I am trying to understand the concept of noise figure and the best achievable SNR.

The noise figure is a measure how much a device degrades the SNR. Say I have a receiver with a NF of 16dB, then the output SNR is smaller by 16dB than the input SNR. But what is the input SNR? Of course, it depends on some parameters.

As an example, let's say the system has a bandwidth of 100 MHz. Furthermore the antenna is ideal with an output impedance of 50 Ohm, so my receiver should have an input resistance of 50 Ohm.

Now suppose I have a hypothetical receiver which requires (due to hypothetical constraints, e.g. linearity) an input amplitude of 50mV but has NF=0dB (noise less). Then the output SNR represents the best achievable SNR in this setting and should be given by:

$$ \rm SNR = 10 \log \left( \frac{\frac{\hat{v}^2}{2R}}{k T B} \right) = 10 \log \left( \frac{50\cdot10^{-3}}{2 k T 50 \cdot 100 \cdot 10^{6}} \right) = 77.98 \mathrm{dB} $$

This sounds a little bit low too me given the fact 50mV is not too small.

The SNR could be improved by making the antenna resistance smaller (but to my understanding this should be 50 Ohm to match it with the impedance of the air).

Is the best achievable SNR with an ideal receiver in a typical setting like this really so low or are there some flaws in my understanding?

  • \$\begingroup\$ Generally SNR is a ratio of power, but it looks like you used voltage instead. If you use Volts to Volts, just change the 10log to 20log. But I don't think you have volts in the denominator.. In any case, where did the 8 come from? \$\endgroup\$ – MikeP Sep 3 '15 at 18:35
  • \$\begingroup\$ No, I use power, I just did not write intermediate results. Let me add this. (EDIT: Added). The 8 is wrong, I just corrected it, it should be 2. \$\endgroup\$ – divB Sep 3 '15 at 18:36
  • \$\begingroup\$ I think for power the numerator should be 50 mV ^2 * 50 ohms or .125 mW. The denominator should just be kTB or k*T*100*10^6, which gives me about 85 dB. \$\endgroup\$ – MikeP Sep 3 '15 at 18:40

I think that you do have to compare powers and also in a certain bandwith. The noise is a power in a certain bandwidth, if you choose a smaller BW the noise will be less !

If your signal is only 1 MHz wide, you don't need your the 100 MHz bandwidth (the LNA could be 100 MHz but after mixing you would filter to a 1 MHz bandwidth and get rid of the extra noise).

So you cannot compare your 50 mV signal to the noise power (in a specified BW). You need a signal power (in a specified BW) and compare that to the noise (in the same BW).

Think of it like: if your signal is only a 50 mV sinewave it will use a very small BW (it is only one frequency). Now compare that to for example a WiFi signal which uses many frequencies in a certain BW. Combined they can also peak at 50 mV but they would contain a lot more power (and information!).

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.