A cascade of two such amplifiers will have noise figure of
a) $6$ dB
b) $8$ dB
c) $12$ dB
d) none of these
Correct Answer: Option D
Solution:
Solution:
Gain of A = $6\;dB$
$10\;log_{10}(G)=6\;dB$
$G=10^{6/10}=10^{0.6}$
$G=3.98$
Noise figure = $3\;dB$
$10\;log_{10}(G)=3\;dB$
$F=10^{3/10}=10^{0.3}$
$F=1.995$
Equivalent noise figure of cascaded amplifiers, $$F_{cas}=F_1+\frac{F_2-1}{G_1}+\frac{F_3-1}{G_1\;G_2}+...$$ $$F_{cas}=1.995+\frac{1.995-1}{3.98}$$ $$F_{cas}=1.995+\frac{0.995}{3.98}$$ $$F_{cas}=1.995+0.25$$ $$F_{cas}=2.245$$ $$F_{cas}(dB)=10\;log_{10}(2.245)$$ $$F_{cas}(dB)=3.512\;dB$$
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