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Mumbai University > Electronics > Sem 7 > Digital image processing

**Marks:** 10 M

**Year:** Dec 2013

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Explain the following examples: (i) Signature (ii) Fourier Descriptor

written 8.1 years ago by | modified 2.5 years ago by |

Mumbai University > Electronics > Sem 7 > Digital image processing

**Marks:** 10 M

**Year:** Dec 2013

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written 8.1 years ago by |

**(i) Signatures:**

The signature of an object is a simple function representation that can be used to describe and reconstruct a boundary. A Simple approach is to plot the distance from the centroid to the boundary as a function of the angle. This is called the polar radii signature of the object.

Point C is the centroid. The phasor r completes 3600, we get a one dimensional representation of r(Φ) will be constant, equal to the radius of the circle. The signature can thus be defined as 1-diemnsional representation of a boundary.

Consider another boundary in the shape of a square. In the case r(Φ) will be given by the formula r(Φ)=x sec θ

**(ii) Fourier Descriptors:**

A common method of describing the contour of an object is by using 1-Dimesnional Fourier Transform.

The figure shows a N-point digital boundary in the spatial Domain. Each of these edges pixels can be defined by its x and y coordinates. Starting at an arbitrary point (x0,y0),(x1,y1),(x2,y2)……..(xn-1,yn-1) points are encountered as we move in the counter clockwise direction. Each of these points can be expressed as xr and yr for r=0,1……N-1. These coordinates values can be used to generate a complex function of the form,

f(n)= x(n)+j y(n) for n=0,1,2,……..,N-1

Hence the x-axis is treated as real axis and y-axis as the imaginary axis. The Fourier transform of this function f(n) yields the frequency components that describe the given edge. The discrete Fourier transform (DFT) of f(n) is

$F(u)= 1/N \sum_{n=0}^{N-1} f(n) e^{-j2πun/N} \ \ \ \ \ for \ u=0,1,2,………...,N-1$

The advantage of using this equation is that it reduces the edge description problem from 2-Diemsnsional to 1-Dimension.

Substituting the value of f(n) we have,

$F(u)= 1/N \sum_{n=0}^{N-1} [x(n)+jy(n)] e^{-j2πun/N} \ \ \ \ \ for \ \ u=0,1,2,…..,N-1$

The coefficients of F(u) are called Fourier descriptors. The inverse discrete fourier transform (IDFT) of F(u) gives back f(n).

$F(n)= \sum_{n=0}^{N-1} F(u) e^{+j2πun/N} \ \ \ \ for \ \ n=0,1,2,3……N-1$

However instead of using all the F(u) coefficients, we only use few of them while remaining terms are made zero,

$f (n)= \sum_{n=0}^{N-1} F(u) e^{+j2πun/N} \ \ \ \ \ for \ \ n=0,1,2,3……N-1$

Although only M terms are used for F(u), f(n) still has 0 to N-1 values. That is the same number of points exist in the new approximated boundary but not as many terms are used in the reconstruction of each point.

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