There are many things I might do enable my station to hear weaker signals that would otherwise be buried in noise. Just a few possibilities:

  • Install a nice LNA
  • Increase antenna gain
  • Reduce feedline losses
  • Buy a better receiver
  • Move to a quieter location

Is there a way I can quantitatively calculate the effect these changes on my station's weak-signal receive performance?


1 Answer 1


Try the Friis noise formula:

$$ F_{eq} = F_1 + {F_2-1 \over G_1} + {F_3-1 \over G_1 G_2} + \cdots \tag 1 $$

$F_n$ is the noise factor of the n-th component, and likewise $G_n$ is the gain. The noise factor $F$ is the linear ratio form of the noise figure which is given in decibels.

For example, the first component may be an LNA, the second component a transmission line from the LNA to the receiver, and the third component the receiver itself.

Example 1: a good LNA on a crappy receiver

I read somewhere an RTL-SDR has a noise figure of 6 dB, or a noise factor of 4. Let's say we have an LNA with a noise figure of 1 dB (1.26) and a gain of 20 dB (100). By equation 1:

$$ 1.26 + {4-1 \over 100} = 1.29 = 1.11\:\mathrm{dB} $$

Thus, the addition of the LNA has brought the noise factor from 6 dB down to 1.11 dB by dividing the RTL-SDR's noise by the LNA's gain.

The additional power into the receiver increases distortion and reduces dynamic range, but that's a different can of worms.

Noise factor of passive components

This is easy: the noise factor is equal to the attenuation, assuming the thermodynamic temperature (as measured by a thermometer) of the component is "room temperature", conventionally 290 K (17 °C or 62 °F). And logically, the noise figure is equal to the loss in decibels.

So, a piece of coax with 4 dB of loss has a noise figure of 4 dB. The noise factor and attenuation are $ 10^{4/10} = 2.51 $.

This is true for any passive component, like a filter, or even antenna inefficiencies.

Example 2: not putting the LNA at the antenna

Same as before, but with a 4 dB loss feedline between the LNA and antenna.

Antenna -> 4 dB loss -> LNA -> receiver

We'll need to know the gain of the coax for the $G_1$ term, which will be less than one. It's the reciprocal of the attenuation: 1/2.51 = 0.398

$$ 2.51 + {1.26 - 1\over 0.398} + {4 - 1\over 0.398 \cdot 100} = 3.24 = 5.10\:\mathrm{dB} $$

The feedline noise dominates, negatively impacting the system in two ways:

  • It adds Johnson-Nyquist noise generated by the resistive losses of the cable, represented by the $F_1$ term, and
  • it reduces the power passed to the next stage, making the noise power added by that next stage more significant, represented by the $G_1$ term.

Example 3: put the LNA at the antenna

Same as before, but now with the LNA at the antenna:

Antenna -> LNA -> 4 dB loss -> RTL-SDR

$$ 1.26 + {2.51-1 \over 100} + {4-1 \over 100 \cdot 0.398} = 1.35 = 1.30\:\mathrm{dB} $$

Almost as good as example 1 where there was no feedline.

How does noise figure relate to SNR in practice?

Noise figure is the reduction in signal to noise ratio (SNR) under an assumption: the input noise is equivalent to the Johnson-Nyquist noise of a resistor at 290 K. 290 K is very quiet: it's the same noise power you'd get from a dummy load at room temperature. Often input noise is more.

The relationship of input SNR to output SNR is related not only to the receiving system's noise, but the input noise:

$$ { \text{SNR}_\text{in} \over \text{SNR}_\text{out} } = 1 + {T_\text{sys} \over T_\text{in}} \tag 2 $$

When $T_\text{sys} \ll T_\text{in}$ the right hand side of the equation approaches 1, thus $ \text{SNR}_\text{in} \approx \text{SNR}_\text{out} $.

To make sense of that brings us to...

Introducing noise temperature

A resistor produces Johnson-Nyquist noise due to thermal motion of the charge carriers within it. The noise power is related to the temperature by:

$$ {P \over B} = kT \tag 3 $$

  • $P$ is the power in watts
  • $B$ is the bandwidth in Hz over which that power is measured
  • $k$ is the Boltzmann constant, approximately 1.381e-23 joules per kelvin
  • $T$ is the temperature of the resistor, in kelvin

It's useful to pretend all noise is generated by a resistor at some temperature, which makes noise temperature a useful unit for "noisiness" that may or may not be directly related to thermodynamic temperature as measured by a thermometer.

Noise factor and noise temperature are related by:

$$ T = 290\,(F-1) \tag 4 $$

$$ F = {T \over 290} + 1 \tag 5 $$

This leads to an alternate version of the Friis noise formula expressed in noise temperature rather than noise factor:

$$ T_{eq} = T_1 + {T_2 \over G_1} + {T_3 \over G_1 G_2} + \cdots \tag 6 $$

What's the input noise temperature?

ITU-R P.372-14 provides some excellent data on environmental noise temperatures. The document is worth a read, but here are a few relevant figures from it:

figure 1

figure 2

figure 3

Each figure covers a different frequency range, noise temperature on the right axes, and equivalent noise figure on the left. The noise temperature doesn't come anywhere near 290 K within HF. At microwave frequencies 290 K might be realistic, provided no man-made noise and an antenna directional enough to avoid the sun.

Example 4: huge loss and a cheap receiver at different frequencies

We'll use an RTL-SDR with 12 dB of loss due to feedline and antenna inefficiencies:

Antenna -> 12 dB loss -> RTL-SDR

The 12 dB noise figure of the feedline is equivalent to a noise temperature of 4306 K. And the 6 dB NF of the RTL-SDR is 865 K.

$$ T_{eq} = 4306\:\mathrm K + {875 \over 10^{-12/10}} = 18174\:\mathrm K $$

By equation 5 that's a noise figure of 18 dB.

Let's say we're using this at 400 MHz in a city. Line A (man-made noise) in figure 3 puts external noise temperature around 3000 K. So by equation 3:

$$ { \text{SNR}_\text{in} \over \text{SNR}_\text{out} } = 1 + {18174\:\mathrm K \over 3000\:\mathrm K} = 7 = 8.49\:\mathrm{dB} $$

Under these conditions, this terrible station will make the SNR 8.49 dB worse. Bad, but not as bad as the 18 dB noise figure would suggest.

Now what if this is a 7 MHz station is a very quiet location? Figure 2 is hard to read precisely since it spans 18 orders of magnitude, but line D (galactic noise) at 7 MHz is at least 300,000 K.

$$ { \text{SNR}_\text{in} \over \text{SNR}_\text{out} } = 1 + {18174\:\mathrm K \over 300000\:\mathrm K} = 1.06 = 0.256\:\mathrm{dB} $$

Attempts to reduce noise with an LNA, better receiver, etc. are futile. The SNR must be improved at the source, for example by increasing antenna directivity.


  • Improving system noise performance only matters if the external noise isn't already dominating.
  • Increasing antenna directivity improves SNR even if external noise dominates.
  • On HF, external noise is probably dominating.
  • Noise figures assume an external noise of 290 K, which is really quiet: about as quiet as things can get for a terrestrial antenna. Noiser environments make the noise figure matter less.
  • Feedlines, filters, antenna inefficiency, or any other passive component that has loss also introduces noise.

Finally a caveat: distortion is a detriment to receive performance also. The gain of an LNA may reduce the significance of the receiver's noise, but it will also increase distortion. Always consider the whole picture.

  • 1
    $\begingroup$ wow, what an impressive answer $\endgroup$ Commented Jan 15, 2019 at 19:57
  • $\begingroup$ For those trying to do the math with NF in dB from datasheets, here is a Noise Factor<->Noise Figure calculator: microwaves101.com/calculators/1339-noise-conversion-calculator. These are the conversions: F = Noise Factor = 10^(Noise Figure/10) ; NF = Noise Figure = 10log_10(Noise Factor). $\endgroup$
    – KJ7LNW
    Commented May 20, 2021 at 0:49
  • 1
    $\begingroup$ Phil, I don't see the "-1" on each F_2-1 / G_1 example. Is there a reason you omitted the -1 or should that be fixed? In Example2 I get ~7.61dB without the -1's, but I get about 5.1dB with them. $\endgroup$
    – KJ7LNW
    Commented May 20, 2021 at 1:29
  • 1
    $\begingroup$ @KJ7LNW it was an error on my part. Thanks for the catch $\endgroup$ Commented May 20, 2021 at 11:56

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