Question

A receiver has a dynamic range of 65 dB. It has a sensitivity of 0.88 nW. The maximum allowable input signal is approximately:

Answer 1

2.783 * 10^-3 watts

Step-by-step explanation:

calculate the maximum allowable input signal

= Dynamic range * sensitivity

Dynamic range is in decibels hence we have to take log

∴ 65 db. =
`10log_(10) (DR)`

=
`10 (65)/(10)` =
`10^(6.5)`
`10^(6.5)`

therefore DR ( dynamic range ) = 31.6223 * 10^5

sensitivity = 0.88 nW = 0.88 * 10 ^-9 watts

therefore the maximum allowable input signal = ( 0.88 * 10^-9 ) * ( 31.6223 * 10^5 ) = 2.783 * 10^-3 watts