2.12 A communication system uses BPSK modulation to transmit data bits b₁, 1 = 0, 1, 2,.... In the transmitter, a sequence of rectan- gular pulses is mixed to a carrier

frequency by multiplying the rectangular pulses by √2P cos(27f₁t): s(t) = √2P bipr(t-IT) cos(2n fit). /-0,1.... At the receiver, the received signal is first mixed down to baseband by multiplying by √2/T cos(27f2t), where f₂ - f₁ = Af is the offset of the two 136 oscillators. After the signal is mixed down, it is filtered with a matched filter, that is h(t) = pr(t). The filter is sampled at time t = iT for i = 1,2,.... In addition, the signal is mixed down by multiplying by -√2/T sin(27f2t). Let ye(iT) denote the first output and y, (iT) denote the second output. Then, ye(it) = = £ $(7) √/²7/ CC cos(21f27)h(iT - t)dt ys(IT) = - = - * S(T) 7) √ sin(2727)h(iT - T)dt. 1-00 (a) In the absence of noise, evaluate the outputs y(iT) and y, (iT) in terms of bi-₁› E = PT. AfT, and į. Ignore double frequency terms in evaluating the output. That is, derive an expression for ye(iT) and y,(iT). (Useful trig identity sin(u) - sin(v) = 2 cos("") sin("").) (b) Assume you buy two crystal oscillators at a 10 MHz nominal frequency that have + 10 PPM accuracy. That is, factual = fnominal (1 ± 10/106). Assume that the data rate is 100 kbps (7 = 10-5), that the data bits are all positive ( b; = 1, i = 0, 1, 2,..., 500), and that E = 1. (i) Are the double frequency terms negligible? (ii) Plot the output of the filters y (iT) and y, (iT) as a function of į for 1 ≤i≤ 500. Assume all the data bits are +1 and f₂-fi = Af = 200 Hz and T = 10-³.