Digital Signal Processing Solutions to Final 2013 Edited by Yang-Ting Justing Chou Confirmed by Prof.…

Digital Signal ProcessingSolutions to Final 2013

Edited by Yang-Ting Justing ChouConfirmed by Prof. Jar-Ferr Kevin Yang

LAB: 92923 R, TEL: ext. 621E-mail: [email protected]

Page of MediaCore: http://140.116.92.181

mailto:[email protected]

mailto:http://140.116.92.181

MediaCore Lab, National Cheng Kung University, Tainan, Taiwan

Announcement

• Final Exam: 繁城 , 9:10~11:30, 01/10

• Course Website: http://140.116.92.181/dsp.htm– Homework– Past Examinations

• TA E-mail: [email protected]

• TA Time: Wednesday, 10-11 am.

• HW#7 & HW#8 Submission: 23:59, 01/14

112/05/14 2

mailto:[email protected]


112/05/14 3

1.1 ( a ) Explanation


112/05/14 4

1.1 ( b ) Explanation


1.1 ( c ) Explanation

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1.1 ( d ) Explanation

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1.2 Auto-Correlation Function (ACF)

Computation: N+2*N*log2N

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假如 pole只有 -j, 則為 BPF, 但因zero中的 BR影響甚大 , 故此filter仍為 BR

1.3 Filters

H1(z):LP H2(z):LPH3(z):BPH4(z):BR

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HPF

BPF

LPF

LPF

BR

HPF

H5(z):LPH6(z):HPH7(z):BRH8(z):BR

H8(z)H4(z)


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1.4 Filters

Filter with real coefficient: Poles and zeros are conjugateLinear phase filter: zeros reciprocally appearMinimum phase filter: poles and zeros are all inside the unit circle (not includes unit circle)All-pass filter: poles and zeros reciprocally appearStable filter: all of poles are inside the unit circle (not includes unit circle)

(a) 1, 2, 3, 4, 6, 7, 8 (b) 3, 6(c) None(d) 3, 4, 5, 6, 7(e) None(f) 5, 8

1 5


(a)

2;25.0

(b)Frequency transformation:

( )

1 1

1

j jj j

j j

jj

j

e ee ee e

eee

Both filters of T1 and T2 are HPF to LPF

11

1, and

1j jzZ Z e z e

z

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2.1 Filter Design

Simplest transformation:

1

111

1)(

z

zzGZk

k


HPF to LPF

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2.1 (c) Filter Design


HFIR(z) = (1+b0z-1-2z-2) (1-3z-1+b1z-2) (1-3z-1+b2z-2+2z-3-b3z-4+z-5) = (1-3z-1+b1z-2+b0z-1-3b0z-2+b0b1z-3-2z-2+6z-3-2b1z-4) (1-3z-1+b2z-2+2z-3-b3z-4+z-5) = (1+(-3+b0) z-1+(b1-3b0-2)z-2+(6+b0b1)z-3-2b1z-4) (1-3z-1+b2z-2+2z-3-b3z-4+z-5)

b2=2; b3=3 b0=6; b1=-0.5;

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2.2 Linear Phase Filter

Linear phase filter: zeros reciprocally appear

?

Type I

M=even, h[n]=h[M-n], 0≤n≤M

Type II

M=odd, h[n]=h[M-n], 0≤n≤M

Type IIIM=even, h[n]=-h[M-n], 0≤n≤M

Type IV

M=odd, h[n]=-h[M-n], 0≤n≤M


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HIIR(z) = 44

321

30

22

11

13

3221)2)(1(

zazzz

zazazaza

44

321

403

3320

232

1311

3221)()2()2(

zazzzzaazaaazaazaaa

=

Kaaaaaaaaaaa

122

23

2 303203231

4

1 , Let K=1

2232

32

31

41

aaaa

aa

8505.5

8546.0

7092.38505.51700.1

4

3

2

1

0

a

a

aaa

2.3 All-Pass Filter

All-pass filter: poles and zeros reciprocally appear

12

30

320

aaaaa


41244224113132313521

],[ mnx

213411312

],[ mnh

213114312

],[~ mnh

(0,0)

(0,0)

(0,0)

y[2, 2] = 10y[-1, 2]= 0y[3, 2] = -14y[3, 4] = 5y[4, 2] = -5

112/05/14 14

2.4 2D - Linear Convolution

y[n, m] = x[n, m] * h[n, m] = Σk Σl x[k, l]∙h[n-k, m-l]


]2[][][][7

0

28

3

04 kXWnxWnzkZ

n

nk

n

nk

(a)

(b)

(c)

(d)

][][))])4[((][(][][ 48

7

08

7

08 kXWkXWnxnxWnwkW k

n

nk

n

nk

][][))])4[((][(][][ 48

7

08

7

08 kXWkXWnxnxWnykY k

n

nk

n

nk

]2[]2[][

]2[][][][][

48

48

kZWkZkW

kXkZkXWkXkW

k

k

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3.1 DFT


2

))][((][][][][Re][* nxnxnxevennykXkY

(a)

(b)

(c)

]5.0,5.3,5.05.1,1,5.05.1,5.3,5.0,2[][ jjjjjjny

][][2

][][2][Im2][][2][ **

nxnxj

nxnxjnxjnwkXOddkW

]4,0,2,0,0,4,0,2[][ jjjjnw

312][8][7

0

27

0

2 n=k=

ny kY

112/05/14 16

3.2 IDFT

]24, 3, 2, 1, 1,24, 3,2{[21][ jjjjny

]} 3,24, 1, 1, 2, 3,24,2[ jjjj

]24, 3, 2, 1, 1,24, 3,2[][ jjjjnw ]24, 3, 2, 1, 1,24, 3,2[ jjjj


112/05/14 17

(a)

]2[][][

]2[])[(

][][][)1(][][22

)1(1even for odd for

,0],[2

0

2

2

NkXkXkG

NkXWnxW

nxWnxnxnxnxnnnx

ng

N

n

nkN

Nn

N

Nn

Nn

n

(b)

)2

)3ln(2

(

][

][

][

][)3()1(][)3(][

][)3(

1

0

)2

)3ln(2

(2

2/)3ln(21

0

2/22

1

0

2)3ln(2/2

1

0

1

0

jNNkX

enx

eeenx

enxee

WnxWnxkY

nxny

N

n

N

jNNknj

jN

NnjN

n

NnNj

Nnkj

N

n

Nnkjj

njN

nNj

N

n

nkN

nnN

n

nkN

n

n

3.3 N-Point DFT

nnjNNnjNn

N eeW )1()2/(2

2

))2

)3ln(21(()

2)3ln((

][

][

][

][)3(][

][)3(

1

0

)2

)3ln((2

2/)3ln(21

0

2

1

0

2)3ln(

1

0

jNkXjNkX

enx

eenx

enxe

WnxkY

nxny

N

n

N

jNknj

jN

NnjN

n

Nnkj

N

n

Nnkjj

nj

N

n

nkN

n

n

OR

j )3ln()3ln(


0000004200210000

],[ mnx

0000426422213211

0000

jjjjrow column

jjjjj

jjjjj

6396335634321213211

4363521

112/05/14 18

3.4 2-D DFT

OR

00630042210021004221

jj

jirowcolumn

jjjjj

jjjjj

6396335634321213211

4363521

(0,0)

(0,0)


Thanks for Your Attention!

112/05/14 19

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Digital Signal Processing Solutions to Final 2013 Edited by Yang-Ting Justing Chou Confirmed by Prof.…