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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up vote
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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?
fft
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up vote
2
down vote
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up vote
2
down vote
favorite
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?
fft
New contributor
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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fft
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edited 4 hours ago
Marcus Müller
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asked 4 hours ago
luffyKun
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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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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
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...
add a comment |
up vote
3
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...
add a comment |
up vote
3
down vote
up vote
3
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...
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...
edited 4 hours ago
answered 4 hours ago
Marcus Müller
11.1k41431
11.1k41431
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