Discussion
Noise power in a bandwidth \( B \) at temperature \( T \) is \( kTB \). For \( T = 290 \ \mathrm{K} \) and \( B = 1 \ \mathrm{MHz} \), the noise power is about
Answer: A
The thermal noise floor is \( -174 \ \mathrm{dBm/Hz} \) at this temperature, so in a megahertz:
\[ \begin{aligned} P &= -174 + 10\log_{10}(10^{6}) \\ &= -174 + 60 = -114 \ \mathrm{dBm} \end{aligned} \]
giving about -114 dBm, a figure worth memorising as the noise floor in one megahertz. -174 dBm is the density per hertz rather than the power in the band. -84 dBm would correspond to a gigahertz of bandwidth. -144 dBm corresponds to a kilohertz.
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