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Solution_3_2007

Solution_3_2007 - ECE 154B Winter 2007 Homework#3 Due...

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ECE 154B Winter 2007 Homework #3 Due: Monday, February 5, 2007 1. Prove that for M a positive even integer, 1 2 +3 2 +…+(M-3) 2 + (M-1) 2 = M(M 2 -1)/6. (Hint: Use induction) Proof by induction: Let M=2. (2-1) 2 =(2)((2) 2 -1)/6=1. Assume true for M. Test for M+2: 1 2 +3 2 +…+(M+2-3) 2 + (M+2-1) 2 = (M+2)((M+2) 2 -1)/6 1 2 +3 2 +…+(M-1) 2 + (M+1) 2 = (M+2)(M 2 +4M+4-1)/6 1 2 +3 2 +…+(M-1) 2 + (M+1) 2 = (M)(M 2 -1 )/6 +(2)(M 2 -1)/6+(M+2)(4M+4)/6 (M+1) 2 = (2M 2 -2+4M 2 +12M+8)/6 (M+1) 2 = (6M 2 +12M+6)/6 (M+1) 2 = (M+1) 2 2. The following applies to M= 4, PAM transmission with unequal a priori probabilities. Assume that one of four signals was sent of the form: {-3A σ (t), -A σ (t), +A σ (t),and +3A σ (t)} with a priori probabilities {1/8, 3/8, 3/8 and 1/8} respectively. σ (t) has duration T and has unit energy. The received signal is equal to the transmitted signal plus additive white Gaussian noise with power spectral density N 0 /2 . Let Z be the output of a correlator-integrator receiver which multiplies by σ (t) and integrates over the duration of each pulse T. (a) Evaluate the average symbol energy, E s , and the average energy per bit, E b . Hint: You need to take into account the a priori probabilities. E 0 = 0 T − 3A   t  2 dt = 9A 2 0 T  t 2 dt = 9A 2 E 1 = 0 T − A   t  2 dt = A 2 , E 2 = E 1 , E 3 = E 0 E s = 1 8 E 0  3 8 E 1  3 8 E 2  1 8 E 3 = 3A 2 E b = E s log 2 4 = 3A 2 2 (b) Find the optimal decision regions for Z. That is find the decision regions that lead to a minimum probability of error.
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N 2 = E [ 0 T
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