Hi, The bandwidth in quadrature modulation is typically assumed to be max(B1,B2) where B1 = bandwidth of I(t) B2 = bw of Q(t) rf signal = I(t) cos wt + Q(t) sin wt = A(t) cos(wt + phi(t) ) w = carrier frequency It seems to me that for a given amount of phase change happening during a given time, constraining the envelope to remain constant increases the bandwidth. If yes , why ? ( Notice that in GMSK-GSM, the maximum phase variation allowed during a symbol is pi/2 which happens if 3 consecutive bits are identical. But in QPSK with same symbol rate, even pi phase change is possible per symbol duration. ) I assume that for same symbol rate, GMSK and QPSK are going to take approximately the same bandwidth. I dont have the data in front of me . Quadrature modulation will result in phase/frequency modulation of the carrier. This should result in consuming more bandwidth than that of I(t) or Q(t). Why this does not happen ? Does bw of cos(phi(t)) give a good estimate of bw of rf signal ? thanks shankar

# linear modulation and cpm

Started by ●September 17, 2003

Reply by ●September 18, 20032003-09-18

kbc32@yahoo.com (kbc) wrote in message news:<a382521e.0309170414.10b3c684@posting.google.com>...> Hi, > > The bandwidth in quadrature modulation is typically > assumed to be max(B1,B2) where > B1 = bandwidth of I(t) > B2 = bw of Q(t) > > rf signal = I(t) cos wt + Q(t) sin wt = A(t) cos(wt + phi(t) ) > w = carrier frequency > > It seems to me that for a given amount of phase change > happening during a given time, constraining the envelope > to remain constant increases the bandwidth. If yes , why ? > > ( Notice that in GMSK-GSM, the maximum phase variation allowed > during a symbol is pi/2 which happens if 3 consecutive bits > are identical. But in QPSK with same symbol rate, even > pi phase change is possible per symbol duration. ) I assume > that for same symbol rate, GMSK and QPSK are going to take > approximately the same bandwidth. I dont have the data in > front of me . > > Quadrature modulation will result in phase/frequency modulation > of the carrier. This should result in consuming more bandwidth > than that of I(t) or Q(t). Why this does not happen ? > > Does bw of cos(phi(t)) give a good estimate of bw of rf signal ?It appears {I(t),Q(t)} is baseband signal. I guess w is the carrier frequency at RF level. From your Eqs, I(t)= A(t)cos(phi(t)) Q(t)= -A(t)sin(phi(t)) So A(t)=sqroot(I(t)^2+ Q(t)^2) phi(t)=arg(-Q(t)/I(t)) From your Eqn. I expect "bw of rf signal/baseband signal" only depends on phi(t)- wonder why you suspect "good estimate"! Secondly "bw of cos(phi(t))" hardly makes any sense! Regards, Santosh> > thanks > shankar

Reply by ●September 19, 20032003-09-19

> > It appears {I(t),Q(t)} is baseband signal. I guess w is the carrier > frequency at RF level. > > From your Eqs, > I(t)= A(t)cos(phi(t)) > Q(t)= -A(t)sin(phi(t)) > > So A(t)=sqroot(I(t)^2+ Q(t)^2) > phi(t)=arg(-Q(t)/I(t)) > > From your Eqn. I expect "bw of rf signal/baseband signal" only > depends on > phi(t)- wonder why you suspect "good estimate"! Secondly "bw of > cos(phi(t))" hardly makes any sense! > > Regards, > Santosh > > >Well, bandwidth at passband is not supposed to depend on carrier frequency value. In cos(wt + phi(t)) , you set carrier freq to zero and you get cos(phi(t)).

Reply by ●September 25, 20032003-09-25