楼主:
djljing (阿劲)
2015-11-14 15:05:21t0=0.15s fc=250Hz 调变指数=0.85=a
绘制出输入讯号,调变过后的讯号,解调后的讯号
mn(t)=m(t)/max|m(t)|
s(t)=Ac[1+ka*m(t)]*cos(2*pi*fc*t)
=Ac[1+a*mn(t)]*cos(2*pi*fc*t)
10sin(20*pi*t) 0 <= t<=1/3t0
m(t)= -5sin(20*pi*t) 1/3<=t<=2/3t0
10sin(20*pi*t) 2/3<=t<=t0
0 otherwise