& n=1,2,\ldots, N_o\end{aligned}$$ where, $o$ and $n$ denote the orientation and scale indexes, respectively. $O$ is the total number of orientations. $N_o$ is the total number of scales at the $o-$th orientation. $\lfloor \rfloor$ is the floor operator. Note that $\lfloor y \rfloor = y$ if $y>0$, otherwise zero. $E_o$ is the local energy at the $o-$th orientation, which is calculated by using the Hilbert transform as follows $$\begin{aligned}
\label{eq:e}
E_o(x)=\sqrt{F_o^2(x)+H_o^2(x)}\end{aligned}$$ where, $F_o(x)$ and $H_o(x)$ are the AC component and Hilbert transform of the image signal $I(x)$ at the $o-$th orientation. However, because the Hilbert transform operator is an improper integral, $F_o(x)$ and $H_o(x)$ are computed by convolving the image signal with a pair of even and odd wavelets filters in quadrature as follows [@Morrone1987; @Venkatesh1990]: $$\begin{aligned}
\label{eq:fs}
&F_o(x)= \sum_{n} I(x) * M_{no}^{e}\\
\label{eq:hs}
&H_o(s)= \sum_{n} I(x) * M_{no}^{o}$$ where, $*$ denotes the convolution operator. $M_{no}^e$ and $M_{no}^o$ are even and odd filters at the $o-$th orientation and $n-$th scale, generated by the logarithmic Gabor function $$\begin{aligned}
\label{eq:lgabor1}
G_{no}(f)=\exp \left(\frac{-\left(\log\frac{f}{\hat{f}_{no}}\right)^2}{2\left( \log \sigma_{no}\right)^2}\right)$$ where, $\hat{f}_{no}$ and $\sigma_{no}$ denote the centre frequency and bandwidth of the $n-$th Gabor filer at the $o-$th iteration. The filters are scaled as follows $$\begin{aligned}
\label{eq:lgabor2}
\hat{f}_{no}=\frac{1}{\lambda_{\min 