Answers to Problems for Adaptive Filters (2nd Edition) - Behrouz Farhang-Boroujeny

Chapter 2
Discrete-Time Signals and Systems
1. (i)
x(n) =

3n n 0
0.7n n ≥ 0
⇒ X(z) =
−1
X
n=−∞
3n
z−n
+
∞
X
n=0
0.7n
z−n
=
∞
X
n=1
(3−1
z)n
+
∞
X
n=0
(0.7z−1
)n
=
z
3 − z
+
z
z − 0.7
And region of convergence is
0.7 |z| 3
(ii)
x(n) =



2n n 0
0.7n 0 ≤ n 5
0.5n n ≥ 5
and thus
X(z) =
∞
X
n=−∞
x(n)z−n
=
−1
X
n=−∞
2n
z−n
+
4
X
n=0
0.7n
z−n
+
∞
X
n=5
0.5n
z−n
=
∞
X
n=1
(2−1
z)n
+
4
X
n=0
(0.7z−1
)n
+
∞
X
n=5
(0.5z−1
)n
=
z
2 − z
+
1 − (0.7z−1)5
1 − 0.7z−1
+
(0.5z−1)5
1 − 0.5z−1
for
0.5 |z| 2
3
s
m
t
b
9
8
@
g
m
a
i
l
.
c
o
m
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4 Contents Chap. 2
2. By method of partial fraction,
H(z) =
1
(1 − 0.5z−1)(1 − 2z−1)
=
A
1 − 0.5z−1
+
B
1 − 2z−1
,
where A = −1
3 and B = 4
3.
(i) When the region of convergence is I, both terms on the right hand side are z-transform of
a left-sided sequence. We thus get
A
1 − 0.5z−1
= −A
−1
X
n=−∞
0.5n
z−n
,
B
1 − 2z−1
= −B
−1
X
n=−∞
2n
z−n
and
h(n) =

0 n ≥ 0
1
3 × 0.5n − 4
3 × 2n n 0.
(ii) When the region of convergence is II, the first corresponds to a right-sided sequence and
the second term to a left-sided sequence. We thus get
A
1 − 0.5z−1
= A
∞
X
n=0
0.5n
z−n
,
B
1 − 2z−1
=
−B × 0.5z
1 − 0.5z
= −B × 0.5z
∞
X
n=0
0.5n
zn
= −B
∞
X
n=1
0.5n
zn
= −B
−1
X
n=−∞
0.5−n
z−n
= −B
−1
X
n=−∞
2n
z−n
and
h(n) =

−1
3 × 0.5n n ≥ 0
4
3 × 2n n 0.
(iii) When the region of the convergence is III, both terms corresponds to a right sided sequence.
We thus get
A
1 − 0.5z−1
= A
∞
X
n=0
0.5n
z−n
,
B
1 − 2z−1
= B
∞
X
n=0
2n
z−n
and
h(n) =

−1
3 × 0.5n + 4
3 × 2n n ≥ 0
0 n 0.

Sec. 2.0 Contents 5
3. From (2.43),
φxx(k) = rxx(k) + |mx|2
.
Taking z-transform from both sides,
Φxx(z) =
∞
X
k=−∞
rxx(k)z−k
+ |mx|2
∞
X
k=−∞
z−k
.
We note that
∞
X
k=−∞
z−k
=
−1
X
k=−∞
z−k
+
∞
X
k=0
z−k
and for this to be convergent, |z| 1 for the first term on the right side, and |z| 1 for the
second term. Clearly there is no region of z for which
∞
P
k=−∞
z−k is convergent.
4. Applying (2.38),
φxx(k) = φvv(k) + E[sin(ω0n + θ) sin(ω0(n − k) + θ)]
= σ2
vδ(k) + E[
1
2
cos(ω0k) −
1
2
cos(ω0(2n − k) + 2θ)] = σ2
vδ(k) +
1
2
cos ω0k
where σ2
v = E[|v(n)|2] Hence,
Φxx(z) = σ2
v +
1
2
∞
X
k=−∞
cos ω0kz−k
and
1
2
∞
X
k=−∞
cos ω0kz−k
=
1
4
[
∞
X
k=−∞
ejω0k
z−k
+
∞
X
k=−∞
e−jω0k
z−k
]
In a similar way to the proof given in Problem 2.5, we can show that here also there is no region
of z-plane in which either of summation on the right side of the above equation may converge.
5. (i) From (2.43),
φxx(k) = rxx(k) + |mx|2
.
Taking z-transform from both sides,
Φxx(z) =
∞
X
k=−∞
rxx(k)z−k
+ |mx|2
∞
X
k=−∞
z−k
.
We note that
∞
X
k=−∞
z−k
=
−1
X
k=−∞
z−k
+
∞
X
k=0
z−k
and for this to be convergent, |z| 1 for the first term on the right side, and |z| 1 for
the second term. Clearly there is no region of z for which
∞
P
k=−∞
z−k is convergent.

6 Contents Chap. 2
(ii) Since x(n) is a stationary process with mx 6= 0, we can separate it into two parts
x(n) = mx + y(n)
In the above equation y(n) is a stationary process with my = 0. Taking a Fourier transform
from the above equation we have
X(ejω
) = mxδ(ω) + Y (ejω
)
y(ejω
) =
∞
X
n=−∞
y(n)e−jωn
In the vicinity of ω = 0, Y (ejω) becomes
lim
ω→0
Y (ejω
) =
∞
X
n=−∞
y(n)
Since my = 0, we can expect that the above equation would also be zero., and thus in
vicinity of ω = 0, we have
X(ejω
) = mxδ(ω)
(iii) As in previous part
x(n) = mx + y(n)
y(n) is a stationary process with my = 0. So we have
φxx(k) = m2
x + φyy(k)
Φxx(ejω
) = m2
xδ(ω) + Φyy(z)
Φyy(z) =
∞
X
k=−∞
φyy(k)e−jωk
In vicinity of ω = 0 we have
Φyy(z) =
∞
X
k=−∞
φyy(k)
=
∞
X
k=−∞
E[y(n)y∗
(n − k)]
= E

y(n)
∞
X
k=−∞
y∗
(n − k)
#
= 0
Accordingly,
Φxx(ejω
) = m2
xδ(ω)

Sec. 2.0 Contents 7
6. Solution to this problem follows the same line of derivations as Problem 2.5, with mx replaced
by Aejω0n.
7. Starting from (2.54), we arrived at (2.55) as follows
Φxx(z) =
∞
X
k=−∞
φxx(k)z−k
= (
∞
X
k=−∞
φ∗
xx(k)(z∗
)−k
)∗
= (
∞
X
k=−∞
φxx(−k)(
1
z∗
)k
)∗
(from (2.48))
= (
∞
X
k=−∞
φxx(k)(
1
z∗
)−k
)∗
(k → −k)
= (Φxx(
1
z∗
))∗
= Φ∗
xx(
1
z∗
)
Equation (2.56) is also proved in the same way.
8. (i) From (2.57)
Φvv(z) = H∗
(
1
z∗
)Φvv(z) = σ2
v × (
1
1 − az∗
)∗
= σ2
v ×
1
1 − a∗z
From (2.80),
Φuu(z) = H(z)H∗
(
1
z∗
)Φvv(z) =
σ2
v
(1 − az−1)(1 − a∗z)
(ii) For |a| 1,
1
1 − a∗z
=
∞
X
k=0
(a∗
)k
ak
=
0
X
k=−∞
(a∗
)−k
z−k
Since φvu(k) is the inverse z-transform of Φvu(z) and φvv(z) =
∞
P
k=−∞
φvu(k)z−k, it follows
that
φvu(k) =

0 k 0
σ2
v(a∗)−k, k ≤ 0
By method of partial fraction,
Φuu(z) =
σ2
v
(1 − a∗z)(1 − az−1) = −(σ2
v/a∗)z
(z−1/a∗)(z−a)
=
A
z − 1/a∗
+
B
z − a

8 Contents Chap. 2
Where
A =

−
σ2
v
a∗
z
z − a

z=1/a∗
= −
σ2
v
a∗
1
1 − |a|2
B =

−
σ2
v
a∗
z
z − 1/a∗

z=a
=
σ2
va
1 − |a|2
The first term on the right hand side corresponds to a left sided sequence,
A
z − 1/a∗
= −
Aa∗
1 − a∗z
=
σ2
v
1 − |a|2
1
1 − a∗z
=
σ2
v
1 − |a|2
∞
X
k=0
(a∗
)k
zk
=
σ2
v
1 − |a|2
0
X
k=−∞
(1/a∗
)k
z−k
and the second term to a right sided sequence
B
z − a
=
Bz−1
1 − az−1
= Bz−1
∞
X
k=0
ak
z−k
=
σ2
v
1 − |a|2
∞
X
k=1
ak−1
z−k
Using the above results, and since Φuu(z) =
∞
P
k=−∞
φuuz−k, we get
φuu(k) =
( σ2
v
1−|a|2 ak−1 k 0
σ2
v
1−|a|2 (a∗)−k k ≤ 0
(iii)
σ2
u = φuu(0) =
σ2
v
1 − |a|2
9. (i) When a 6= b,
Φvu(z) = H∗
(
1
z∗
)Φvv(z) =
σ2
v
(1 − a∗z)(1 − b∗z)
Φuu(z) = H(z)H∗
(
1
z∗
)Φvv(z) =
σ2
v
(1 − az−1)(1 − bz−1)(1 − a∗z)(1 − b∗z)
When a = b,
Φvu(z) = H∗
(
1
z∗
)Φvv(z) =
σ2
v
(1 − a∗z)2
Φuu(z) = H(z)H∗
(
1
z∗
)Φvv(z) =
σ2
v
(1 − az−1)2(1 − a∗z)2

Sec. 2.0 Contents 9
(ii) When a 6= b
Φvu(z) =
σ2
v
a∗ − b∗

a∗
1 − a∗z
−
b∗
1 − b∗z

Thus, we have
φvu(k) =
(
0 k 0
σ2
v
a∗−b∗

(a∗)−k+1 − (b∗)−k+1

k ≤ 0
Similarly,
Φuu(z) =
A
(z − a)
+
B
(z − b)
+
C
(z − 1/a∗)
+
D
(z − 1/b∗)
Where
A =

σ2
vz2
a∗b∗(z − b)(z − 1/a∗)(z − 1/b∗)

z=a
=
σ2
va2
(a − b)(1 − |a|2)(1 − ab∗)
B =

σ2
vz2
a∗b∗(z − a)(z − 1/a∗)(z − 1/b∗)

z=b
=
−σ2
va2
(a − b)(1 − |b|2)(1 − ba∗)
C =

σ2
vz2
a∗b∗(z − a)(z − b)(z − 1/b∗)

z=1/a∗
=
−σ2
v
(a∗ − b∗)(1 − |a|2)(1 − ba∗)
D =

σ2
vz2
a∗b∗(z − a)(z − b)(z − 1/a∗)

z=1/b∗
=
σ2
va2
(a∗ − b∗)(1 − |b|2)(1 − ab∗)
Hence,
Φuu(z) =
Az−1
(1 − az−1)
+
Bz−1
(1 − bz−1)
−
Ca∗
(1 − a∗z)
−
Da∗
(1 − a∗z)
=
∞
X
k=1
Aak−1
z−k
+
∞
X
k=1
Bak−1
z−k
−
0
X
k=−∞
C(a∗
)−k+1
z−k
−
0
X
k=−∞
D(b∗
)−k+1
z−k
and
φvu(k) =

Aak−1 + Bbk−1 k 0
−C(a∗)−k+1 − D(b∗)−k+1 k ≤ 0
When a = b,
φvu(k) =
1
2πj
I
C
Φvu(z)zk−1
dz =
1
2πj
I
C
σ2
vzk−1
(1 − a∗z)2
dz =
1
2πj
I
C
σ2
vzk−1
(a∗)2(z − 1/a∗)2
dz
Where C is a contour in |a| |z| |a|−1. For k ≥ 1, there is no poles inside C implies
that φvu(k) = 0. For k 1, substituting z = 1/p and noting that p = a∗ is the only pole
in C,
φvu(k) =
1
2πj
I
C
Φvu(1/p)p−k−1
dp =
1
2πj
I
C
σ2
vp−k+1
(p − a∗)2
dp

10 Contents Chap. 2
= Residue

σ2
vp−k+1
(p − a∗)2

p=a∗
=

d
dp
σ2
vp−k+1

p
= a∗
= σ2
v(1 − k)(a∗
)−k
Hence,
φvu(k) =

0 k 0
σ2
v(1 − k)(a∗)−k k ≤ 0
Similarly, applying the residue method to find inverse z-transform of Φuu(z),
φuu(k) =
( σ2
v
(1−|a|2)3 [(1 − |a|2)(1 + k) + 2|a|2]ak k ≥ −1
σ2
v
(1−|a|2)3 [(1 − |a|2)(1 − k) + 2|a|2](a∗)−k k −1
(iii)
σ2
u = φuu(0)
10. (i) Using (2.75),
Φvu(z) = H∗
(1/z∗
)Φvv(z) =
N−1
X
n=0
h(n)(z∗
)n
!∗
σ2
v
σ2
v
N−1
X
n=0
h∗
(n)zn
= σ2
v
0
X
n=−(N−1)
h∗
(−n)z−n
Using (2.80),
Φuu(z)H(z)H∗
(1/z∗
)Φvv(z) = σ2
v
N−1
X
n=0
h(n)z−n
N−1
X
m=0
h∗
(m)zm
σ2
v
N−1
X
n=0
N−1
X
m=0
h(n)h∗
(m)z−(n−m)
With n − m = k, we obtain
Φuu(z) = σ2
v
N−1
X
k=−(N−1)
αh(k)z−k
,
where αh(k) =
min(N−1,N−1−k)
P
m=max(0,−k)
h(m + k)h∗(m).
(ii) From the result of Part (a), we immediately obtain
φvu(k) =

σ2
vh∗(−k) −(N − 1) ≤ k ≤ 0
0 otherwise
and
φuu(k) =

σ2
vαh(k) −(N − 1) ≤ k ≤ N − 1
0 otherwise
= σ2
vαh(k) (since αh(k) = 0 for k ≤ −N and k ≥ N)

Sec. 2.0 Contents 11
11. From (2.65),
E[|xN (ejω
)|2
] =
N
X
n=−N
N
X
m=−N
E[x(n)x∗
(m)]e−jω(n−m)
=
N
X
n=−N
N
X
m=−N
φxx(n − m)ejω(n−m)
Next, we note that k = n − m can take values in the range −2N to 2N and for any k = n − m,
there are 2N + 1 − |k| similar terms added together. Thus we obtain
E[|xN (ejω
)|2
] =
2N
X
k=−2N
(2N + 1 − |k|)φxx(k)e−jωk
Dividing both sides of the result by 2N + 1, gives (2.66).
12. From (2.75), φxy(z) = H∗(1/z∗)Φxx(z). Using (2.55) and (2.56), we get
Φ∗
yx(1/z∗
) = H∗
(1/z∗
)Φxx(z)
Hence, Φyx(z) = H(z)Φ∗
xx(1/z∗) = H(z)Φxx(z) From (2.90), Φdy(z) = H∗(1/z∗)Φdx(z). Again
using (2.56), we get
Φ∗
yd(1/z∗
) = H∗
(1/z∗
)Φ∗
xd(1/z∗
)
Hence, Φyd(z) = H(z)Φxd(z).
13. Using ? to denote convolution,
(i) φyx(n) = h(n) ? φxx(n),
(ii) φxy(n) = h∗(−n) ? φxx(n),
(iii) φyd(n) = h(n) ? φxd(n),
(iv) φdy(n) = h∗(−n) ? φdx(n)
14. Let y(n) = y1(n) + y2(n), where y1(n) and y2(n) are the outputs of H(z) and G(z), respectively.
(i) Since u(n) and v(n) are uncorrelated,
φuy(n) = φuy1(n) + φuy2(n) = φuy1(n)
Hence,
Φuy(z) = Φuy1(z) = H∗
(1/z∗
)Φuu(z) = σ2
uH∗
(1/z∗
).
(ii) Similar to Part (b), Φvy(z) = σ2
vG∗(1/z∗).
(iii) Noting that y1(n) and y2(n) are originating from uncorrelated source, u(n) and v(n),
respectively, we obtain Φyy(z) = Φy1y1(z)+Φy2y2(z) = σ2
uH(z)H∗(1/z∗)+σ2
vG(z)G∗(1/z∗).

12 Contents Chap. 2
15. (i) Using Φxx(z) = H(z)H∗(1/z∗)Φvv(z), with H(z) = 1
1−0.5z−1 and Φvv(z) = 1, we obtain
Φxx(z) =
1
1 − 0.5z−1

1
1 − 0.5(1/z∗)−1
∗
=
1
(1 − 0.5z−1)(1 − 0.5z)
(ii) Since Φyy(z) = G(z)G∗(1/z∗)Φzz(z), where G(z) = 1 − 2z−1, we obtain
Φyy(z) =
1 − 2z−1
1 − 0.5z−1
×

1 − 2(1/z∗)−1
1 − 0.5(1/z∗)−1
∗
Φvv(z) =
1 − 2z−1
1 − 0.5z−1
×
1 − 2z
1 − 0.5z
=
1 − 2(z + z−1) + 4
1 − 0.5(z + z−1) + 1/4
= 4
(iii) Using Φvy(z) = H∗(1/z∗)G∗(1/z∗)Φvv(z),
Φvy(z) =

1 − 2(1/z∗)−1
1 − 0.5(1/z∗)−1
∗
Φvv(z) =
1 − 2z
1 − 0.5z
.
16. (i) Since x(n) =
P
k
h(k)v(n − k) and y(n) =
P
k
g(k)v(n − k),
φxy(m) = E[x(n)y∗
(n − m)] = E[
X
l
X
k
h(k)g∗
(l)v(n − k)v∗
(n − m − l)]
=
X
l
X
k
h(k)g∗
(l)E[v(n − k)v∗
(n − m − l)] =
X
l
X
k
h(k)g∗
(l)δ(−k + l + m).
But δ(−k + l + m) = 1 for k = l + m and zero elsewhere. Using this in the above result,
we obtain φxy(m) =
P
l
h(l + m)g∗(l).
(ii) Since Φxy(z) is the z-transform of φxy(m),
Φxy(z) =
X
m
φxy(m)z−m
=
X
m
X
l
h(l + m)g∗
(l)z−m
X
m
X
l
h(l + m)z−(m+l)
g∗
(l)zl
=
X
l
g∗
(l)zl
X
m
h(l + m)z−(m+l)
= G∗
(1/z∗
)H(z) = H(z)G∗
(1/z∗
).
17. (i) Since u(n) = x(n) + y(n), we have
φuu(k) = E[(x(n) + y(n))(x∗
(n + k) + y∗
(n + k))]
= E[x(n)x∗
(n + k)] + E[y(n)y∗
(n + k)] + E[x(n)y∗
(n + k)] + E[y(n)x∗
(n + k)]
= φxx(k) + φyy(k) + φxy(k) + φyx(k)
Hence,
Φuu(z) = Φxx(z) + Φyy(z) + Φxy(z) + Φyx(z) = Φxx(z) + Φyy(z) + Φxy(z) + Φ∗
xy(1/z∗
)

Sec. 2.0 Contents 13
(ii) Since u(n) is the output of a system with input v(n) and transfer function H(z) + G(z),
Φuu(z) = (H(z) + G(z))(H∗
(1/z∗
) + G∗
(1/z∗
))Φvv(z)
= H(z)H∗
(1/z∗
)Φvv(z)+G(z)G∗
(1/z∗
)Φvv(z)+H(z)G∗
(1/z∗
)Φvv(z)+H∗
(1/z∗
)G(z)Φvv(z)
Since v(n) in a zero-mean unit-variance white noise process,
Φvv(z) = 1
and we have
Φuu(z) = Φxx(z) + Φyy(z) + Φxy(z) + Φyx(z) = Φxx(z) + Φyy(z) + Φxy(z) + Φ∗
xy(1/z∗
)

Answers to Problems for Adaptive Filters (2nd Edition) - Behrouz Farhang-Boroujeny

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