TEST BANK FOR Adaptive Filter Theory 4th Edition By Simon Haykin (Solution manual only)

1.1 Let
(1)
(2)
We are given that
(3)
Hence, substituting Eq. (3) into (2), and then using Eq. (1), we get
1.2 We know that the correlation matrix R is Hermitian; that is
Given that the inverse matrix R-1 exists, we may write
where I is the identity matrix. Taking the Hermitian transpose of both sides:
Hence,
That is, the inverse matrix R-1 is Hermitian.
1.3 For the case of a two-by-two matrix, we may
ru(k) = E[u(n)u*(n – k)]
ry(k) = E[y(n)y*(n – k)]
y(n) = u(n + a) – u(n – a)
ry(k) = E[(u(n + a) – u(n – a))(u*(n + a – k) – u*(n – a – k))]
= 2ru(k) – ru(2a + k) – ru(– 2a + k)
RH = R
R –1RH = I
RR –H = I
R –H R –1 =
Ru = Rs + Rν
2
For Ru to be nonsingular, we require
With r12 = r21 for real data, this condition reduces to
Since this is quadratic in , we may impose the following condition on for nonsingularity
of Ru:
where
1.4 We are given
This matrix is positive definite because
r11 r12
r21 r22
σ2 0
0 σ2
= +
r11 σ2 + r12
r21 r22 σ2 +
=
det(Ru) r11 σ2 ( + ) r22 σ2 = ( + ) – r12r21 > 0
r11 σ2 ( + ) r22 σ2 ( + ) – r12r21 > 0
σ2 σ2
σ2 1
2
--(r11 + r22) 1
4Δr
(r11 + r22)2 – 1
– --------------------------------------
 
 
 
>
Δr r11r22 r12
2 = –
R 1 1
1 1
=
aTRa [a1,a2] 1 1
1 1
a1
a2
=
a1
2 2a1a2 a2