HW8Solutions - ECEN 661 Modulation Theory Homework#8...

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ECEN 661 - Modulation Theory Homework #8 Solutions 1. (a) As described in class, comparing with is the same as comparing with . (b) When a binary “0” is sent, the two consecutive transmitted waveforms are the same. Hence the received samples are of the form: , , where . Then , . Let and . Note that and are both complex Gaussian with zero means, and variance . The covariance is . Hence the statistics and are statistically independent. Since is the magnitude of a non-zero mean, complex Gaussian random variable it has a Rician distribution. Similarly, since is the magnitude of a zero mean, complex Gaussian random variable it has a Rayleigh distri- bution. The receiver will make a decision error if > (that is, the Rayleigh is bigger than z 0 z 1 z 0 2 z 1 2 z 0 2 > < z 1 2 r n r n 1 + 2 > < r n r n 1 2 r n 2 r n 1 2 2 Re r n r n 1 * [ ] + + > < r n 2 r n 1 2 2 Re r n r n 1 * [ ] + 2 Re r n r n 1 * [ ] > < 2 Re r n r n 1 * [ ] Re r n r n 1 * [ ] > < 0 r n 1 E p e j φ n k 1 + = r n E p e j φ n k + = 1 2 -- E n n 2 [ ] 1 2 -- E n n 1 2 [ ] σ 2 N o E p = = = z 0 r n 1 r n + 2 E p e j φ n n 1 n n + + = = z 1 r n r n 1 n n n n 1 = = u n n 1 n n + = v n n n n 1 = u v 1 2 -- E u 2 [ ] 1 2 -- E v 2 [ ] 1 2 -- E n n 2 [ ] 1 2 -- E n n 1 2 [ ] + 2 σ 2 = = = 1 2 -- E uv * [ ] 1 2 -- E n n n n 1 + ( ) n n * n n 1 * ( ) [ ] 1 2 -- E n n 2 [ ] 1 2 -- E n n 1 2 [ ] 0 = = = z 0 z 1 z 0 z 1 z 1 z 0
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the Rician). If a binary “1” was sent, the two consecutive transmitted signals are opposite. In that case, following similar steps, we find that is Rayleigh and is Rician and an error will be made if < . The two Rayleigh and Rician distributions involved work out to be: (Rayleigh) (Rician) (c) The probability that the Rayleigh is bigger than the Rician is: . To evaluate this integral, we manipulate the integrand into the form of a valid Rician PDF as fol- lows. First, let . Then, . Next, let and so that: . Now the integrand is a properly normalized Rician PDF and hence the integral is 1. The resulting error probability is then . Finally, noting that we get . 2. In coherent BPSK, suppose a data bit of “-1” was sent so that the transmitted waveform (over one symbol interval) is z 0 z 1 z 1 z 0 f z ( ) z 2 σ 2 --------- z 2 4 σ 2 --------- U z ( ) exp = f z ( ) z 2 σ 2 --------- z 2 4 E p 2 + 4 σ 2 -------------------- I 0 E p z σ 2 -------- U z ( ) exp = Pr Rayleigh>Rician ( ) z 2 σ 2 --------- z 2 4 E p 2 + 4 σ 2 -------------------- I 0 E p z σ 2 -------- y 2 σ 2 --------- y 2 4 σ 2 --------- exp y d z exp z d 0 = z 2 σ 2 --------- z 2 4 E p 2 + 4 σ 2 -------------------- I 0 E p z σ 2 -------- z 2 4 σ 2 --------- exp exp z d 0 = z 2 σ 2 --------- 2 z 2 4 E p 2 + 4 σ 2 ----------------------- I 0 E p z σ 2 -------- exp z d 0 = u 2 2 z 2 = Pr Rayleigh>Rician ( ) 1 2 -- u 2 σ 2 --------- u 2 4 E p 2 + 4 σ 2 --------------------- I 0 E p u 2 σ 2 ------------- exp u d 0 = A 2 2 E p 2 = s 2 2 σ 2 = Pr Rayleigh>Rician ( ) 1 2 -- A 2 2 s 2 ------- exp u s 2 ---- u 2 A 2 + 2 s 2 ----------------- I 0 Au s 2 ------ exp u d 0 = Pr Rayleigh>Rician ( ) 1 2 -- A 2 2 s 2 ------- exp 1 2 -- E p 2 2 σ 2 --------- exp 1 2 -- E p 2 N 0 --------- exp = = = E p
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