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First, multiply (-174dBm/Hz) by 180000, then convert the result into watt? Because the decibel is a logarithmic unit, you need to be careful here. $-174\mathrm{dBm}$ means $10^{-17.4} \mathrm{mW}$. Common usage of the decibel tends to mix operations, so $-174\mathrm{dBm/Hz}$ means $10^{-17.4} \mathrm{mW/Hz}$ You can really complicate your units by adding $... 3 Let's model the data as: $${y}_{i} = {x}_{i} + {n}_{i}$$ So the the$ i $-th pixel in the noisy image$ Y $is composed by the noiseless image data and additive IID noise. Now assume we have 2 images:$ {Y}^{1} $and$ {Y}^{2} $: $${y}^{j}_{i} = {x}_{i} + {n}^{j}_{i}, j = 1, 2$$ Indeed, noise wise, when we combine data temporarily (Between 2 images) and ... 2 I'm just making this up on the fly. There's got to be at least one paper out there on this, or sections in Kalman filtering books. Supposing that Δtk is distributed normally, does this just come out naturally as disturbance? i.e. could I account for this with my Kalman gain Nope. Take the continuous-time model of the system:$\begin{aligned} \dot x(t) &... 2 You would use the second method suggested by the OP. dBm uses a power ratio relative to 1 mW (specifically as10\log_{10}(P_{in}/1\text{mW})$. To convert -174 dBm/Hz into Watts/Hz, you could either subtract 30 dB to get -204 dBW/Hz (where dBW uses a ratio of Watts) and then reverse the dB by dividing by 10 and raising to the 10th power, or equivalently ... 2$E_s\$ would be the energy of the channel symbols at the output of the matched filter as scaled by the path loss between transmitter and receiver as long as all timing and carrier offsets have been corrected for and channel equalization completed - consider how after matched filtering with carrier and timing offsets removed we use just one sample per symbol ...