Restoration by Spatial Differentiation

To simplify the problem we assume:

• The image is blurred by linear motion:
  
  $$g(x)=\int_0^Tf(x-x_d(t)) dt = \int_0^T f(x-vt) dt$$
  
  
  where $v$ is the constant speed of the motion and $L=vT$ is the distance traveled during the exposure time $T$.
• The width of the image $W$ is a multiple of $L$:
  
  $$W=KL$$
  
  

We next introduce a new variable $x'=x-vt$, and have $t=(x-x')/v$ and $dt=-dx'/v$. Moreover, the integral limits $0$ and $T$ for $t$ become, respectively, $x$ and $x-vT=x-L$ for $x'$. Now the image becomes

$$g(x)=\int_0^T f(x-vt) dt = -\frac{1}{v} \int_x^{x-L} f(x') dx' =\frac{1}{v}[F(x)-F(x-L)]$$

where

$$F(x)=\int f(x') dx'$$

For convenience, we will ignore the constant factor $1/v$.

As the motion distortion is essentially an integration $g(x)=\int f(x') dx'$, to restore $f(x)$ from $g(x)$, we can simply differentiate $g(x)$:

$$g'(x)=\frac{d}{dx}g(x)=f(x)-f(x-L)$$

and restore the original signal $f(x)$ as

$$f(x)=g'(x)+f(x-L) \;\;\;\;\;\;\;\mbox{for $0 \leq x \leq L$}$$

Note that above equation only recovers $f(x)$ inside the interval $0 \leq x \leq L$.

To recover the rest of $f(x)$, we replace $x$ by $x+mL$ for $m=0,1,\cdots,K-1$ and apply the above relationship recursively

$$f(x+mL) = g'(x+mL)+f(x+(m-1)L)$$

$$= g'(x+mL)+g'(x+(m-1)L)+f(x+(m-2)L)$$

$$= \cdots \cdots$$

$$= g'(x+mL)+g'(x+(m-1)L)+\cdots+g'(x)+f(x-L)$$

$$= \sum_{i=0}^m g'(x+iL)+f(x-L) \;\;\;\;\;\;\;\;\;\;(m=0,1,\cdots,K-1)$$

Here $f(x-L)$ represents the segment of signal of length $L$ that moves from outside the image into the image during the exposure time $T$. If $f(x-L)$ is known, for example, if we can assume $f(x-L)=constant$ (e.g., uniform background), then the original signal $f(x)$ over the entire interval $0 \leq x \leq KL=W$ can be obtained by evaluating the above equation at $0 \leq x \leq L$ for all $m=0,1,\cdots,K-1$.

However, if we cannot assume $f(x-L)=constant$, it need be estimated. As the above equation is valid for $m=0,1,\cdots,K-1$, we actually have $K$ equations which can be added up to give

$$\sum_{m=0}^{K-1} f(x+mL) = \sum_{m=0}^{K-1} \sum_{i=0}^m g'(x+iL) +\sum_{m=0}^{K-1} f(x-L)$$

which can be solved for $f(x-L)$

$$f(x-L)=\frac{1}{K}\sum_{m=0}^{K-1}f(x+mL)- \frac{1}{K}\sum_{m=0}^{K-1} \sum_{i=0}^m g'(x+iL)$$

The first term on the right is an average of $f(x)$ over the entire range of the image and can be estimated by the average of $g(x)$.