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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.G.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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About the Author: 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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