If the locally generated vector is modeled as i=[]=[h%i1,h62hk] (5.49) Then the filter output at the kth sample is identical to the dot-product output of the correlator y, i=0,1 (5.50) Matched j=1 If the underlying noise process is white,then filter (5.51) UESTC [h]=[,4k-w] 16
16 UESTC If the locally generated vector is modeled as 0 0 0,1 0,2 0, [ ] [ , , ] (5.49) k h h h h h = = * 0 h Then the filter output at the kth sample is identical to the dot-product output of the correlator , 1 , 0,1 (5.50) k j i j j y h i = = If the underlying noise process is white, then 2 , , 1 ,1 1 [ ] [ , , ] (5.51) i i k i k i h u u u = − Matched filter
36 Produces T T largest 41,1 SNR! Re( T Ts 立∑M Choose largest 2立∑M Re ( Figure 5.5.Functional block diagram of optimal matched-filter receiver in AWGN UESTC 17
17 UESTC Produces largest SNR!
Example 5.4 Consider binary phase-shift keying(PSK),where the transmitted signal can be modeled as cos[2zf.(t-(i-l)T)+dJ.(i-ITst<iT (5.52) where d,is a digital random variable taking on value of +1 with equal probability.It can be assumed for simplicity,that f=m/T.The transmitted signal model is simplified to cos2πf1+d7]=Ree4e'2x],i-T≤i<i0(5.53) UESTC 18
18 UESTC Example 5.4 Consider binary phase-shift keying (PSK) , where the transmitted signal can be modeled as cos[2 ( ( 1) ) ], ( 1) (5.52) 2 c i f t i T d i T t iT − − + − where is a digital random variable taking on value of with equal probability. It can be assumed for simplicity, that . The transmitted signal model is simplified to i d 1 c f m T = / 2 2 cos[2 ] Re[ ], ( 1) (5.53) 2 i c Jd J f t c i f t d e e i T t iT + = −
A baseband equivalent received signal is y(t)=em2+z(t) (5.54) where z(t)represents a complex equivalent white noise process.From this equation,we recognize u (t)=e'"R =J, (i-1)T≤t<i江 (5.55 4,()=eJ2=-J, (i-l)T≤t<i讧 (5.56 UESTC 19
19 UESTC A baseband equivalent received signal is / 2 ( ) ( ) (5.54) i Jd y t e z t where represents a complex equivalent white noise process. From this equation, we recognize z t( )2 1 ( ) , ( 1) (5.55) J u t e J i T t iT = = − 2 0 ( ) , ( 1) (5.56) J u t e J i T t iT − = = − −
/9 It follows that a matched filter in white noise has constant tap coefficients.Figure 5.6 shows the FIR matched filter Ts Output Figure 5.6.FIR matched filtero for Example 5.4 UESTC 20
20 UESTC It follows that a matched filter in white noise has constant tap coefficients. Figure 5.6 shows the FIR matched filter . 0 h