Review:NP theorem W-1 pxHo)=了 N-1 px:H1)=7 Follow the NP rule W-1 L(x)= {三w} m=0 a=…{可空-宫]}， W-1 W-1 0 whxiong@uestc.edu.cn 11
whxiong@uestc.edu.cn Review: NP theorem 11 p(x; H1) = 1 (2¼¾2) N=2 exp ( ¡ 1 2¾2 N X¡1 n=0 [x(n) ¡ s(n)] 2 ) p(x; H0) = 1 (2¼¾2) N=2 exp ( ¡ 1 2¾2 N X¡1 n=0 [x(n)] 2 ) Follow the NP rule L(x) = p(x; H1) p(x; H0) = exp ( ¡ 1 2¾2 N X¡1 n=0 £ (x[n] ¡ s[n]) 2 ¡ x 2 [n] ¤ ) L(x) = exp ( ¡ 1 2¾2 " ¡ N X¡1 n=0 2s[n]x[n] + N X¡1 n=0 s 2 [n] #) > °
Review:NP theorem N-1 :Ho)= N-1 p(x;H)= Follow the NP rule () {三w到} W-1 m=0 w-{[4三小} N-1 N-1 m=0 - 1 W-1 ∑ rijzin>7+a∑s2m m=0 m=0 whxiong@uestc.edu.cn 12
whxiong@uestc.edu.cn Review: NP theorem 12 p(x; H1) = 1 (2¼¾2) N=2 exp ( ¡ 1 2¾2 N X¡1 n=0 [x(n) ¡ s(n)] 2 ) p(x; H0) = 1 (2¼¾2) N=2 exp ( ¡ 1 2¾2 N X¡1 n=0 [x(n)] 2 ) Follow the NP rule L(x) = p(x; H1) p(x; H0) = exp ( ¡ 1 2¾2 N X¡1 n=0 £ (x[n] ¡ s[n]) 2 ¡ x 2 [n] ¤ ) L(x) = exp ( ¡ 1 2¾2 " ¡ N X¡1 n=0 2s[n]x[n] + N X¡1 n=0 s 2 [n] #) > ° 1 ¾2 N X¡1 n=0 s[n]x[n] > ln °+ 1 2¾2 N X¡1 n=0 s 2 [n]
Matched Filter (1) x(n] y(n] h(n] T(x) whxiong@uestc.edu.cn
whxiong@uestc.edu.cn Matched Filter (1)
Matched Filter (1) x(n] y[] h(n] T(x) What is the filter's impulse response,such that when sampled at NT is T(x) whxiong@uestc.edu.cn 14
whxiong@uestc.edu.cn Matched Filter (1) 14 What is the filter’s impulse response, such that when sampled at NT is T(x)
Matched Filter (1) x(n] y[] h(n] T(x) What is the filter's impulse response,such that when sampled at NT is T(x) N-1 N-1 m]=∑x[hn-[W-1=∑x[h[N-1- k三0 k=0 whxiong@uestc.edu.cn 15
whxiong@uestc.edu.cn Matched Filter (1) 15 What is the filter’s impulse response, such that when sampled at NT is T(x) y[n] = N X¡1 k=0 x[k]h[n ¡ k] y[N ¡ 1] = N X¡1 k=0 x[k]h[N ¡ 1 ¡ k]