Bandwidth-Noise-SNR relationship

Let’s say we have the noise figure F of a system of devices. We also have the bandwidth improvement coefficient ( as defined by my teacher ) $$BI=10log(\frac{B_{in}}{B_{out}})$$ where B1 is the bandwidth at the input and B2 the bandwidth at the output of the system. Having these two, my teacher claims that the following stands : $$SNR_{out}=SNR_{in} - F + BI$$

Everything in the expression above is in decibels.

I really don’t understand why the BI is in the equation. Even Wikipedia’s definition of noise figure doesn’t include that term.

The equation is supposedly derived from the definition of thermal noise : $$N_{th}(dBm)=-174 dBm +10logB_1$$ where B is the bandwidth and the definition of noise figure as the amount of noise, produced by a device,which is above the thermal noise : $$N(dBm)=-174dBm +10logB_2+F(dB)$$

I can see that $$N-N_{th}=-F(dB)+ BI(dB)$$ This is as far as I can get .. Moreover, if BI is greater than F , then SNR_out will be greater than SNR_in. Is that even possible ?

Note : This is all based on an exercise on superheterodyne receiver.

• Noise power is proportional to bandwidth (noise voltage is proportional to squareroot of bandwidth). If you drop the bandwidth by 4:1, the noise power drops 4:1, or (3 + 3) dB. – analogsystemsrf Nov 20 '18 at 3:33

If you have a large bandwidth with a lot of white(thermal) noise and decrease this BW, your SNR will go up, assuming your signal is still intact. Therefore the BI factor will increase the $$\SNR_{out}\$$ when you decrease the BW through the system.