## Appendix B  
By definition, the equivalent rectangular bandwidth of a filter, \(H_{\text{ERB}}\)[58], is defined as:  
\[H_{\text{ERB}}=\frac{\int_{-\infty}^{+\infty}\abs{H(f)}^{2}df}{2\abs{H(f_{0}) }^{2}}, \tag{15}\]  
where \(\abs{H(f_{0})}^{2}\) is the maximum value of the power spectrum, which occurs at \(\pm f_{0}\). By using the [PERSON]'s theorem, we can write:  
\[\int_{-\infty}^{+\infty}\abs{H(f)}^{2}df=\int_{-\infty}^{+\infty}\abs{h(f)}^{ 2}dt. \tag{16}\]  
Hence, the equivalent rectangular bandwidth can be expressed as:  
\[H_{\text{ERB}}=\frac{\int_{-\infty}^{+\infty}\abs{h(t)}^{2}dt}{2\abs{H(f_{0}) }^{2}}, \tag{17}\]  
Let us assume, \(\tilde{h}(t)=\abs{h(t)}^{2}\); hence, the equivalent rectangular bandwidth can be simplified as:  
\[H_{\text{ERB}}=\frac{\int_{-\infty}^{+\infty}\tilde{h}(t)dt}{2\abs{H(f_{0})}^{ 2}}, \tag{18}\]  
From the definition of the Fourier transform of \(\tilde{h}(t)\), we can write:  
\[\tilde{H}(f)=\int_{-\infty}^{+\infty}\tilde{h}(t)e^{-j2\pi ft}dt, \tag{19}\]We can find the D.C. component of \(\tilde{\rm H}({\rm f})\) by substituting \({\rm f}=0\) in Equation (A15) as:  
\[\tilde{\rm H}(0)=\int_{-\infty}^{+\infty}\tilde{\rm h}({\rm t}){\rm dt}.\] (A16)  
Hence, Equation (A14) can be written as:  
\[{\rm H}_{\rm ERB}=\frac{\tilde{\rm H}(0)}{2|{\rm H}({\rm f}_{0})|^{2}},\] (A17)  
Now, using Equation (A4), we can find the expression of \(\tilde{\rm h}({\rm t})\) as:  
\[ \begin{array}{ll}\tilde{\rm h}({\rm t})&=[{\rm c.r}({\rm t})c({\rm t})]^{2},\\ &={\rm c}^{2}{\rm r}^{2}({\rm t}){\rm c}^{2}({\rm t}),\end{array} \] (A18)  
\[ \begin{array}{ll}&={\rm c}^{2}\tilde{\rm r}({\rm t})\tilde{\rm c}({\rm t}). \end{array} \] (A19)  
\[ \begin{array}{ll}&={\rm t}^{2n-2}{\rm e}^{-4\pi{\rm bt}}{\rm u}({\rm t}). \end{array} \] (A20)  
\[ \begin{array}{ll}\tilde{\rm c}({\rm t})&={\rm c}^{2}({\rm t}),\\ &={\rm cos}^{2}(2\pi{\rm f}_{0}{\rm t}+\varphi).\end{array} \] (A21)  
By taking Fourier transform of both sides of Equation (A19), we can write:  
\[\tilde{\rm H}({\rm f})={\rm c}^{2}.\big{[}\tilde{\rm R}({\rm f})*\tilde{\rm C }({\rm f})\big{]}.\] (A22)  
where \(\tilde{\rm R}({\rm f})\) is the Fourier transform of \(\tilde{\rm r}({\rm t})\), and \(\tilde{\rm C}({\rm f})\) is the Fourier transform of \(\tilde{\rm c}({\rm t})\). Now, we need to find the Fourier transform of \(\tilde{\rm r}({\rm t})\) and \(\tilde{\rm c}({\rm t})\). The Fourier transform of \(\tilde{\rm r}({\rm t})\) can be found as:  
\[\tilde{\rm R}({\rm f})=(2n-2)!(4\pi{\rm b})^{-(2n-1)}\left[1+{\rm j}\frac{{ \rm f}}{2{\rm b}}\right]^{-(2n-1)}.\] (A23)  
The Fourier transform of \(\tilde{\rm c}({\rm t})\) can be expressed as:  
\[ \begin{array}{ll}\tilde{\rm C}({\rm f})=\left[\frac{1}{2}\delta({\rm f})+ \frac{1}{4}{\rm e}^{{\rm j}\varphi}\delta({\rm f}-2{\rm f}_{0})+\frac{1}{4}{ \rm e}^{-{\rm j}\varphi}\delta({\rm f}+2{\rm f}_{0})\right].\end{array} \] (A24)  
Finally, substituting \(\tilde{\rm R}({\rm f})\) and \(\tilde{\rm C}({\rm f})\) in Equation (A22), we find the expression of \(\tilde{\rm H}({\rm f})\) as:  
\[ \begin{array}{ll}\tilde{\rm H}({\rm f})={\rm c}^{2}(2n-2)!&(4\pi{\rm b})^{-( 2n-1)}\left[1+{\rm j}\frac{{\rm f}}{2{\rm b}}\right]^{-(2n-1)}\\ &\ast\left[\frac{1}{2}\delta({\rm f})+\frac{1}{4}{\rm e}^{{\rm j}\varphi} \delta({\rm f}-2{\rm f}_{0})+\frac{1}{4}{\rm e}^{-{\rm j}\varphi}\delta({\rm f }+2{\rm f}_{0})\right],\end{array} \] (A25)  
We can further simplify Equation (A25) as:  
\[ \begin{array}{ll}\tilde{\rm H}({\rm f})={\rm c}^{2}(2n-2)!(4\pi{\rm b})^{-( 2n-1)}\left\{\left[\frac{1}{2}\big{(}1+{\rm j}\frac{{\rm f}}{2{\rm b}}\big{)} \right]^{-(2n-1)}+\frac{1}{4}{\rm e}^{{\rm j}2\varphi}\left[1+{\rm j}\frac{({ \rm f}-2{\rm f}_{0})}{2{\rm b}}\right]^{-(2n-1)}+\frac{1}{4}{\rm e}^{-{\rm j} 2\varphi}\left[1+{\rm j}\frac{({\rm f}+2{\rm f}_{0})}{2{\rm b}}\right]^{-(2n-1 )}\right\},\end{array} \] (A26)  
By substituting \({\rm f}=0\) in Equation (A26), we can find the expression of \(\tilde{\rm H}(0)\) as:  
\[