By qpsk,  i mean  filtered qpsk obviously.

               shankar

> 
> 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)).

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

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