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).