Recently i watched a video which talks about noise and filtering.That why i am posting this.After searching the same in wiki, i didn't see a detailed derivation of the formula. so I thought to write this and post.
Wiener deconvolution is a frequency domain technique which helps to remove the noise if we know the degradation function(H(t)) in advance. The degradation function degrades the image quality. It can be blurring like Gaussian blur or it can be motion blur.
X(x,y) -> Passes through degradation function -> Some noise gets added here(N) -> our final signal Y(x,y)
If we multiply the Y with Wiener filter then it will provide an approximation of X. Lets call this approximation as X$ .Now am going to write down the full derivation. a part of this is available in wiki. But those who are not remembering complex number maths may find it difficult.
Mean square error between approximation and original data is = E | X(x,y) - X$(x,y) | ^2
but we know X$ is can be obtained by G(x,y) * Y(x,y) [ * is convolution operator ]
= E( | X(x,y) - G(x,y) * Y(x,y) |^2)
But if convert all to frequency domain convolution becomes multiplication, ie a(t)*b(t) = a(f)b(f)
Also we know Y is X * H + N
= E(| X - G ( XH + N)) |^2)
= E(| X - GXH - GN) |^2)
= E(| (1 - G H)X - GN) |^2)
if A and B are two complex numbers | A B |^2 = (A B)(AB)* , here * is complex conjugate.
= E( ((1-GH)X - GN)((1-GH)X - GN)* )
= E( ((1-GH)X - GN)((1-G*H*)X* - G*N*)) , here all * means complex conjugate
Expanding this is a painful thing. but i am not giving up. So are you ?
= E( ((1-GH)X(1-GH)* XX* - (1-GH)XG*N* - GN(1-GH)*X* + GNG*N* )
= ( (1-GH)(1-GH)* E{|X|^2} - (1-GH)G* E{XN*} - G(1-GH)* E{NX*} + GG* E{|N|^2}
wooha,
Lets assume noise and input signal has no relation,That why it is noise. So we can just ignore the terms E{XN*} ,E{NX*}.I think this is because at a particular frequency we can't find any relation between input data and noise. So the probability is null here. That why taking E( expectation) which sums the probability of values. here since we can't find any probability we can assume it 0.
We can denote S(f) = E{|X|^2} , it is the power spectrum of the complex values of the input data
We can denote N(f) = E{|N|^2} , it is the power spectrum of the complex values of the noise data
so final equation becomes
minError(f) = (1-GH)(1-GH)* S + GG*N
Now we need to find the minimum error which we can obtain for G. So following the basic calculus (like how we will find the min value of a function ) . careful about G* ,it is not G. So we can treat that as constant.
d(minError(f)/dG = -H(1-GH)*S + G*N = 0 ,
Now we need to find G from this. we will find it in a minute.
That is G*N - H[1-GH]*S = 0
G*N = HS[1-GH]*
G*N = HS[1-G*H*]
G*N = S[H-G*HH*]
G*N = S[H-G*|H|^2 ]
G*N = SH-SG*|H|^2 ]
G*N = (G*) [ (1/G*)SH-S|H|^2 ]
N = [ (1/G*)SH - S|H|^2 ]
N + S|H|^2 = SH/G*
G*/SH = 1/ (N+ S|H|^2 )
G* = SH / ( N+ S HH* ). Take out s
G* = SH / S( N/S+ HH* )
N/S is noise to signal ratio, We can replace it with a constant K
G* = H / (K+ HH* )
G = H*/ (K+H*H), H*H is |H|^2|
G = H*/ (K+|H|^2)
Yes we reached the final answer , G = H*/(K+|H|^2).
if you know H, then you can find G, Then just multiply this with Fourier transformed input data. you will get a better picture. Thus you can remove some noise from the input data. In next post i will try to post some samples after applying G or wiener filter. Be ready for it.
![X_k =
\sum_{n=0}^{N-1} x_n \cos \left[\frac{\pi}{N} \left(n+\frac{1}{2}\right) k \right] \quad \quad k = 0, \dots, N-1.](http://upload.wikimedia.org/math/9/b/9/9b912066d54e929387a28251790b9c4e.png)
