I'm confused by the following:

  1. What's the noise figure of an attenuator with 10 dB loss at very high temperature, having a noise power of like 80 dBm/Hz?
  2. How about if the attenuator at very low temperature having a noise power of -500 dbm/Hz?

The attenuators are 50 Ω matched and purely passive.

Now, we connect a 50 Ω source with room temperature (noise power -173 dBm/Hz) to these two attenuators. What is the noise power coming out? How about if the attenuators are not 50 Ω matched, rather 70 Ω or 100 Ω with mismatching, what's the noise power at the output now?

Could anyone help? I have read many articles, but not a clear explanation.

PS: please do not think much about the non-realistic values, i took the large values to show the problem. 'Noise figre of attenuator equals to its loss' seems not correct in some cases.

  • 3
    \$\begingroup\$ -500 dBm is a ridiculous figure. \$\endgroup\$
    – Andy aka
    Jan 21, 2021 at 16:04
  • 2
    \$\begingroup\$ Determine the NF of a 50 ohm, 10 dB attenuator at room temperature. Derive the formula for the NF. Fill in the noise of a resistor, what do you get? That can be used to answer your questions. My point: Instead of looking for formulas, derive them yourself. If you understand noise then this isn't that hard. \$\endgroup\$ Jan 21, 2021 at 16:18
  • \$\begingroup\$ @Bimpelrekkie thanks! i know the NF of a 50Ohm and 10 dB attenuator at room temperrature is 10 dB, i can also derive it. to be more clear, if the the source and attenautor are not at same temperature. like in case 1, source at roome temperature and attenautor at high temperature like with 80dBm/Hz if posssible. now what's the noise power at output, should it be 80dBm/Hz? i think yes, the input noise is way low than anttenuator internal noise. then what's the NF now? 80+173= 253dB. rather than 10 dB(attenuator loss). \$\endgroup\$
    – Xin
    Jan 21, 2021 at 16:27
  • \$\begingroup\$ @Andy aka just a number for showing that the noise power from source is very low at low temperature. more important i want to show or ask: it seems that 'NF of attenuator is equal to its loss' is no more correct if source and attenuator are not at same temperature. \$\endgroup\$
    – Xin
    Jan 21, 2021 at 16:32
  • \$\begingroup\$ 80dBm/Hz is 1MW in 10Hz or 1GW in 10kHz. I'd recommend you don't use an active volcano or a Tokamak as an attenuator. \$\endgroup\$
    – user16324
    Jan 21, 2021 at 16:39

1 Answer 1


The key is to know how power is defined and what is actually being measured.

Imagine a 10V signal passed through a 10 ohm source resistor and then a 10 ohm attenuator

enter image description here

The input signal power to the 10 ohm attenuator is measured by multiplying the RMS voltage and current over a load if the attenuator were to be removed and replaced with a load matched to the source (10 ohms). In this case pretend RMS voltage is 10V and not the amplitude. The current will be 0.5A and RMS voltage 5V, giving 2.5W.

The output signal power from the attenuator is the voltage and current over the load if the load is matched to the source + attenuator. In this case 20 ohm load. The current will be 0.25A and RMS voltage over the load will be 5V. Output power is therefore 1.25W. This is a loss of 2, i.e. a gain of 0.5.

The input noise power to the attenuator is the Johnson–Nyquist noise from the source resistor.that is dissipated in a matched load. In this example the source has a noise temperature of 290K, and the noise power dissipated in a matched load will be kT0B. The output noise is the noise power from the source and attenuator across a matched load (20 ohms). This is determined from the equation for the noise of 2 resistors in series:

$$\frac{R_ST_A + R_ST_A}{R_S+R_A}$$

Because the resistance of the source and the attenuator is the same, the noise into a matched load is

$$k\frac{T_S + T_A}{2}B$$

If we give the physical temperature of the attenuator to be 290K then you get 290kB. If you give it the physical temperature of 350K then it will be 320kB.

$$F=\frac{S_i/N_i}{S_o/N_o} ​$$ $$ = \frac{ 2.5(W) / 290kB(W)}{1.25(W) / 320kB(W)}$$ $$= 2.206$$

This is the same as the similar definition

$$ F = \frac{GT_i+T_a} {GT_i}$$ $$ = 1 + \frac{T_a} {GT_i}$$ $$ = 1 + \frac{LT_a} {T_i}$$ $$ = 1 + \frac{T_e} {T_i}$$ $$ = 1 + \frac{(L-1)T_A} {T_i}$$ $$ = 1 + \frac{(2-1)350} {290}$$ $$ = 2.206$$

So L=F only when source and physical temperature of attenuator is the same

enter image description here

Temperature is referenced to a specific stage in the component chain.

The equivalent temperature of each component in the chain (if the input noise temperature to the chain is T0 is


The equivalent temperature referenced to the beginning of the chain is the sum of the equivalent temperatures divided by the gain of the components before them (in order to mathematically achieve the Tsub>e at the actual input of the component, which is the noise added by the component referred to the input).

If the whole chain is referred to the output, then the output temperature is the sum of the equivalent input temperatures multiplied by the gain of all the components after each individual component

The noise figure is

$$1+(F_1-1) + \frac{F_2-1}{G_1}+\frac{F_3-1}{G_1G_2}...$$ $$=1+\frac{T_e}{T_0}$$

Where Te is the input referred temperature to the whole chain


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