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Suppose I have two time series $A:=[0,0,0,4,5,6]$ and $B:=[1,2,3,4,5,6]$. I implement the following filter (with initial value at the first element):

$$ f(t) = f(t-1) + \alpha*(f(t)-f(t-1))$$

with $\alpha \in [0,1]$.

Is there any way I can use filtered values of $B$ to recover the filtered values of $A$? For example suppose I can store all historical filtered values of $B$ for any choice of $\alpha$ in memory, is it possible to reconstruct what the filtered values for $A$ would have been for some choice of $\alpha$?

Intuitively this would involve subtracting the some filtered value(s) of $B$ at time step $3$ from the filtered value(s) of $B$ at time step $6$, to recover an approximation of $A$.

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Yes. Your filter $f(t)$ is an LTI system, so you can use that to your advantage. Let:

$$ C(t) = B(t) - A(t) $$

Then, if $y_A(t)$, $y_B(t)$, and $y_C(t)$ are the system's responses to $A(t)$, $B(t)$, and $C(t)$ as inputs, then you can straightforwardly do what you want:

$$ A(t) = B(t) - C(t) \rightarrow y_A(t) = y_B(t) - y_C(t) $$

So, pass $B(t)$ through the filter, then pass $C(t) = [1, 2, 3, 0, 0, 0]$ through the filter and find the difference between the two filter outputs. That's exactly what you would get if you passed $A(t)$ through the filter on its own.

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  • $\begingroup$ Obviously correct, but I just ask myself why one would want that. I mean if you have to filter two functions ($B(t)$ and $C(t)$) to get the response to $A(t)$, why not filter $A(t)$ directly? $\endgroup$ – Matt L. May 29 '14 at 17:36
  • $\begingroup$ Agreed; I'm not sure what the practical application of this would be. The OP might just be asking the wrong question; a more fundamental change of approach might be needed. $\endgroup$ – Jason R May 29 '14 at 17:52
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    $\begingroup$ i can't really decode the OP's question anyway. but something about this appears similar to the issue of Truncated IIR filters. see ccrma.stanford.edu/~jos/tiirts/tiirts.pdf and ccrma.stanford.edu/~jos/pdf/tiir.pdf . if you want an exponentially damped filter but you want it to totally forget about the input samples some $L$ samples ago, what you really have is an FIR, but like a moving average filter, this exponentially-damped FIR can be efficiently implemented as a truncated IIR filter. i have a pdf on how to design the coefficients if anyone is interested lemme know. $\endgroup$ – robert bristow-johnson May 29 '14 at 18:07

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