Discussion
A receiver has a \( 200 \ \mathrm{kHz} \) noise bandwidth and a \( 6 \ \mathrm{dB} \) noise figure. Its noise floor is about
Answer: C
Building it up from the reference density:
\[ \begin{aligned} P &= -174 + 10\log_{10}(2\times10^{5}) + 6 \\ &= -174 + 53 + 6 = -115 \ \mathrm{dBm} \end{aligned} \]
so about -115 dBm. -121 dBm omits the noise figure. -108 dBm double-counts it. And -174 dBm is the density per hertz before any bandwidth or noise figure is applied. Assembling a noise floor from density, bandwidth and noise figure is the standard three-term calculation of every link budget.
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