Electronic – Calculating FFT for only part of full frequency range

The maximum sample rate of the device is 2Mhz, this is just something else that I don't understand, how can I detect 100Mhz+ signal if the maximum sample rate is 2Mhz ?

You can't, at least not unambiguously if there are other signals that are of a lower frequency in the input.

Anyway, if I understand correctly, the maximum frequency in FFT is 1Mhz, via nyquist, my question is, how can I find the bin that corresponds to the 148Mhz signal ? I can see a spike that corresponds to the signal but I want to be able to index into the bins and find it.

Probably reason why you see a spike that corresponds to the signal is because you're experiencing aliasing. There's no way to unambiguously detect the \$148.369\text{ MHz}\$ signal without a sufficiently high sample rate. It could actually be a much lower frequency signal.

For example, if I try to sample a \$2\text{ kHz}\$ sine wave signal with a sampling rate of \$1.5\text{ kHz}\$, the ADC reconstruction of the signal might end up actually being \$0.5\text{ kHz}\$, so there's no way to tell the difference between a \$2\text{ kHz}\$ or a \$0.5\text{ kHz}\$ signal using only the data acquired by an ADC. That's why ADCs are often used with an analog low-pass antialiasing filter to avoid the problem.

If you want to detect this \$148.369\text{ MHz}\$ signal, you need to either sample the signal at more than twice that frequency (the more, the better), or use an alternative strategy that does not involve directly looking for the \$148.369\text{ MHz}\$ signal.

For example, you can mix the signal with another (sine wave) signal generated by a local oscillator and look for the beat frequencies instead of the \$148.369\text{ MHz}\$ signals. You can then use an ADC and antialiasing filter filter to look for them. A proper choice of the local oscillator frequency would create beat frequencies much less than the sampling rate of the system.

This is actually the technique used in some (all?) radios to tune to a specific frequency, called heterodyning.

There is a direct, and actually quite simple, relationship between all the figures.

Let's start with the sample size. The numbers of bins (or "buckets") is equal with half of the samples in your set. For instance, if you have 1024 samples, then you get 512 bins. As simple as that.

Now for the sample rate. The maximum frequency is half the sample rate (see Nyquist-Shannon sampling theorem). So if you have a sample rate of 2.67ksps then your frequency range is 0-1.335kHz.

Again pretty straight forward - just another divide by two.

Now the bins are spread evenly over the frequency range - so your 512 bins, over 1355Hz is 2.607421875 Hz per bin.

For 0.5Hz per bin up to 300Hz you want 600 bins. Your target sample rate would be 600Hz. How you down-sample to that would be up to you.

Most FFT code I have seen works on 2ⁿ sample sizes, so 600 bins isn't a nice number. That would be 1200 samples - not a 2ⁿ. So you'd probably want to round it up to 2¹¹, or 2048 samples. That would give you 1024 bins, and you'd want a target sample rate of 1024sps for 0-512Hz range.