problem 2.63 review
g = e^(-2t)u(t)
PSIg(tau) = int<inf>(g(t)g(t-tau)dt)
tau>0 -> = e^(2tau)int<tau,inf>(e^(-4t)dt) = e^(2tau)*(-1/4)*e^(-4t)|(t -> tau,inf)
tau<=0 -> = e^(2tau)int<0,inf>(e^(-4t)dt)


have a sawtooth with T=2pi trough/peak at (0,0) and min/max at 0/1
can write sig as infinite sum s = sum<inf>(Dn*e^(j2ntpi*f0))
Dn = 1/T0 * int<T0>(s(t)e^(-j2ntpi*f0)dt) = 1/2pi * int<0,2pi>(tdt/2pi * e^(-jnt)) = 1/4pipi * e^(-jnt)/(-nn) * (-jnt - 1)|(t -> 0,2pi)
 = -1/4nnpipi * (e^(-jn2pi) * (-jn2pi-1) + 1) = j/n2pi = Dn = mag(Dn)e^(j*phase(Dn)), D0 = 1/2 = 1/T0 * int<T0>(s(t)dt)

2.8a
trig form an = 4/npi * sin(npi/2), bn = 0 -> Cn = sqrt(anan + bnbn) = abs(an), thetan = arctan(-bn/an) = 0

get one page of 8.5x11 handwritten

int(xe^(ax)dx) = e^(ax)/aa * (ax-1), apply above

going over quiz 2....