Full text: Proceedings of the Symposium "From Analytical to Digital" (Part 1)

  
  
  
Difficulties can also arise in connection with the inversion of the 
matrix H. Andrews and Hunt (1977) and Jansson (1984) mentioned 
problem as i) limitations in primary computer memory when H is very 
large, ii) if the inverse exists and is unique, it may be ill 
conditioned which may cause the term H^!1n in (4-4) to dominate the 
result, iii) roundoff error may become à serious problem. To iilu- 
strate the computer storage problem, consider the two-dimensional case. 
I£ g has the dimension N x N then the matrix H has the dimension N^ x 
N2,. .Thus, for.a rather small image the H | matrix becomes difficult to 
solve with conventional methods. Andrews and Hunt (1977) illustrate the 
problem with noise by deriving S/N between data g and solution fund 
g/N for the data is alpha and snr for the solution is beta then the 
ratio is defined as alpha/beta = norm(H 1) which is the Euclidian norm 
of the matrix Hi, Usually normQr 1) >> 1 which leads to the solution 
being dominated by noise. Again, this arises from the difficulties of 
solving Wi with regard to the above mentioned ill conditioning. 
Equation (4-2) can be solved using least square techniques. We have 
n=g-Hf (4-5) 
Minimize the norm of n. If H is square and has full rank 
€ = (qiTin-1gfg - iig (4-6) 
which has the same solution as the approximate inverse solution eq. 
(4-3). : 
4,1.2 Fourier Approach 
The Fourier transform of eq. (4-2) gives 
G(s) = H(8)F(g) + N(s) (4-7) 
As in the matrix approach, an approximate solution of the unknown 
undegraded image f may be obtained by an inverse computation. By 
multiplying G with a filter transfer function Y(s) - 1/H(s) we obtain 
F(s) = Y(s)G(s) = F(a) + Y(2)N(s) (4-8) 
which is an approximation of the original function. In optical science 
literature it is common to describe the inverse filtering procedure 
with the true solution on the left hand side of the equation. Equation 
(4-7) would then be written in the form F(e)=Y(s)G(s)-Y(sIN(8) = 
G(s)/H(s)-N(s)/H(S). For frequencies where H approaches zero the filter 
Y in eq. (4-8) becomes very large. This is usually the case for higher 
frequencies. Since the expected amplitude of the noise is independent 
of the frequency it follows that the signal dominates the spectrum at 
low frequencies and the noise dominates at high frequencies (Fig 4.1). 
That is one reason for using a filter which only restores frequencies 
with high S/N (Castleman 1979, Rosenfeld and Kak 1976) so that the 
second term in eq. (4-8) will never dominate the result. This suggests 
a filter similar to the filter Y but with a given cut-off frequency. On 
the basis of resolution it is desirable to have the cut-off frequency 
f. as large as possible. On the other hand as fc increases the filter Y 
yields a large additional noise. Assume that the object image f and 
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