How large FFTs can pull signals out of the noise floor?











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I am trying to detect unknown RF tones around -140 dBm and my scan BW is 5 MHz, going through the Noise power calculations the signal is below the thermal noise based on the scan BW. I read that using large FFTs can help to pull signals out of the noise floor. My question is how large FFTs can accomplish this?










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    I am trying to detect unknown RF tones around -140 dBm and my scan BW is 5 MHz, going through the Noise power calculations the signal is below the thermal noise based on the scan BW. I read that using large FFTs can help to pull signals out of the noise floor. My question is how large FFTs can accomplish this?










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    luffyKun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      I am trying to detect unknown RF tones around -140 dBm and my scan BW is 5 MHz, going through the Noise power calculations the signal is below the thermal noise based on the scan BW. I read that using large FFTs can help to pull signals out of the noise floor. My question is how large FFTs can accomplish this?










      share|improve this question









      New contributor




      luffyKun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I am trying to detect unknown RF tones around -140 dBm and my scan BW is 5 MHz, going through the Noise power calculations the signal is below the thermal noise based on the scan BW. I read that using large FFTs can help to pull signals out of the noise floor. My question is how large FFTs can accomplish this?







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      edited 7 hours ago









      Marcus Müller

      11.2k41431




      11.2k41431






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      asked 8 hours ago









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          If you look at the formula of a single DFT bin



          $$X[k] = sum_{n=0}^{N-1}x[n]e^{-j2pi kfrac nN}text,$$



          you'll notice that his is essentially a correlation of $x$ with the complex sinusoid $e^{-j2pi kfrac nN}$.



          That means the DFT can just be understood as a filter bank of matched filters for single tones that fall in the DFT "raster".



          Hence, you simply get FFT length-based processing gain: The length of the sum.



          But: you probably don't have perfect knowledge of the exact frequency of the signal you're trying to detect¹! So, you can't put things into that perfect DFT raster.



          Now, the larger you choose the FFT length $N$, the finer that raster will get, but also, the longer your observation has to be, and the more compute power you'll need.



          At some point, the DFT stops being the best possible tone detector, and superresolution techniques become relevant. In this case (weak tone, you're sure that you've only got exactly one tone in your signal), the ESPRIT algorithm with a long observation period leading to the autocovariance matrix estimate that it takes as input, would probably work very nicely.




          ¹ There's inevitably frequency error in your receiver, and in your transmitter. Papers that start with We assume perfect synchronization typically skip the hard part of making a system work...




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          • the matched filter is out performed by ESPRIT? I don’t think so
            – Stanley Pawlukiewicz
            47 mins ago











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          1 Answer
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          up vote
          4
          down vote













          If you look at the formula of a single DFT bin



          $$X[k] = sum_{n=0}^{N-1}x[n]e^{-j2pi kfrac nN}text,$$



          you'll notice that his is essentially a correlation of $x$ with the complex sinusoid $e^{-j2pi kfrac nN}$.



          That means the DFT can just be understood as a filter bank of matched filters for single tones that fall in the DFT "raster".



          Hence, you simply get FFT length-based processing gain: The length of the sum.



          But: you probably don't have perfect knowledge of the exact frequency of the signal you're trying to detect¹! So, you can't put things into that perfect DFT raster.



          Now, the larger you choose the FFT length $N$, the finer that raster will get, but also, the longer your observation has to be, and the more compute power you'll need.



          At some point, the DFT stops being the best possible tone detector, and superresolution techniques become relevant. In this case (weak tone, you're sure that you've only got exactly one tone in your signal), the ESPRIT algorithm with a long observation period leading to the autocovariance matrix estimate that it takes as input, would probably work very nicely.




          ¹ There's inevitably frequency error in your receiver, and in your transmitter. Papers that start with We assume perfect synchronization typically skip the hard part of making a system work...




          share|improve this answer























          • the matched filter is out performed by ESPRIT? I don’t think so
            – Stanley Pawlukiewicz
            47 mins ago















          up vote
          4
          down vote













          If you look at the formula of a single DFT bin



          $$X[k] = sum_{n=0}^{N-1}x[n]e^{-j2pi kfrac nN}text,$$



          you'll notice that his is essentially a correlation of $x$ with the complex sinusoid $e^{-j2pi kfrac nN}$.



          That means the DFT can just be understood as a filter bank of matched filters for single tones that fall in the DFT "raster".



          Hence, you simply get FFT length-based processing gain: The length of the sum.



          But: you probably don't have perfect knowledge of the exact frequency of the signal you're trying to detect¹! So, you can't put things into that perfect DFT raster.



          Now, the larger you choose the FFT length $N$, the finer that raster will get, but also, the longer your observation has to be, and the more compute power you'll need.



          At some point, the DFT stops being the best possible tone detector, and superresolution techniques become relevant. In this case (weak tone, you're sure that you've only got exactly one tone in your signal), the ESPRIT algorithm with a long observation period leading to the autocovariance matrix estimate that it takes as input, would probably work very nicely.




          ¹ There's inevitably frequency error in your receiver, and in your transmitter. Papers that start with We assume perfect synchronization typically skip the hard part of making a system work...




          share|improve this answer























          • the matched filter is out performed by ESPRIT? I don’t think so
            – Stanley Pawlukiewicz
            47 mins ago













          up vote
          4
          down vote










          up vote
          4
          down vote









          If you look at the formula of a single DFT bin



          $$X[k] = sum_{n=0}^{N-1}x[n]e^{-j2pi kfrac nN}text,$$



          you'll notice that his is essentially a correlation of $x$ with the complex sinusoid $e^{-j2pi kfrac nN}$.



          That means the DFT can just be understood as a filter bank of matched filters for single tones that fall in the DFT "raster".



          Hence, you simply get FFT length-based processing gain: The length of the sum.



          But: you probably don't have perfect knowledge of the exact frequency of the signal you're trying to detect¹! So, you can't put things into that perfect DFT raster.



          Now, the larger you choose the FFT length $N$, the finer that raster will get, but also, the longer your observation has to be, and the more compute power you'll need.



          At some point, the DFT stops being the best possible tone detector, and superresolution techniques become relevant. In this case (weak tone, you're sure that you've only got exactly one tone in your signal), the ESPRIT algorithm with a long observation period leading to the autocovariance matrix estimate that it takes as input, would probably work very nicely.




          ¹ There's inevitably frequency error in your receiver, and in your transmitter. Papers that start with We assume perfect synchronization typically skip the hard part of making a system work...




          share|improve this answer














          If you look at the formula of a single DFT bin



          $$X[k] = sum_{n=0}^{N-1}x[n]e^{-j2pi kfrac nN}text,$$



          you'll notice that his is essentially a correlation of $x$ with the complex sinusoid $e^{-j2pi kfrac nN}$.



          That means the DFT can just be understood as a filter bank of matched filters for single tones that fall in the DFT "raster".



          Hence, you simply get FFT length-based processing gain: The length of the sum.



          But: you probably don't have perfect knowledge of the exact frequency of the signal you're trying to detect¹! So, you can't put things into that perfect DFT raster.



          Now, the larger you choose the FFT length $N$, the finer that raster will get, but also, the longer your observation has to be, and the more compute power you'll need.



          At some point, the DFT stops being the best possible tone detector, and superresolution techniques become relevant. In this case (weak tone, you're sure that you've only got exactly one tone in your signal), the ESPRIT algorithm with a long observation period leading to the autocovariance matrix estimate that it takes as input, would probably work very nicely.




          ¹ There's inevitably frequency error in your receiver, and in your transmitter. Papers that start with We assume perfect synchronization typically skip the hard part of making a system work...





          share|improve this answer














          share|improve this answer



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          edited 7 hours ago

























          answered 7 hours ago









          Marcus Müller

          11.2k41431




          11.2k41431












          • the matched filter is out performed by ESPRIT? I don’t think so
            – Stanley Pawlukiewicz
            47 mins ago


















          • the matched filter is out performed by ESPRIT? I don’t think so
            – Stanley Pawlukiewicz
            47 mins ago
















          the matched filter is out performed by ESPRIT? I don’t think so
          – Stanley Pawlukiewicz
          47 mins ago




          the matched filter is out performed by ESPRIT? I don’t think so
          – Stanley Pawlukiewicz
          47 mins ago










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