presence of two different clusters with each
one consisting of a large number of scatterers, and (2) the absence of line—of—sight paths in the
wireless environment. Deﬁne h(t) = iz(z)ej¢1 9 = ¢2‘ 4’1
We may then rewrite Eq. (1) in the form
h(t) = a15(t — 1:) + aze'jesa — 'c) as stated in the problem. (a) (i) The transfer function of the model is 447 H(f) Flh(t)l —j27rfr1 ~j(2rtfrz+0)
ale +a26 (ii) The power—delay proﬁle of the model is P, = E[Iz(t)2] 1 = E[a,5(t — 1,) + age—1950 — 1:2)(a15(z— 1)) + azejoaa — 12)] ll E[a§52(z — 1,) + (13520  12) + a1a2c0365(t — mesa — 12)] E[af152(z—rl)+E[a§]52(t—rz) (l) (b) The magnitude response of the model is ll —j21tfi:l ~j21rf(12+ 0)
ale + aze H(f) af + a: + 2ala2cos(21tf('tz — 1,) + e) which exhibits frequency selectivity due to two factors: (1) variations in the coefﬁcients a1 and a2.
and (2) variations inthe delay difference 1211. Problem 8.16 The multipath inﬂuence on a communication system is usually described in terms of two effects 
selective fading and intersymbol interference. In a Rake receiver, selective fading is mitigated by
detecting the echo signals individually, using a correlation method, and adding them algebraically
(with the same sign) rather than vectorially, and intersymbol interference is dealt with by reinseiting different delays into the various detected echoes so that they fall into step again. Making each correlator perform at its assigned value of delay can be done by inserting the right
amount of delay in either the reference (called the delayedreference) or received signals (called
the delayedsignal). Independent of the form of the reference signals employed, the output SNR
from the integrating ﬁlters is substantially the same for both conﬁgurations, under the assumption that the length of the delay Td is signiﬁcantly smaller than the symbol duration T. Each integrating
ﬁlter responds to signals only within about il/T of the frequency f. Therefore, the noises adding 448 shorter than T, regardless of the form of reference signal. The only difference between the tap
circuit contributions of the delayedsignal scheme and those of the delayed—reference scheme is
that the latter are staggered in time by various fractions of Td, and since such staggering is small compared to the signiﬁcant ﬂuctuation period of the contributions, we conclude that the noise
outputs of the two conﬁgurations are equivalent. However, there are three practical advantages of the delayed—signal scheme over the delayed—
reference scheme. First, one delay line instead of two is required. Second, in the latter conﬁguration. corresponding taps in the mark (symbol 0) and space (symbol 1) lines would have
to be adjusted and be kept in phase coincidence. Third, coherent intersymbol interference (eliminated in the delayedsignal scheme) is still present in the latter scheme, (Price and Green
1958), see the Bibliography.
Problem 8.17 (a) The output of the linear combiner is given by N
x(t) = 209ij)
j=l N
ZOLJ(zij) +nj(t)) j=1 N N
ZOszJmU) + Zulu/(t)
j=l j=1
W—J g—V—J
signal noise The output signal—to—noise ratio is therefore Average signal power
Average noise power  (SNR)0 N 2
E[2 Otj(zjm(t)):l j=1 II N 2
E[20tjnj(t):
j=1 449 Etn,<t)nk(r)] = { N N
[2 Z “j“kzjzkm2(t)]
J: = E
lkl
E H N N
: OLJOtknj(t)nk(t):
j=lk=1 NN ZEULJUckzjzkE[m2(l)]
= i=1k=1 N N
2 Z Otjock E[nj(t)nk(t)] j=lk=l Using the following expectations E[m2(t)] = 1 for all t (i.e., unit message power) ojfork=j 0 for k¢j we ﬁnd that Eq. (1) simpliﬁes to (b) Equation (2) can be rewritten in the equivalent form 450 (1) (2) = JEL—_ (3) where uj = 09 oj and (SNR)J = 23/03. We now invoke the Schwarz inequality, which, in discrete form for the problem at hand, is stated as follows N 2 N N 2
(ZuJ(SNR);/2] s[2uf][2((SNR)}/2)] (4)
j=l j=l j=1
Hence, inserting this inequality into the righthand side of Eq. (3), we may write N
(SNR)0 s 2(SNR)J.
j=1 which proves the formula under subpart (i). To prove subpart (ii), we recall that the Schwarz inequality of Eq. (4) is satisﬁed with the
equality sign if (except for a scaling factor) (SNR)J‘./2 = uj or, equivalently, Z.
_J = . .
0‘10] I That is, 451 Z.
_ J
1 Problem 8.18
For the problem at hand, we have M = N = 2. Therefore,
M  N+l = l and so the weight subspace W is one—dimensional. We thus have the following representation for
the action of the antenna array: C2
(Interferer)
One—dimensional
0 ’/sub—space W
ﬂ—J
Weight
Problem 8.19 (a) The cost function is 2 = le[nle*lnl 1
J — §e[n] 2 where the error signal is M eln] = dlnl—ZWZInlka] k=l Let
wk[n] = ak[n] +jbk[n] Hence
{)1 1 ae*[n] 1 * cae[n]
3—6—1; — 2e[n] aak +26 M] aak : —%e[n]x:[n] — 56*[n1xkh’l]
1 *
= —§Re{xk[n]e [11]} 61 1 ae*[n] 1* defnl
[n] abk +26 [n] abk abk 26 : —Ze[n]x:[n] + 13*[n]xk[n] 2 2 = —Im{xk[n](e*[n])} The adjustment applied to the kth weight is therefore Awk[n] = Aak[n] +jAbk[n] _ a_J_ . 2i
Maak ”319k where H is the stepsize parameter. Substituting Eqs. (1) and (2) into (3), Awk[n] * While [11} (b) The complex LMS algorithm is described by the following pair of relations: wk[n+1] = wk[n]+Awk[n] = wk[n] + uxk[n]e*[n], k = 1, 2, ~ M
e(n) = d[n]—§E\v;xk[n]
k=l 453 uRe{xk[nie*[n1} + uIm{xklznie*[n1} ,M (1) (2) (3) Problem 8.20 (a) We are told that the speed of response of the weights in the LMS algorithm is proportional to
the average signal power at the antenna array input. Conversely, we may say that the average
signal power at the array input is proportional to the speed of response of the weights in the LMS algorithm. Moreover, the maximum speed of response of the LMS weights is
proportional to Rb/fmax, where Rb is the bit rate and fmax is the maximum fade rate in Hz. It follows therefore that the dynamic range of the average signal power at the antenna array
input is proportional to Rb/fmax, as shown by max watts (1;) max where CC is the proportionality constant. (b) For CL = 0.2, Rb = 32 x 103 b/s, andfmax = 70 Hz, the use of Eq. (1) yields Pmax = 0.2 x 32 x 103/70 = 640/7
= 91 watts
which is somewhat limited in value. Problem 8.21 (a) According to the Wiener ﬁlter, derived for the case of complex data, the optimum weight
vector is deﬁned by R w = rxd (l) x 0
where Rx = correlation matrix of the input signal vector x[n] = E[x[n]xH[n]] (2) rxd = cross—correlation vector between X[n] and desired response d[n] = E[x[n]d*[n]] (3) wo = optimum weight vector. 454 Note that the formulation of Eq. (1) is based on the premise that the array output is deﬁned as the inner product wa[n]. The Wiener ﬁlter for real data is a special case of Eq. (1), where the
Hermitian transpose H in Eq. (2) is replaced by ordinary transposition and the complex conjugation in Eq. (3) is omitted. Assuming that the input x[n] and desired response d[n] are
jointly ergodic, we may use the following estimates for Rx and rxd: K
A 1 H
Rx : [—(2x[n]x [u] (4)
k=1
1 K *
fxd = szmd [n] (5)
k=l where K is the total number of snapshots used to train the antenna array. Correspondingly, the
estimate of the optimum weight vector wo is computed as A_[ w = Rx rm (6) —1 . .
where Rx is the inverse of Rx. (b) The DMI algorithm for computing the estimate W may now proceed as follows: 1. Collect K snapshots of data denoted by {X[k], d[kl }f=. where K is sufﬁciently large for w to approach wo and yet small enough to ensure
stationarity of the data. 2. Use Eqs. (4) and (5) to compute the correlation estimates ﬁx and 1"“.
3. Invert the correlation matrix Rx and then use Eq. (6) to compute the weight estimate w. For an antenna array consisting of M elements, the matrix ﬁx is an M—by—M matrix and f‘xd is an M—by—l vector. Therefore, the inversion of ﬁx and its multiplication by EM requires multiplications and additions on the order of M3. 455 ...
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