m(t) has frequency range 5 kHz to 15 kHz Now it is amplitude modulated f(t)=A(1+m(t))cos2πfct where fc=6000kHz ∴ AM signal will have highest frequency =fc+fm(max) =600+15=615kHz And AM signal will have lowest frequency =fc−fm(max) =600−15=585kHz It is a band pass signal so we use bandpass sampling fs=1.2×
2fH
k
K=
fH
fH−fL
=
615
615−585
K=20.5 We select K=20 ∴fs=1.2×
2×615
20
∴fs=73.8kHz Now L=256 And 2n=L=256 ∴n=8 Bitrate =Rb=nfs ∴Rb=8×73.8×103 ∴Rb=0.59Mbps