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Introduction to Digital Filters
    Matrix Filter Representations
       State Space Realization
          State Space Filter Realization Example

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State Space Filter Realization Example

The digital filter having difference equation

$\displaystyle y(n) = u(n-1) + u(n-2) + 0.5\, y(n-1) - 0.1\, y(n-2) + 0.01\, y(n-3) $

can be realized in state-space form as follows:F.5
$\displaystyle \left[\begin{array}{c} x_1(n+1) \\ [2pt] x_2(n+1) \\ [2pt] x_3(n+1)\end{array}\right]$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 0 & 1 & 0\\ [2pt] 0 & 0 & 1\\ [2pt] 0.01... ...\right] + \left[\begin{array}{c} 0 \\ [2pt] 0 \\ [2pt] 1\end{array}\right] u(n)$  
$\displaystyle \underline{y}(n)$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 0 & 1 & 1\end{array}\right] \left[\begin{array}{c} x_1(n) \\ [2pt] x_2(n) \\ [2pt] x_3(n)\end{array}\right] \protect$ (F.5)

Thus, $ {\underline{x}}(n) = [x_1(n), x_2(n), x_3(n)]^T$ is the vector of state variables at time $ n$, $ B = [0,0,1]^T$ is the state-input gain vector, $ C = [0,1,1]$ is the vector of state-gains for the output, and the direct-path gain is $ D=0$.

This example is repeated using matlab in §G.7.8 (after we have covered transfer functions).

A general procedure for converting any difference equation to state-space form is described in §G.7. The particular state-space model shown in Eq.$ \,$(F.5) happens to be called controller canonical form, for reasons discussed in Appendix G. The set of all state-space realizations of this filter is given by exploring the set of all similarity transformations applied to any particular realization, such as the control-canonical form in Eq.$ \,$(F.5). Similarity transformations are discussed in §G.8, and in books on linear algebra [58].

Note that the state-space model replaces an $ N$th-order difference equation by a vector first-order difference equation. This provides elegant simplifications in the theory and analysis of digital filters. For example, consider the case $ B=C=I$, and $ D=0$, so that Eq.$ \,$(F.4) reduces to

$\displaystyle \underline{y}(n+1) = A \underline{y}(n) + \underline{u}(n), \protect$ (F.6)

where $ A$ is the $ N\times N$ transition matrix, and both $ \underline{u}(n)$ and $ \underline{y}(n)$ are $ N\times 1$ signal vectors. (This filter has $ N$ inputs and $ N$ outputs.) This vector first-order difference equation is analogous to the following scalar first-order difference equation:

$\displaystyle y(n+1) = a y(n) + u(n) $

The response of this filter to its initial state $ y(0)$ is given by

$\displaystyle y(n) = a^n y(0), \quad n=0,1,2,3,\ldots\,. $

(This is the zero-input response of the filter, i.e., $ u(n)\equiv 0$.) Similarly, setting $ \underline{u}(n)=0$ to in Eq.$ \,$(F.6) yields

$\displaystyle \underline{y}(n) = A^n \underline{y}(0), \quad n=0,1,2,3,\ldots\,. $

Thus, an $ N$th-order digital filter ``looks like'' a first-order digital filter when cast in state-space form.

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Previous: State Space Realization
Next: Time Domain Filter Estimation
written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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