With the help of neat block diagram explain the basic of filtering in frequency domain. Give the reasons of shifting the origin.

Importance of Fourier Transform and Frequency domain tools:

i. The Fourier transform is one of the most important transforms that is used in image processing

ii. It is used to convert the image from time domain to frequency domain, so that frequency domain tools can be used for image enhancement.

iii. Fourier image analysis has several useful properties, for instance convolution in the spatial domain corresponds to multiplication in the frequency domain. This is important because multiplication is a simpler mathematical operation than convolution. As with one-dimensional signals, this property enables FFT convolution and various deconvolution techniques.

iv. The idea of blurring an image by reducing its high frequency components, or sharpening an image by increasing the magnitude of its high frequency components is intuitively easy to understand.

v. Understanding frequency domain concepts and using the frequency domain tools are important, as it leads to enhancement techniques that might not have been thought of by restricting attention to the spatial domain.

Basics of filtering in frequency domain

i. Filtering techniques in the frequency domain are based on modifying the fourier transform to achieve a specific objective and then computing the inverse DFT to get us back to the image domain.

ii. Filtering in the frequency domain consists of modifying the Fourier transform of an image and then computing the inverse transform to obtain the processed result.

iii. Thus given a digital image, f(x,y) of size M x N, the basic filtering equation in which we are interested has the form : $g(x,y)= G^{-1}[H(u,v)F(u,v)]\space\space\space\space \text {(Equation 1)}$

where $G^{-1}$ is the IDFT

F(u,v) is the DFT of the input image f(x,y)

H(u,v) is a filter function (also called simply the filter or Filter transfer function)

g(x,y) is the filtered (output) image

Figure 1. Filtering in frequency domain

iv. Functions F, H and G are arrays of size $M \times N$, the same as the input image. The product H(u,v)F(u,v) is formed using array multiplication

v. The filter function modifies the transform of the input image to yield a processed output, g(x,y). Specifications of H(u,v) is simplified considerably by using functions that are symmetric about their center, which requires that F(u,v) be centered i.e. its origin needs to be shifted. This shift of origin is explained below.

Shifting the origin :

i. The periodicities of the transform and its inverse are important issues in the implementation of OFT-based algorithms. Consider the 1-D spectrum in Fig. 2(a).the transform data in the interval from 0 to $M_1$ consists of two back-to-back half periods meeting at point $M_{12}. $

ii. For display and filtering purposes, it is more convenient to have in this interval a complete period of the transform in which the data are contiguous, as in Fig. 2(b).

$f(x)e^{j2π(u0x/M)} = \gt F(u - uo)$

iii In other words, multiplying f(x) by the exponential term shown shifts the data so that the origin, F(O), is located at uo. If we let $U_o= M/2,$ the exponential term becomes $ej^{πx}$ which is equal to $(-1)^x$ because x is an integer. In this case,

That is, multiplying f(x) by $(-1)^x$ shifts the data so that F(O) is at the center of the interval [0, M - 1], which corresponds to Fig. 1(b), as desired.

iv. In 2-D, the principle is same, Fig. 2(c) shows. Instead of two half periods, there are now four quarter periods meeting at the point (M /2, N /2). The dashed rectangles correspond the infinite number of periods of the 2-D DFT. As in the I-D case, visualization is simplified if we shift the data so that F(0,0) is at (M/2, N/2).

We get the equation

$f(x,y)(-1)^{x+y} \lt=\gt F(u – M/2,v N/2)$

v. Using this equation shifts the data so that F(O, 0) is at the center of the frequency rectangle defined by the intervals [0, M -- 1] and [0, N - 1], as desired. Figure 2(d) shows the result.