lecture notes 07 Fall 19 complete

Lecture 7: Goals Lecture Notes 7 Goals Be able to determine the error probability in a system with two signals and a receiver filter in the presence of additive white Gaussian noise. EECS 455 (Univ. of Michigan) Fall 2019 October 28, 2019 1 / 142 BPSK Modulator Lecture Notes 7 Model 1.5 200 1 100 0.5 0 0 −0.5 −1 −1.5 0 1 2 3 4 5 6 7 8 9 10 time 20 10 0 −10 −20 −30 −40 −50 −10 −8 −6 −4 −2 0 2 4 6 8 10 frequency −100 −200 0 2 4 6 8 time 30 20 10 0 −10 −20 −15 −10 −5 0 5 10 15 20 frequency Data Source b(t ) s(t ) √P EECS 455 (Univ. of Michigan) Fall 2019 October 28, 2019 2 / 142 Channel Lecture Notes 7 Model 200 100 0 −100 −200 0 2 4 6 8 time 300 200 100 0 −100 −200 −300 0 1 2 3 4 5 6 7 8 9 10 time (s) 30 30 20 20 10 10 0 0 −10 −20 −15 −10 −5 0 5 10 15 20 frequency ( ) −10 −20 −15 −10 −5 0 5 10 15 20 frequency (Hz) ( ) 200 0 −200 0 1 2 3 4 5 6 7 8 9 10 time n(t ) 30 20 10 0 −10 −20 −15 −10 −5 0 5 10 15 20 frequency 20 10 0 −400 −300 −200 − 400 i EECS 455 (Univ. of Michigan) Fall 2019 October 28, 2019 3 / 142 Pulse Waveform Lecture Notes 7 Model b(t) = Σ∞ l =−∞ bl pT (t − lT ), bl ∈ {+1, −1}. pT (t) pT (t) = 1, 0 ≤ t ≤ T 1 0, otherwise. T t s(t) = √P b(t). The signal has power P. The energy of the transmitted bit is E = PT . EECS 455 (Univ. of Michigan) Fall 2019 October 28, 2019 4 / 142 Lecture Notes 7 Model s(t) t EECS 455 (Univ. of Michigan) Fall 2019 October 28, 2019 5 / 142 BPSK Demodulator Lecture Notes 7 Model 1 T EECS 455 (Univ. of Michigan) Fall 2019 October 28, 2019 6 / 142 Demodulator Lecture Notes 7 Model The output of the filter is used to decide the data bit transmitted. We let bˆi denote the decision as to which bit was transmitted (during the interval (iT , (i + 1)T ]). If bˆi = bi then the correct decision was made. If bˆi /= bi then an error was made. EECS 455 (Univ. of Michigan) Fall 2019 October 28, 2019 7 / 142 Matched Filter Lecture Notes 7 Model The filter is “matched” to the baseband signal being transmitted. For simple rectangular type signals this is just a rectangular pulse of duration T . The impulse response is h(t) = pT (T − t) = pT (t). EECS 455 (Univ. of Michigan) Fall 2019 October 28, 2019 8 / 142 Lecture Notes 7 Performance Analysis Demodulator Output Decomposition: Signal Component The output of the filter due only to the transmitted signal is given by sˆ(t) = = ∫ ∞ √ 1/Th(t τ)r (τ)d τ −∞ ∞ √ 1/TpT (t τ)r (τ)d τ −∞ ∫ t = t−T √ 1/Tr (τ)d τ That is, the filter is essentially a sliding integrator, integrating over the last T seconds. EECS 455 (Univ. of Michigan) Fall 2019 October 28, 2019 9 / 142 Lecture Notes 7 Performance Analysis Matched Filter Output: Single Shot Case Consider first transmitting just a sing√le pulse (beginning at t = 0 and ending at t = T ) with amplitude P b0, b0 ∈ {+1, −1}. The output of the filter would have the form √ ∫ t sˆ(t) = P/T t−T b0 pT (τ)d τ That is, the filter is essentially integrating a square wave over T second intervals. The output would be maximum when the filter integrated from 0 to T√, i.e. at t = T√ in which case the output would be ( P/T )T = E . In general the output would look like a triangular function shown below (for b0 = +1). EECS 455 (Univ. of Michigan) Fall 2019 October 28, 2019 10 / 142 Lecture Notes 7 Performance Analysis Matched Filter Outputs ...............................................continued...........................................