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Let us assume that we have a narrowband signal $s[n]$, at frequency $f_1$, and let us further assume that this $f_1$ is just below $\frac{f_s}{2}$, where $f_s$ is the sampling rate.

Furthermore, let us also assume that we have a fixed number of time-domain samples of this signal, $N$, and that our FFT length is also $N$.

The objective is to detect this signal $s[n]$'s presence in the frequency domain. (Further assume that we know to expect this signal in or around that frequency $f_1$).

My question is simple: With all things being equal, which is better from a detection perspective:

  • Simply do the FFT of this signal and look at the peak at that frequency bin

Or

  • BPF a band of where we expect the signals' frequency $f_1$ to be, (eg, BPF this band $f_1 - \epsilon \text{ to} f_1 + \epsilon$). Then, mix this band down such that $f_1 + \epsilon$ is guaranteed to be just above DC, and LPF that entire band. Now, take the FFT, and detect energy.

Would the second method be better from a detection point of view because we have more samples per cycle, whereas the first one has just over 2 samples per cycle?

Thanks,

TLDR: Is it better from a detection standpoint to FFT a signal or down-mix it and then FFT it (with appropriate filtering)?

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    $\begingroup$ Filtering could be beneficial, whether at baseband or passband, but I don't think mixing it down to baseband help from a theoretical standpoint. I think it is done for practical considerations, like being able to reduce the sample rate. $\endgroup$
    – Jim Clay
    Commented Mar 22, 2012 at 22:43
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    $\begingroup$ I don't quite understand the second method. The dual filter operation seems redundant (or could be combined into one) and I don't see the purpose of the FFT. $\endgroup$
    – Hilmar
    Commented Mar 23, 2012 at 11:24
  • $\begingroup$ @Hilmar The purpose of the second filtering (LPF) was simply to remove the double-frequency harmonics from the down-mixing operation. $\endgroup$
    – Spacey
    Commented Mar 23, 2012 at 15:47
  • $\begingroup$ @Hilmar Actually, now that I think about it, they might not exist as double-frequencies since I am so close to Fs/2 anyway... $\endgroup$
    – Spacey
    Commented Mar 23, 2012 at 17:06

2 Answers 2

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There is no difference in performance of the optimum detector in either case. Translating a signal in frequency before performing an FFT cannot improve the optimum probability of detection due to the principle of reversibility in detection theory; since shifting a signal in frequency is a reversible operation, then the detection performance of optimum detectors cannot vary based upon the signal's center frequency.

Stated differently, the number of samples per cycle of a particular DFT bin frequency has no bearing whatsoever upon the "quality" of its corresponding output.

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  • $\begingroup$ Where I am mixing up in thinking that the more samples we have per cycle, the more noise immune we can be if we integrate over that cycle, and this situation here? $\endgroup$
    – Spacey
    Commented Mar 22, 2012 at 23:26
  • $\begingroup$ In a word, no. The "integration time" inherent in the DFT is driven only by the length of the transform. Think about this: each output bin of a DFT can equivalently be generated by mixing down by a complex sinusoid at the bin's corresponding frequency, then integrating the resulting samples over a duration equal to the DFT's length. This is the "filter bank interpretation" of the DFT. $\endgroup$
    – Jason R
    Commented Mar 23, 2012 at 2:16
  • $\begingroup$ Hmm...yes... that makes sense... so in a way the DFT is already doing this second method so to speak. $\endgroup$
    – Spacey
    Commented Mar 23, 2012 at 15:49
  • $\begingroup$ @Mohammad: Exactly. $\endgroup$
    – Jason R
    Commented Mar 23, 2012 at 16:02
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If you are doing are doing this in both the analog and digital domain, then there may be significant differences due to trade-offs between the noise sources in the analog stage(s) in the filters and the mixer oscillator frequency, and the corresponding sampling jitter and quantization noise in converting to digital.

Even strictly in the digital domain, the bandpass and lowpass filter responses may have different effects on how potential stop-band noise might affect your detection accuracy, compared to how windowing (rectangular or otherwise) affects the continuous frequency response of each FFT bin.

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  • $\begingroup$ This would be in the digital domain, however it is my understanding that windowing doesnt really reduce noise per se, (just gives you a bazaar of options in relation to dynamic range and main-lobe-widths/side-lobe-height ratios), so in this case, why would it affect detection accuracy in this example? $\endgroup$
    – Spacey
    Commented Mar 23, 2012 at 18:37
  • $\begingroup$ @Mohammed : That's a big enough topic for new question by itself. The "bazaar of options" can change the shape of the noise floor for each bin, depending on the characteristics of the noise. Change the floor, and the statistical reliability of your detection metric may change. $\endgroup$
    – hotpaw2
    Commented Mar 23, 2012 at 23:05

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