Finding the Eigenstructure of A Next  |  Prev  |  Up  |  Top  |  Index  |  JOS Index  |  JOS Pubs  |  JOS Home  |  Search

### Finding the Eigenstructure of A

Starting with the defining equation for an eigenvector and its corresponding eigenvalue ,

we get

 (E.23)

We normalized the first element of to 1 since is an eigenvector whenever is. (If there is a missing solution because its first element happens to be zero, we can repeat the analysis normalizing the second element to 1 instead.)

Equation (E.23) gives us two equations in two unknowns:

 (E.24) (E.25)

Substituting the first into the second to eliminate , we get

Thus, we have found both eigenvectors

They are linearly independent provided and finite provided .

We can now use Eq. (E.24) to find the eigenvalues:

Assuming , the eigenvalues are

 (E.26)

and so this is the range of corresponding to sinusoidal oscillation. For , the eigenvalues are real, corresponding to exponential growth and decay. The values yield a repeated root (dc or oscillation).

Let us henceforth assume . In this range is real, and we have , . Thus, the eigenvalues can be expressed as

Equating to , we obtain , or , where denotes the sampling rate. Thus the relationship between the coefficient in the digital waveguide oscillator and the frequency of sinusoidal oscillation is expressed succinctly as

We see that the coefficient range (-1,1) corresponds to frequencies in the range , and that's the complete set of available digital frequencies.

We have now shown that the system of Fig.E.3 oscillates sinusoidally at any desired digital frequency rad/sec by simply setting , where denotes the sampling interval.

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