Solutions Manual
for
Digital Communications, 5th Edition
(Chapter 10) 1
Prepared by
Kostas Stamatiou
January 15, 2008
2
Problem 10.1
(a)
F(z) = 4
5+3
5z−1⇒X(z) = F(z)F∗(z−1) = 1 + 12
25 z+z−1
Hence :
Γ=
112
25 0
12
25 112
25
012
25 1
ξ=
3/5
4/5
0
−0.456
(b) The eigenvalues of the matrix Γare given by :
|Γ−λI|= 0 ⇒
1−λ0.48 0
0.48 1 −λ0.48
= 0 ⇒λ= 1,0.3232,1.6768
The step size ∆ should range between :
(c) Following equations (10-3-3)-(10-3-4) we have :
ψ=
1 0.48
0.48 0.64
, ψ
c−1
c0
=
0.6
0.8
⇒
Problem 10.2
(a)
(b) From (10-1-31) :
J∆= ∆2Jmin
3
X
k=1
λ2
k
1−(1 −∆λk)2≈1
2∆Jmin
3
X
k=1
λk
1 + N0
(c) Let C′=VtC, ξ′=Vtξ, where Vis the matrix whose columns form the eigenvectors of the
covariance matrix Γ(note that Vt=V−1).Then :
C(n+1) = (I−∆Γ)C(n)+ ∆ξ⇒
C(n+1) =I−∆VΛV−1C(n)+ ∆ξ⇒
which is a set of three de-coupled difference equations (de-coupled because Λ is a diagonal matrix).
Hence, we can write :
k=ξ′
λk
or going back to matrix form :
C′=Λ−1ξ′⇒
which agrees with the result in Probl. 9.49(a).
Problem 10.3
Suppose that we have a discrete-time system with frequency response H(ω); this may be equalized
by use of the DFT as shown below :
an
System H(ω)
yn
Equalizer
output
N−1
X
n=0
N−1
X
n=0
Let :
E(ω) = A(ω)Y∗(ω)
|Y(ω)|2
Then by direct substitution of Y(ω) we obtain :
If the sequence {an}is sufficiently padded with zeros, the N-point DFT simply represents the
values of E(gw) and H(ω) at ω=2π
Nk=ωk,for k= 0,1, …N −1 without frequency aliasing.
Therefore the use of the DFT as specified in this problem yields E(ωk) = 1
Problem 10.4
The MSE performance index at the time instant kis
J(ck) = E
N
X
n=−N
ck,nvk−n−Ik
2
5
then its l−th element is
Gk,l =ϑJ(ck)
2ϑck,l
=1
2E“2 N
X
ck,nvk−n−Ik!v∗
k−l#
Problem 10.5
The tap-leakage LMS algorithm is :
C(n+ 1) = wC(n) + ∆ǫ(n)V∗(n) = wC(n) + ∆ (ΓC(n)−ξ) = (wI−∆Γ)C(n)−∆ξ
and since ∆ >0,the convergence criterion is :
Problem 10.6
The estimate of g can be written as : ˆg=h0x0+… +hM−1xM−1=xTh,where x,hare column
vectors containing the respective coefficients. Then using the orthogonality principle we obtain the
optimum linear estimator h:
6
or :
hopt =R−1
xx c
where we have used the fact that gand ware independent, and that E[g] = 0.Also, the column
Problem 10.7
(a) The time-update equation for the parameters {Hk}is :
H(n+1)
k=H(n)
k+ ∆ǫ(n)y(n)
k
where nis the time-index, kis the filter index, and y(n)
kis the output of the k-th filter with transfer
function : 1−z−M/1−ej2πk/M z−1as shown in the figure below :
– –
?66
y0(n)
H0
z−1
+X
+
–
– –
y1(n)
+X
+
– –
?66
yM−1(n)
HM−1
z−1
+X
+
–
ej2π(M−1)/M
CCCCCCCCCCCC
1/M z−M
.
.
Parallel Bank of Single −Pole Filters
7
(b) It is straightforward to prove that the transfer function of the k-th filter in the parallel bank
has a resonant frequency at fk= 2πk
Problem 10.8
(a) The gradient of the performance index Jwith respect to his : dJ
dh = 2h+ 40.Hence, the time
update equation becomes :
hn+1 =hn−1
2∆(2hn+ 40) = hn(1 −∆) −20∆
(b) We note that Jhas a minimum at h=−20,with corresponding value : Jmin =−372.To
illustrate the convergence of the algorithm let’s choose : ∆ = 1/2.Then : hn+1 =hn/2−10,and,
using induction, we can prove that :
X
k=0 1
where h0is the initial value for h. Then, as n→ ∞,the dependence on the initial condition h0
vanishes and hn→ −10 1
1−1/2=−20,which is the desired value. The following plot shows the
expression for Jas a function of n, for ∆ = 1/2 and for various initial values h0.
h0=−25
h0=−30
h0=0
0 2 4 6 8 10 12 14 16 18 20
−400
−350
−250
−200
−150
−100
−50
0
50
Iteration n
J(n)
PROPRIETARY MATERIAL. c
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Problem 10.9
The linear estimator for xcan be written as : ˆx(n) = a1x(n−1)+a2x(n−1) = [x(n−1) x(n−2)]
a1
a2
.
Using the orthogonality principle we obtain :
E
x(n−1)
x(n−2)
ǫ
= 0 ⇒E
x(n−1)
x(n−2)
x(n)−[x(n−1) x(n−2)]
a1
a2
= 0
Problem 10.10
In Probl. 10.9 we found that the optimum (MSE) linear predictor for x(n),is ˆx(n) = bx(n−1).
Problem 10.11
9
The system C(z) = 1
1−0.9z−1has an impulse response : c(n) = (0.9)n, n ≥0.Then, we write the
input y(n) to the adaptive FIR filter :
y(n) = ∞
X
k=0
c(k)x(n−k) + w(n)
E
y(n)
y(n−1)
ǫ
= 0 ⇒E
y(n)
y(n−1)
x(n)−[y(n)y(n−1)]
b0
b1
= 0
b0
b1
y(n)y(n)y(n)y(n−1)
y(n−1)y(n)y(n−1)y(n−1)
−1
y(n)x(n)
y(n−1)x(n)
The various correlations are as follows :
E[y(n)x(n)] = E“∞
X
c(k)x(n−k)x(n) + w(n)x(n)#=c(0) = 1
1−0.81 +σ2
0.19 +σ2
and :
E[y(n)y(n−1)] = E
∞
X
k=0
∞
X
j=0
c(k)c(j)x(n−k)x(n−1−j)
X
X
10
Hence :
b0
b1
1
0.91
0.19
1
0.19 + 0.1
−1
1
0
0.85
−0.75
It is interesting to note that in the absence of noise (i.e when the term σ2
w= 0.1 is missing from the
diagonal of the correlation matrix), the optimum coefficients are : B(z) = b0+b1z−1= 1 −0.9z−1,
Problem 10.12
(a) If we denote by Vthe matrix whose columns are the eigenvectors {vi}:
V= [v1|v2|…|vN]
then its conjugate transpose matrix is :
V∗t=
v∗t
1
v∗t
2
…
v∗t
N
(b) To compute Γ1/2,we first determine V,Λ(i.e the eigenvalues and eigenvectors of the correlation
matrix). Then :
Γ1/2=
N
X
i=1
λ1/2
iviv∗t
i=VΛ1/2V∗t