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• 1. TELE3113 Analogue and Digital Communications &#x2013; Detection Theory (2) Wei Zhang w.zhang@unsw.edu.au School of Electrical Engineering and Telecommunications The University of New South Wales7 Oct. 2009 TELE3113 1
• 2. Integrate-and-Dump detector Integrate-and-Dump detectorr(t)=si(t)+n(t) &#xF8F1; s (t ) = + A 0&#x2264;t &#x2264;T for 1 si (t ) = &#xF8F2; 1 &#xF8F3;s2 (t ) = &#x2212; A 0&#x2264;t &#x2264;T for 0 t 0 +T &#xF8F1; a1 (t ) + no for 1 Output of the integrator: z (t ) = &#x222B; [si (t ) + n(t )]dt = &#xF8F2; t 0 +T t0 &#xF8F3;a2 (t ) + no for 0 where a1 = &#x222B; Adt = AT t0 t 0 +T a2 = &#x222B; (&#x2212; A)dt = &#x2212; AT t0 t 0 +T no = &#x222B; n(t )dt t07 Oct. 2009 TELE3113 2
• 3. Integrate-and-Dump detector no is a zero-mean Gaussian random variable. &#xF8F1;t0 +T &#xF8F4; &#xF8FC; t 0 +T &#xF8F4; E{no } = E &#xF8F2; &#x222B; n(t )dt &#xF8FD; = &#x222B; E{n(t )}dt = 0 &#xF8F4; t0 &#xF8F3; &#xF8F4; t0 &#xF8FE; &#xF8F1; &#xF8EE; t 0 +T &#xF8F9; &#xF8FC; 2 &#xF8F4; { } &#x3C3; no = Var{no } = E no = E &#xF8F2;&#xF8EF; &#x222B; n(t )dt &#xF8FA; &#xF8FD; 2 2 &#xF8F4; &#xF8F4; &#xF8EF; t0 &#xF8F3;&#xF8F0; &#xF8FA; &#xF8F4; &#xF8FB; &#xF8FE; t o +T t 0 +T = &#x222B; &#x222B; E{n(t )n(&#x3B5; )}dtd&#x3B5; t0 t0 t o +T t 0 +T &#x3B7; = &#x222B; &#x222B; &#x3B4; (t &#x2212; &#x3B5; )dtd&#x3B5; t0 t0 2 t 0 +T &#x3B7; &#x3B7;T = &#x222B; t0 2 d&#x3B5; = 2 1 1 pdf of no: f n (&#x3B1; ) = &#x2212;&#x3B1; 2 /( 2&#x3C3; no ) 2 2 e = e &#x2212;&#x3B1; /(&#x3B7;T ) o 2&#x3C0; &#x3C3; no &#x3C0;&#x3B7;T7 Oct. 2009 TELE3113 3
• 4. Integrate-and-Dump detector &#xF8F1; s1 (t ) = + A 0&#x2264;t &#x2264;T for 1 As si (t ) = &#xF8F2; &#xF8F3;s 0 (t ) = &#x2212; A 0&#x2264;t &#x2264;T for 0 s0 s1 We choose the decision threshold to be 0. 0 &#x2212; AT + AT Two cases of detection error: (a) +A is transmitted but (AT+no)&lt;0 no&lt;-AT (b) -A is transmitted but (-AT+no)&gt;0 no&gt;+AT Error probability: Pe = P (no &lt; &#x2212; AT | A) P ( A) + P (no &gt; AT | A) P (&#x2212; A) &#x2212; AT 2 &#x221E; 2 e &#x2212;&#x3B1; /(&#x3B7;T ) e &#x2212;&#x3B1; /(&#x3B7;T ) = P ( A) &#x2212;&#x221E; &#x222B; &#x3C0;&#x3B7;T d&#x3B1; + P (&#x2212; A) &#x222B; AT &#x3C0;&#x3B7;T d&#x3B1; &#x221E; 2 e &#x2212;&#x3B1; /(&#x3B7;T ) &#xF8EB; 2 A2T &#xF8F6; d&#x3B1; [P( A) + P (&#x2212; A)] &#x221E; 2 e &#x2212;u / 2 = &#x222B; &#x3C0;&#x3B7;T Thus, Pe = Q&#xF8EC; &#xF8EC; &#xF8F7; Q Q(x ) = &#x222B; du AT &#xF8ED; &#x3B7; &#xF8F7; &#xF8F8; x 2&#x3C0; &#x221E; 2 e &#x2212;u / 2 2&#x3B1; &#xF8EB; 2 Eb &#xF8F6; = &#x222B; T du Qu = 2&#x3C0; &#x3B7;T = Q&#xF8EC; &#xF8EC; &#x3B7; &#xF8F7; &#xF8F7; Q Eb = &#x222B; A2 dt 2 A2T &#x3B7; &#xF8ED; &#xF8F8; 07 Oct. 2009 TELE3113 4
• 5. Integrate-and-Dump detector Consider two signal symbols s1 and s2. Let Ed be the energy of the difference signal (s1- s2), s2 s1 T &#xF8EB; Ed &#xF8F6; signal symbol energy=si2 Ed = &#x222B; [s1 (t ) &#x2212; s2 (t )] dt Pe = Q&#xF8EC; &#xF8F7; 2 i.e. &#xF8EC; 2&#x3B7; &#xF8F7; 0 &#xF8ED; &#xF8F8; &#xF8F1; s (t ) = + A 0&#x2264;t &#x2264;T for 1 For example: If si (t ) = &#xF8F2; 1 &#xF8F3;s2 (t ) = &#x2212; A 0&#x2264;t &#x2264;T for 0 T T Thus Ed = &#x222B; [s1 (t ) &#x2212; s2 (t )] dt = &#x222B; [A &#x2212; (&#x2212; A)] dt = 4 A T 2 2 2 0 0 &#xF8EB; 4 A2T &#xF8F6; &#xF8EB; 2 A2T &#xF8F6; &#x21D2; Pe = Q&#xF8EC; &#xF8F7; = Q&#xF8EC; &#xF8F7; &#xF8EC; 2&#x3B7; &#xF8F7; &#xF8EC; &#x3B7; &#xF8F7; &#xF8ED; &#xF8F8; &#xF8ED; &#xF8F8;7 Oct. 2009 TELE3113 5
• 6. Integrate-and-Dump detector Example: In a binary system with bipolar binary signal which is a +A volt or &#x2013;A volt pulse during the interval (0,T), the sending of either +A or &#x2013;A are equally probable. The value of A is 10mV. The noise power spectral density is 10-9 W/Hz. The transmission rate of data (bit rate) is 104 bit/s. An integrate-and-dump detector is used. (a) Find the probability of error, Pe. (b) If the bit rate is increased to 105 bit/s what value of A is needed to attain the same Pe, as in part (a). &#x3B7; Solution: (a) With = 10 &#x2212;9 , P(+A)=P(-A)=0.5 , bit interval T=10-4 seconds, A=10mV 2 &#xF8EB; 2 A 2T &#xF8F6; &#xF8EB; &#x2212;4 &#xF8F6; ( ) &#x2212;3 2 Pe = Q&#xF8EC; &#xF8F7; = Q&#xF8EC; 2(10 &#xD7; 10 ) (10 ) &#xF8F7; = Q 10 = 7.8 &#xD7; 10 &#x2212; 4 &#xF8EC; &#x3B7; &#xF8F7; &#xF8EC; 2 &#xD7; 10 &#x2212;9 &#xF8F7; &#xF8ED; &#xF8F8; &#xF8ED; &#xF8F8; &#xF8EB; &#xF8F6; ( ) 2 (b) Bit interval T=10 -5 seconds, for the same P , i.e. P = Q&#xF8EC; 2 A T &#xF8F7; = Q 10 e e &#xF8EC; &#x3B7; &#xF8F7; &#xF8ED; &#xF8F8; 2 A 2T = 10 &#x2192; A= ( 10 2 &#xD7; 10 &#x2212;9 ) = 31.62mV &#x3B7; 2 10 &#x2212;5 ( )7 Oct. 2009 TELE3113 6
• 7. Optimal Detection Threshold Pe = P (detect s 2 | s1 ) P( s1 ) + P (detect s1 | s 2 ) P( s 2 ) &#x3BB; &#x221E; = P ( s1 ) &#x222B; f (r | s1 )dr + P( s 2 ) &#x222B; f (r | s 2 )dr &#x2212;&#x221E; &#x3BB; &#x3BB; &#xF8EE; &#x3BB; &#xF8F9; = P ( s1 ) &#x222B; f (r | s1 )dr + P( s 2 ) &#xF8EF;1 &#x2212; &#x222B; f (r | s 2 )dr &#xF8FA; &#x2212;&#x221E; &#xF8F0; &#x2212;&#x221E; &#xF8FB; &#x3BB; = P( s 2 ) + &#x222B; [P(s ) f (r | s ) &#x2212; P(s &#x2212;&#x221E; 1 1 2 ) f (r | s 2 )]dr7 Oct. 2009 TELE3113 7
• 8. Optimal Detection Threshold To find a threshold &#x3BB;o which minimizes Pe, we set dPe = 0 d&#x3BB; gives Taking ln(.) on both sides, then P( s1 ) f (&#x3BB;o | s1 ) = P ( s 2 ) f (&#x3BB;o | s 2 ) &#x3BB;o (s1 &#x2212; s2 ) s12 &#x2212; s2 2 P(s ) f (&#x3BB;o | s1 ) P ( s 2 ) &#x2212; = ln 2 = &#x3C3;n 2 2&#x3C3; n2 P(s1 ) f (&#x3BB;o | s 2 ) P( s1 ) o o s +s &#x3C3;n 2 P(s ) &#x2212; ( &#x3BB;o &#x2212; s1 ) 2 /( 2&#x3C3; no ) 2 &#x3BB;o &#x2212; 1 2 = o ln 2 f (&#x3BB;o | s1 ) e P( s 2 ) 2 s1 &#x2212; s2 P(s1 ) = &#x2212;( &#x3BB; &#x2212; s ) 2 /( 2&#x3C3; 2 ) = s1 + s2 &#x3C3; no 2 f (&#x3BB; o | s 2 ) e o 2 no P( s1 ) &#x3BB;o = + P(s ) ln 2 2 s1 &#x2212; s2 P(s1 ) 2 2 2 2 &#x3BB;o ( s1 &#x2212; s2 ) / &#x3C3; no &#x2212; ( s1 &#x2212; s 2 ) /( 2&#x3C3; no ) P( s 2 ) e = If P ( s1 ) = P ( s 2 ) &#x21D2; &#x3BB;o = s1 + s 2 P ( s1 ) 27 Oct. 2009 TELE3113 8
• 9. Correlator Receiver r r 2 Recall ML decision criterion: minimize r &#x2212; si r r = &#x222B; [r (t ) &#x2212; si (t )] dt = &#x222B; r 2 (t )dt + &#x222B; &#x2212; 2 &#x222B; r (t ) si (t )dt 2 With r &#x2212; si 2 si2 (t )dt T T T T 1 24 4 3 1 24 4 3 constant energy of i - th signal r r &#x222B; si2 (t )dt &#x2212; 2 &#x222B; r (t ) si (t )dt 2 minimize r &#x2212; si minimize T T Let &#x3BE; i denotes the energy of si(t). ML decision criterion becomes Detected Find i to maximize signal &#x222B; r (t ) si (t )dt &#x2212; 1 &#x222B; si2 (t )dt symbol 2 T T 1 24 4 3 &#x3BE;i Or simply: Find i to maximize &#x222B; r (t )s (t )dt T i if all signal symbols have the same energy Correlation Receiver7 Oct. 2009 TELE3113 9
• 10. Matched Filter The multiplying and integrating in correlation receiver can be reduced to a linear filtering. Consider the received signal r(t) passes through a filter hi(t): i.e. r (t ) &#x2217; hi (t ) = &#x222B; r (&#x3C4; )hi (t &#x2212; &#x3C4; )d&#x3C4; T Let hi (&#x3C4; ) = si (T &#x2212; &#x3C4; ) &#x21D2; &#x222B; r (&#x3C4; )h (t &#x2212; &#x3C4; )d&#x3C4; = &#x222B; r (&#x3C4; )s (t &#x2212; T + &#x3C4; )d&#x3C4; i i for 0 &#x2264; t &#x2264; T T T Then we sample the filter output at t=T, thus &#x222B; r (&#x3C4; )h (t &#x2212; &#x3C4; )d&#x3C4; T i = &#x222B; r (&#x3C4; ) si (t &#x2212; T + &#x3C4; )d&#x3C4; T = &#x222B; r (&#x3C4; ) si (&#x3C4; )d&#x3C4; T (correlation) t =T t =T ML decision criterion: &#x222B; r (t ) s (t )dt &#x2212; &#x222B; s 1 2 Find i to maximize i 2 i(t )dt T T 1 24 4 3 becomes Find i to maximize &#x3BE;i Detected signal symbol &#x222B; r (&#x3C4; )h (t &#x2212; &#x3C4; )d&#x3C4; T i &#x2212; 1 &#x222B; si2 (t )dt 2 T t =T 1 24 4 3 &#x3BE;i7 Oct. 2009 TELE3113 10 Matched-Filter Receiver
• 11. Matched Filter Consider the matched filter hi (t ) = si (T &#x2212; t ) si(-t) t t t si(t) si(-t) hi(t) 0 T t -T 0 t 0 T t Equivalence of matched filter and correlator: Matched filter receiver &#x2264; &#x2264; Correlator receiver7 Oct. 2009 TELE3113 11
• 12. Detection of PAM Detection of M-ary PAM (one-dimension)symbol s1 s2 sM/2-1 sM/2 sM/2+1 sM/2+2 sM-1 sM &#x2026; &#x2026;amplitude &#x2212;3 E &#x2212; E 0 E 3 E ( M &#x2212; 1) E 2Let si = E Ai where i = 1,2 ,...,M and Ai = (2i &#x2212; 1 &#x2212; M ) , energy of si is siFor equally probable signals, i.e. P( si ) = 1 / M for i = 1,2,...,M 1 M 2 E M 2 E M E M (M 2 &#x2212; 1) &#xF8EB; M 2 &#x2212; 1 &#xF8F6;average energy &#x3B5; av = &#x2211; si = &#x2211; Ai = &#x2211; (2i &#x2212; 1 &#x2212; M ) = =&#xF8EC; &#xF8EC; 3 &#xF8F7;E 2 M i =1 M i =1 M i =1 M 3 &#xF8F7; &#xF8ED; &#xF8F8;Received signal r = si + n = E Ai + n where n = 0, &#x3C3; n = &#x3B7; 2 2 Average Prob(symbol error) Pe = M &#x2212;1 M ( P r &#x2212; si &gt; E ) &#x221E; M &#x2212;1 2 &#x3C0;&#x3B7; &#x222B;E 2 = e &#x2212; x &#x3B7; dx M 2( M &#x2212; 1) &#xF8EB; 2 E &#xF8F6; &#x221E; x 2 M &#x2212;1 2 &#x222B; Q&#xF8EC; &#xF8F7; 2 let y = = e &#x2212; y 2 dy = &#xF8EC; &#x3B7; &#xF8F7; &#x3B7; M 2&#x3C0; 2 E /&#x3B7; M &#xF8ED; &#xF8F8;7 Oct. 2009 TELE3113 12
• 13. Detection of QAM &#x2026; Detection of M-ary QAM (two-dimension) : M=2k 2E For one-dimension M -ary PAM: 2E Average Prob(symbol error) Pe ( M &#x2212; PAM ,1D ) &#x2026; &#x2026; 2( M &#x2212; 1) &#xF8EB; 2 E &#xF8F6; = Q&#xF8EC; &#xF8EC; &#x3B7; &#xF8F7; &#xF8F7; M &#xF8ED; &#xF8F8; For two-dimension M-ary QAM: Average Prob(correct decision) Pc ( M &#x2212;QAM , 2 D ) = 1 &#x2212; Pe ( ( ) 2 &#x2026; M &#x2212; PAM ,1D ) Average Prob(symbol error) Pe ( M &#x2212;QAM , 2 D ) = 1 &#x2212; Pc ( M &#x2212;QAM , 2 D ) &#xF8EB; 2E &#xF8F6; ( = 1 &#x2212; 1 &#x2212; Pe ( M &#x2212; PAM ,1D ) ) 2 For M=4, Pe ( 4 &#x2212; PAM ,1D ) = Q&#xF8EC; &#xF8EC; &#x3B7; &#xF8F7; &#xF8F7; &#xF8ED; &#xF8F8; Average Prob(symbol error) Pe ( 4&#x2212;QAM , 2 D ) = 1 &#x2212; 1 &#x2212; Pe ( ( 4 &#x2212; PAM ,1D ) ) 2 2 &#xF8EE; &#xF8EB; 2 E &#xF8F6;&#xF8F9; &#xF8EB; 2E &#xF8F6;&#xF8EE; &#xF8EB; 2 E &#xF8F6;&#xF8F9; = 1 &#x2212; &#xF8EF;1 &#x2212; Q&#xF8EC; &#xF8EC; &#x3B7; &#xF8F7; &#xF8F7; &#xF8FA; = Q&#xF8EC; &#xF8EC; &#x3B7; &#xF8F7; &#xF8F7; &#xF8EF; 2 &#x2212; Q&#xF8EC; &#xF8EC; &#x3B7; &#xF8F7;&#xF8FA; &#xF8F7; &#xF8EF; &#xF8F0; &#xF8ED; &#xF8F8;&#xF8FA;&#xF8FB; &#xF8ED; &#xF8F8;&#xF8EF;&#xF8F0; &#xF8ED; &#xF8F8;&#xF8FA;&#xF8FB;7 Oct. 2009 TELE3113 13