# Closest two frequencies that can be resolved

Forgive me if I misuse the terminology.

I wish to transmit two slowly changing analogue signals by amplitude modulating two higher frequency sinusoidal carrier waves. I will superimpose the two AM signals, send them down a wire, and then demodulate them at the other end using a digital signal processor.

To demodulate them, I will simply use two DFT bins.

I want to select two carrier frequencies as close together as possible. What is the closest I can get two carrier frequencies without them interfering?

• Welcome to SE.DSP! How long can you wait? The longer the amount of data you collect, the closer the frequencies you can resolve. – Peter K. Jul 19 '17 at 14:30
• Can you elaborate on how you plan to use the DFT to demodulate the signals? I can see how to use the DFT to detect them. – MBaz Jul 19 '17 at 14:42
• @MBaz - That might be a little too much detail for this question. But imagine that both carrier waves can fit an integer number of cycles into 120 samples. I would take 120 samples, and multiply those samples by a sin wave table that's in phase with the first carrier wave (by magic). That should pick out the first carrier wave, and ignore the second. Am I correct? – Rocketmagnet Jul 19 '17 at 16:15
• @Rocketmagnet You would still need to filter out the second carrier, which will be shifted in frequency by the multiplication with the first carrier. Also, you seem to be describing a time-domain process; what do you need the DFT for? – MBaz Jul 19 '17 at 16:44
• @MBaz - It's hard for me to talk about this, because I'm not totally sure I'm using the terminology correctly. – Rocketmagnet Jul 19 '17 at 17:17

There is no issue of frequency resolution in a typical modulation and demodulation scheme. (unless you want to implement some frequency/signal detectors)

Putting two message signals in adjacent channels requires that their spectrums do not overlap. In your case you should be able to specify the associated bandwidths of those two slowly changing analogue signals.

Assuming that associated (baseband) bandwidths of those signals are $F_1$ and $F_2$ Hz, then the carrier frequencies $f_{c1}$ and $f_{c2}$ have the relation: $$f_{c2} - f_{c1} > F_1 + F_2$$

That's the minimum distance allowed for classical AM modulation, you could have used, instead, a single side band AM (SSB-AM) which would reduce the minimum distance to : $$f_{c2} - f_{c1} > \min(F_1,F_2)$$

But definitely that would require a more complex and difficult modulator and demodulator in terms of implementation.

Assuming that you have met this criterion (and assuming that you have selected the simple classical AM option), after down-converting, demodulating you should sample the signals satifying their respective nyquist rates.

• Thank you for your answer. Please could you clarify the meaning of detector. Is this a function with a Boolean output, or an 'analogue' output? – Rocketmagnet Jul 19 '17 at 14:56
• A detector is the technical name given to the statistical hypothesis tesing operation wose results are yes or no, therefore binary. Frequency resolution is a concept in spectral analysis in which you practically take DFT of a windowed segment of input signal and try to deduce the presence of sinusoidal components. Also you can use periodogram estimators for this. Finally you would then have afrequency resolution based on window type and window length. – Fat32 Jul 19 '17 at 15:42

The frequency resolution depends on number of DFT points used and sampling frequency. The resolution(frequency range per bin) could be written as
$$\textrm{res} = \frac{f_s}{\textrm{DFT points}}.$$

So, if you want to use two consecutive bins, above is the resolution. Also ensure that $f_s$ and $\textrm{DFT points}$ are chosen in such a way that input frequencies don't span across bins.