If these 2 signals, that form this duplex, have ranges of 15-30 [kHz] and 30-45 [kHz] using:

$NoiseFloor_{dBm} =10\log _{10}(k\times T_{0}\times 1000)+NF+10\log _{10}(BW)$ (formula from wikipedia)

and if $k,T,NF$ and $\Delta f$ are given, only thing left to figure out is the $BW$ (bandwidth). Is the $ BW=2(\Delta f+f_{m})$ where $f_{m}=2\times 15=30 [kHz]$ correct formula for the bandwidth of this duplex?


Yes, the 30 kHz are correct, but only assuming your receiver has filters that filter out any noise outside these bands. That might or might not be the case – if your receiver in fact observes 0 – 45 kHz, for example, then the noise energy it'll see is defined by that 45 kHz bandwidth.

I'd also recommend you not just peruse the formula from Wikipedia (which I'd argue is in terrible shape), but try to understand how this noise floor calculation comes to be, and then, after that, apply the conversion to dBm.

Namely, you get Johnson-Nyquist thermal noise. This noise has a power density of $k_B T$, that is, Boltzmann's constant times the temperature of your system. This constant represents Energy per temperature above absolute zero (eg. Joules per Kelvin), so $k_B T$ is Energy.

In signal-processing systems we commonly care about Power more than energy (because, for example, if we know how much power gets dissipated by a resistor, we know the voltage across it), so we divide the Energy by time, or, multiply with bandwidth/frequency (which is the inverse of time):

$$P_\text{N, thermal}=k_B T \Delta f$$

Real-world active systems, especially receiver's amplifiers, add more than just the noise present due to thermal noise – the factor between the noise going in and coming out is the Noise Figure $\text{NF}$. Therefore:

$$P_\text{N, thermal}=k_B T \Delta f\, \text{NF}$$

Because we like our $\text{dB}$-based units, let's cancel out all units (I'm doing this in-line here just for clarity) and then do the dB-conversion. Then we get

$$P_\text{N, thermal}[\text{dBW}]=10 \log_{10}\left((k_B \cdot \frac KJ)( T \cdot \frac 1K)(\Delta f \cdot s) \, \text{NF}\right)$$

You wanted $\text{dBm}$, but a milliwatt is simply $\frac1{1000}$ of a Watt, so

$$\begin{align} P_\text{N, thermal}[\text{dBm}]=10 \log_{10}\left((k_B \cdot \frac KJ)( T \cdot \frac 1K)(\Delta f \cdot s) \, \text{NF}\,\cdot 1000\right) \end{align}$$

Let's apply a few logarithm rules here:

$$\begin{align} P_\text{N, thermal}[\text{dBm}]&=10 \log_{10}\left((k_B \cdot \frac KJ)( T \cdot \frac 1K)\right)&+&10 \log_{10}\left(\Delta f \cdot s \right)&+&10 \log_{10}\left( \text{NF}\right) &+& 10\log_{10}\left(1000\right)\\ &=10 \log_{10}\left((k_B \cdot \frac KJ)( T \cdot \frac 1K)\right)&+&10 \log_{10}\left(\Delta f \cdot s \right)&+&\text{NF}_\text{dB} &+& 30 \end{align}$$

If you drop in the values of Boltzmann's constant and room temperature, you'll end up with the classical -174dBm/Hz; simply add your bandwidth (in dBHz) and your Noise Figure (in dB), and you get your observed noise power.

  • $\begingroup$ $f_m$ is 30kHz not the BW, right? $\endgroup$ – Desperado Jan 14 '17 at 20:58
  • $\begingroup$ I don't know – all I know is that from your description, you've got two 15 kHz ranges, and the total bandwidth of that is 30 kHz. $f_m$ is used in your formula, but I don't know where it comes from. $\endgroup$ – Marcus Müller Jan 14 '17 at 21:33
  • $\begingroup$ That's a maximum frequency of the modulating signal. Carson's rule $\endgroup$ – Desperado Jan 14 '17 at 22:39
  • $\begingroup$ Well, it doesn't really matter at all for the question. The question is how wide the input filters of your receiver are, not how wide your signal of interest is, @Desperado. $\endgroup$ – Marcus Müller Jan 15 '17 at 9:10

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