I need help solving the following blurring function question.Assume an image $f(x,y)$ is moving in front of a camera so that $π‘₯_0(𝑑)$ and $𝑦_0(𝑑)$ are the time-varying components of motion in the x- and y- directions respectively. The camera’s shutter opens at 𝑑 = 0 and closes at 𝑑 = 𝑇. Assume that the shutter opening and closing operations are instantaneous. The blurred image 𝑔(π‘₯, 𝑦) captured by the camera is calculated as:

$$ g(x,y) = \int_{0}^{T} f( x - x_0(t), y - y_0(t)) dt $$

Calculate the blurring transfer function $H(u,v) = G(u,v) / F(u, v)$ in frequency domain with respect to $π‘₯_0(𝑑)$ and $𝑦_0(𝑑)$, where $𝐺(𝑒,𝑣)$ and $𝐹(𝑒,𝑣)$ are the Fourier transforms of $𝑔(π‘₯,𝑦)$ and $𝑓(π‘₯,𝑦)$ respectively and then calculate the blurring transfer function for the case where the image $𝑓(π‘₯, 𝑦)$ moves in only x- direction with a constant speed $\frac{a}{T}$.

Attempt: I really have no idea how to set up this blurring function.It is clear that the speed, $v$, is

$$ v = x_0(t)i + y_0(t)j $$

and also that

$$ \mathcal F \{ f(x - x_0, y - y_0) \} = \text{exp}( (-2\pi i (x_0u + y_0v) ) )F(u,v) $$

$$ G(u,v) = \mathcal F \{ g(x,y) \} = \mathcal F \left\{\int_{0}^{T} f( x - x_0(t), y - y_0(t))\right\} $$

but I don't know how to proceed from there.Any help appreciated


Your start looks right, although I think your speed equation is wrong. To continue, the FT has no time dependance so you can take the integral out.

$$G(u, v) = \int_0^T \mathcal F\{f(x-x_0, y-y_0)\} dt $$

$$ = F(u,v) \int_0^T e^{-2\pi i(x_0u + y_0v)} dt$$

Divide by $F$ to get the transfer function.

If I understand the question correctly $x_0(t)$ and $y_0(t)$ are positions such that $x_0(t) = x_0(0) + v_x(t)t$ and similar for $y$. You can then put in an appropriate version of the speed given and do the integration.

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