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Problem 3.1 (Invariance of coding gain)
(a) Show that in the power-limited regime the nominal coding gain γc(A) of (5.9),
the UBE (5.10) of Pb(E), and the effective coding gain γeff (A) are invariant to
scaling, orthogonal transformations and Cartesian products.
In the power-limited regime, the nominal coding gain is defined as
Scaling A by α>0 multiplies both dmin(A)and Eb(A)by α2, and therefore leaves
γc(A) unchanged. Orthogonal transformations of A do not change either
dmin(A)or Eb(A). As we have seen in Problem 2.1, taking Cartesian products also
does not change either 2 2 min(A)or Eb(A). Therefore γc(A) is invariant under all
these operations. The UBE of Pb(E)involves γc(A)and Kb(A)= Kmin(A)/(| log |A|).
Kmin(A)isalso obviously unchanged under scaling or orthogonal transformations.
Problem 2.1 showed that Kmin(A)increases byafactorof K under a K-fold Cartesian
product, but so does log |A|,so Kb(A) is also unchanged under Cartesian products.
The effective coding gain is a function of the UBE of Pb(E), and therefore it is
invariant also.
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(b) Show that in the bandwidth-limited regime the nominal coding gain γc(A) of
(5.14), the UBE (5.15) of Ps(E), and the effective coding gain γeff (A) are invariant
to scaling, orthogonal transformations and Cartesian products.
In the bandwidth-limited regime, the nominal coding gain is defined as
Scaling A by α>0 multiplies both dmin(A)and Es(A)by α2 and does not change
ρ(A), and therefore leaves γc(A) unchanged. Orthogonal transformations of A do
not change 2 min(A), Eb(A)or ρ(A). As we have seen in Problem 2.1, taking
Cartesian products also does not change d2 min(A), E b (A)or ρ(A). Therefore
γc(A) is invariant under all these operations.
The UBE of Ps(E)involves γc(A)and Ks(A)=(2/N)Kmin(A). Kmin(A) is also
obviously unchanged under scaling or orthogonal transformations. Problem 2.1
showed that min(A)increases byafactorof K under a K-fold Cartesian product, but
so does N,so Ks(A) is also unchanged under Cartesian products. The effective
coding gain is a function of the UBE of Ps(E), and therefore it is invariant also.

Problem 3.2 (Orthogonal signal sets)
An orthogonal signal set is a set A = {aj, 1 ≤ j ≤ M } of M orthogonal vectors in RM
with equal energy E(A); i.e., aj, aj = E(A)δjj’ (Kronecker delta).
(a) Compute the nominal spectral efficiency ρ of A in bits per two dimensions.
Compute the average energy Eb per information bit.
The rate of A is log2 M bits per M dimensions, so the nominal spectral efficiency is
ρ =(2/M )log2 M bits per two dimensions.
The average energy per symbol is E(A), so the average energy per bit is
(b) Compute the minimum squared distance d2 min(A). Show that every signal has
K min(A)= M − 1 nearest neighbors.
The squared distance between any two distinct vectors is
||aj − aj’ ||2 = ||aj||2 − 2aj, aj + ||aj ||2 = E(A) − 0+ E(A)=2E(A),

so d2 min(A)=2E(A), and every vector has all other vectors as nearest neighbors,
so Kmin(A)= M − 1.
(c) Let the noise variance be σ2 = N0/2 per dimension. Show that the probability
of error of an optimum detector is bounded by the UBE
Pr(E) ≤ (M − 1)Q (E(A)/N0).
The pairwise error probability between any two distinct vectors is
Pr{aj → aj } = Q (||aj − aj ||2/4σ2)= Q (2E(A)/2N0)= Q (E(A)/N0).
By the union bound, for any aj ∈A,
so the average Pr(E) also satisfies this upper bound.
(d) Let M →∞ with Eb held constant. Using an asymptotically accurate upper
bound √ for the Q (·) function (see Appendix), show that Pr(E) → 0 provided that
Eb/N0 > 2ln2 (1.42 dB). How close is this to the ultimate Shannon limit on
Eb/N0? What is the nominal spectral efficiency ρ in the limit?

By the Chernoff bound of the Appendix, Q (x2) ≤ e −x2/2 . Therefore
Pr(E) ≤ (M − 1)e −E(A)/2N0 <e (ln M) e −(Eb log2 M)/2N0
Since ln M =(log2 M )(ln 2), as M →∞ this bound goes to zero provided that
Eb/2N0 > ln 2,
or equivalently Eb/N0 > 2ln2 (1.42 dB).
The ultimate Shannon limit on Eb/N0 is Eb/N0 > ln 2 (-1.59 dB), so this shows
that we can get to within 3 dB of the ultimate Shannon limit with orthogonal
signalling. (It was shown in 6.450 that orthogonal signalling can actually achieve
Pr(E) → 0 for any Eb/N0 > ln 2, the ultimate Shannon limit.)
Unfortunately, the nominal spectral efficiency ρ =(2 log2 M )/M goes to 0 as M
→∞
Problem 3.3 (Simplex signal sets)
Let A be an orthogonal signal set as above. (a) Denote the mean of A by m(A).
Show that m(A) = 0, and compute ||m(A)||2 .
By definition,

Therefore, using orthogonality, we have
By the strict non-negativity of the Euclidean norm, ||m(A)||2 = 0.
The zero-mean set A = A− m(A) (as in Exercise 2) is called a simplex signal set.It is
universally believed to be the optimum set of M signals in AWGN in the absence of
bandwidth constraints, except at ridiculously low SNRs.
(b) For M =2, 3, 4, sketch A and A .
For M =2, 3, 4, A consists of M orthogonal vectors in M -space (hard to sketch for
M =4). For M =2, A consists of two antipodal signals in a 1-dimensional subspace
of 2-space; for M =3, A consists of three vertices of an equilateral triangle in a
2dimensional subspace of 3-space; and for M =4, A consists of four vertices of a
regular tetrahedron in a 3-dimensional subspace of 4-space.
(c) Show that all signals in A have the same energy E(A). Compute E(A). Compute
the inner products aj , aj for all aj , aj ∈A .

The inner product of m(A)withany aj is
The energy of aj = aj − m(A) is therefore
For j = j, the inner product aj , aj is
In other words, the inner product is equal to M −1 E(A)if j = j and − 1 E(A)for j = j.
M M
(d) [Optional]. Show that for ridiculously low SNR, a signal set consisting of M − 2
zero signals and two antipodal signals {±a} has a lower Pr(E) than a simplex signal
set. [Hint: see M. Steiner, “The strong simplex conjecture is false,” IEEE
Transactions on Information Theory, pp. 721-731, May 1994.]
See the cited article.

Problem 3.4 (Biorthogonal signal sets)
The set A = ±A of size 2M consisting of the M signals in an orthogonal signal set A
with symbol energy E(A) and their negatives is called a biorthogonal signal set.
(a) Show that the mean of A is m(A)= 0, and that the average energy is E(A). The
mean is
and every vector has energy E(A).
(b) How much greater is the nominal spectral efficiency ρ of A than that of A?
The rate of A is log2 2M = 1+log2 M bits per M dimensions, so its nominal spectral
efficiency is ρ =(2/M )(1 + log2 M )b/2D,whichis2/M b/2D greater than for A. This
is helpful for small M , but negligible as M →∞.
(c) Show that the probability of error of A is approximately the same as that of an
orthogonal signal set with the same size and average energy, for M large.
Each vector in A has 2M − 2 nearest neighbors at squared distance 2E(A), and one
antipodal vector at squared distance 4E(A). The union bound estimate is therefore
Pr(E) ≈ (2M − 2)Q (E(A)/N0) ≈|A|Q (E(A)/N0),

which is approximately the same as the estimate Pr(E) ≈ (2M − 1)Q (E(A)/N0) ≈
|A|Q (E(A)/N0) for an orthogonal signal set A of size |A| =2M .
(d) Let the number of signals be a power of 2: 2M =2k . Show that the nominal
spectral efficiency is ρ(A)=4k2−k b/2D, and that the nominal coding gain is γc(A)=
k/2. Show that the number of nearest neighbors is Kmin(A)=2k − 2.
If M =2k−1, then the nominal spectral efficiency is
ρ(A”)=(2/M )(1 + log2 M )=22−k k =4k2−k b/2D.
We are in the power-limited regime, so the nominal coding gain is
The number of nearest neighbors is Kmin(A)=2M − 2=2k − 2.
Problem 3.5 (small nonbinary constellations)
(a) For M =4, the (2 × 2)-QAM signal set is known to be optimal in N =2
dimensions. Show however that there exists at least one other inequivalent two-
dimensional signal set

A’ with the same coding gain. Which signal set has the lower “error coefficient”
Kmin(A)?
The 4-QAM signal set A with points {(±α, ±α)} has b =2, d2 min(A)=4α2 and
E(A)= 2α2,so A has Eb = E(A)/2= α2 and γc(A)= d2 min(A)/4Eb =1 The 4-point
hexagonal signal set A with points at {(0, 0), (α, 3α), (2α, 0), (3α, 3α)} √ has mean
m =(3α/2, 3α)/2) and average energy E(A)=5α2 . If we translate A to A” = A − m to
remove the mean, then E(A)= E(A) −||m||2 =5α2 − 3α2 =2α2 . Thus A has the
same minimum squared distance, the same average energy, and thus the same
coding gain as A.
In A, each point has two nearest neighbors, so Kmin(A)=2. In A, two points have
two nearest neighbors and two points have three nearest neighbors, so
Kmin(A)=2.5. (This factor of 1.25 difference in error coefficient will cost about
(1/4) · (0.2) = 0.05 dB in effective coding gain, by our rule of thumb.)
[Actually, all parallelogram signal sets with sides of length 2α and angles between
60◦ and 90◦ have minimum squared distance 4α2 and average energy 2α2, if the
mean is removed.]
(b) Show that the coding gain of (a) can be improved in N =3 dimensions. [Hint:
consider the signal set A = {(1, 1, 1), (1, −1, −1), (−1, 1, −1), (−1, −1, 1)}.] Sketch A .
What is the geometric name of the polytope whose vertex set is A?

The four signal points in A are the vertices of a tetrahedron (see Chapter 6, Figure
1). The minimum squared distance between points in A is 2 · 4 = 8, and the average
energy is E(A)=3, so Eb =3/2. Thus the coding gain of A is γc(A)= d2 min(A)/4Eb
=4/3, afactorof4/3 (1.25 dB) better than that of A.
However, the nominal spectral efficiency ρ of A is only 4/3 b/2D, compared to ρ =2
b/2D for A; i.e., A is less bandwidth-efficient. Also, each point in A has Kmin(A)=3
Since each signal sends 2 bits, the UBE for Pb(E)is 1 √ nearest neighbors, which
costs about 0.1 dB in effective coding gain.
(c) Give an accurate plot of the UBE of the Pr(E) for the signal set A of (b). How
much is the effective coding gain, by our rule of thumb and by this plot?
The UBE for Pr(E) is
Since each signal sends 2 bits, the UBE for Pb(E) is ½ Pr(E): Pb(E) ≈ 1.5Q (2 4
Eb/N0). An accurate plot of the UBE may be obtained by moving the baseline
curve Pb(E) ≈ 3 Q (2Eb/N0) to the left by 1.25 dB and up by a factor of 1.5, as
shown in Figure 1. This shows that the effective coding gain is about γeff (A) ≈ 1.15
dB at Pb(E) ≈ 10−5.Our √ rule of thumb gives approximately the same result, since
1.5 is equal to about

Figure 1. Pb(E) vs. Eb/N0 for tetrahedron (4-simplex) signal set.
(d) For M =8 and N =2, propose at least two good signal sets, and determine
which one is better. [Open research problem: Find the optimal such signal set,
and prove that it is optimal.]

Possible 8-point 2-dimensional signal sets include:
(i) 8-PSK. If the radius of each signal point is r, then the minimum distance is
dmin = 2r sin 22.5◦, so to achieve dmin = 2 requires r =1/(sin 22.5◦)=2.613,
oranenergyof 6.828 (8.34 dB).
(ii) An 8-point version of the V.29 signal set, with four points of type (1, 1) and
four points of type (3, 0). The average energy is then 5.5 (7.40 dB), about 1 dB
better than 8-PSK. Even better, the minimum distance can be maintained at
dmin = 2 if the outer √ points are moved in to (1 + 3, 0), which reduces the
average energy to 4.732 (6.75 dB).
(iii) Hexagonal signal sets. One hexagonal 8-point set with dmin = 2 has 1 point
at the origin, 6 at squared radius 4, and 1 at squared radius 12, for an average
energy of √ √ 36/8=4.5 (6.53 dB). The mean m has length 12/8, so removing
the mean reduces the energy further by 3/16 = 0.1875 to 4.3125 (6.35 dB).
Another more symmetrical hexagonal signal set (the “double diamond”) has
points
at √ (±1, 0), (0, ± 3) and (±2, ± 3). This signal set also has average energy
36/8=4.5
(6.53 dB), and zero mean.

Problem 3.6 (Even-weight codes have better coding gain) Let C be an (n,k,d)
binary linear code with d odd. Show that if we append an overall parity check p =
i xi to each codeword x, then we obtain an (n +1,k,d +1) binary linear code C with
d even. Show that the nominal coding gain γc(C) is always greater than γc(C) if k>
1. Conclude that we can focus primarily on linear codes with d even. The new code
C has the group property, because the mod-2 sum of two codewords (x1,... ,xn,p =
i xi)and (x1,... ,xn,p = i xi) is
another codeword in C . Its length is n = n+ 1, and it has the same number of
codewords (dimension). Since the parity bit p is equal to 1 for all odd-weight
codewords in C,the weight of all odd-weight codewords is increased by 1, so the
minimum nonzero weight becomes d = d+ 1. We conclude that C is a binary linear
(n+1,k,d+ 1) block code.
The nominal coding gain thus goes from dk . Since
if d<n, the nominal coding gain strictly increases unless d = n— i.e., unless C is a
repetition code with k = 1— in which case it stays the same (namely 1 (0 dB)).

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