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

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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.

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(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.