Linear predictive coefficients of signal y are defined as the best $k$ coefficients $a_i, i = 1, \ldots, k$, that will approximate $y_n$ by $-\sum_{i=1}^k{a_iy_{n-i}}$. (Best approximation is that which minimizes the sum of squared errors.)

How are LPCs defined when this problem has multiple solutions, e.g. when y$$ is constant zero?

  • $\begingroup$ your example doesn't work: in case of a constant, only the first $a_1$ will be nonzero, the rest would be zero. Find a better example! $\endgroup$ – Marcus Müller Jun 9 '18 at 22:32
  • $\begingroup$ Can you clarify what you are asking for? A definition or methods to compute optimal coefficients? $\endgroup$ – Jonas Schwarz Jun 10 '18 at 19:17
  • $\begingroup$ Sorry, you are right, fixed it. I am asking for a definition. $\endgroup$ – seed Jun 11 '18 at 9:14
  • $\begingroup$ In a real case I believe that it is very unlikelly that you get a full sequence of zeros. Even if this happens, it is very easy to detect this situation. $\endgroup$ – Filipe Pinto Jun 26 '19 at 11:29
  • $\begingroup$ Filipe, agreed, I just did not immediately realize that full zero sequence is the only problematic one. $\endgroup$ – seed Jul 1 '19 at 9:52

Constant zero sequence is the only finite sequence for which $a_i$ are not uniquely defined.

Indeed, $a_i$ are a solution to the equation $$ \begin{bmatrix} R_0 && R_1 && \ldots && R_{k-1} \\ R_1 && R_0 && \ldots && R_{k-2} \\ \vdots && \vdots && \ddots && \vdots \\ R_{k-1} && R_{k-2} && \ldots && R_0 \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_k \end{bmatrix} = - \begin{bmatrix} R_0 \\ R_1 \\ \vdots \\ R_{k-1} \end{bmatrix} $$ To prove their uniqueness, it is sufficient to prove that the autocorrelation matrix R is nondegenerate.

If $y_0, y_1, \ldots, y_n$ is the initial sequence, $y_0 \ne 0$, and $$ Y = \begin{bmatrix} y_0 && \ldots && y_n && 0 && \ldots && 0 \\ 0 && y_0 && \ldots && y_n && \ldots && 0 \\ \vdots && \ddots && \ddots && \vdots && \ddots && \vdots \\ 0 && \ldots && 0 && y_0 && \ldots && y_n \end{bmatrix}, $$ then $$ R = YY^T, $$ so R is the Gram matrix of the rows of Y. A Gram matrix is nondegenerate iff the vectors are linearly independent, which they obviously are.

To deal with the constant zeros problem, I should probably just trim silence at the beginning and end of the audio.

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The least square solution is strictly convex. So you only have one minimum, although in cases, several combinations of inputs yield the same minimum.

Let us go back to a simpler situation: assume that you are given two numbers, $a$ and $b$, what is their best estimate? "Best" requires some more information:

  • How best (distance)?
  • What do variables relates (model)?
  • Where do $a$ and $b$ dwell (probability)?

If how best is a least-squares error loss function, a quadratic measure can be minimized. If the model is a polynomial function, its degree should be fixed). If $a$ and $b$ have known properties, they should be assumed.

The case of constant zero values renders most loss functions useless and reduces the space of variation to nil. In other words, if $a$ and $b$ are equal, and you don't did fancy behavior, one answer is, for any $\alpha$,

$$ \alpha a + (1-\alpha)b$$

but you could as well:

$$ (a^\alpha b ^ \beta)^{\frac{1}{\alpha+\beta}}$$

and many others, depending on the model you wish. So, the proper "best estimate" is either:

  • totally determined by your deterministic situation, and you ought to check input values and determine a predefined output based on logical testing (not based on stochastic optimization)
  • set by "other" rules, imposed on the model ( sparse model on, $\alpha$ or $\beta$ for instance), that can give you a more "singular" solution.

So, in your case, the simplest model could be a solution. So what's the simplest?

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