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

$$\left[ \begin{array}{c} y_0 \\ [2pt] y_1 \\ [2pt] y_2 \end{array}\right] = \left[ \begin{array}{ccc} h_{00} & h_{01} & h_{02} \\ [2pt] h_{10} & h_{11} & h_{12} \\ [2pt] h_{20} & h_{21} & h_{22} \end{array}\right] \left[ \begin{array}{c} x_0 \\ [2pt] x_1 \\ [2pt] x_2 \end{array}\right].$$

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

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

Consider the non-causal time-varying filter defined by

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

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

$$\begin{eqnarray*} C_3(0)&=&\frac{1}{3}\left[ \begin{array}{ccc} 1 & 1 & 1 \\ [2pt] 1 & 1 & 1 \\ [2pt] 1 & 1 & 1 \end{array}\right], \\ [5pt] C_3(1)&=&\frac{1}{3}\left[ \begin{array}{ccc} 1 & e^{j2\pi\frac{1}{3}} & e^{j2\pi\frac{2}{3}} \\ [2pt] 1 & e^{j2\pi\frac{1}{3}} & e^{j2\pi\frac{2}{3}} \\ [2pt] 1 & e^{j2\pi\frac{1}{3}} & e^{j2\pi\frac{2}{3}} \end{array}\right] =\frac{1}{3}\left[ \begin{array}{ccc} 1 & e^{j\frac{2\pi}{3}} & e^{-j\frac{2\pi}{3}} \\ [2pt] 1 & e^{j\frac{2\pi}{3}} & e^{-j\frac{2\pi}{3}} \\ [2pt] 1 & e^{j\frac{2\pi}{3}} & e^{-j\frac{2\pi}{3}} \end{array}\right],\\ [5pt] C_3(2)&=&\frac{1}{3}\left[ \begin{array}{ccc} 1 & e^{j2\pi\frac{2}{3}} & e^{j2\pi\frac{4}{3}} \\ [2pt] 1 & e^{j2\pi\frac{2}{3}} & e^{j2\pi\frac{4}{3}} \\ [2pt] 1 & e^{j2\pi\frac{2}{3}} & e^{j2\pi\frac{4}{3}} \end{array}\right] =\frac{1}{3}\left[ \begin{array}{ccc} 1 & e^{-j\frac{2\pi}{3}} & e^{j\frac{2\pi}{3}} \\ [2pt] 1 & e^{-j\frac{2\pi}{3}} & e^{j\frac{2\pi}{3}} \\ [2pt] 1 & e^{-j\frac{2\pi}{3}} & e^{j\frac{2\pi}{3}} \end{array}\right]. \end{eqnarray*}$$

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

$$x_0\isdef \left[ \begin{array}{c} 1 \\ [2pt] 1 \\ [2pt] 1 \end{array}\right],\qquad x_1\isdef \left[ \begin{array}{c} 1 \\ [2pt] e^{j\frac{2\pi}{3}} \\ [2pt] e^{-j\frac{2\pi}{3}} \end{array}\right],\qquad x_2\isdef \left[ \begin{array}{c} 1 \\ [2pt] e^{-j\frac{2\pi}{3}} \\ [2pt] e^{j\frac{2\pi}{3}} \end{array}\right].$$

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

$$C_3(k)X = C_3(k)\sum_{i=0}^2\alpha_ix_i = \sum_{i=0}^2\alpha_iC_3(k)x_i = \alpha_k\left[ \begin{array}{c} 1 \\ [2pt] 1 \\ [2pt] 1 \end{array}\right].$$

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

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