Yes Alberto, of course you are right. The average length of the rotating I/Q vector is the DC component of course = half the "P.E.P. voltage" without modulation. I forgot about that...
I'm still totally astonished by my new "toy" :-) :-) :-) 73 Johan SM6LKM ---- i2phd wrote: > --- In [email protected], "Johan H. Bodin" <[EMAIL PROTECTED]> wrote: >>> 1-An algorithm for demodulating AM, in detail >> Output = (I^2 + Q^2)^0.5 > > > Correct Johan, but you must not forget to place a highpass > filter after that. In practice, the digital equivalent of a DC > blocking capacitor... otherwise you will have a substantial offset > from zero in your demodulated signal. What I use in Winrad is this : > > // 3rd order elliptic filter, fs=11025 Hz, > // hi-pass, fcut = 150 Hz, ripple = 2dB > > static float HPFcoef_11k[12] = > { 1., -1.9999374905129277, 1., 1., -1.9571775980131627, > 0.96526631928081519, > 1., -1., 0., 1., -0.79282675867370578, 0. > }; > > > 73 Alberto I2PHD
