bandwidth usage doesn't change for modulated signals, but band location changes
mod sigs written as m(t)*cos()
PSD gets shifted up

B = 1/Tb for the primary BW
total usage = 2/Tb for shifted freqs

PHIpsk(t) = m(t)cos(wt)
m = sum(akp(t-kTb))
p(t) = rect((t-Tb/2)/Tb)
Sm = Tb*sinc^2(fpi*Tb) = mag(P(f))^2/Tb
Sphi_psk = Tb/4 * (sinc^2(f+fc) + sinc^2(f-fc))

Bbaseband = 1/Tb = Rb

Bpsk = Fc+1/Tb-(fc-1/Tb) = 2/Tb = 2Rb

FSK
PHIfsk = sum(akp(t-kTb)cos(wct)) + sum((1-ak)p(t-kTb)cos(wct))
ak is 0 or 1
FSK is a superposition of 2 ASK sigs with different freqs

Bfsk = fc1-fc0+2/Tb = 2Rb+fc1-fc0

homeworks 5,6,7,8
maybe a little from 4

multiple channels
all channels with 300Hz bw
3 channels
inputs are sampling freq and number of bits per sample
samples combined into a frame
more bits per frame, same frequency
calc bits per frame, channels*bits/chan + overhead
Rb = (3n+x) bits/frame * fs frames/sec = fs(3n+x)bits/sec

no raised cosine fun

M-ary modulation techniques
inc data rate: reduce time to transmit bits, transmit multiple bits in Tb sec
reducing Tb incs BW
increasing bits per symbol either keeps bw the same, but requires more power or increases bw need but requires less power depending on which modulation scheme is used
M-ASK
simple extension of binary ASK
sigs are of form PSI = ai*cos(wct)
ai is chosen from a set ai in (0,A,2A,...,(M-1)A)
more efficient to center them around 0