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Thermal Noise (Johnson–Nyquist)

  1. Nov 5, 2008 #1
    Could someone explain the equation describing Thermal Noise?

    The equation is given by

    [tex]
    \overline{\left(\delta U\right)^{2}_{kT}} = 4RkT\Delta f
    [/tex]

    Specifically I dont understand what the [tex]\Delta f[/tex] is. Some online resources describe [tex]\Delta f[/tex] as:
    - "frequency bandwidth within which information is sought"
    - "bandwidth being considered"
    - "bandwidth over which noise is measured"
    These descriptions are a bit vague. Could someone explain [tex]\Delta f[/tex] in more detail.
    thanks.
     
  2. jcsd
  3. Nov 5, 2008 #2

    f95toli

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    It is usually just the bandwidth of the total system: i.e. measurement system+sample.

    The equation essentially tells you what voltage you would see if you measured a resistor with resistance R held at a temperature T using using a measurement system with a bandwidth [itex]\Delta f[/itex].
     
  4. Nov 5, 2008 #3
    Thank you for your reply.

    When you say "...measurement system with bandwidth [tex]\Delta f [/tex]", does this mean that the voltage fluctuates in a range [tex]\Delta f [/tex]?


    So for example
    if we want to see how much of the voltage fluctuations (due to thermal noise) lie in a range 20Hz, then it is this 20Hz that is [tex]\Delta f [/tex]?
     
  5. Nov 6, 2008 #4

    f95toli

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    No, the spectral density is the same for all frequencies; there is as much noise around 20 Hz as there is around 2 MHz or 2 GHz, the spectrum of thermal noise is "white" and does not depend on the frequency.

    [itex]\Delta f [/itex] is just the electrical bandwidth. If you e.g. have a measurement systems which behaves like a low-pass filter with a cut-off frequency of 1 MHz (i.e. it will let all frequencies DC-1 MHz), [itex]\Delta f=1e6 [/itex] Hz
    If the system instead behaves like a band-pass filter [itex]\Delta f [/itex] is just the bandwidth; i.e. the voltage does not depend on the centre frequency.
     
  6. Nov 11, 2008 #5
    Agree.

    That is, the narrower your receiver, the less noise it picks and creates. I built a magnetic receiver at 457kHz that had a noise bandwidth of about 1Hz (triple heterodyne), its sensitivity was about -170dBm: enough to pick the 10.000th harmonics of the mains. It worked like an ARVA (no idea how you call it in English: it locates the magnetic transmitter hold by a victim entrapped in an avalanche) but mine had a range of 100m.

    The basic reason is that the physical fluctuation is an energy (kT). If you observe this fluctuation over a broader bandwidth, that is a shorter time, the corresponding power is bigger.

    Now, it is difficult to now precisely what the noise bandwidth is. It is not the -3dB bandwidth, but the integral of the power response of whatever filters the frequencies. For instance, some spectrum analyser's doc give the noise bandwidths of the IF filters; other analysers have software to convert to dBm/Hz (should be : per log(Hz)...); bad ones have nothing.

    If you think of a spread spectrum receiver, you first have a broad bandwidth with much noise, and after pulse compression (call it despreading if your job wants it) a narrow bandwidth with hopefully less noise. Well, sometimes. Because then, the signal-to-noise ratio before pulse compression would generally be negative, and few receivers still work under these conditions - which means that spread spectrum often has a poor sensitivity.

    Several effects are less obvious. For instance in a heterodyne (not a Q-I one), noise at the image frequency, picked at the antenna or created by the preamplifier is filtered out at the input or output of the preamplifier, but noise added by the mixer is not. So mixers have a bad noise figure.

    You may understand an optimum demodulator as a means of reducing the noise bandwidth.

    Also think of radio-astronomy correlation receivers. They use two or more receivers and preferable antennas. Instead of adding the signals, they integrate their product over a (very) long time but keep the input bandwidth large. This filters out very efficiently the uncorrelated noises produced by the receivers. The equivalent noise bandwidth is much smaller than the input bandwith, but is not the inverse of the integration time - one wins like sqrt (F*T). This gets complicated because the process isn't linear.

    Nothing obvious here. Noise is subtle anyway.
     
  7. Apr 6, 2010 #6
    Ok .. I agree with everything that you guys said. I have a question however. Assume that you have a transducer that behaves like first order active low pass filter with the pole being determined by an RC resistor circuit. If i wanted to calculated the Thermal (Johnson) noise of the R component, what Δf should I use?? From what you guys said I should use the bandwidth of the transducer (equivalent to a low pass filter) which will be 1/(2*π*RC). According to the formula of Johnson Noise the noise will be reduced when the resistance is reducing. However, changing R you change as well the bandwidth. Increasing the bandwidth, the noise is increased even though the resistance is low.
    So does that mean the value of the resistor will not be changing the output noise unless the bandwidth of the overall system is kept constant i.e. limit the frequency spectrum with a bandpass filter at the output of the transducer.
     
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