I have a question regarding the signal to noise degradation after a device with a known noise figure.

I must be wrong in some of the assumptions I'm making. As far as I know, the S/N degradation should be equal to the noise figure, but if I calculate it using noise temperatures, I get very different values.


SNidB: Signal no noise at the input, in dB
SNidB: Signal to noise at the output, in dB
Te: Excess temperature of the device at the input
T0: Reference temperature = 290 ºK
Ti: Noise temperature at the input = 10 ºK (e.g. an antenna pointing at the sky)
Nf: Noise figure of the device = 0.8 dB
Ps: Power of the input signal = -120 dBm
Pni: Noise power at the input
Pno: Noise power at the output
B: Bandwidth = 2.4 kHz

The noise power at the input is:

Pni = 10log10 (kTi*B / 10^-3) = -154.8 dBm

So, the signal to noise at the input:

SNidB = -120 + 154.8 = 34.8 dBm

If we calculate the noise power after the device in terms of it's noise temperature contribution:

Knowing that:


Nf = 10*log(1+Te/T0)

Then, with our given Nf of 0.8 dB:

Te = 58.7K

And the noise power at the output is

Pno = 10log10(k(Ti+Te)*B / 10^-3) = -146.4 dBm

So, the signal to noise at the output:

SNodB = -120 + 146.4 = 26.4 dB

Which gives us a degradation of 34.8 - 26.4 = 8.4 dB

But, according to the noise figure definition

** Nf = SNidB - SNodB, which should be our given 0.8 dB != 8.4 dB

¿What am I doing wrong?

Thanks in advance.

  • \$\begingroup\$ SNodB calculation omits the gain of this system. \$\endgroup\$
    – user16324
    Commented Dec 10, 2020 at 13:32

1 Answer 1


In plain language, the problem is that you warmed your signal up without adding any gain.

Unless you actually chill that first stage to the same temperature as your source (it's been done), then if you don't add enough gain to overcome the noise temperature of 290K, your theoretic noise figure is meaningless.

Eg. if we run your math for a .8 dB noise figure relative to 10K, then we get a 2K noise temperature.

But your system runs at room temperature, so you followed standard practice and calculated relative to 290K, and got a noise temperature of 58.7K. That's reality. You then tried to figure out the output noise power, and came out to the truth for the temperature of your system, which is to say quite a bit worse than expected.

The conclusion is that if you want to actually see your theoretic noise figure, you have to add enough gain that the amplified input noise dominates over the thermal noise of the electronics (at whatever temperature they operate at).

  • \$\begingroup\$ Thank you Chris. That's the key I was missing. The noise figure based calculation yields the same results only when the reference temperature is matched with that of the system. Otherwise, it's like warming the signal just before entering the device. Likewise, if I was pointing the antenna at a 400K source, the degradation would be only 0.6dB if calculated the same, not the 0.8dB expected. \$\endgroup\$
    – Angel
    Commented Dec 10, 2020 at 19:18

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