Determine (a) the peak frequency deviation, (b) minimum bandwidth, and (c) baud for a binary FSK signal with a mark frequency of 38 kHz, a space frequency of 40 kHz, and an input bit rate of 4 kbps.
a)
2Δ f=∣fm−fs∣2\Delta\>f=|fm-f_s|2Δf=∣fm−fs∣
Δ f=(40−38)2=1kHz\Delta\>f=\frac{(40-38)}{2}=1kHzΔf=2(40−38)=1kHz
b)
Bmin=2(Δ f+fb)B_{min}=2(\Delta\>f+f_b)Bmin=2(Δf+fb)
=2(1+4)=2(1+4)=2(1+4)
=10kHz=10kHz=10kHz
c)
Baud for a binary FSK =fbN=\frac{f_b}{N}=Nfb where N=1N=1N=1
=40001=4000=\frac{4000}{1}=4000=14000=4000
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