written 6.2 years ago by | • modified 6.2 years ago |
A1=6dB A2=15dB A3=10d
NF1=10dB NF2=6dB NF3=10dB
Mumbai University > Electronics > Sem 4 > Priniciples of Communication Engineering
written 6.2 years ago by | • modified 6.2 years ago |
A1=6dB A2=15dB A3=10d
NF1=10dB NF2=6dB NF3=10dB
Mumbai University > Electronics > Sem 4 > Priniciples of Communication Engineering
written 6.2 years ago by |
Given :
Power gain →A1=6dB,A2=15dB,A3=10dB
Noise factor →NF1=10dB,NF2=6dB,NF3=10dB
To Find : - Overall Noise factor ( FT) and Nosie figure = NFT
Solution : -
FT(dB)=F1+F2−B1Alin1+F3−1Alin1.Alin2
NF=10logF→F= antilog (NF10)
F1=antilog(10dB10)=10′=10;A1=6dB→6dB=10logAlin1
Hence, Alin1=antilog(6dB10)=3.98
F2=antilog(6dB10)=100.6=3.98;A2=15dB→A2=10logAlin2
Therefore , 15 = 10logAlin2
i.e Alin2=antilog(15dB10) = 31.623
F3=antilog(10dB10)=10′=10;A3=10dB→A3=10logAlin3
Therefore , 10 = 10logAlin3
i.e Alin3=antilog(10dB10) = 10
Noise Factor
FT=10+3⋅98−13⋅98+10−1(3.98)(31⋅623)=10.820
Noise Figure
NFT=10log(10⋅820)=10⋅342