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I need to enhance the visibility of veins on dorsal hand vein images in my project. I use two different even-symmetric Gabor filters bank improve vein visibility.

First bank consists of these gabor functions: $$G^\mathit{e}_\mathit{mk}(x,y)=\dfrac{\gamma}{2\pi\sigma^2}\exp\Bigg\{-\frac{1}{2}\left(\dfrac{x_\mathit{\theta}+\gamma^2y_\mathit{\theta}^2}{\sigma^2}\right)\Bigg\}\times \left(\cos(2\pi f_\mathit{0}x_\mathit{\theta})-\exp(-\dfrac{\upsilon^2}{2})\right)$$

Second bank consists of these ones:

$$G^\mathit{e}_\mathit{mk}(x,y)=\exp\Bigg\{-\frac{1}{2}\left(\dfrac{x_\mathit{\theta}+\gamma^2y_\mathit{\theta}^2}{\sigma^2}\right)\Bigg\}\times \cos(2\pi f_\mathit{0}x_\mathit{\theta})$$

where $m$ is the scale index, $k$ is the orientation index, $f_\theta$ is the filter center frequency, $\sigma$ is the standard deviation (often called scale), $\gamma$ is the aspect ratio of the elliptical Gaussian envelope, $\upsilon$ is the factor determining DC response, $x_\theta=(x\cos\theta+y\sin\theta)$ and $y_\theta=(-x\sin\theta+y\cos\theta)$ are rotated versions of the $x$ and $y$ coordinates.

I have coded these filters in MATLAB, I do not have any problem on coding. But I cannot understand the underlying difference between these two gabor functions.

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  • $\begingroup$ How is v determined ? $\endgroup$
    – vini
    Jan 21, 2012 at 16:34
  • $\begingroup$ Sorry for late response. $\upsilon=\sqrt{2ln2/\beta}$ where $\beta=(2^{\Delta\omega}-1)/(2^{\Delta\omega}+1)$. $\Delta\omega$ represents the frequency bandwith in octaves which is proposed that there is a significant range in bandwidth ($\Delta\omega(\in [1, 1.5])$) $\endgroup$
    – saglamp
    Mar 19, 2012 at 8:53

2 Answers 2

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Depending on the location of the peak and the scale of the two axes of the Gaussian envelope the filter may have a large DC response. A popular approach to get a zero DC response is to subtract the output of a low-pass Gaussian filter, which is what the first of these two does. In the case of images, if the DC response is not removed, the filter will respond to the absolute intensity of the image.

The tutorial, Javier R. Movellan - Tutorial on Gabor Filters, gives a bit more detail.

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  • $\begingroup$ Thank you for the answer an the tutorial. I have read the tutorial but i am still confused about "the absolute intensity of the image". I need more information about the difference between DC response subtracted Gabor filter and not subtracted one. For example, I wonder that if we look at the convolution of both filters on the same image, what will be dissimilar in these results? $\endgroup$
    – saglamp
    Jan 13, 2012 at 15:21
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In addition to the DC component difference mentioned (where typically v^2 = sigma^2). The first formula has a normalized gaussian because of the first coefficient, though I'm not sure how much use normalizing part of a wavefunction is, since it does not involve probability functions.

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