HW7sol - ENEE 630 Homework#7 Solution to 1 Recall that m =...

ENEE 630 Homework #7
Solution to 1 Recall that
Γ
m
= -
Δ
m − 1
P
m − 1
where
Δ
m − 1
=
m − 1
summationdisplay
k =0
a
m − 1 ,k
r ( k - m )
We have
r ( m ) = - Γ
∗
m
P
m − 1
-
m − 1
summationdisplay
k =1
a
∗
m − 1 ,k
r ( m - k )
or,
Γ
m
= -
r
∗
( m )
P
m − 1
-
1
P
m − 1
m − 1
summationdisplay
k =1
a
m − 1 ,k
r
∗
( m - k )
We also need,
a
m,k
= a
m − 1 ,k
+ Γ
m
a
∗
m − 1 ,m − k
, k = 0 , 1 , . . . , m
P
m
= P
m − 1
(1 - | Γ
m
|
2
)
(a)
P
0
= r (0) = 1
Γ
1
= -
r
∗
(1)
P
0
= - 0 . 8
P
1
= P
0
(1 - | Γ
1
|
2
)
= 0 . 36
Γ
2
= -
r
∗
(2)
P
1
-
r
∗
(1)
P
1
Γ
1
=
1
9
P
2
= P
1
(1 - | Γ
2
|
2
)
= 0 . 356
Also,
a
22
= Γ
2
a
21
= Γ
1
+ Γ
2
Γ
∗
1
= - 0 . 889
Δ
2
= - 0 . 0444
Γ
3
= -
Δ
2
P
2
= 0 . 125
1

(b) See Figure 1.
-1 z
-1 z -1 z
-0.8
-0.8
0.111
0.281
0.281
0.111
u[n]
Figure 1: P-1
(c) P
3
= P
2
(1 - | Γ
3
|
2
) = 0 . 35. See Figure 2
P
m
m
1
1 2 3
Figure 2: P-1
Solution to 2 (a) (i) The tap weights of a prediction error filter is
a
m
=
parenleftbigg
1
- w
0
parenrightbigg
where w
0
= R
− 1
M
r
M
and
r
M
= E [ u
M
( n - 1) u
∗
1
( n )]
= σ
2
α





e
− jw
e
− j 2 w
.
.
.
e
− jMw





= σ
2
α
e
− jw
s
M
( w )
2

with
s
M
( w )
△
=





1
e
− jw
.