woooo....4/20
getting the test back....greeeaat
ganked a 83 out of that shit, kick ass!

performance criterion
xmitter input chosen from finite set of possible symbols
rx obj: accurately det which symbol was tx
figure of merit: prob of error in decision
	often interested in prob of bit error or bit error rate (BER)
	
general rxer:
	binary sig txed
	using gaussians simplifies equations
	sig txed through additive white gaussian noise (AWGN) channel where noise N(t) is added to sig so that the rxed sig is R(t) = si(t) + N(t)
	N is gaussian noise process with mean of 0, Sn(f) = n/2, PSD is const for all freqs
	the rxed sig is filtered to produce Y(t) = conv(si(t),h(t)) + conv(N(t),h(t)) = ^si(t) + n(t)
	output of filter is sampled at time tm to produce the decision statistic, Z = ^si(tm) + n(tm)
	Z is a gaussian RV
	mean of Z is mui = ^si(tm), depends only on signal component
	variance(Z) = n/2 * int<inf>(h^2(t)dt), variance does not depend on which sig was sent
	denote cond decision stats as Z0 (decision stat, conditioned on s0(t) being sent) and Z1 (decision stat, conditioned on s1(t) being sent)
	gaussian rand vars have density f()s of f0(z) = 1/sqrt(2pi*sigma^2) * e^(-(z-mu0)^2/(2sigma^2))
	f1(z) = 1/sqrt(2pi*sigma^2) * e^(-(z-mu1)^2/(2sigma^2))
	assume mu1>mu0, rxer will comp Z with thresh gamma and if Z>=gamma, s1 sent, Z<gamma, s0 sent
	choice of gamma will affect prob of error
	to min prob of error, thresh should be gamma = (mu0 + mu1)/2
	if sigs are antipodal (s0 = -s1), gamma=0
	
ex11-1
find Pe0 = P(error given s0 sent) = P(z>=gamma|s0 sent) = P(z0>=gamma), z0 is gaussian
find Pe1 = P(error given s1 sent) = P(z>=gamma|s1 sent) = P(z1<gamma), z1 is gaussian
Pe1 = 1-P(z1>=gamma) = 1-Q((gamma-mu1)/sigma) = Q((mu1-gamma)/sigma)

if a<b, Q(a)<Q(b)

ex11-2
s1(t) = rect((t-T/2)/T)
s0(t) = -rect((t-T/2)/T)
h(t) = rect((t-T/2)/T)
mui = conv(si,h)
mu1 = tri((t-T)/T)
mu0 = -tri((t-T)/T)
max at T, so sample at T to get widest gap between means
variance = n/2 * int<inf>(h^2(t)) = nT/2
gamma = (mu1+mu0)/2 = 0
Pe0 = Q((gamma-mu0)/sigma) = Q(sqrt(2T/n))
Pe0 = Q((gamma-mu1)/sigma) = Q(sqrt(2T/n)) = Pe (because they're the same)