Machine-translated by Google. Questions are authored in English.
Top

Discussion

The available thermal noise power in a bandwidth \( B \) is

  • A.kT over B
  • B.kTB
  • C.4kTBR
  • D.2qIB

Answer: B

The available power is \( kTB \), independent of the resistance value, which is why a noise floor can be quoted in dBm per hertz without naming an impedance. \( 4kTBR \) is the mean square open-circuit voltage, which does depend on resistance. \( 2qIB \) is shot noise power spectral density in current terms. And dividing by bandwidth would give a density rather than a power. Knowing which expression gives a power and which a voltage prevents most errors in noise arithmetic.

No comment is present. Be the first to comment.
Loading…

Post your comment

Please log in or register to post a comment.