Digital Signal Processing Solutions to Final 2013

1. Digital Signal ProcessingSolutions to Final 2013 Edited by Yang-Ting Justing Chou Confirmed by Prof. Jar-Ferr Kevin Yang LAB: 92923 R, TEL: ext. 621 E-mail: yangting115@gmail.com Page of MediaCore: http://140.116.92.181

2. 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: yangting115@gmail.com • TA Time: Wednesday, 10-11 am. • HW#7 & HW#8 Submission: 23:59, 01/14

3. 1.1 ( a ) Explanation

4. 1.1 ( b ) Explanation

5. 1.1 ( c ) Explanation

6. 1.1 ( d ) Explanation

7. 1.2 Auto-Correlation Function (ACF) Computation: N+2*N*log2N

8. 1.3 Filters H5(z):LP H6(z):HP H7(z):BR H8(z):BR H1(z):LP H2(z):LP H3(z):BP H4(z):BR BPF HPF LPF H4(z) H8(z) BR 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 LPF HPF

9. 1.4 Filters Filter with real coefficient: Poles and zeros are conjugate Linear phase filter: zeros reciprocally appear Minimum phase filter: poles and zeros are all inside the unit circle (not includes unit circle) All-pass filter: poles and zeros reciprocally appear Stable filter: all of poles are inside the unit circle (not includes unit circle) 1 5 (a) 1, 2, 3, 4, 6, 7, 8 (b) 3, 6 (c) None (d) 3, 4, 5, 6, 7 (e) None (f) 5, 8

10. 2.1 Filter Design (a) Simplest transformation: (b) Frequency transformation: Both filters of T1 and T2 are HPF to LPF

11. 2.1 (c) Filter Design HPF to LPF

12. 2.2 Linear Phase Filter ? Linear phase filter: zeros reciprocally appear Type I Type II Type III Type IV 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) M=even, h[n]=h[M-n], 0≤n≤M M=odd, h[n]=h[M-n], 0≤n≤M M=even, h[n]=-h[M-n], 0≤n≤M M=odd, h[n]=-h[M-n], 0≤n≤M b2=2; b3=3 b0=6; b1=-0.5;

13. 2.3 All-Pass Filter All-pass filter: poles and zeros reciprocally appear HIIR(z) = = , Let K=1

14. 2.4 2D - Linear Convolution y[n, m] = x[n, m] * h[n, m] = ΣkΣlx[k, l]∙h[n-k, m-l] (0,0) (0,0) (0,0) y[2, 2] = 10 y[-1, 2]= 0 y[3, 2] = -14 y[3, 4] = 5 y[4, 2] = -5

15. 3.1 DFT (a) (b) (c) (d)

16. 3.2 IDFT (a) (b) (c)

17. 3.3 N-Point DFT (a) (b) OR

18. 3.4 2-D DFT row column (0,0) OR column row (0,0)

19. Thanks for Your Attention!