【導(dǎo)讀】London:Prentice-Hall,3rded.,2020.Wiley-Interscience,N.J.,2020.EngineeringPress,Massachusetts,2020.38?h[n]=0forn<0andn?N. 2-BFIRFilterDesign. h[-1]h[0]h[1]h[k-2]h[k-1]h[k]h[k+1]h[k+2]h[N-2]h[N-1]h[N]h[N+1]. r[-k-1]r[-k]r[-k+1]r[-2]r[-1]r[0]r[1]r[2]r[k-1]r[k]r[k+1]r[k+2]. (a)r[n]=h[n+k],wherek=(N?1)/2.Specially,whenh[n]isevensymmetrich[n]=h[N?1?n]. (b)s[0]=r[0],s[n]=2r[n]for0<n?k.h[n](h[n]?0for0?n?N?1). r[n]=h[n+k],k=(N?1)/2(r[n]?0for-k?n?k,seepage41). n].Sets[0]=r[0],s[n]=2r[n]forn?0.(F=f?2jFkHFeRF???RFsnnF?2jFn. HFhne???dfMaxHfHf?45?dfMaxHfHf?[]d

　　

【正文】 –, h[1] = h[7] = s[3]/2 = , h[0] = h[8] = s[4]/2 = . 66 ? Example 2: Design a 7length digital filter in the minimax sense ideal filter: Hd(F) = 1 for 0 ? F , Hd(F) = 0 for F ? , transition band: F weighting function: W(F) = 1 for 0 ? F ? , W(F) = for ? F ? , (Step 1) Since N = 7, k = (N1)/2 = 3, k+2 = 5, ? Choose 5 extreme frequencies (., F0 = , F1 = , F2 = , F3 = , F4 = ) 67 (Step 2) ? s[0] = , s[1] = , s[2] = , s[3] = , e = 1 1 [ 0] 11 1 [ 1 ] 11 2 [ 2] 01 2 [ 3 ] 01 1 1 1 2 0sssse? ? ? ? ? ?? ? ? ? ? ?? ? ?? ? ? ? ? ????? ? ? ? ? ?? ? ? ? ? ?? ? ? ? ? ?? ? ? ? ? ???? ? ? ? ? ?R(F) 68 After Step 2, (Step 3) err(F) (Step 4) extreme points: , , , , . (Step 5) E0 = Max[|err(F)|] = , return to Step 2. ? ? ? ?? ? ? ? ? ?[ 0 .4 6 3 8 0 .6 3 2 7 c o s 2 0 .0 8 0 9 c o s 40 .1 6 0 8 c o s 6 ]dFFF H F W F???? ? ???Iteration 1 2 3 4 5 6 7 Max[|err(F)|] 69 After 7 times of iteration s[0] = , s[1] = , s[2] = , s[3] = , e = (Step 5): h[3] = , h[2] = h[4] = s[1]/2 = , h[1] = h[5] = s[2]/2 = , h[0] = h[6] = s[3]/2 = , h[n] = 0 for n 0 and n 6 0 0 . 0 5 0 . 1 0 . 1 5 0 . 2 0 . 2 5 0 . 3 0 . 3 5 0 . 4 0 . 4 5 0 . 5 0 . 500 . 511 . 5R ( F )dH ( F )70 附錄二： Spectrum Analysis for Sampled Signals ? ? ? ?1 2/0N j nm NnX m x n e ?? ??? ?簡(jiǎn)單的規(guī)則： 把間隔由 1 換成 fs /N where fs = 1/ ?t 已知 x[n] 是由一個(gè) continuous signal y(t) 取樣而得 x[n] = y(n?t) DFT: FT: Q: x[n] 的 DFT 和 y(t) 的 Fourier transform 之間有什麼關(guān)係？ ? ? ? ?2j f tY f e y t d t?? ???? ?(學(xué)信號(hào)處理的人一定要會(huì)的基本常識(shí) ) sffmN? (Very important) 71 ? ? st fX m Y m N???? ????fs = 1/ ?t for m ? N/2 ? ? () st fX m Y m N N??? ? ?????for N/2 證明： ? ? ? ?2j f tY f e y t d t?? ???? ?? ? ? ? ? ?22f t f tj m n j m nf t t tnnY m e y n e x n??? ? ? ? ? ?? ? ? ? ? ???用 t = n?t, f = m?f 代入 當(dāng) 1tf N? ? ? ., 1 sftfNN? ? ??? ?? ?? ?2 mnjsNtntfY m e x nND F T x n?????????????72 Example： 已知 y() = , y() = , y(0) = , y() = , y() = (Step 1) Choose N (N ? 5), N is the number of points of the DFT 如果已知 y() = 0 for k 2 and k 2 (也就是 y(t) 在無限多個(gè)點(diǎn)上的值是已知的 )， 則 N 選得越大越好 如果不確定 y() 在 k 2 或 k 2 的時(shí)候的值， (也就是 y(t) 只有 5 個(gè)點(diǎn)值是已知的 ) 則選擇 N = 5 如何用 DFT 來出 y(t) 的頻譜？ 0 . 8 0 . 6 0 . 4 0 . 2 0 0 . 2 0 . 4 0 . 6 0 . 80123y ( t )73 (Step 2) Set ? ? ? ?0 .1x n y n? for n = 0, 1, 2 ? ? ? ?0 . 1x n N y n?? for n = 2, 1 例如，當(dāng) N = 16 時(shí) ? ? 0xn? otherwise Q: 為什麼要把 y(), y() 移到 x[N2], x[N1] 而非移到 x[2], x[1]？ A: 因?yàn)? x[n] 是 DFT 的 input DFT input 的 index 範(fàn)圍必需是 0 ? n ? 15 0 2 4 6 8 10 12 14 160123x [ n ]74 (Step 3) Perform the DFT for x[n] ? ? ? ?1 2/0N j nm NnX m x n e ?? ??? ?(Step 4) ? ?stfY m X mN?? ??????? ?() s tfY m N X mN??? ? ?????for m ? N/2 for m N/2 以這個(gè)例子而言 11 2516 tfNN? ? ???0 2 4 6 8 10 12 14 1602468X [ m ]75 ? ? ? ?0 . 6 2 5 0 . 1Y m X m?for m ? 8 ? ? ? ?0 . 6 2 5 ( 1 6 ) 0 . 1Y m X m??for m 8 5 4 3 2 1 0 1 2 3 4 500 . 20 . 40 . 60 . 8Y ( f )faxis