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Choice of Output Signal and Initial Conditions

The two most natural choices of output signal are

\begin{eqnarray*}
y_1(n) &=& [1, 0] \underline{x}(n) = [1, 0] \left[\begin{array}{cc} 1 & 1 \\ [2pt] \eta & -\eta \end{array}\right]\tilde{\underline{x}}(n)\\ [10pt]
&=& [1, 1]\tilde{\underline{x}}(n) = \lambda_1^n\,\tilde{x}_1(0) + \lambda_2^n\,\tilde{x}_2(0)\\ [20pt]
y_2(n) &=& [0, 1] \underline{x}(n) = [0, 1] \left[\begin{array}{cc} 1 & 1 \\ [2pt] \eta & -\eta \end{array}\right]\tilde{\underline{x}}(n)\\ [10pt]
&=& [\eta, -\eta]\tilde{\underline{x}}(n) = \eta\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{x}_2(0)\\
&=& e^{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 \tilde{\underline{x}}(0) = \mathbf{E}^{-1}\left[\begin{array}{c} 1 \\ [2pt] 0 \end{array}\right] = \frac{-1}{2e}\left[\begin{array}{cc} -e & -1 \\ [2pt] -e & 1 \end{array}\right]\left[\begin{array}{c} 1 \\ [2pt] 0 \end{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)
$

Similarly $ y_2(n)$ is proportional to $ \sin(\omega n T)$
(``phase quadrature'' output).


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``Introduction to State Space Models'', by Julius O. Smith III, (From Lecture Overheads, Music 420).
Copyright © 2014-03-24 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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