Digital Signal ProcessingSolutions to Final 2011

Edited by Shih-Ming HuangConfirmed by Prof. Jar-Ferr Yang

 LAB: 92923 R, TEL: ext. 621

E-mail: smhuang@video5.ee.ncku.edu.twPage of MediaCore: http://mediawww.ee.ncku.edu.tw

• 1-1• (a)

],[

0000

0000

0000

09600

02400

02400

02400

02400

6666

6666

6666

6666

],[

3operationcolumn

operationrow

3

kX

mnx

3

0

3

044],[],[

n m

mnkWWmnxkX

We could compute 2D DFT by row-column method:

e.g.

• 1-1• (b)

• 1-1• (c)

• 1-2• (a)

• 1-2• (b)

• 1-2• (c)

• 1-3• H1 :LP• H2 :LP• H3 :BP• H4 :HP• H5 :HP• H6 :HP• H7 :BR• H8 :LP

2-1

(a)

1

1

1, and

1j jz

Z Z e z ez

11

11

1 0.2

( ) ( ) zLP Zz

H z H Z

0.2 (b) Frequency transformation:

1 1

j j

j j

j j

e ee e

e e

11

1

221

( ) ( ) zLP Zz

H z H Z

2

1

1

1, and

1j jz

Z Z e z ez

( )

1 1

1

j j

j j

j j

j

j

j

e ee e

e ee

ee

LPF to LPF LPF to HPF

(c)

2-1

(c)

)( jeH

H(e

jω)

ω

2ππ

-π

1

11

1

z

zZ

,05.0

LPF to HPF2-1

• 2-2

1 1 2 1 2 3 40 1 2 3

0 1 2 3

0 1 2 2

( ) (2 )(3 2 )(1 2 2 )

2, 3, 0, 1

2, 3, 0, 1

FIRH z b z z b z z b z z b z

b b b b

b b b b

or

• 2-3

To make the above formula be a all-pass filter, we should set:

a4a1= 0

a3a1 + a4 = -5k

a4a2 + a3= 3k

a4a0 + a3a2 = -2k

a3a0= 1

All-pass filter: poles and zeros reciprocally appear

1 1 2 34 3 1 2 0

1 2 3

1 2 3 44 1 3 1 4 4 2 3 4 0 3 2 3 0

1 2 3 4

( )( )( )

1 2 3 5

( ) ( ) ( )

1 2 3 5 0

IIR

a a z a z a z a zH z

z z z

a a a a a z a a a z a a a a z a a z

z z z z

0,3,3

2,

3

5,

3

143210 aaaaa

• 2-4

y[0,1]=7, y[-1, 2]=0, y[5, 2]=24,y[3, 6]=0, y[6, 9]=0

(0,0)

(0,0)

1 2 2 3 1

3 2 3 2 3[ , ]

3 5 2 2 2

2 4 2 1 2

x n m

2 1 3

[ , ] 2 2 4

1 1 5

h n m

5 1 1

[ , ] 4 2 2

3 1 2

h n m

(0,0)

• 3-1• (a)

• (b)

• (c)

• (d)

• (e)

15 154

16 16 160 0

( ) [ ] [ ] 16 [4] 64k k

k k

j X k W X k x

16( ( [ ])) 16 [ ]

[80 0 0 0 0 0 0 0 0 32 -48 64 -64 48 -16 32]

DFT DFT x n x n

15 158

16 16 16 160 0

[8] [ ] ( 1) [ ] 14n n

n n

X W x n x n

7 70

8 8 8 80 0

[0] [ ] [ ] 8n

n n

X W x n x n

7 74

8 8 8 80 0

[4] [ ] ( 1) [ ] 14n n

n n

X W x n x n

• 3-2• (a)

• (b)

• (c)

*

*8

[ ] [ ][ ] Re{ [ ]}

21

[ ] ( [ ] [(( )) ]2

x n x ny n x n

Y k X k X k

*

*

[ ] { [ ]} ( [ ] [ ]) / 2

[ ] ( [ ] [ ]) / 2

w n Even x n x n x n

W k X k X k

7 72 /8

00 00

[ ] ( [ ] ) [ ] [0]j nk

kn nk

W k w n e w n W

j]-1.5j,-2.51.5j,11.5j,1,2.5-1.5j,2.5-j,1[-3,-2.5

],1,4,-2,-3[-3,-2,4,1

=-3

• 3-3• (a)

• (b)

• (c)

2 3

[ ] 2 [(( 2)) ] 3 [(( 3)) ]

[ ] 2 [ ] 3 [ ]

N N

k kN N

w n x n x n

W k X k W X k W

[ ], 1 ( 1)[ ] [ ] 0.5 [ ] 0.5( 1) [ ]

0, 2

[ ] 0.5 [ ] 0.5 [ ]2

nnx n n even

g n x n x n x nn odd

NG k X k X k

[ ] ( 1) [ ]

[ ] [ ]2

ny n x n

NY k X k

• 3-4

X(0,0)(0,0) (0,0)

1111

11

1111

11

0031

0042

0000

0000

1111

11

1111

11

jj

jj

jj

jj

jj

jjjj

jj

jjjj

3432

3625524

718716

5562324

(0,0)

(0,0)