# Maximum achievable SNR (SNR before LNA)

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?

• 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? – MikeP Sep 3 '15 at 18:35
• 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. – divB Sep 3 '15 at 18:36
• 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. – MikeP Sep 3 '15 at 18:40