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Does anyone know a mathematical proof of this concept? Are there limitations of some kind or is this a general statement for all passive elements?

Thanks in advance.

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3 Answers 3

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The proof is physical, not mathematical, and based on thermodynamics. this is a reasonable description.

The noise temperature (hence noise figure) of the attenuator depends on the temperature of the dissipative elements that convert the signal to heat. If these are within the attenuator, then with the attenuator at ambient temperature the noise figure is the attenuation. As you lower the attenuator temperature you will lower it's noise figure. If it were held at absolute zero, the noise figure would be 1 (0dB).

One way I like to understand this is to 'make' the attenuator from a lossless coupler and a termination, e.g. for a 3dB attenuator think of a 3dB pad with the other ports terminated in matched loads. The output of the coupler consists of the signal from the input port attenuated by 3dB, along with the noise from one of the terminations attenuated by 3dB. For the termination at ambient temperature, this corresponds to a 3dB noise figure. Now cool the termination that is contributing to the output noise and you can, ideally, reduce this to near zero. This would reduce the noise figure to near 1 (0dB).

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If you Google "Attenuator noise figure" you will find

http://analog.intgckts.com/noise-figure-of-attenuator/

which proves your theorem.

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  • \$\begingroup\$ In that link there is an unclear step for me. \$N_{in}=kT_0B\$ is the thermal noise at the attenuator input. If the attenuator was ideal (if it introduces no additional noise) the noise power at the output would be \$N_{out_{ideal}}=kT_0B/L\$. At this point he says that the power difference \$N_{in}-N_{out_{ideal}}=kT_0B(1-1/L)\$ is the same as the additional noise power introduced by the attenuator itself: why? \$\endgroup\$
    – Nameless
    Feb 23, 2020 at 8:58
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Noise figure is defined as SNR in - SNR out.

As the attenuator attenuates the signal, and generates its own noise consistent with the impedance it's being used in, which replaces the input noise that was attenuated, the noise figure drops straight out of the definition.

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