Processing 47-53 MHz signal sampled at 96000 Hz

Question

I am fairly new to the topic of DSP and FFT and I have hit a mental block. For a project, I am trying to pick out frequency's with amplitude peaks in a 47-53 MHz signal. This signal is being put through a 24 bit ADC with a sample rate of 96000 Hz. In order to successfully measure a sample, don't you have to have a sample rate >= to twice the desired frequency (Nyquist rate)? How can you get an accurate representation of a 47-53 MHz signal with such a low sample rate?

The second block I have hit involves the FFT of this signal. From my limited knowledge of FFT, the first "bin" (in the real part) represents 0 Hz and the last "bin" (in the real part) represents: $$ F=i\times \frac{Fs}{N} $$ Where $i$=Bin Index, $Fs$=Sample Rate, $N$=FFT Size

In my case with size 1024 and above sample rate, $511 \times 96000/1024 = 47906.25$ Hz, right? How can this range ($0$ Hz-$47906.25$ Hz) possibly represent my $47-53$ MHz signal? I get the feeling that I am missing something really crucial but I just can't figure it out.

Sorry if this is a really basic question and thanks for the help!

Answer 1

In short, what you're trying to do is impossible. The technique is known as band-pass sampling, or undersampling, but for it to work, you need that the signal's bandwidth be less than half your sampling frequency.

When you sample a signal with spectrum $S(f)$ and bandwidth $B$ at sampling frequency $f_s$, the sampled signal has spectrum $$S_s(f)=k\sum_i S(f-if_s)$$ where $k$ is a constant. If $B>f_s/2$ then the replicas $S(f-if_s)$ overlap and you end up with a distorted spectrum.

You are correct that the highest frequency bin in your FFT is 47906.25 Hz.