Foundations of MIMO Communication 1st Edition Heath Solutions Manual

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Foundations of MIMO
Communication 1st Edition Heath
Solutions Manual
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51 Problems
Problems
1.1 Show that, for sto be proper complex, its in-phase and quadrature components must
be uncorrelated and have the same variance.
Solution
As defined in Appendix C.1.4, proper complexity entails E[s2] =E[s]2. Letting
s=si+ jsq, this corresponds to
E[s2
i]E[s2
q] + 2jE[sisq] =E[si]2E[sq]2+ 2jE[si]E[sq]; (1.257)
which can be split into twin conditions for the real and imaginary parts, respectively
E[s2
i]E[s2
q] =E[si]2E[sq]2(1.258)
and
E[sisq] =E[si]E[sq]: (1.259)
Condition (1.258) can be rearranged into var[si] = var[sq]while condition (1.259)
amounts tosiandsqbeing uncorrelated.
Ifswere a vector, then the above conditions would generalize to Rsi=Rsqand
Rsisq=RT
sisq. The real and imaginary vectors need not be uncorrelated, only
for each entry do the real and imaginary parts need to be uncorrelated (because the
diagonal ofRsisqdoes need to be zero for the condition Rsisq=RT
sisqto hold).
1.2 Letsconform to a 3-PSK constellation defined by s0=1p
2(1j),s1=1p
2(1j),
ands2= j. Is this signal proper complex? Is it circularly symmetric?
Solution
This zero-mean ternary signal is not proper complex because E[s]2= 0 whereas
E[s2] =1=3. Referring to Problem 1.1, with s=si+ jsq, in the ternary signal
at hand the in-phase and quadrature components are uncorrelated, but var[si] = 1=3
while var[sq] = 2=3.
The signal is not circularly symmetric either because a rotation thereof will result
in a constellation with differently positioned points. Only rotations of 120degrees
would yield an equivalent constellation.
1.3 Letsconform to a ternary constellation defined by s0=1,s1= 0, and s2= 1. Is
this signal proper complex? Is it circularly symmetric?TESTBANKSELLER.COM
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52 A primer on information theory and MMSE estimation
Solution
This zero-mean ternary signal is not proper complex because E[s]2= 0 whereas
E[s2] = 2=3. Referring to Problem 1.1, in the ternary signal at hand, var[si] = 2=3
while var[sq] = 0 .
The signal is not circularly symmetric either because a rotation thereof will result
in a constellation with differently positioned points. Only a rotation of 180degrees
would yield an equivalent constellation.
1.4 Give an expression for the minimum distance between neighboring points in a one-
dimensional constellation featuring Mpoints equidistant along the real axis.
Solution
SuchM-PAM constellation (with BPSK as special case for M= 2) is described by

(2m+ 1M)dmin
2
m= 0;:::;M1; (1.260)
from which, by imposing that the variance be unity,
dmin= 2r
3
M21: (1.261)
1.5 Letxbe a discrete random variable and let y=g(x)withg(
)an arbitrary function.
IsH(y)larger or smaller than H(x)?
Solution
SinceH(
)quantities the uncertainty in a random variable, H(y) H (x)with
equality if every value of xmaps to a distinct value of y, and with strict inequality if
multiple values of xmap to the same value of y.
1.6 Express the entropy of a discrete random variable xas a function of the information
divergence between xand a uniformly distributed counterpart.
Solution
LetDbe the information divergence between xsome other another random variable
that takesMvalues equiprobably. Then, from (1.34),
D=X
xpx(x) log2px(x)X
xpx(x) log21
M(1.262)
=H(x) + log2M: (1.263)TESTBANKSELLER.COM
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53 Problems
It follows that
H(x) = log2MD: (1.264)
1.7 Express the differential entropy of a real Gaussian variable xN(;2).
Solution
Invoking the PDF in (C.13),
h(x) =E
log2fx(x)
(1.265)
=E(x)2
22
log2elog21p
2(1.266)
=1
2log2e+1
2log2(22) (1.267)
and thus
h(x) =1
2log2(2e2): (1.268)
1.8 Compute the differential entropy of a random variable that takes the value 0with
probability 1=3and is otherwise uniformly distributed in the interval [1;1].
Solution
The differential entropy of a discrete random variable (or, as in this case, a mixed
random variable having a discrete component) is 1. To see that, notice that the
density of discrete mass points can be represented by delta functions. Since a delta
function can be obtained from a uniform random variable by allowing the support
to vanish, we can refer to Example 1.4 and let b!0, which makes the differential
entropy grow unboundedly negative.
As advanced in the text, care must be exercised when dealing with differential
entropies.
1.9 Calculate the differential entropy of a random variable xthat abides by the exponen-
tial distribution
fx(x) =1
ex=: (1.269)TESTBANKSELLER.COM
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