Take the 2-minute tour ×
Signal Processing Stack Exchange is a question and answer site for practitioners of the art and science of signal, image and video processing. It's 100% free, no registration required.

Suppose that there is a FIR filter F and a signal S. The filtered signal is the convolution of F and S, F * S.

The problem: how to calculate a signal S' such that F * S' = S' (the filtered version is the same as itself, up to a fixed time shift), and S' minimizes norm(S, S') (where norm is some kind of distance metric, for example sum of squared differences).

Loosely: how to calculate the most similar signal to a given signal which is not affected by a filter?

EDIT: The original problem may be over constrained because there may be very few (or no) signals for which F*S = S exactly. An alternate formulation is to find S' which minimizes the combined error alpha*norm( F*S' - S') + beta*norm( S' - S ), for some weights alpha and beta.

EDIT: The specific filter I have in this case is F=[ 0.00097656, -0.00976562, 0.06347656, -0.15625 , 0.18554688, 0.83203125, 0.18554688, -0.15625 , 0.06347656, -0.00976562, 0.00097656] which is a fairly modest lowpass filter. I was interested in the general case as well though.

share|improve this question
1  
A handwaving answer: It's easy to prove that $\mathcal{X} = \{S: F*S = S\}$ is a subspace. You are looking for a projection of an arbitrary signal onto $\mathcal{X}$. If you know more about F (eg. are there sinusoids that F filters out with gain 1 and zero phase shift?) then this projection simply the sum over appropriate inner products with such "unit-gain-no-phase-shift" signals. –  Atul Ingle Feb 10 '13 at 16:17
1  
Are you truly looking for $F * S' = S'$? Seems to me that that would be impossible, unless $F$ is the trivial system where its output always equal to its input. I wasn't sure if you were looking for something like $F * S' = S$ instead. –  Jason R Feb 10 '13 at 19:12
2  
If it's an LTI system, then any pure sinusoid not at a zero or pole in the frequency response will be unchanged except in amplitude and phase. If the frequency response of the FIR filter does not include a point with a gain of 1.0, then a solution does not exist. –  hotpaw2 Feb 10 '13 at 21:06
add comment

1 Answer

If a filter with transfer function $H(f)$ passes a signal $x(t)$ with Fourier transform $X(f)$ unchanged, then $H(f)X(f) = X(f)$ for all $f$, and so $H(f)$ must have the property that $$H(f) = 1 ~ \text{for all frequencies}~ f~ \text{for which}~ X(f) \neq 0.$$ Thus, an arbitrary filter will not enjoy this property for very many signals, perhaps for no signal at all. Indeed, if $H(f_0) = H(-f_0) = 1$ for some $f_0$, then the filter will pass the sinusoid $A\cos(2\pi f_0t + \theta), -\infty < t < \infty$ unchanged, but it will not pass a sinusoidal pulse such as $$x(t) = \begin{cases} A\cos(2\pi f_0t + \theta), & -T < t < T,\\ 0, &\text{otherwise,}\end{cases}$$ without distorting it. Of course, it is possible that $H(f) \neq 1$ for all $f$ and so even passing a monochromatic sinusoid is not possible (cf. @hotpaw's comment). In short, the subspace referred to by @AtulIngle might well be the trivial subspace of zero dimension containing only the zero signal. It is also worth noting that if $H(f)$ has constant nonzero value over an interval, then the filter is physically unrealizable for essentially the same reason that an ideal low-pass filter is unrealizable.

So, the OP's question is likely to have no answer at all because the signals $S^\prime$ that he seeks to find may not exist at all for the given filter. Perhaps something else was intended to be asked?

share|improve this answer
    
Please modified version of problem above, to minimize both filter-induced distortion (which will likely be nonzero for the reasons you explain) and the difference from original signal. –  Alex I Feb 11 '13 at 0:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.