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

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ECEN 661 - Modulation Theory Homework #8 Solutions 1. Let the BPSK signal be written as over the bit interval . (a) The received will base its decision on a statistic of the form . This decision statistic will be Gaussian with mean of and variance of . Therefore the probability of error is . The loss in SNR is . Hence we will experience 0.5dB loss when . (b) For the case of QPSK we can proceed in the same manner except now the transmitted signal is going to be of the form . We form the statistic as before to make a decision on the I-channel bit resulting in and the variance is the same as before. In this case the probability of error then becomes To find to tolerable phase error, note that in the ideal situation (no phase error), the system will require =6.79dB to achieve . Hence, with 0.5dB loss, our system will operate at =7.29dB. With this value of SNR, the desired will be achieved when . Hence the QPSK needs much greater phase estimation. 2. (a) The decision statistic can be written as where and are zero mean Gaussain random variables with variance . In order to simplify the calculations, form a rotated statistic and then the phase error will be simply the angle of s t ( ) A ± ω c t θ + ( ) cos = 0 T s , ( ) z 2 r t ( ) ω c t φ + ( ) cos t d 0 T s = E z [ ] A ± T s θ ( ) cos = σ 2 N o T s = P e Q E z [ ] ( ) 2 σ 2 ------------------ Q A 2 T s N o ----------- θ ( ) cos Q 2 E b N o --------- θ ( ) cos = = = loss(dB) 10log 10 cos 2 θ ( ) ( ) = θ cos 1 10 0.05 ( ) 19.25 ° = = s t ( ) A ± ω c t θ + ( ) A ± ω c t θ + ( ) sin cos = z E z [ ] A ± T s θ ( ) AT s θ ( ) sin + cos = P e 1 2 -- Q A 2 T s N o ----------- θ ( ) θ ( ) sin + cos ( ) 2 1 2 -- Q A 2 T s N o ----------- θ ( ) θ ( ) sin cos ( ) 2 + = 1 2 -- Q 4 E b N o --------- θ π 4 -- cos 2 1 2 -- Q 4 E b N o --------- θ π 4 -- + cos 2 + = E b N o P e 10 3 = E b N o P e 10 3 = θ
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