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Wikipedia says:

In a wireless communications receiver, the equivalent input noise temperature \$T_{eq}\$ would equal the sum of two noise temperatures:

$$T_{eq} \ = \ T_{ant} \ + \ T_{sys}$$

I understand that these values of \$T\$ are related to temperatures, but they are not themselves actual temperatures that one measures with a thermometer.

If I put a 273K ice cube in my 357K coffee (no pun intended) I'd get cooler coffee, not 630K coffee. The same applies if they are two streams of fluid mixing rather than static objects.

In another setting; at given frequency, the noise power of an external radio source, like a blackbody source, would scale as the fourth power of temperature, not linearly.

I need help understanding why noise temperatures are simply added, even though in the real world the last thing we'd think of doing is adding two temperatures together.

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I think this is a case of just confusingly worded Wikipedia articles. That passage would seem to suggest that if you ever have two noise temperatures then you can just add them together. As you've aptly explained, that doesn't make any sense.

Rather, \$T_{sys}\$ is a figure of merit that's calculated by measuring the noise added by some component. The input noise is \$T_{ant}\$. After passing through some component, the noise will be \$T_{eq}\$, which must be equal to or greater than \$T_{ant}\$. And the difference is \$T_{sys}\$, by definition.

If \$T_{sys} = 0\$, you have an ideal component which adds no noise.

If \$T_{sys} \ll T_{ant}\$, you have a realistic component which adds only negligible noise, and the signal to noise ratio (SNR) is not significantly decreased. Like a good LNA.

Thus \$T_{sys}\$ makes a convenient figure of merit: by comparing it with the input noise temperature it's easy to see how relevant the noise added by this component will be. If the input noise is already high there's not much reason to spend more money on components with a lower \$T_{sys}\$.

By using a LNA with a very low \$T_{sys}\$, the signal and the noise can be amplified with a minimal decrease in SNR. Once that amplification is done, the input noise (\$T_{ant}\$) is much higher (because all the noise power was amplified), so now all the components that follow can have a much higher \$T_{sys}\$ (and thus lower cost) without having an unacceptable impact on SNR.

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  • \$\begingroup\$ This makes a lot of sense to me, thanks! So the numerical values "look like" temperatures (e.g. 300K, 50K...) because they are referenced to noise levels that would be produced by devices that are at those temperatures? \$\endgroup\$ – uhoh Apr 27 '17 at 7:53
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    \$\begingroup\$ I think part of it is that, and another part is just convention. Random noise of one kind is like any other kind, so it makes sense to just pick one kind and call all random noise that. I guess since temperature is ubiquitous and unavoidable (and in many but not all cases, the most significant), it became the canonical kind of noise. \$\endgroup\$ – Phil Frost Apr 27 '17 at 8:01
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    \$\begingroup\$ Or maybe I misunderstood your comment...the answer is yes, whatever the origin of the noise, we just pretend it's all thermal noise equivalent to some thing at some temperature. \$\endgroup\$ – Phil Frost Apr 27 '17 at 8:03
  • \$\begingroup\$ I see statements that \$T_{ant}\$ for example represent the temperature of a hypothetical \$50\Omega\$ resistor at the input that would make an equivalent amount of noise. Noise temperatures are usually in the 10K to 1000K range. You don't see 0.000001K or 1,000,000K in electronics that often. The numbers "look like" temperatures that might actually occur in receivers, roughly speaking. \$\endgroup\$ – uhoh Apr 27 '17 at 8:07
  • \$\begingroup\$ I've finally gotten to the bottom of this, and it turns out that it does make sense after all! \$\endgroup\$ – uhoh Sep 1 '18 at 12:59
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I've since written a longer answer elsewhere which I will just summarize here.

The key is that most of of the time we work in the Rayleigh-Jeans regime where the energy associated with the working temperature is much higher than that of photons of the frequency of interest.

For example, at room temperature \$k_B T \approx \$ 4E-21, whereas even at 32 GHz the energy associated with photons (or quanta) is only 2E-23.

So while the Plank distribution has a strong temperature dependence at the higher end,

$$B_{\nu}(T) = \frac{2 h \nu^3}{c^2} \frac{1}{\exp(h \nu / k_BT)-1}$$

at the low end the behavior can be written as

$$B_{\nu}(T) = \frac{2 \nu^2}{c^2} k_B T$$

This is easily seen in the plot below, where all curves have a slope of 1 well below the maximum.

Since in the Rayleigh-Jeans regime the power per unit bandwidth is indeed proportional to temperatures, adding two values of power together is effectively adding two equivalent temperatures together. You are not really adding two temperatures, you're adding noise figures, expressed conveniently in terms of temperature.


below: From New Jersey Institute of Technology's Dr. Dale Gary's nicely-written Physics 728 Radio Astronomy; Lecture #1 notes:

Rayleigh-Jeans vs Planck

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