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(ELEC317)2003_midtermSol.pdf
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Elec317 midterm solution
Q.1
a) [14 180 181 182 17 10 3 2 1]
b)
Let

vv
y1 = x1 . h
vv
y2 = x2 . h
vv vv
(ax + bx ) . h = a(x . h) + b(x . h)
12 12
vv
= ay1 + by2
c) [20 21 23 24 23 4 3 2 1]
d)
Input1/Output1
[0 910] MedianFilter [0 9 9]
Input2/Output2
[100 0 50] MedianFilter [050 0]

Sum of Inputs
[0 910] + [100 0 50] = [100 9 60]

Sum of Outputs

[0 9 9] + [050 0] = [0 59 9]
Apply median filter on sum of inputs
[100 9 60] MedianFilter [9 60 9] [0 59 9]
Since sum of input does not give sum of output, so median filter is not linear
e) median filter f) median filter g) use a length 5 median filter
Q.2 a)
P.1 = PH
or
.1*
P = PT
b)
Energy of x is

Ex =

[]2 +
x 0
[]2 +... +
x 1
x[N .1]2
H
=xx Energy of y is
Ey =yHy =
[]2 +... +
y 0 y[N .1]2
=[][]Ax H Ax
=xH AHAx H AHA
=xx =Ex (Note =I ) .Energy is preserved.
v
v rv v
c) Rx =E[(x .E[x])(x .E[x])H ]
vv
d) E[ y] =AE[x] e)
vv
Let x = E[x]
vv
y = E[ y]
vv vv
Ry = E[( y .y )( y .y )H ]
vv vv
= E[(Ax .Ax )(Ax .Ax )H ]
v vv
= E[A(x .x )(x .x )H AH ]
v vv
= AE[(x .x )(x .x )]AH
= ARx AH
f)
KL Transform/Eigen Decomposition/Diagonalization

v
Let ti and i be corresponding eigenvector and eigenvalues of Rx
vv v
.1 H
T =[tt K t ] , T = T
12 N
.10 K 0 . ..
0 M
. 2 .=
.MO 0 . ..
.0 K 0 N .
Rt v =t v
xi i
RxT = T
T H RxT = Put A=TH R = AR AH = T HRT =
yx x
Since the covariance matrix is diagonal Cov( yi , yj ) = 0 (for i j, . i,j[1,N])
so, the elements of y are uncorrelated
g)
Rx .I
= 0
5 . 3

= 0
35 .
(5 .)2 . 9 = 0
( . 2)( . 8)= 0
= 2,8

for =2 .
.
.. 1 .. = 2.. 1 ..
.53..u ..u .
35 uu

. .. 2 .. 2 .
.5u + 3u = 2u

12 1
.
.3u1 + 5u2 = 2u2
u1 =. u2

. 1 .
eigen vector=
..
.1
..
for =8
.53.. v1 .. v1 . .
.
.. .. = 8.. ..
35 vv
. .. 2 .. 2 . 5v + 3v = 8v
. 12 1
.
3v + 5v = 8v
. 12 2 v = v
12
.1.
eigen vector=
.
.

1
..
1 . 11.
T =

.
.

.11
2 .. 1 .1 .1..20.
A =
, Ry =
.
..
.

11 08
2 .. ..
Alternative method for 2g)
By trial and error, try different values of A, and prove ARx AH =

3a)

91 61
. .
.
0 25

.

32

.

.

A

.
.=
.
.

.

7 39

.

38

100 129

.

F

.
.=
91 140

0 255 0

.

.

Y

.
.=
.
.

0 255 0

100

91 61
0 f (i, j) <= 42
. .
.
.

126

.

E

.
.=
91

.

115

3b) modify equation 3c
. ...

85 42

f (i, j
) <= 127

<

y(i, j)
=

170
. ..
127

f (i, j
) <= 212

<

255 f (i, j)
212

>

3c) use length 3 low-pass filter
.. . ..
1 1 1
5757 57
.. . .

h
=

.

.

X.

.
.=
.
.

5757 57

other filter such as triangular filter is also accepted.