Ex1
P_2=\frac{1}{2} P_110 log \frac{P_2}{P_1} = 10 log \frac{0.5P_1}{P_1}=10 log (0.5)=-3dbEx2
10 log \frac{10P_1}{P_1} =10 log10 =10dbEx3


P_2=\frac{1}{2} P_110 log \frac{P_2}{P_1} = 10 log \frac{0.5P_1}{P_1}=10 log (0.5)=-3db10 log \frac{10P_1}{P_1} =10 log10 =10db
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