Determine $f(x)$ knowing $f(x)+f(x+\varepsilon)$

Question

I encountered a problem at work that, in my opinion, has a fundamental mathematical reasoning to determine its solvability.

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Due to an unwitting software configuration, my associate recorded the audio of our meeting twice, resulting in the audio sounding like the sum of the original signal combined with a slightly delayed copy of itself.

The final audio file can be thought as a time-dependent signal $g(t)$ that is related to the original audio signal, $f(t)$, by the following equation

$$g(t) = f(t) + f(t+\varepsilon)$$

Knowing $g(t)$ and $\varepsilon$ is it possible to approximate $f(t)$, either analytically or numerically?

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My primary intuition was that it would be possible, but now it feels like trying to unscramble two eggs after they were mixed in a bowl.

Answer 1

A possible avenue, mathematically, may be the Fourier transform.

Let $$ \newcommand{\FF}{\mathcal{F}} \newcommand{\dd}{\mathrm{d}} \FF[f](\xi) := \int_{-\infty}^\infty f(x) e^{-2\pi i \xi x} \, \dd x $$ denote the usual one-dimensional transformation for $f$.

We know two properties:

• $\FF$ is linear, in the sense that given $\alpha,\beta \in \mathbb{C}$, we have $\FF[\alpha f + \beta g] = \alpha \FF[f] + \beta \FF[g]$

• $\FF$ satisfies certain time-shift properties. In particular, $$ \FF[f(x-x_0)](\xi) = e^{2\pi i x_o \xi} \FF[f](\xi) $$

Consequently, if $g(x) = f(x) + f(x+\varepsilon)$, then $$\begin{align*} \FF[g](\xi) &= \FF[f(x) + f(x+\varepsilon)](\xi) \\ &= \FF[f](\xi) + e^{-2\pi i \varepsilon \xi} \FF[f](\xi) \\ &= \FF[f](\xi) \Big( 1 + e^{-2\pi i \varepsilon \xi} \Big) \end{align*}$$

Therefore, assuming we know $g,\varepsilon$, $$ \FF[f](\xi) = \frac{\FF[g](\xi) }{ 1 + e^{-2\pi i \varepsilon \xi}} $$ To recover $f$, one uses the similarly-defined inverse Fourier transform, $$ \FF^{-1}[f](x) := \int_{-\infty}^\infty f(\xi) e^{2\pi i \xi x} \, \dd \xi $$ or, more appropriately, $$ f(x) = \FF^{-1}\Big[ \FF[f] \Big](x) := \int_{-\infty}^\infty \FF[f](\xi) e^{2\pi i \xi x} \, \dd \xi $$

However, there are no guarantees it would be easy to calculate this inverse, and depending on the $\varepsilon,\xi$ involved it might not even be well-defined (say, if the exponential in the denominator evaluated to $-1$). Based on the other comments, issues may even arise for certain types of $f$.

Answer 2

Υοur second feelings are correct from a purely mathematical point of view. For a given $\varepsilon$, there are many functions $f$ such that $f(t) = - f(t + \varepsilon)$ for all $t$. For such an $f$, $g(t) = 0$ for all $t$, and so $g$ gives no information about $f$ apart from that fact that $f(t) = f(t + \varepsilon)$ for all $t$.

It may be possible to retrieve some information about $f$, if you make some assumptions about it based on the physics of the situation (which make my above contrived example highly unlikely). Probably some kind of machine learning approach would be a good place to start. MSE is probably not the best forum to ask if you are interested in following that up.

Answer 3

Assume that $f$ is (locally approximated by) a straight line, $f(t) = at + b$. This gives us:

$$f(t + \epsilon) = a(t + \epsilon) + b = at + b + a\epsilon$$ $$g(t) = 2at + 2b + a\epsilon$$

We now have two coefficients to solve for, so assume that we know $g(t_1)$ and $g(t_2)$, giving us the system of equations:

$$g(t_1) = (2t_1 + \epsilon)a + 2b$$ $$g(t_2) = (2t_2 + \epsilon)a + 2b$$

Solving for $a$ and $b$ gives:

$$a = \frac{g(t_2) - g(t_1)}{2(t_2 - t_1)}$$ $$b = \frac{(2t_2 + \epsilon)g(t_1) - (2t_1 + \epsilon)g(t_2)}{4(t_2 - t_1)}$$

And plugging these into the formula for $f$ gives:

$$f(t) = \frac{(2(t - t_1) - \epsilon)g(t_2) - (2(t - t_2) - \epsilon)g(t_1)}{4(t_2 - t_1)}$$

If we choose $t_1 = t$ and $t_2 = t + \epsilon$, then this simplifies to:

$$\boxed{f(t) = \frac{3g(t) - g(t + \epsilon)}{4}}$$

If $f$ is expressible as a Taylor series, then the maximum error in this approximation is $\frac{1}{4}\epsilon^2 |f''(t)|$.

Answer 4

You problem is one of deconvolution:

$$ g(t) = f(t) \ast \left( \delta(t) + \delta(t+\epsilon) \right) = f(t) \ast h(t), $$

where $h(t) = \delta(t) + \delta(t+\epsilon)$. In the frequency domain, this is

$$ G(\omega) = F(\omega) \cdot H(\omega). $$

Thus, to recover $f(t)$, we deconvolve $g(t)$ with $h(t)$:

$$ f(t) = \mathcal{F}^{-1} \left( { G(\omega) \over H(\omega) } \right), $$

where $\mathcal{F}$ is the inverse Fourier transform. For the system to be solvable, $H(\omega) \neq 0$.

There are better ways to do deconvolution that are more stable, but the above is the gist of it and where I would suggest you start.