State-Space Analysis of Waveguide Oscillator

We will now use state-space analysis^C.19[452] to determine Equations (C.154-C.157).

From Equations (C.149-C.153),

$$x_1(n+1) = c[g x_1(n) + x_2(n)] - x_2(n) = c\,g x_1(n) + (c-1) x_2(n)$$

and

$$x_2(n+1) = g x_1(n) + c[g x_1(n) + x_2(n)] = (1+c) g x_1(n) + c\,x_2(n)$$

In matrix form, the state time-update can be written

$$\left[\begin{array}{c} x_1(n+1) \\ [2pt] x_2(n+1) \end{array}\right] = \underbrace{\left[\begin{array}{cc} gc & c-1 \\ [2pt] g(c+1) & c \end{array}\right]}_\mathbf{A} \left[\begin{array}{c} x_1(n) \\ [2pt] x_2(n) \end{array}\right] \protect$$ (C.157)

or, in vector notation,

$$\underline{x}(n+1) = \mathbf{A}\underline{x}(n) + \mathbf{B}u(n)$$ (C.158)

$$y(n) = \mathbf{C}\underline{x}(n)$$ (C.159)

where we have introduced an input signal $u(n)$ , which sums into the state vector via the $2\times 1$ (or $2\times N_u$ ) vector $\mathbf{B}$ . The output signal is defined as the $1\times 2$ vector $\mathbf{C}$ times the state vector $\underline{x}(n)$ . Multiple outputs may be defined by choosing $\mathbf{C}$ to be an $N_y \times 2$ matrix.

A basic fact from linear algebra is that the determinant of a matrix is equal to the product of its eigenvalues. As a quick check, we find that the determinant of $A$ is

$$\det{\mathbf{A}} = c^2g - g(c+1)(c-1) = c^2g - g(c^2-1) = c^2g - gc^2+g = g. \protect$$ (C.160)

When the eigenvalues ${\lambda_i}$ of $\mathbf{A}$ (system poles) are complex, then they must form a complex conjugate pair (since $\mathbf{A}$ is real), and we have $\det{\mathbf{A}} = \vert{\lambda_i}\vert^2 = g$ . Therefore, the system is stable if and only if $\vert g\vert<1$ . When making a digital sinusoidal oscillator from the system impulse response, we have $\vert g\vert=1$ , and the system can be said to be ``marginally stable''. Since an undriven sinusoidal oscillator must not lose energy, and since every lossless state-space system has unit-modulus eigenvalues (consider the modal representation), we expect $\left\vert\det{A}\right\vert=1$ , which occurs for $g=1$ .

Note that $\underline{x}(n) = \mathbf{A}^n\underline{x}(0)$ . If we diagonalize this system to obtain $\tilde{\mathbf{A}}= \mathbf{E}^{-1}\mathbf{A}\mathbf{E}$ , where $\tilde{\mathbf{A}}=$   diag$[\lambda_1,\lambda_2]$ , and $\mathbf{E}$ is the matrix of eigenvectors of $\mathbf{A}$ , then we have

$$\tilde{\underline{x}}(n) = \tilde{A}^n\,\tilde{\underline{x}}(0) = \left[\begin{array}{cc} \lambda_1^n & 0 \\ [2pt] 0 & \lambda_2^n \end{array}\right]\left[\begin{array}{c} \tilde{x}_1(0) \\ [2pt] \tilde{x}_2(0) \end{array}\right]$$

where $\tilde{\underline{x}}(n) \isdef \mathbf{E}^{-1}\underline{x}(n)$ denotes the state vector in these new ``modal coordinates''. Since $\tilde{A}$ is diagonal, the modes are decoupled, and we can write

$$\begin{eqnarray*} \tilde{x}_1(n) &=& \lambda_1^n\,\tilde{x}_1(0)\\ \tilde{x}_2(n) &=& \lambda_2^n\,\tilde{x}_2(0) \end{eqnarray*}$$

If this system is to generate a real sampled sinusoid at radian frequency $\omega$ , the eigenvalues $\lambda_1$ and $\lambda_2$ must be of the form

$$\begin{eqnarray*} \lambda_1 &=& e^{j\omega T}\\ \lambda_2 &=& e^{-j\omega T}, \end{eqnarray*}$$

(in either order) where $\omega$ is real, and $T$ denotes the sampling interval in seconds.

Thus, we can determine the frequency of oscillation $\omega$ (and verify that the system actually oscillates) by determining the eigenvalues $\lambda_i$ of $A$ . Note that, as a prerequisite, it will also be necessary to find two linearly independent eigenvectors of $A$ (columns of $\mathbf{E}$ ).