1
$\begingroup$

Let $x[n]$ be a time-series, and two filters:

  • $A[n] = a_0 x[n] + a_1 x[n-1] + ... a_q x[n-q]$
  • $B[n] = b_0 x[n] + b_1 x[n-1] + ... b_r x[n-r]$

I think the answer is No, but is there a way to have this problem (sign of 2 filters should be both positive):

$$A[n] > 0 \quad AND \quad B[n] > 0$$

expressed in terms of the study of the sign of a single linear filter

$$C[n] = c_0 x[n] + c_1 x[n-1] + ... c_s x[n-s]$$

?

If not possible, is there a way to have nearly equivalence (that holds for most values of $n$)?


Application:

A[n] = x[n] - x[n-1]                 # derivative  (or a smoothed version with moving average)
B[n] = x[n] - 2 * x[n-1] + x[n-2]    # acceleration (idem)

I'd like to find when both derivative and acceleration of $x[n]$ are positive by studying the sign of only 1 linear filter.

Or said in another way: is it possible to create a linear filter $C[n] = c_0 x[n] + c_1 x[n-1] + ... c_s x[n-s]$ such that $C[n] > 0$ if and only if derivative and acceleration of $x[n]$ are both positive?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.