P_{n}=4kTBR is the minimum NOISE power, so your signal-to-noise ratio is P_{s}/P_{n} for a signal power of P_{s}. The only way to reduce the noise power is to reduce T (temperature), B (bandwidth), or R (resistance).Thanks Bob,
Sowhy do we need the relationship
P = 4.k.t.R.B
If we already know the power in the circuit?
4 k T R B is the variance of the voltage (which you could call "the power" but of the voltage signal, "power" not being physical power, but "square of the signal").Thanks Bob,
Sowhy do we need the relationship
P = 4.k.T.R.B
If we already know the power in the circuit?
Because you cannot extract power without a load. A better way to think of the physics here is to think of a black-body radiator. A black-body radiator is a device that is in thermal equilibrium, it emits the same amount of power as it absorbs from its environment. The circuit that you are given is undergoing the same process, it is absorbing radiation (thermal, visible, RF, what have you) from the surrounding environment. Normally, it would just spit it back out since it is at equilibrium. However, if you have a load in your circuit, then part of the absorbed power is dissipated by the load which gives rise to the noise signal.Thanks vanesch, so
p=k*t*b is the real power.
p=4*k*t*r*b is the signal power.
But if the power is independent of conductor length, what is to stop me from cutting a conductor into *infininte* (read lots) of peices and getting lots of power?