Combined invariants to blur and rotation using Zernike moment descriptors

Abstract

Moment invariants that are not affected by geometric transform have been utilized as pattern features in a number of applications. But in most cases, images are processed subject to blur degradations. The traditional blur invariant sets were constructed using geometric moments, central moments or complex moments. However, these non-orthogonal moments are generally considered as a disadvantage over orthogonal moments, such as Zernike, pseudo-Zernike, and Legendre moments, in decreasing information redundancy and sensitivity to noises. To solve this problem, this paper addresses a method for recognizing objects in an image in a way that is invariant to images’ blur and rotation transformations to improve the robustness to noises. The proposed method is based on Zernike descriptors which are orthogonal over a unit circle, and is invariant to a central symmetric blur, such as linear motion or out-of-focus blur. We present a mathematical framework of obtaining the Zernike moments of blurred images, and a framework of deriving the combined blur and rotation invariants. The classification experimental results are presented to confirm the proposed method outperforms other similar ones in the presence of various blur-degraded and rotation-transformed images.

Appendices

Appendix A: Proof of Theorem 1

In this appendix, we detail a derivation of Theorem 1. According to (8), we obtain the following expression.

$$ \begin{aligned} \frac{\pi }{p + 1}Z_{pq}^{g} & = \sum\limits_{k = q}^{p} {B_{pqk} D_{kq}^{g} } \\ & = \sum\limits_{k = q}^{p} {B_{pqk} } \int\limits_{x} {\int\limits_{y} {(x - iy)^{(k + q)/2} (x + iy)^{(k - q)/2} g(x,y)\hbox{d}x\hbox{d}y} } \\ & = \sum\limits_{k = q}^{p} {B_{pqk} } \int\limits_{x} {\int\limits_{y} {(x - iy)^{(k + q)/2} (x + iy)^{(k - q)/2} } \left[ {\iint_{{R^{2} }} {h(a,b)f(x - a,y - b)dadb}} \right]} \hbox{d}x\hbox{d}y \\ & = \sum\limits_{k = q}^{p} {B_{pqk} } \iint_{{R^{2} }} {h(a,b)}\left[ {\int\limits_{x} {\int\limits_{y} {(x - iy)^{(k + q)/2} (x + iy)^{(k - q)/2} f(x - a,y - b)\hbox{d}x\hbox{d}y} } } \right]dadb \\ & = \sum\limits_{k = q}^{p} {B_{pqk} } \iint_{{R^{2} }} {h(a,b)\left[ {\int\limits_{x} {\int\limits_{y} {[(x + a) - i(y + b)]^{(k + q)/2} [(x + a) + i(y + b)]^{(k - q)/2} } } f(x,y)\hbox{d}x\hbox{d}y} \right]}dadb \\ & = \sum\limits_{k = q}^{p} {B_{pqk} } \iint_{{R^{2} }} {h(a,b)\left[ {\int\limits_{x} {\int\limits_{y} {\sum\limits_{m = 0}^{{\frac{k + q}{2}}} {\left( \begin{gathered} \frac{k + q}{2} \\ m \\ \end{gathered} \right)(x - iy)^{m} (a - ib)^{{\frac{k + q}{2} - m}} } \sum\limits_{n = 0}^{{\frac{k - q}{2}}} {\left( \begin{gathered} \frac{k - q}{2} \\ n \\ \end{gathered} \right)(x + iy)^{n} (a + ib)^{{\frac{k - q}{2} - n}} f(x,y)\hbox{d}x\hbox{d}y} } } } \right]dadb} \\ & = \sum\limits_{k = q}^{p} {\sum\limits_{m = 0}^{{\frac{k + q}{2}}} {\sum\limits_{n = 0}^{{\frac{k - q}{2}}} {\left( \begin{gathered} \frac{k + q}{2} \\ m \\ \end{gathered} \right)\left( \begin{gathered} \frac{k - q}{2} \\ n \\ \end{gathered} \right)} } B_{pqk} } \iint_{{R^{2} }} {(a - ib)^{{\frac{k + q}{2} - m}} (a + ib)^{{\frac{k - q}{2} - n}} h(a,b)dadb}\int\limits_{x} {\int\limits_{y} {(x - iy)^{m} (x + iy)^{n} f(x,y)\hbox{d}x\hbox{d}y} } \\ & = \sum\limits_{k = q}^{p} {\sum\limits_{m = 0}^{{\frac{k + q}{2}}} {\sum\limits_{n = 0}^{{\frac{k - q}{2}}} {\left( \begin{gathered} \frac{k + q}{2} \\ m \\ \end{gathered} \right)\left( \begin{gathered} \frac{k - q}{2} \\ n \\ \end{gathered} \right)} } B_{pqk} } D_{k - m - n,q - m + n}^{h} D_{m + n,m - n}^{f} \\ \end{aligned} $$

(15)

From (15) and (8), we can obtain the following expression of radial moments after blurring degradation.

$$ D_{pq}^{g} = \sum\limits_{m = 0}^{{\frac{p + q}{2}}} {\sum\limits_{n = 0}^{{\frac{p - q}{2}}} {\left( \begin{gathered} \frac{p + q}{2} \\ m \\ \end{gathered} \right)\left( \begin{gathered} \frac{p - q}{2} \\ n \\ \end{gathered} \right)} } D_{p - m - n,q - m + n}^{h} D_{m + n,m - n}^{f} $$

(16)

Appendix B: Proof of theorem 2

Proof of $Z_{pp}^{g} = Z_{pp}^{f}$

According to (15), if p = q, Zernike moments $Z_{pq}^{g}$of an image after Gaussian blurring becomes

$$ Z_{pp}^{g} = \frac{p + 1}{\pi }\sum\limits_{m = 0}^{p} {\left( \begin{gathered} p \hfill \\ m \hfill \\ \end{gathered} \right)B_{ppp} D_{p - m,p - m}^{h} D_{mm}^{f} } $$

(17)

From (7), we can obtain the radial moments in PSF form as

$$ \begin{aligned} D_{p - m,p - m}^{h} & = \int\limits_{x} {\int\limits_{y} {h(x,y)(x - iy)^{p - m} \hbox{d}x\hbox{d}y} } \\ & = \int\limits_{x} {\int\limits_{y} {h(x,y)} } \sum\limits_{t = 0}^{p - m} {\left( \begin{gathered} p - m \\ t \\ \end{gathered} \right)( - i)^{t} x^{p - m - t} y^{t} \hbox{d}x\hbox{d}y} \\ & = \sum\limits_{t = 0}^{p - m} {\left( \begin{gathered} p - m \\ t \\ \end{gathered} \right)( - i)^{t} } \int\limits_{x} {\int\limits_{y} {\frac{1}{{2\pi \sigma^{2} }}e^{{ - \frac{{x^{2} + y^{2} }}{{2\sigma^{2} }}}} x^{p - m - t} y^{t} \hbox{d}x\hbox{d}y} } \\ & = \sum\limits_{t = 0}^{p - m} {\left( \begin{gathered} p - m \\ t \\ \end{gathered} \right)( - i)^{t} g_{p - m - t} g_{t} } \\ \end{aligned} $$

(18)

where g _i g _j is the central moment of Gaussian PSF, according to the conclusion drawn by Flusser and Suk [14] and Liu and Zhang [18] we have

$$ g_{i} g_{j} = \int\limits_{ - \infty }^{ + \infty } {\int\limits_{ - \infty }^{ + \infty } {(x - 0)^{i} (y - 0)^{j} \frac{1}{{2\pi \sigma^{2} }}e^{{ - \frac{{x^{2} + y^{2} }}{{2\sigma^{2} }}}} \hbox{d}x\hbox{d}y} } $$

(19)

and

$$ g_{i} = \frac{1}{{\sqrt {2\pi } \sigma }}\int\limits_{ - \infty }^{ + \infty } {x^{i} e^{{ - \frac{{x^{2} }}{{2\sigma^{2} }}}} \hbox{d}x} = \left\{ \begin{gathered} 0\quad {\text{if }}i{\text{ is an odd number}} \hfill \\ 1.3 \ldots (i - 1)\sigma^{i} \quad {\text{if }}i{\text{ is an even number}} \hfill \\ \end{gathered} \right. $$

(20)

Then, from (18–20) we have the following conclusions:

• (1) If p − m is odd, whether t is odd or even, $D_{p - m,p - m}^{h} = 0$.

• (2) If p − m is even and t is odd, $D_{p - m,p - m}^{h} = 0$.

• (3) If p − m is even and t is even

Let t = 2k, then

$$ \begin{aligned} D_{p - m,p - m}^{h} & = \sum\limits_{k = 0}^{(p - m)/2} {\left( \begin{aligned} p - m \\ 2k \\ \end{aligned} \right)( - i)^{2k} g_{p - m - 2k} g_{2k} } \\ & = \sum\limits_{k = 0}^{(p - m)/2} {\left( \begin{aligned} p - m \\ 2k \\ \end{aligned} \right)( - 1)^{k} g_{p - m - 2k} g_{2k} } \\ \end{aligned} $$

(21)

where

$$ g_{2k} = 1 \cdot 3 \cdot 5 \cdots (2k - 1)\sigma^{2k} = \frac{(2k)!}{{2^{k} k!}}\sigma^{2k} $$

(22)

$$ g_{p - m - 2k} = \frac{(p - m - 2k)!}{{2^{{\frac{p - m}{2} - k}} \left( {\frac{p - m}{2} - k} \right)!}}\sigma^{p - m - 2k} $$

(23)

Substitute (22) and (23) into (21), we get

$$ \begin{aligned} D_{p - m,p - m}^{h} & = \sum\limits_{k = 0}^{(p - m)/2} {\frac{{( - 1)^{k} (p - m)!}}{(2k)!(p - m - 2k)!}\frac{(2k)!}{{2^{k} k!}}\frac{(p - m - 2k)!}{{2^{{\frac{p - m}{2} - k}} (\frac{p - m}{2} - k)!}}\sigma^{2k} \sigma^{p - m - 2k} } \\ & = \sigma^{p - m} \frac{(p - m)!}{{2^{{\frac{p - m}{2}}} \left( {\frac{p - m}{2}} \right)!}}\sum\limits_{k = 0}^{(p - m)/2} {\left( \begin{gathered} (p - m)/2 \\ k \\ \end{gathered} \right)} ( - 1)^{k} (1)^{{\left( {\frac{p - m}{2} - k} \right)}} \\ & = \sigma^{p - m} \frac{(p - m)!}{{2^{{\frac{p - m}{2}}} \left( {\frac{p - m}{2}} \right)!}}( - 1 + 1)^{{\frac{p - m}{2}}} = 0 \\ \end{aligned} $$

Due to the fact that imaging system is energy preserving, i.e.,$\iint_{{R^{2} }} {h(x,y)\hbox{d}x\hbox{d}y = 1}$, thus, we have

$$ Z_{00}^{h} = \frac{1}{\pi }B_{000} D_{00}^{h} = \frac{1}{\pi } $$

(24)

While p = q, $p \ne m$,$D_{p - m,p - m}^{h} = 0$. Substituting $D_{p - m,p - m}^{h} = 0$ into (17), we get

$$ \begin{aligned} Z_{pp}^{g} & = \frac{p + 1}{\pi }\left( \begin{gathered} p \hfill \\ p \hfill \\ \end{gathered} \right)B_{ppp} D_{00}^{h} D_{pp}^{f} \\ & = \frac{p + 1}{\pi }D_{pp}^{f} \\ & = Z_{pp}^{f} \\ \end{aligned} $$

(25)

Therefore, when p = q, and $Z_{pp}^{g} = Z_{pp}^{f}$