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Choice of Output Signal and Initial Conditions

Recalling that $ {\tilde x}= E\underline{{\tilde x}}$, the output signal from any diagonal state-space model is a linear combination of the modal signals. The two immediate outputs $ x_1(n)$ and $ x_2(n)$ in Fig.E.3 are given in terms of the modal signals $ {\tilde x}_1(n) = \lambda_1^n{\tilde x}_1(0)$ and $ {\tilde x}_2(n)=
\lambda_2^n{\tilde x}_2(0)$ as

\begin{eqnarray*}
y_1(n) &=& [1, 0] {\underline{x}}(n) = [1, 0] \left[\begin{arr...
...\lambda_1^n {\tilde x}_1(0) - \eta \lambda_2^n\,{\tilde x}_2(0).
\end{eqnarray*}

The output signal from the first state variable $ x_1(n)$ is

\begin{eqnarray*}
y_1(n) &=& \lambda_1^n\,{\tilde x}_1(0) + \lambda_2^n\,{\tilde...
...{j\omega n T} {\tilde x}_1(0) + e^{-j\omega n T}{\tilde x}_2(0).
\end{eqnarray*}

The initial condition $ {\underline{x}}(0) = [1, 0]^T$ corresponds to modal initial state

$\displaystyle \underline{{\tilde x}}(0) = E^{-1}\left[\begin{array}{c} 1 \\ [2p...
...nd{array}\right] = \left[\begin{array}{c} 1/2 \\ [2pt] 1/2 \end{array}\right].
$

For this initialization, the output $ y_1$ from the first state variable $ x_1$ is simply

$\displaystyle y_1(n) = \frac{e^{j\omega n T} + e^{-j\omega n T}}{2} = \cos(\omega n T).
$

A similar derivation can be carried out to show that the output $ y_2(n) = x_2(n)$ is proportional to $ \sin(\omega nT)$, i.e., it is in phase quadrature with respect to $ y_1(n)=x_1(n)$). Phase-quadrature outputs are often useful in practice, e.g., for generating complex sinusoids.


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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (August 2006 Edition).
Copyright © 2007-02-02 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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