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Noise power paradox

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Explain why the noise power across a resistor in thermal equilibrium is limited in contradiction to the formula for the noise [tex]V_{RMS} = \sqrt{4RkT\Delta f}[/tex]
    which states that if we measure with infinite bandwidth we have infinitely large power


    2. Relevant equations
    [tex]V_{RMS} = \sqrt{4RkT\Delta f}[/tex]


    3. The attempt at a solution
    So the formula states that if a system measures the noise power of a resistor with infinite bandwidth is the noise power is infinitely large. But the noise power cannot exceed the internal energy of the resistor, which is not infinitely large. But I don't know where and why the formula can't hold anymore.

    Thanks in advance
     
  2. jcsd
  3. Oct 14, 2009 #2

    gabbagabbahey

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    Hi The Alchamist, welcome to PF!:smile:

    Hmmm... well, how was the formula derived? What assumptions was the derivations based on?
     
  4. Oct 14, 2009 #3
    Thanks,

    Well the power spectral density function of thermal noise is [tex] \phi = 4RkT [/tex]
    So the power is the integral of the power spectral density function over the bandwidth.
    [tex]W = \int\limits_{bandwidth} \phi \mathrm{d}f = \int_{f_{1}}^{f_{2}}4RkT \mathrm{d}f = 4RkT\Delta f [/tex]

    With [tex]\Delta f = f_{2}-f_{1} [/tex]
    This means that
    [tex]V_{RMS} = \sqrt{W} = \sqrt{4RkT \Delta f}[/tex]

    So probably there is something wrong with the assumption of the boundary condition of the integral that f2 can't go to infinity. But I have no clue why there is such condition it should be possible to take an infinite bandwidth. So maybe the phi is wrong :?. confusing
     
  5. Oct 14, 2009 #4

    gabbagabbahey

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    There are certain assumptions that underline the claim that [itex]\phi = 4RkT[/itex]....what are they?
     
  6. Oct 15, 2009 #5
    Well, the PSD is uniform over an infinite range of frequencies but it is proportional to the temperature of the resistor. What you want is a PSD that is cut-off at a certain frequency to limit the power. But I don't know why the PSD should have such a frequency.
     
  7. Oct 15, 2009 #6

    gabbagabbahey

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    The point I'm hinting towards is that the PSD is valid only for an Ideal resistor, something which is inherently unphysical. Real resistors will always possess some small stray Capacitance. If you call the effective capacitance of the resistor [itex]C[/itex], the effect of it is to modify the power spectral density to

    [tex]\phi=\frac{4RkT}{\left(1+(2\pi fRC)^2\right)}[/tex]

    Which you should derive yourself by considering a circuit consisting of an ideal resistor, a capacitor and a noise source....

    What happens when you integrate this distribution over all frequencies? Why is [itex]\phi=4RkT[/itex] a good approximation for most frequencies?
     
  8. Oct 17, 2009 #7
    Ok I put a capacitor parallel to the resistor and derive the new impedance which is indeed
    [tex]
    Z = \frac{R}{\left(1+(2\pi fRC)^2\right)}
    [/tex]
    So [tex]
    \phi= 4ZkT = \frac{4RkT}{\left(1+(2\pi fRC)^2\right)}
    [/tex]

    Integrating over df gives:
    [tex]
    \frac{2kT}{ \pi C} \arctan{2 \pi C f R}
    [/tex]
    filling in f with limits from 0 to infinity gives:
    [tex]
    \frac{2kT}{ \pi C} \frac{\pi}{2} - 0
    = \frac{kT}{C}
    [/tex]
    which is of course finite with constant temperature.
    The idea that phi is a good approximation up to a certain frequency is because the capacitance of the parallel capacitor is very small so the term [tex](2 \pi f R C)^2[/tex] is negligible and phi is the "original" phi for the ideal resistor.

    That is what I can think of and seems correct,
    Thanks
     
  9. Oct 17, 2009 #8

    gabbagabbahey

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    Looks good to me!:approve:
     
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