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Tele3113 wk11wed
Tele3113 wk11wed
Tele3113 wk11wed
Tele3113 wk11wed
Tele3113 wk11wed
Tele3113 wk11wed
Tele3113 wk11wed
Tele3113 wk11wed
Tele3113 wk11wed
Tele3113 wk11wed
Tele3113 wk11wed
Tele3113 wk11wed
Tele3113 wk11wed
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Tele3113 wk11wed

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  • 1. TELE3113 Analogue and Digital Communications – 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)  s (t ) = + A 0≤t ≤T for 1 si (t ) =  1 s2 (t ) = − A 0≤t ≤T for 0 t 0 +T  a1 (t ) + no for 1 Output of the integrator: z (t ) = ∫ [si (t ) + n(t )]dt =  t 0 +T t0 a2 (t ) + no for 0 where a1 = ∫ Adt = AT t0 t 0 +T a2 = ∫ (− A)dt = − AT t0 t 0 +T no = ∫ n(t )dt t07 Oct. 2009 TELE3113 2
  • 3. Integrate-and-Dump detector no is a zero-mean Gaussian random variable. t0 +T   t 0 +T  E{no } = E  ∫ n(t )dt  = ∫ E{n(t )}dt = 0  t0   t0    t 0 +T   2  { } σ no = Var{no } = E no = E  ∫ n(t )dt   2 2    t0      t o +T t 0 +T = ∫ ∫ E{n(t )n(ε )}dtdε t0 t0 t o +T t 0 +T η = ∫ ∫ δ (t − ε )dtdε t0 t0 2 t 0 +T η ηT = ∫ t0 2 dε = 2 1 1 pdf of no: f n (α ) = −α 2 /( 2σ no ) 2 2 e = e −α /(ηT ) o 2π σ no πηT7 Oct. 2009 TELE3113 3
  • 4. Integrate-and-Dump detector  s1 (t ) = + A 0≤t ≤T for 1 As si (t ) =  s 0 (t ) = − A 0≤t ≤T for 0 s0 s1 We choose the decision threshold to be 0. 0 − AT + AT Two cases of detection error: (a) +A is transmitted but (AT+no)<0 no<-AT (b) -A is transmitted but (-AT+no)>0 no>+AT Error probability: Pe = P (no < − AT | A) P ( A) + P (no > AT | A) P (− A) − AT 2 ∞ 2 e −α /(ηT ) e −α /(ηT ) = P ( A) −∞ ∫ πηT dα + P (− A) ∫ AT πηT dα ∞ 2 e −α /(ηT )  2 A2T  dα [P( A) + P (− A)] ∞ 2 e −u / 2 = ∫ πηT Thus, Pe = Q   Q Q(x ) = ∫ du AT  η   x 2π ∞ 2 e −u / 2 2α  2 Eb  = ∫ T du Qu = 2π ηT = Q  η   Q Eb = ∫ A2 dt 2 A2T η   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  Ed  signal symbol energy=si2 Ed = ∫ [s1 (t ) − s2 (t )] dt Pe = Q  2 i.e.  2η  0    s (t ) = + A 0≤t ≤T for 1 For example: If si (t ) =  1 s2 (t ) = − A 0≤t ≤T for 0 T T Thus Ed = ∫ [s1 (t ) − s2 (t )] dt = ∫ [A − (− A)] dt = 4 A T 2 2 2 0 0  4 A2T   2 A2T  ⇒ Pe = Q  = Q   2η   η     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 –A volt pulse during the interval (0,T), the sending of either +A or –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). η Solution: (a) With = 10 −9 , P(+A)=P(-A)=0.5 , bit interval T=10-4 seconds, A=10mV 2  2 A 2T   −4  ( ) −3 2 Pe = Q  = Q 2(10 × 10 ) (10 )  = Q 10 = 7.8 × 10 − 4  η   2 × 10 −9        ( ) 2 (b) Bit interval T=10 -5 seconds, for the same P , i.e. P = Q 2 A T  = Q 10 e e  η    2 A 2T = 10 → A= ( 10 2 × 10 −9 ) = 31.62mV η 2 10 −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 ) λ ∞ = P ( s1 ) ∫ f (r | s1 )dr + P( s 2 ) ∫ f (r | s 2 )dr −∞ λ λ  λ  = P ( s1 ) ∫ f (r | s1 )dr + P( s 2 ) 1 − ∫ f (r | s 2 )dr  −∞  −∞  λ = P( s 2 ) + ∫ [P(s ) f (r | s ) − P(s −∞ 1 1 2 ) f (r | s 2 )]dr7 Oct. 2009 TELE3113 7
  • 8. Optimal Detection Threshold To find a threshold λo which minimizes Pe, we set dPe = 0 dλ gives Taking ln(.) on both sides, then P( s1 ) f (λo | s1 ) = P ( s 2 ) f (λo | s 2 ) λo (s1 − s2 ) s12 − s2 2 P(s ) f (λo | s1 ) P ( s 2 ) − = ln 2 = σn 2 2σ n2 P(s1 ) f (λo | s 2 ) P( s1 ) o o s +s σn 2 P(s ) − ( λo − s1 ) 2 /( 2σ no ) 2 λo − 1 2 = o ln 2 f (λo | s1 ) e P( s 2 ) 2 s1 − s2 P(s1 ) = −( λ − s ) 2 /( 2σ 2 ) = s1 + s2 σ no 2 f (λ o | s 2 ) e o 2 no P( s1 ) λo = + P(s ) ln 2 2 s1 − s2 P(s1 ) 2 2 2 2 λo ( s1 − s2 ) / σ no − ( s1 − s 2 ) /( 2σ no ) P( s 2 ) e = If P ( s1 ) = P ( s 2 ) ⇒ λo = s1 + s 2 P ( s1 ) 27 Oct. 2009 TELE3113 8
  • 9. Correlator Receiver r r 2 Recall ML decision criterion: minimize r − si r r = ∫ [r (t ) − si (t )] dt = ∫ r 2 (t )dt + ∫ − 2 ∫ r (t ) si (t )dt 2 With r − 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 ∫ si2 (t )dt − 2 ∫ r (t ) si (t )dt 2 minimize r − si minimize T T Let ξ i denotes the energy of si(t). ML decision criterion becomes Detected Find i to maximize signal ∫ r (t ) si (t )dt − 1 ∫ si2 (t )dt symbol 2 T T 1 24 4 3 ξi Or simply: Find i to maximize ∫ 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 ) ∗ hi (t ) = ∫ r (τ )hi (t − τ )dτ T Let hi (τ ) = si (T − τ ) ⇒ ∫ r (τ )h (t − τ )dτ = ∫ r (τ )s (t − T + τ )dτ i i for 0 ≤ t ≤ T T T Then we sample the filter output at t=T, thus ∫ r (τ )h (t − τ )dτ T i = ∫ r (τ ) si (t − T + τ )dτ T = ∫ r (τ ) si (τ )dτ T (correlation) t =T t =T ML decision criterion: ∫ r (t ) s (t )dt − ∫ s 1 2 Find i to maximize i 2 i(t )dt T T 1 24 4 3 becomes Find i to maximize ξi Detected signal symbol ∫ r (τ )h (t − τ )dτ T i − 1 ∫ si2 (t )dt 2 T t =T 1 24 4 3 ξi7 Oct. 2009 TELE3113 10 Matched-Filter Receiver
  • 11. Matched Filter Consider the matched filter hi (t ) = si (T − 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 ≤ ≤ 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 … …amplitude −3 E − E 0 E 3 E ( M − 1) E 2Let si = E Ai where i = 1,2 ,...,M and Ai = (2i − 1 − 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 − 1)  M 2 − 1 average energy ε av = ∑ si = ∑ Ai = ∑ (2i − 1 − M ) = =  3 E 2 M i =1 M i =1 M i =1 M 3   Received signal r = si + n = E Ai + n where n = 0, σ n = η 2 2 Average Prob(symbol error) Pe = M −1 M ( P r − si > E ) ∞ M −1 2 πη ∫E 2 = e − x η dx M 2( M − 1)  2 E  ∞ x 2 M −1 2 ∫ Q  2 let y = = e − y 2 dy =  η  η M 2π 2 E /η M  7 Oct. 2009 TELE3113 12
  • 13. Detection of QAM … Detection of M-ary QAM (two-dimension) : M=2k 2E For one-dimension M -ary PAM: 2E Average Prob(symbol error) Pe ( M − PAM ,1D ) … … 2( M − 1)  2 E  = Q  η   M   For two-dimension M-ary QAM: Average Prob(correct decision) Pc ( M −QAM , 2 D ) = 1 − Pe ( ( ) 2 … M − PAM ,1D ) Average Prob(symbol error) Pe ( M −QAM , 2 D ) = 1 − Pc ( M −QAM , 2 D )  2E  ( = 1 − 1 − Pe ( M − PAM ,1D ) ) 2 For M=4, Pe ( 4 − PAM ,1D ) = Q  η     Average Prob(symbol error) Pe ( 4−QAM , 2 D ) = 1 − 1 − Pe ( ( 4 − PAM ,1D ) ) 2 2   2 E   2E   2 E  = 1 − 1 − Q  η    = Q  η    2 − Q  η          7 Oct. 2009 TELE3113 13

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