text: # 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. Nov 20, 2018 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.
explanation: The text extract discusses the relationship between bandwidth, noise, and signal-to-noise ratio (SNR) in a system, particularly in the context of a superheterodyne receiver. It includes a specific equation that incorporates the noise figure (F) and the bandwidth improvement coefficient (BI) to determine the output SNR. The user expresses confusion about the inclusion of the BI term in the equation and questions its validity, especially when BI is greater than F.
  
  The text does not contain a direct question or a question-and-answer format. Instead, it presents a detailed explanation and a note from a comment, which provides additional context but does not form a structured question or answer.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # 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. Nov 20, 2018 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.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract discusses the relationship between bandwidth, noise, and signal-to-noise ratio (SNR) in a system, particularly in the context of a superheterodyne receiver. It includes a specific equation that incorporates the noise figure (F) and the bandwidth improvement coefficient (BI) to determine the output SNR. The user expresses confusion about the inclusion of the BI term in the equation and questions its validity, especially when BI is greater than F.
  
  The text does not contain a direct question or a question-and-answer format. Instead, it presents a detailed explanation and a note from a comment, which provides additional context but does not form a structured question or answer.
  
  Label: no-question