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According to the research paper Multidirectional Scratch Detection and Restoration in Digitized Old Images, we have,

$$H(u, v) = \frac{1}{1 + 0.414 {. \sqrt[{2n}]{\frac {u^*}{D_u}+\frac {v^*}{D_v}}}} \tag{1}$$

where:

\begin{align} u^* &= \cos \theta . (u + t_x) + \sin \theta . (v + t_y)\\ v^* &= -\sin\theta . (u + t_x) + \cos \theta . (v + t_y)\\ t_x &= \mbox{center}_x \times \cos \theta\\ t_y &= \mbox{center}_y \times \sin \theta \end{align}

The above formula generated a kernel of the following data:

The convolution operation is generating a complete black image. My guess is, the kernel values are too small.

Is there any technique to magnify these data?

N.B.1. I have actually corrected formula $(1)$

N.B.2. I have used an 8-bit-indexed gray-scale image and a 32x32 mask. I padded the mask. Then converted both the image and the mask to complex 2d array. Then I applied Fourier transform to both of them. Then I multiplied them. Then did the Inverse transform. I have tested with $\theta=0.9$ and $radian=0.9$. The results are not coming as expected.

According to the research paper Multidirectional Scratch Detection and Restoration in Digitized Old Images, we have,

$$H(u, v) = \frac{1}{1 + 0.414 {. \sqrt[{2n}]{\frac {u^*}{D_u}+\frac {v^*}{D_v}}}} \tag{1}$$

where:

\begin{align} u^* &= \cos \theta . (u + t_x) + \sin \theta . (v + t_y)\\ v^* &= -\sin\theta . (u + t_x) + \cos \theta . (v + t_y)\\ t_x &= \mbox{center}_x \times \cos \theta\\ t_y &= \mbox{center}_y \times \sin \theta \end{align}

N.B. I have used an 128x128 8-bit-indexed gray-scale image and a 32x32 mask. I padded the mask. Then converted both the image and the mask to complex 2d array. Then I applied Fourier transform to both of them. Then I multiplied them. Then did the Inverse transform. 

I have tested with values:

• $\theta=0.9$ and $radian=0.9$
• $D_u=2, D_v=2$
• $CenterX=16, CenterY=16$ and $CenterX=-16, CenterY=-16$

The results are not coming as expected.

The convolution operation is generating a complete black image. My guess is, the kernel values are too small.

Is there any technique to magnify these data?

P.S I have actually corrected formula $(1)$

According to the research paper Multidirectional Scratch Detection and Restoration in Digitized Old Images, we have,

$$H(u, v) = \frac{1}{1 + 0.414 {. \sqrt[{2n}]{\frac {u^*}{D_u}+\frac {v^*}{D_v}}}} \tag{1}$$

N.B. I have actually corrected formula $(1)$

where:

\begin{align} u^* &= \cos \theta . (u + t_x) + \sin \theta . (v + t_y)\\ v^* &= -\sin\theta . (u + t_x) + \cos \theta . (v + t_y)\\ t_x &= \mbox{center}_x \times \cos \theta\\ t_y &= \mbox{center}_y \times \sin \theta \end{align}

The above formula generated a kernel of the following data:

The convolution operation is generating a complete black image. My guess is, the kernel values are too small.

Is there any technique to magnify these data?

According to the research paper Multidirectional Scratch Detection and Restoration in Digitized Old Images, we have,

$$H(u, v) = \frac{1}{1 + 0.414 {. \sqrt[{2n}]{\frac {u^*}{D_u}+\frac {v^*}{D_v}}}} \tag{1}$$

where:

\begin{align} u^* &= \cos \theta . (u + t_x) + \sin \theta . (v + t_y)\\ v^* &= -\sin\theta . (u + t_x) + \cos \theta . (v + t_y)\\ t_x &= \mbox{center}_x \times \cos \theta\\ t_y &= \mbox{center}_y \times \sin \theta \end{align}

The above formula generated a kernel of the following data:

The convolution operation is generating a complete black image. My guess is, the kernel values are too small.

Is there any technique to magnify these data?

N.B.1. I have actually corrected formula $(1)$

N.B.2. I have used an 8-bit-indexed gray-scale image and a 32x32 mask. I padded the mask. Then converted both the image and the mask to complex 2d array. Then I applied Fourier transform to both of them. Then I multiplied them. Then did the Inverse transform. I have tested with $\theta=0.9$ and $radian=0.9$. The results are not coming as expected.

According to the research paper Multidirectional Scratch Detection and Restoration in Digitized Old Images, we have,

$H(u, v) = \frac{1}{1 + 0.414 {. \sqrt[{2n}]{\frac {u^*}{D_u}+\frac {v^*}{D_v}}}} ... ... ... ... ... ... ... (1)$

N.B. I have actually, corrected the formula$\ldots(1)$

where:

$t_x = \mbox{center}_x \times \cos \theta\,,$

$t_y = \mbox{center}_y \times \sin \theta\,,$

$u^* = \cos \theta . (u + t_x) + \sin \theta . (v + t_y)\,,$

$v^* = -\sin\theta . (u + t_x) + \cos \theta . (v + t_y)\,.$

The above formula generated a kernel of the following data:

The convolution operation is generating a complete black image.

  My guess is, the kernel values are too small.

Is there any technique to magnify these data?

According to the research paper Multidirectional Scratch Detection and Restoration in Digitized Old Images, we have,

$$H(u, v) = \frac{1}{1 + 0.414 {. \sqrt[{2n}]{\frac {u^*}{D_u}+\frac {v^*}{D_v}}}} \tag{1}$$

N.B. I have actually corrected formula $(1)$

where:

\begin{align} u^* &= \cos \theta . (u + t_x) + \sin \theta . (v + t_y)\\ v^* &= -\sin\theta . (u + t_x) + \cos \theta . (v + t_y)\\ t_x &= \mbox{center}_x \times \cos \theta\\ t_y &= \mbox{center}_y \times \sin \theta \end{align}

The above formula generated a kernel of the following data:

The convolution operation is generating a complete black image. My guess is, the kernel values are too small.

Is there any technique to magnify these data?