Solution_6_2007

Solution_6_2007 - ECE 154B Homework #6 Solutions 1. Making...

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ECE 154B Homework #6 Solutions 1. Making the usual assumptions of equally likely signals in AWGN and an optimal receiver, it can be shown that the probability of symbol error for M/2 orthogonal signals of energy E and their negatives is given as: P[symbol error] = 1 - 1 2 0 exp − 1 2 x 2 E s N 0 2  1 2Q x  M 2 2 dx A special case of this for M=4 is the case of QPSK where the 4 signals are: + sin( ω t), -sin( ω t), + cos( ω t), and - sin( ω t). For this case, we have derived a formula in class for the probability of symbol error. Show how the above formula reduces to the formula we derived in class. In class P s : P s = 1 − 1 Q E s N 0  2 M/2 Orthogonal Signals and their negatives, M=4: P c = 0 1 2 e 1 2 x 2 E s N 0 2 1 2Q x  dx = 0 1 2 e 1 2 x 2 E s N 0 2 x x 1 2 e 1 2 y 2 dydx = 0 x x 1 2 e 1 2  x 2 E s N 0 2 y 2 dydx The 2-D integration is over the area: We can apply a 45 degree rotation to the axes so that the integration in both dimensions go to infinity: x y x' y'

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Rotation matrix: [ cos sin sin cos ] [ x' y' ] = [ cos 4 sin 4 sin 4 cos 4 ] [ x y ] = [ x y 2 x y 2 ] P c = 0 x x 1 2 e 1 2 x 2 y 2 2 2 E s N 0 x 2 E s N 0 dydx = 0 0 1 2 e 1 2 x' 2 y' 2 2 2 E s N 0 x' y' 2  2 E s N 0 dy' dx' = 0 0 1 2 e 1 2 x' 2 2 E s N 0 x' E s N 0 y' 2 2 E s N 0 x' E s N 0 ∂ x ,y ∂ x' ,y' dy' dx' = 0 0 1 2 e 1 2  x' E s N 0 2  y' E s N 0 2 1 dy' dx' = 1 Q E s N 0  2 P s = 1 − 1 Q E s N 0  2 2. For the case of 4 orthogonal signals and their negatives (i.e., M=8) how should we map 3 binary digits to the 8 waveforms to minimize the probability of binary error at high signal to noise ratio? The minimum distance is between orthogonal signals. The maximum distance is between
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This note was uploaded on 11/11/2009 for the course ECE 670377 taught by Professor Wolf during the Winter '07 term at UCSD.

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Solution_6_2007 - ECE 154B Homework #6 Solutions 1. Making...

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