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Appendix A: Polynomial form of PWD

If all the coefficients of $B(Z_t)$ are polynomials of $p$, equation 4 is also a polynomial of $p$, and the plane-wave destruction equation becomes in turn a polynomial equation of $p$. The problem is to design a $2N+1$ points filter $B(Z_t)$ with polynomial coefficients such that the allpass system $H(Z_t)=\frac{B(1/Z_t)}{B(Z_t)}$ can approximate the phase-shift operator $Z_t^p=e^{j\omega p}$. Denoting the phase response of the system as $\theta(\omega)$, that is $H(e^{j\omega})=e^{j\theta(\omega)}$, the group delay of the system is

\begin{displaymath}
\tau(\omega)=\frac{\partial \theta(\omega)}{\partial \omega}.
\end{displaymath} (24)

The maximally flat criteria designs a filter with a smoothest phase response. There are $2N$ unknown coefficients in $H(Z_t)$, so we can add $2N$ flat constraints for the first $2N-$th order deviratives of the phase response. It becomes (Zhang, 2009, equation 7)
\begin{displaymath}
\left\{\begin{array}{l}
\tau(\omega)=p \\
\displaystyle{\fr...
...partial \omega^n}}=0
\qquad n=1,2,\dots,2N
\end{array}\right.,
\end{displaymath} (25)

which is equivalent to the following linear maximally flat conditions (Thiran, 1971):
\begin{displaymath}
\sum_{k=-N}^N (d-k)^{2n+1}b_k =0,
\end{displaymath} (26)

where $n=0,1,\dots,2N-1$ and $d=p/2$ is the fractional delay of $B(1/Z_t)$ or $1/B(Z_t)$.

In order to solve $b_k$ from the above equations, Thiran (1971) used an additional condition $b_0=1$, which leads $b_k(k\neq 0)$ to be a fractional function of $p$. Differently from that, we use the following condition,

\begin{displaymath}
\sum_{k=-N}^N b_k =1,
\end{displaymath} (27)

where $b_k$ can be proved to be polynomials of $p$.

Let vector $\mathbf b=[b_0,b_N,\dots,b_1,b_{-1},\dots,b_{-N}]^\textrm T$. Combining equations A-3 and A-4, we rewrite them into the following matrix form:

\begin{displaymath}
\left[\begin{array}{c\vert cccccc}
1 & 1 & \dots & 1 & 1 & \...
...begin{array}{c}
1 \\ 0 \\ 0 \\ \vdots \\ 0
\end{array}\right].
\end{displaymath}

The matrix on the left side, denoted as $\tensor V$, can be split into four blocks $\left[\begin{array}{c\vert c}
\tensor A & \tensor B \\ \hline \tensor C & \tensor D
\end{array}\right]$ as shown above. Following the lemma of matrix inversion,

\begin{displaymath}
\tensor V^{-1}=\left[\begin{array}{cc}
(\tensor A-\tensor B\...
...\tensor C)^{-1}\tensor B\tensor D^{-1} \\
\end{array}\right],
\end{displaymath} (28)

therefore the coefficients
\begin{displaymath}
\mathbf b=\tensor V^{-1}[1,0,\dots,0]^\textrm T=
\left[\begi...
... A-\tensor B\tensor D^{-1}\tensor C)^{-1}
\end{array}\right].
\end{displaymath} (29)

Let subindex $i=-N,-N+1,\dots,-1,1,2,\dots,N$ and $x_i=d+i$. Submatrix $\tensor D$ can be expressed as

\begin{eqnarray*}
\tensor D &=& \tensor E\tensor X \\
&=&
\left[\begin{array}{c...
...}\\ \vdots \\ x_{-1} \\ x_1 \\ \vdots \\ x_N
\end{array}\right],
\end{eqnarray*}

so $\tensor D^{-1}=\tensor X^{-1}\tensor E^{-1}$. Denoting $\tensor U=\tensor E^{-1}$ with elements $u_{ij},~j=1,2,\dots,2N$, as $\tensor E$ is a Vandermonde matrix, $u_{ij}$ and Lagrange intepolating polynomials have the following relationship:
\begin{displaymath}
\sum_{j=1}^{2N}u_{ij}x^{2j-2}=\ell_i(x),
\end{displaymath} (30)

where $i=-N,\dots,-1,1,\dots,N$, and $\ell_i(x)$ is the Lagrange polynomial related to the basis $d+i$,
\begin{displaymath}
\ell_i(x)=
\prod_{-N\leq m\leq N}^{m\neq i,m\neq 0}
\frac{x^2-(d+m)^2}{(d+i)^2-(d+m)^2}.
\end{displaymath} (31)

Substituting the above equation, $u_{ij}$ and $x$ into equation A-7, we can prove equation A-7. It follows that

\begin{displaymath}[\tensor E^{-1}\tensor C]_i=d\ell_i(d),
\end{displaymath} (32)


\begin{displaymath}[\tensor D^{-1}\tensor C]_i=
[\tensor X^{-1}\tensor E^{-1}\tensor C]_i=
\frac{d}{d+i}\ell_i(d),
\end{displaymath} (33)

with
\begin{displaymath}
\ell_i(d)=\frac{(-1)^{i+1}N!N!}{(N+i)!(N-i)!}
\frac{d+i}{d}
\prod_{m=-N}^N \frac{2d+m}{2d+m+i}.
\end{displaymath} (34)

Thus hence

$\displaystyle \tensor A-\tensor B\tensor D^{-1}\tensor C$ $\textstyle =$ $\displaystyle \sum_{i=-N}^N
(-1)^{i}\frac{N!N!}{(N+i)!(N-i)!}
\prod_{m=-N}^N \frac{p+m}{p+m+i}$  
  $\textstyle =$ $\displaystyle \frac{(4N)!N!N!}
{(2N)!(2N)!}
\frac{1}{
\displaystyle{\prod_{m=N+1}^{2N}(m^2-p^2)}}$ (35)

and
\begin{displaymath}[\tensor A-\tensor B\tensor D^{-1}\tensor C]^{-1} =
\frac{(2N)!(2N)!}{(4N)!N!N!}
\prod_{m=N+1}^{2N}(m^2-p^2).
\end{displaymath} (36)

It is the coefficient $b_0$, a $2N$-th degree polynomial of $p$. Substituting it into equation A-6, the coefficients at $k=\pm 1,\pm 2,\dots \pm N$ are expressed as

$\displaystyle b_k$ $\textstyle =$ $\displaystyle -[\tensor D^{-1}\tensor C]_k
[\tensor A-\tensor B\tensor D^{-1}\tensor C]^{-1}$  
  $\textstyle =$ $\displaystyle \frac{(2N)!(2N)!}{(4N)!(N+k)!(N-k)!}
\prod_{m=0}^{N-1-k}(m-2N+p)
\prod_{m=0}^{N-1+k}(m-2N-p).$ (37)

With the additional condition A-4 in $2N+1$ points approximation, all the coefficients are polynomials of $p$ of $2N$-th degree. Thus the plane-wave destruction equation 6 therefore is proved to be a polynomial equation of $2N$-th degree.


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Next: Bibliography Up: Chen, Fomel & Lu: Previous: Acknowledgments

2013-07-26