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Derivation

For notational simplicity, we restrict exposition to the three-dimensional case. The general linear digital filter equation $ Y=HX$ is written in three dimensions as

\begin{displaymath} \left[ \begin{array}{c} y_0 \\ [2pt] y_1 \\ [2pt] y_2 \end{a... ...in{array}{c} x_0 \\ [2pt] x_1 \\ [2pt] x_2 \end{array}\right]. \end{displaymath}

where $ x_i$ is regarded as the input sample at time $ i$, and $ y_i$ is the output sample at time $ i$. The general causal time-invariant filter appears in three-space as

\begin{displaymath} H=\left[ \begin{array}{ccc} h_0 & 0 & 0 \\ [2pt] h_1 & h_0 & 0 \\ [2pt] h_2 & h_1 & h_0 \end{array}\right]. \end{displaymath}

Consider the non-causal time-varying filter defined by

\begin{displaymath} C_3(k)={1/3}\left[ \begin{array}{ccc} 1 & W_3^1(k) & W_3^2(k... ...3^2(k) \\ [2pt] 1 & W_3^1(k) & {W_3^2(k)} \end{array}\right]. \end{displaymath}

We may call $ C_3(k)$ the collector matrix corresponding to the $ k^{th}$ frequency.We have

\begin{eqnarray*} C_3(0)&=&\frac{1}{3}\left[ \begin{array}{ccc} 1 & 1 & 1 \\ [2p... ... e^{-j\frac{2\pi}{3}} & e^{j\frac{2\pi}{3}} \end{array}\right]. \end{eqnarray*}

The top row of each matrix is recognized as a basis function for the order three DFT (equispaced vectors on the unit circle). Accordingly, we have the orthogonality and spanning properties of these vectors. So let us define a basis for the signal space $ \{x_0,x_1,x_2\}$ by

\begin{displaymath} x_0\isdef \left[ \begin{array}{c} 1 \\ [2pt] 1 \\ [2pt] 1 \e... ...rac{2\pi}{3}} \\ [2pt] e^{j\frac{2\pi}{3}} \end{array}\right]. \end{displaymath}

Then every component of $ C_3(k)x_k = 1$ and every component of $ C_3(k)x_j=0$ when $ k\neq j$. Now since any signal $ X$ in $ \Re ^3$ may be written as a linear combination of $ \{x_1,x_2,x_3\}$, we find that

\begin{displaymath} C_3(k)X = C_3(k)\sum_{i=0}^2\alpha_ix_i = \sum_{i=0}^2\alp... ...[ \begin{array}{c} 1 \\ [2pt] 1 \\ [2pt] 1 \end{array}\right]. \end{displaymath}

Consequently, we observe that $ C_N(k)$ is a matrix which annihilates all input basis components but the $ k^{th}$. Now multiply $ C_N(k)$ on the left by a diagonal matrix $ D(k)$ so that the product of $ D(k)$$ C_N(k)$ times $ x_k$ gives an arbitrary column vector $ (d_1,d_2,d_3)$. Then every linear time-varying filter $ G$ is expressible as a sum of these products as we will show below. In general, the decomposition for every filter on $ \Re ^N$ is simply

$\displaystyle G=\sum_{k=0}^{N-1}D(k)C_N(k). \protect$ (H.1)

The uniqueness of the decomposition is easy to verify: Suppose there are two distinct decompositions of the form Eq.$ \,$(H.1). Then for some $ k$ we have different D(k)'s. However, this implies that we can get two distinct outputs in response to the $ k^{th}$ input basis function which is absurd.

That every linear time-varying filter may be expressed in this form is also easy to show. Given an arbitrary filter matrix of order N, measure its response to each of the N basis functions (sine and cosine replace $ e^{j\omega t}$) to obtain a set of N by 1 column vectors. The output vector due to the $ k^{th}$ basis vector is precisely the diagonal of $ D(k)$.

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