14-A Orthogonal and Dual Orthogonal Y = A X

Transcription

1 489 XIV. Orthogonal Transform and Multiplexing 14-A Orthogonal and Dual Orthogonal Any M N discrete linear transform can be expressed as the matrix form: N N N 1 y[0] x[0] y[1] x[1] y[2] x[2] ym [ 1] [ 1] M10 M 11 M 12 M 1N 1 xn Y = A X N 1 ym xn [ ], [ n] xn [ ] [ n] m inner product n0 m

2 Orthogonal: N 1 k h k h n0 [ n], [ n] [ n] [ n] 0 when k h 490 orthogonal transforms 的例子 : discrete Fourier transform discrete cosine, sine, Hartley transforms Walsh Transform, Haar Transform discrete Legendre transform discrete orthogonal polynomial transforms Hahn, Meixner, Krawtchouk, Charlier

3 為什麼在信號處理上, 我們經常用 orthogonal transform? Orthogonal transform 最大的好處何在? 491

4 If partial terms are used for reconstruction for orthogonal case, perfect reconstruction: N 1 1 xn Cm ym [ ] m[ n] m0 492 partial reconstruction: 由於 K xnx n 2 N1 N1 n0 mk 1 1 K 1 1 K m m m0 x n C y[ m] [ n] reconstruction error of partial reconstruction C 1 m y[ m] 2 1 m N1 N1 N1 1 1 m m m 1 m n0 mk m K N1 N1 N1 1 1 m m 1 m m mk m K n0 N1 N1 mk m K C y[ m] [ n] m 1 1 K < N C y[ m] [ n] C y [ m ] [ n] C y[ m] C y [ m ] [ n] [ n] C N m y[ mc ] m y [ m 1 1] Cm m m1 Cm y[ m] mk 一定是正的, 可以保證 K 越大, reconstruction error 越小

5 For non-orthogonal case, perfect reconstruction: partial reconstruction: N 1 xn Bnmy, [ m] m0 K 1 xk n Bn, my[ m] m0 B = A 1 K < N 493 reconstruction error of partial reconstruction N1 N1 2 xn x n K n0 mk N1 N1 N1 1 Bnmym, [ ] n0 mk m K N1 N1 N1 ym [ ] y [ m1] B n, m B n, m1 mk m K n0 N Bnmy, [ m] B n, m y [ m ] 1 1 由於 ymy [ ] [ m] Bnm, B nm, 不一定是正的, 1 1 n0 無法保證 K 越大, reconstruction error 越小

6 14-B Frequency and Time Division Multiplexing 494 傳統 Digital Modulation and Multiplexing: 使用 Fourier transform Frequency-Division Multiplexing (FDM) N 1 exp 2 z t X j f t X n = 0 or 1 n0 n n X n can also be set to be 1 or 1 When (1) t [0, T] (2) f n = n/t N 1 exp 2 nt n z t X j n0 T it becomes the orthogonal frequency-division multiplexing (OFDM).

7 Furthermore, if the time-axis is also sampled t = mt/n, m = 0, 1, 2,.., N 1 N 1 T X exp 2 nm n j z m N n0 N t [0,T] sampling for t-axis 495 then the OFDM is equivalent to the transform matrix of the inverse discrete Fourier transform (IDFT), which is one of the discrete orthogonal transform. Modulation: N 1 T, Ym z m A m n X N m j j j 1 e e e j j j A 1 e e e j j j 1 e e e 2 4 2( N 1) N N N 4 8 4( N 1) N N N 2( N1) 4( N1) 2( N1)( N1) N N N n

8 Modulation: Y m N 1 Am, nx m0 n 496 N 1 1 X n A m ny N Demodulation:, m0 m Example: N = 8 X n = [1, 0, 1, 1, 0, 0, 1, 1] (n = 0 ~ 7)

9 Time-Division Multiplexing (TDM) 497 T T T z 0 X, z X, z 2 X,, z ( N 1) XN N N N N 1 ym zmt Am, nx N m0 n A (also a discrete orthogonal transform)

10 498 思考 : 既然 time-division multiplexing 那麼簡單 那為什麼要使用 frequency-division multiplexing 和 orthogonal frequency-division multiplexing (OFDM)?

11 14-C Code Division Multiple Access (CDMA) 499 除了 frequency-division multiplexing 和 time-division multiplexing, 是否還有其他 multiplexing 的方式? 使用其他的 orthogonal transforms 即 code division multiple access (CDMA) CDMA is an important topic in spread spectrum communication 參考資料 [1] M. A. Abu-Rgheff, Introduction to CDMA Wireless Communications, Academic, London, 2007 [2] 邱國書, 陳立民譯, CDMA 展頻通訊原理, 五南, 台北, 2002.

12 CDMA 最常使用的 orthogonal transform 為 Walsh transform 500 y x1 y x 2 y x3 y x 4 y x 5 y x6 y x 7 y x 8 8 channel 1 channel 2 channel 3 channel 4 channel 5 channel 6 channel 7 channel 8 channel 1

13 501 當有兩組人在同一個房間裡交談 (A 和 B 交談 ), (C 和 D 交談 ), 如何才能夠彼此不互相干擾? (1) Different Time (2) Different Tone (3) Different Language

14 502 CDMA 分為 : (1) Orthogonal Type (2) Pseudorandom Sequence Type Orthogonal Type 的例子 : 兩組資料 [1, 0, 1] [1, 1, 0] (1) 將 0 變為 1 [1, 1, 1] [1, 1, 1] (2) 1, 1, 1 modulated by [1, 1, 1, 1, 1, 1, 1, 1] (channel 1) [1, 1, 1, 1, 1, 1, 1, 1, -1, -1, -1, -1, -1, -1, -1, -1, 1, 1, 1, 1, 1, 1, 1, 1] 1, 1, 1 modulated by [1, 1, 1, 1, -1, -1, -1, -1] (channel 2) [1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, -1, -1, -1, -1, 1, 1, 1, 1] (3) 相合 [2, 2, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, -2, -2, -2, -2, 0, 0, 0, 0, 2, 2, 2, 2]

15 demodulation [2, 2, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, -2, -2, -2, -2, 0, 0, 0, 0, 2, 2, 2, 2] [1, 1, 1, 1, 1, 1, 1, 1] [1, 1, 1, 1, 1, 1, 1,1] [1, 1, 1, 1, 1, 1, 1, 1] 503 內積 = 8

16 504 注意 : (1) 使用 N-point Walsh transform 時, 總共可以有 N 個 channels (2) 除了 Walsh transform 以外, 其他的 orthogonal transform 也可以使用 (3) 使用 Walsh transform 的好處

17 Orthogonal Transform 共通的問題 : 需要同步 synchronization R 1 = [ 1, 1, 1, 1, 1, 1, 1, 1] R 2 = [ 1, 1, 1, 1, 1, 1, 1, 1] R 5 = [ 1, 1, 1, 1, 1, 1, 1, 1] R 8 = [ 1, 1, 1, 1, 1, 1, 1, 1] 505 但是某些 basis, 就算不同步也近似 orthogonal <R 1 [n], R 1 [n]> = 8, <R 1 [n], R k [n]> = 0 if k 1 <R 1 [n], R k [n1]> = 2 or 0 if k 1. 這裡的 shift 為 circular shift

18 Pseudorandom Sequence Type 不為 orthogonal,capacity 較少 但是不需要同步 (asynchronous) 506 Pseudorandom Sequence 之間的 correlation C() -axis b 1 p(t+ 1 ) + b 2 p(t + 2 ) recovered: ( ) ( ) ( ) 0 bpt bpt pt dtbc bc b ( 若 C(0) = 1, C( 2 1 ) 0) 1, 2 不必一致

19 507 CDMA 的優點 : (1) 運算量相對於 frequency division multiplexing 減少很多 (2) 可以減少 noise 及 interference 的影響 (3) 可以應用在保密和安全傳輸上 (4) 就算只接收部分的信號, 也有可能把原來的信號 recover 回來 (5) 相鄰的區域的干擾問題可以減少

20 相鄰的區域, 使用差距最大的 語言, 則干擾最少 508 B 區 A 區 假設 A 區使用的 orthogonal basis 為 k [n], k = 0, 1, 2,, N 1 B 區使用的 orthogonal basis 為 h [n], h = 0, 1, 2,, N 1 設法使 max [ n], [ n] k h [ n], [ n] k h 為最小 k = 0, 1, 2,, N 1, h = 0, 1, 2,, N 1

21 附錄十五 Conventional Machine Learning Methods 509 雖然近年來有許多準確度很高的 advanced machine learning 的方法, 但一些較傳統的方法, 由於架構簡單, 不需大量且價格昂貴的 GPU, 且 training time 較少, 仍被產業界廣泛的使用 Conventional machine learning methods for classification: (i) Nearest Neighbor, k Nearest Neighbors (knn) (ii) Support Vector Machine (SVM) (iii) Neural Network Deep learning and the convolutional neural network (CNN) can be viewed as the improved versions of the neural network. Andrew Ng, Machine Learning, Stanford Cousera, available from

22 510 A. k Nearest Neighbors (knn) k Nearest Neighbors (knn) ( 最簡單的方法 ) 選擇 k 個在特徵平面上最近的點 再投票 例子 : 左圖當中,input ( 綠色 ) 要被分類為紅色還是藍色? 3 Nearest Neighbors (solid lines) 藍色 1 票, 紅色 2 票, 判斷結果為紅色 5 Nearest Neighbors (dash lines) 藍色 3 票, 紅色 2 票, 判斷結果為藍色 Each axis corresponds to a feature.

23 511 Weighted k Nearest Neighbors (W-kNN) f 2 Some features are more important than others. Assign larger weights for the features of more importance. f 1 distance w ( f ( a) f ( b)) w ( f ( a) f ( b)) w ( f ( a) f ( b)) m m m w1 w2 w m,,, : weights

24 B. SVM (support vector machine) 512 SVM (support vector machine, 支持向量機 ): 找出適當的分界線, 將資料分為 Class 1, Class 2 兩群

25 Nonlinear SVM: x: inputs, L 1, L 2, L 3,., L m : 標竿 513 f 1 = similarity(x, L 1 ) L 2 L 7 f 2 = similarity(x, L 2 ) L 1 L 3 L 6 f 3 = similarity(x, L 3 ).. f m = similarity(x., L m ) L 4 L 5 L m 以 f 1, f 2, f 3,., f m 為 features SVM 程式下載 C. C. Chang and C. J. Lin, LIBSVM -- A library for SVM, available from

26 C. Neural Network 514 Neural Network: y k 判斷結果 2-Layer Neural Network x j f i 特徵 (features) x g w w f j 0, j i, j i yk gv0, k vj, kxj i j g z 1 1 e z 兩層以上的 Neural Network: Deep Learning

27 期末的勉勵 515 人生難免會有挫折, 最重要的是, 我們面對挫折的態度是什麼 長遠的願景要美麗, 短期的目標要務實

28 516 祝各位同學暑假愉快! 各位同學在研究上或工作上, 有任何和 digital signal processing 或 time frequency analysis 方面的問題, 歡迎找我來一起討論