Pt t pt t 1 0 t t 0 otherwise 1 6 t eecs

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Unformatted text preview: 55 (Univ. of Michigan) ˆ |s0 (t )|2 dt Fall 2012 September 6, 2012 80 / 130 Lecture Notes 5 Up and Down Conversion Up-Down Conversion Summary Two independent low-pass signals of bandwidth W can be up-converted to a signal of frequency fc with bandwidth 2W and then individually recovered by mixing down to baseband. We can represent the two low-pass signals as a single complex-lowpass signal. Phase offset and frequency offset between the two oscillators can be corrected for by addition circuitry or signal processing. EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 81 / 130 Lecture Notes 5 BPSK Modulation: Revisited Binary Phase Shift Keying (BPSK) b (t ) √ '$ '$ s(t ) r (t ) Ed E E d &% &% T T 2P cos(2π fc t ) n(t ) Modulator EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 82 / 130 Lecture Notes 5 BPSK Modulation: Revisited BPSK Signals b (t ) = ∞ l =−∞ bl pT (t − lT ), bl ∈ {+1, −1}. pT (t ) pT (t ) = 1, 0 ≤ t ≤ T 0, otherwise. 1 6 - T EECS 455 (Univ. of Michigan) Fall 2012 t September 6, 2012 83 / 130 Lecture Notes 5 BPSK Modulation: Revisited BPSK Signals s(t ) = = √ √ 2P ∞ l =−∞ bl cos(2π fc t )pT (t − lT ) 2P b (t ) cos(2π fc t ) = √ 2P cos(2π fc t + φ(t )) where φ(t ) is the phase waveform. The signal has power P . The energy of the transmitted bit is E = PT . The phase of a BPSK signal can take on one of two values as shown below. EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 84 / 130 Lecture Notes 5 BPSK Modulation: Revisited Signals for BPSK Modulation b(t) 1 0 -1 0 1 2 3 time 4 5 6 1 2 3 time 4 5 6 1 2 3 time 4 5 6 phi(t) 4 2 0 0 s(t) 1 0 -1 0 EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 85 / 130 Lecture Notes 5 BPSK Modulation: Revisited Demodulator r (t ) '$ Ed E LPF d &% T § d c t = iT X (iT ) E > 0 dec bi −1 = +1 < 0 dec bi −1 = −1 2/T cos(2π fc t ) EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 86 / 130 Lecture Notes 5 BPSK Modulation: Revisited Demodulator Output: Signal and Noise The low pass filter (LPF) is a filter “matched” to the baseband signal being transmitted. For BPSK this is just a rectangular pulse of duration T . The impulse response is h(t ) = pT (t ). X (t ) = X (iT ) = ∞ −∞ ∞ −∞ iT 2/T cos(2π fc τ )h(t − τ )r (τ )d τ 2/T cos(2π fc τ )pT (iT − τ )r (τ )d τ 2/T cos(2π fc τ )[s(τ ) + n(τ )]d τ = (i −1)T ˆ = si + ηi EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 87 / 130 Lecture Notes 5 BPSK Modulation: Revisited Demodulator: Signal Term The signal term is iT ˆ si 2/T cos(2π fc τ ) = √ 2P b (τ ) cos(2π fc τ ) d τ (i −1)T iT = (i −1)T =2 = = 2 P /T bi −1 cos(2π fc τ ) cos(2π fc τ )d τ P /T bi −1 P /T bi −1 iT cos(2π fc τ ) cos(2π fc τ )d τ (i −1)T iT [1 + cos(2π 2fc τ )]d τ (i −1)T P /T bi −1 [τ |iT−1)T + (i EECS 455 (Univ. of Michigan) sin(2π 2fc τ ) iT |(i −1)T ] 2π 2fc Fall 2012 September 6, 2012 88 / 130 Lecture Notes 5 BPSK Modulation: Revisited Demodulator: Signal Term ˆ si sin(2π 2fc iT ) sin(2π 2fc (i − 1)T ) − 2π 2fc T 2π 2fc T √ cos(2π fc (2i − 1)T ) sin(2π fc T ) PT bi −1 [1 + = 2π fc T = ( P /T )Tbi −1[1 + If 2π fc T = 2π n for some integer n then the last term is zero. If 2π fc T >> 1 then the last term is much smaller than the first term and can be neglected. EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 89 / 130 Lecture Notes 5 BPSK Modulation: Revisited Statistics of the Noise iT ηi = 2/T n(τ ) cos(2π fc τ )d τ (i −1)T iT 2/T n(τ ) cos(2π fc τ )d τ ] E [ηi ] = E [ iT (i −1)T 2/T E [n(τ )] cos(2π fc τ )d τ = (i −1)T iT 2/T (0) cos(2π fc τ )d τ = (i −1)T =0 EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 90 / 130 Lecture Notes 5 BPSK Modulation: Revisited Statistics of the Noise iT E [ηi2 ] = E [ iT 2/T n(τ ) cos(2π fc τ )d τ (i −1)T iT 2/T n(s) cos(2π fc s)ds] (i −1)T iT = (2/T )E [n(τ )n(s)] cos(2π fc τ ) cos(2π fc s)dsd τ (i −1)T (i −1)T iT iT (2/T ) = (i −1)T (i −1)T iT = (4/T ) (i −1)T = (1/T ) N0 2 N0 δ (τ − s) cos(2π fc τ ) cos(2π fc s)dsd τ 2 N0 cos2 (2π fc τ )d τ 2 iT (1 + cos(2π 2fc τ ))d τ (i −1)T N0 2 The last equality follows from the assumption that 2π fc T = 2π n or fc >> 1/T . Thus ηi is a Gaussian random variable, with mean 0 and variance N0 /2. = EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 91 / 130 Lecture Notes 5 BPSK Modulation: Revisited Signal and Noise X (iT ) = EECS 455 (Univ. of Michigan) √ √ PT bi −1 + ηi = E bi −1 + ηi . Fall 2012 September 6, 2012 92 / 130 Lecture Notes 5 BPSK Modulation: Revisited Filter Output with Noise (E /N...
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