Reply by robert bristow-johnson●November 24, 20062006-11-24
ma wrote:
>
> When can I asssume that Q/I are small?
it's not the smallness of Q/I that you need to worry about but the
smallness in the rate of change of
omega[n] = arg{ I[n] + j*Q[n] } - arg{ I[n-1] + j*Q[n-1] }
that needs to be small. if |omega[n] - omega[n-1]| is less than pi/2,
you can use a trig identity:
arctan(y) - arctan(x) = arctan( (y-x)/(1 + y*x) )
as long as |arctan(y) - arctan(x)| <= pi/2
and get:
omega[n] = arctan( Q[n]/I[n] ) - arctan( Q[n-1]/I[n-1] )
omega[n] = arctan( (Q[n]*I[n-1] - I[n]*Q[n-1])
/ (I[n]*I[n-1] + Q[n]*Q[n-1]) )
that's how to do FM demod (or group delay if n is frequency, not time)
without having to worry about quadrants and discontinuities and such.
r b-j
> It is not small if there is a carrier
> frequncy and it is not small if I bring the signal to base band or anywhere
> in between.
>
> Regards
Reply by ma●November 24, 20062006-11-24
"Mirko Liss" <mirko.liss@web.de> wrote in message
news:20061124012401.354.4.NOFFLE@mliss.myfqdn.de...
> ma:
>> I found that it is believed that if we have an FM modulation then we can
>> demodulate it by calculating (Q/I)' But why? How (sin/cos)' relates to
>> frequency ?
>
> You need (arctan(Q/I))' to demodulate FM signals.
>
> Since (arctan(x))' = 1/(1+x^2)
> you get:
> (arctan(Q/I))' = 1/(1+(Q/I)^2) * (Q/I)'
>
>
> If you can make sure that Q/I < 0.1 then 1/(1+(Q/I)^2) > 0.990.
>
> So replacing (arctan(Q/I))' with (Q/I)' causes an error of <1%.
> And that's no big deal.
When can I asssume that Q/I are small? It is not small if there is a carrier
frequncy and it is not small if I bring the signal to base band or anywhere
in between.
Regards
Reply by Mirko Liss●November 23, 20062006-11-23
ma:
> I found that it is believed that if we have an FM modulation then we can
> demodulate it by calculating (Q/I)' But why? How (sin/cos)' relates to
> frequency ?
You need (arctan(Q/I))' to demodulate FM signals.
Since (arctan(x))' = 1/(1+x^2)
you get:
(arctan(Q/I))' = 1/(1+(Q/I)^2) * (Q/I)'
If you can make sure that Q/I < 0.1 then 1/(1+(Q/I)^2) > 0.990.
So replacing (arctan(Q/I))' with (Q/I)' causes an error of <1%.
And that's no big deal.
Reply by Mirko Liss●November 23, 20062006-11-23
ma:
> I found that it is believed that if we have an FM modulation then we can
> demodulate it by calculating (Q/I)' But why? How (sin/cos)' relates to
> frequency ?
You need (arctan(Q/I))' to demodulate FM signals.
Since (arctan(x))' = 1/(1+x^2)
you get:
(arctan(Q/I))' = 1/(1+(Q/I)^2) * (Q/I)'
If you can make sure that Q/I < 0.1 then 1/(1+(Q/I)^2) < 0.990.
So replacing (arctan(Q/I))' with (Q/I)' causes an error of <1%.
And that's no big deal.
Reply by naebad●November 23, 20062006-11-23
ma wrote:
> Hello,
> I found that it is believed that if we have an FM modulation then we can
> demodulate it by calculating (Q/I)' But why? How (sin/cos)' relates to
> frequency ?
>
> Regards
It's actually d/dt {atan(Q/I)} ie the rate of change of phase which is
being calculated.
Naebad
Reply by ma●November 23, 20062006-11-23
Hello,
I found that it is believed that if we have an FM modulation then we can
demodulate it by calculating (Q/I)' But why? How (sin/cos)' relates to
frequency ?
Regards