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Johnson noise power distribution?

  1. Aug 5, 2009 #1
    I've been reading through the Feynman lectures (almost done with Volume 1), and I have been trying to prove everything to myself. I have a bit of a problem now though, in Chapter 41 on Brownian motion, Feynman shows what the voltage in an LRC circuit is due to thermal noise, and then says that the power distribution (with respect to angular velocity) is the following:
    P(w)dw=(2/pi)ktdw
    Now I have two problems. First, I do not understand where this comes from. At that point he says merely that it will be proven later, and the later proof is unhelpful and vague. The second problem is that the above formula does not check out with dimensional analysis. Power is in terms of J/s, kt is in terms of J, dw is /s, and cancels on both sides. Leaving
    J/s=J
    Is it wrong or am I missing something?
    The later explanation that is given is that the noise generator in a circuit with resonance (adjustable) can be described as the signal received from an antenna due to thermal radiation emitted from the surrounding environment. This he shows (and I follow the explanation for this part) to be proportional to I(w)=w^2*(kt)/(pi^2c^2), and the correction for high temperature or low frequency is also shown, where I is the intensity of the radiation. It seems to me that Power=Area*Intensity, and so it would depend on the size of the antenna. Can anyone explain this?
     
  2. jcsd
  3. Aug 6, 2009 #2
  4. Aug 6, 2009 #3
    Ok, I think I understand it. I found an explanation using two resistors in series and standing voltage waves. What I don't understand now is his derivation of P from the spectral distribution for a black body radiator. As far as I understand it, he says this:
    1. The black body radiator can be considered to be an individual atom, containing an electron that is allowed to vibrate as a harmonic oscillator. This atom must gain and lose energy at the same rate in order for it to be at thermal equilibrium in a box full of gas.
    2. The equation for the root mean square voltage in an LRC circuit is w^2*LkT
    Both of those statements, as well as the subsequent derivation for a black body radiator, I agree with. The problem is how the radiator relates to the thermal noise.
     
  5. Aug 8, 2009 #4
    Neither inductors nor capacitors can be sources of thermal noise. Only resistors can be the source of noise, but inductances and capacitors can define bandwidth. So it is better to write the noise voltage with an "R" in it.
     
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