A1=6dB A2=15dB A3=10d

NF1=10dB NF2=6dB NF3=10dB

Mumbai University > Electronics > Sem 4 > Priniciples of Communication Engineering

Given :

Power gain $ \rightarrow A_1 = 6dB , A_2 = 15dB , A_3 = 10dB$

Noise factor $ \rightarrow NF_1 = 10dB , NF_2 = 6dB , NF_3 = 10dB$

To Find : - Overall Noise factor ( $F_T$) and Nosie figure = $N_{FT}$

Solution : -

$ F_{T}(d B)=F_{1}+\frac{F_{2}-B1}{A_{lin1}} + \frac{F_{3}-1}{A_{lin1}.A_{lin2}} $

$ N F=10 \log F \rightarrow F= $ antilog $ \left(\frac{N F}{10}\right) $

$ F_{1}=\operatorname{antilog}\left(\frac{10 d B}{10}\right)=10^{\prime}=10 ; A_{1}=6 d B \rightarrow 6 d B=10 \log A_{lin1} $

Hence, $A_{lin1} = \operatorname{antilog}\left(\frac{6 d B}{10}\right) = 3.98 $

$ F_{2}=\operatorname{antilog}\left(\frac{6 dB}{10}\right)=10^{0.6}=3.98 ; A_{2}=15 d B \rightarrow A_{2} = 10 \log A_{lin2} $

Therefore , 15 = $10 \log A_{lin2}$

i.e $ A_{lin2}=\operatorname{antilog}\left(\frac{15 dB}{10}\right) $ = 31.623

$ F_{3}=\operatorname{antilog}\left(\frac{10 dB}{10}\right)=10^{\prime}=10 ; A_{3}=10 d B \rightarrow A_{3} = 10 \log A_{lin3} $

Therefore , 10 = $10 \log A_{lin3}$

i.e $ A_{lin3}=\operatorname{antilog}\left(\frac{10 dB}{10}\right) $ = 10

Noise Factor

$ F_{T}=10+\frac{3 \cdot 98-1}{3 \cdot 98}+\frac{10-1}{(3.98)(31 \cdot 623)}=10.820 $

Noise Figure

$ N _{FT}=10 \log (10 \cdot 820)=10 \cdot 342 $