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Log-gabor are filters defined similarly as gabor filters in the sense that their envelope consist in a Gaussian in Fourier space. This is advantageous because this makes them optimal with respect to the compromise between localization (in space) and detection (of the mean frequency).

The difference is that log-gabor (as their name implies) are defined in the log-space frequency domain. This makes sense as this relevant feature (frequency) may for some applications better optimized when the precision in frequency is proportional to the mean frequency. In perception, this is often the case: for instance the human ear is in a large range sensitive to the relative increments of frequency (this is called the Fechner-Weber rule). This is also used in vision, see for instance this computational neuroscience paper

  • This makes sense as this relevant feature (frequency) may for some applications better optimized when the precision in frequency is proportional to the mean frequency.
  • this has the drawback that they have no simple analytical formulation in the space domain as simple Gabor filters,
  • In perception, this is very advantageous: for instance the human ear is in a large range sensitive to the relative increments of frequency (this is called the Fechner-Weber rule). This is also used in vision, see for instance the filters obtained in this this computational neuroscience paper mimicking the receptive field of simple cells in the primary visual cortex:

example 2D log-gabor filters

To specifically answer your question, log-Gabor filters have indeed the property of having a zero DC offset. To show this consider that this offset is the value of the spectrogram at the origin (for null frequencies). In log-frequencies, the DC corresponds to the limit of the log function at zero, that is minus infinite. Necessarily, this implies that this value is zero for a finite bandwidth of the filter.:

In (simplified) mathematical terms, the envelope is $f \propto \exp( - \log(f)^2 / B^2 /2$ so that for $B>0$, we have $f(0) \propto \lim \exp( - \inf^2 / B^2 /2) = 0$.

  • To show this consider that this offset is the value of the spectrogram at the origin (for null frequencies). In log-frequencies, the DC corresponds to the limit of the log function at zero, that is minus infinite. Necessarily, this implies that this value is zero for a finite bandwidth of the filter. In (simplified) mathematical terms, the envelope is : $$f \propto \exp( - \log(f)^2 / B^2 /2$$ so that for $B>0$, we have $f(0) \propto \lim_{f \rightarrow \infty} \exp( - \log(f)^2 / B^2 /2) = 0$.

  • This is relevant in perception. Indeed, remember that the DC component represent the average luminance value in the image, that is some global configuration (daylight vs nightlight, some specific exposure setting in your camera/eyeball) more relevant to specific tasks such as keeping the circadian rhythm (= syncing with the day rhythm). When doing a perceptually relevant task, such as detecting the shape of an object or tracking the motion of an object, these features are independent of the global configuration: e. g. a cat at night has the same shape as a cat during day!

Log-gabor are filters defined similarly as gabor filters in the sense that their envelope consist in a Gaussian in Fourier space. This is advantageous because this makes them optimal with respect to the compromise between localization (in space) and detection (of the mean frequency).

The difference is that log-gabor (as their name implies) are defined in the log-space frequency domain. This makes sense as this relevant feature (frequency) may for some applications better optimized when the precision in frequency is proportional to the mean frequency. In perception, this is often the case: for instance the human ear is in a large range sensitive to the relative increments of frequency (this is called the Fechner-Weber rule). This is also used in vision, see for instance this computational neuroscience paper

To specifically answer your question, log-Gabor filters have indeed the property of having a zero DC offset. To show this consider that this offset is the value of the spectrogram at the origin (for null frequencies). In log-frequencies, the DC corresponds to the limit of the log function at zero, that is minus infinite. Necessarily, this implies that this value is zero for a finite bandwidth of the filter.

In (simplified) mathematical terms, the envelope is $f \propto \exp( - \log(f)^2 / B^2 /2$ so that for $B>0$, we have $f(0) \propto \lim \exp( - \inf^2 / B^2 /2) = 0$.

Log-gabor are filters defined similarly as gabor filters in the sense that their envelope consist in a Gaussian in Fourier space. This is advantageous because this makes them optimal with respect to the compromise between localization (in space) and detection (of the mean frequency).

The difference is that log-gabor (as their name implies) are defined in the log-space frequency domain:

  • This makes sense as this relevant feature (frequency) may for some applications better optimized when the precision in frequency is proportional to the mean frequency.
  • this has the drawback that they have no simple analytical formulation in the space domain as simple Gabor filters,
  • In perception, this is very advantageous: for instance the human ear is in a large range sensitive to the relative increments of frequency (this is called the Fechner-Weber rule). This is also used in vision, see for instance the filters obtained in this this computational neuroscience paper mimicking the receptive field of simple cells in the primary visual cortex:

example 2D log-gabor filters

To specifically answer your question, log-Gabor filters have indeed the property of having a zero DC offset:

  • To show this consider that this offset is the value of the spectrogram at the origin (for null frequencies). In log-frequencies, the DC corresponds to the limit of the log function at zero, that is minus infinite. Necessarily, this implies that this value is zero for a finite bandwidth of the filter. In (simplified) mathematical terms, the envelope is : $$f \propto \exp( - \log(f)^2 / B^2 /2$$ so that for $B>0$, we have $f(0) \propto \lim_{f \rightarrow \infty} \exp( - \log(f)^2 / B^2 /2) = 0$.

  • This is relevant in perception. Indeed, remember that the DC component represent the average luminance value in the image, that is some global configuration (daylight vs nightlight, some specific exposure setting in your camera/eyeball) more relevant to specific tasks such as keeping the circadian rhythm (= syncing with the day rhythm). When doing a perceptually relevant task, such as detecting the shape of an object or tracking the motion of an object, these features are independent of the global configuration: e. g. a cat at night has the same shape as a cat during day!

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Log-gabor are filters defined similarly as gabor filters in the sense that their envelope consist in a Gaussian in Fourier space. This is advantageous because this makes them optimal with respect to the compromise between localization (in space) and detection (of the mean frequency).

The difference is that log-gabor (as their name implies) are defined in the log-space frequency domain. This makes sense as this relevant feature (frequency) may for some applications better optimized when the precision in frequency is proportional to the mean frequency. In perception, this is often the case: for instance the human ear is in a large range sensitive to the relative increments of frequency (this is called the Fechner-Weber rule). This is also used in vision, see for instance this computational neuroscience paper

To specifically answer your question, log-Gabor filters have indeed the property of having a zero DC offset. To show this consider that this offset is the value of the spectrogram at the origin (for null frequencies). In log-frequencies, the DC corresponds to the limit of the log function at zero, that is minus infinite. Necessarily, this implies that this value is zero for a finite bandwidth of the filter.

In (simplified) mathematical terms, the envelope is $f \propto \exp( - \log(f)^2 / B^2 /2$ so that for $B>0$, we have $f(0) \propto \lim \exp( - \inf^2 / B^2 /2) = 0$.