# Blurring transfer function of image

I need help solving the following blurring function question.Assume an image $f(x,y)$ is moving in front of a camera so that $𝑥_0(𝑡)$ and $𝑦_0(𝑡)$ are the time-varying components of motion in the x- and y- directions respectively. The camera’s shutter opens at 𝑡 = 0 and closes at 𝑡 = 𝑇. Assume that the shutter opening and closing operations are instantaneous. The blurred image 𝑔(𝑥, 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 $𝑥_0(𝑡)$ and $𝑦_0(𝑡)$, where $𝐺(𝑢,𝑣)$ and $𝐹(𝑢,𝑣)$ are the Fourier transforms of $𝑔(𝑥,𝑦)$ and $𝑓(𝑥,𝑦)$ respectively and then calculate the blurring transfer function for the case where the image $𝑓(𝑥, 𝑦)$ 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

$$G(u, v) = \int_0^T \mathcal F\{f(x-x_0, y-y_0)\} dt$$
$$= F(u,v) \int_0^T e^{-2\pi i(x_0u + y_0v)} dt$$
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.