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Noise power in conductor

by Rudibot
Tags: conductor, noise, power
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Rudibot
#1
Dec13-09, 09:50 PM
P: 10
The noise power in a conductor is given by
P = k.t.B

Does this mean the noise power is independant of the conductor length?
Also, does this only apply for a conductor with no resistance?

Thnks,
Rudi
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Bob S
#2
Dec13-09, 10:01 PM
P: 4,663
KTB is just the thermal power in a circuit at temperature T and bandwidth B, not considering any other sources of noise. KTB is about -114 dBm per MHz at 273 kelvin. It has nothing to do with conductor length. See
http://en.wikipedia.org/wiki/Johnson...3Nyquist_noise
Bob S
Rudibot
#3
Dec13-09, 10:37 PM
P: 10
Thanks Bob,

Sowhy do we need the relationship

P = 4.k.t.R.B

If we already know the power in the circuit?

Bob S
#4
Dec13-09, 10:46 PM
P: 4,663
Noise power in conductor

Quote Quote by Rudibot View Post
Thanks Bob,

Sowhy do we need the relationship

P = 4.k.t.R.B

If we already know the power in the circuit?
Pn=4kTBR is the minimum NOISE power, so your signal-to-noise ratio is Ps/Pn for a signal power of Ps. The only way to reduce the noise power is to reduce T (temperature), B (bandwidth), or R (resistance).
Bob S
vanesch
#5
Dec14-09, 12:29 AM
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Quote Quote by Rudibot View Post
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").

BTW, if you look at the "current signal", then the "power" is 4 k T B / R

The physical power you can *extract* from a voltage source with voltage variance V2, and internal resistance R, is by connecting a load-resistor R to it.

The current flowing in the circuit is then sqrt(V2)/(2 R), and the power dissipated in the load resistor is then (I^2 R) or (sqrt(V2) / (2 R))^2 R = 4 k T R B / (4 R) = k T B.

(btw, that's with an un-physical ideal resistor with no noise as a load - in reality, the load resistor pumps just as much power back into the "source" resistor as the other way around, so that there is no net power flux if both resistors are at same temperature).
Rudibot
#6
Feb7-10, 10:44 PM
P: 10
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?
Born2bwire
#7
Feb8-10, 02:29 AM
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Quote Quote by Rudibot View Post
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?
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.

You can cut up your conductors all you want (within a classical limit), each one will be absorbing energy from the environment and emitting it back out. It isn't until there is a resistive load (and a complete circuit) that we interrupt the flow here and redirect some of the absorbed energy as a signal in the circuit.

So we need to have a complete circuit involved here. Cutting up conductor lengths does not make sense since now we just have pieces of wire.


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