Sharpening

As the opposite of smoothing operations, image sharpening has the goal to enhance the details (the high spatial frequency components) of the image. A high-pass filtered image can be obtained as the difference between the original image and its low-pass filtered version. The original image can be considered as all-pass filtered by a delta function kernel:

$$W_{all-pass}=\left[ \begin{array}{ccc} 0 & 0 & 0 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right]$$

as the image convolved with this delta function is not changed:

$$W_{allpass} * I_{original} = I_{original}$$

Therefore we have:

$$I_{high-pass} = I_{all-pass}-I_{low-pass} =W_{allpass}*I_{original}-W_{lowpass}*I_{original}$$

$$= (W_{allpass}-W_{lowpass})*I_{original} =W_{highpass}*I_{original}$$

where $W_{highpass}=W_{allpass}-W_{lowpass}$ is the high-pass kernel corresponding to the low-pass kernel $W_{lowpass}$. Equivalently the frequency domain filtering is shown in the figure:

We can therefore get a high-pass filter kernel corresponding to each low-pass filter kernel by subtracting the low-pass kernel from the all-pass one:

• 
  
  $$W_{highpass}=W_{allpass}-W_{lowpass} =\left[ \begin{array}{c... ...0 & -1 & 0 -1 & 4 & -1 \\ 0 & -1 & 0 \end{array} \right]$$
  
  

• 
  
  $$W_{highpass}=W_{allpass}-W_{lowpass} =\left[ \begin{array}{c... ... -1 & -1 -1 & 8 & -1 \\ -1 & -1 & -1 \end{array} \right]$$
  
  

• 
  
  $$W_{highpass}=W_{allpass}-W_{lowpass} =\left[ \begin{array}{cc... ...-2 & -1 -2 & 12 & -2 \\ -1 & -2 & -1 \end{array} \right]$$
  
  

These are various forms of the Laplace operators.

Note that the sum of all elements of the resulting high-pass filter is always zero. When such a high-pass kernel is convolved with a region of an image where all pixels have same gray level (constant or DC component), the result is zero, i.e., the zero spatial frequency component is totally suppressed by the high-pass filter.

Similar band-pass filters can be obtained by finding the difference between two low-pass filters of different cut-off frequencies.

The following figure shows the original image, a panda (left), and its low-pass (middle) and high-pass (right) filtered versions.