Hi, everyone. When I did a question on Wiener filter, I do not know how to deal with it.
The question is as follows.
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It is known that a certain object g(x,y) of size 256X256 is filtered by h(x,y) and then is corrupted by white additive Gaussian noise n(x,y) to become i(x,y). We observe i(x,y), and our goal is to recover g(x,y). We intend to use Wiener filter.
It is defined that Φ
g(f
x,f
y) is power spectral density of g(x,y) and Φ
n(f
x,f
y) is the power spectral density of the noise n(x,y) and assume that Φ
n(f
x,f
y)÷Φ
g(f
x,f
y) ≈ 0.1 at all frequencies.
In this particular problem, we have some prior measurement for h(x,y) and found that it can be represented as h(x,y) = k
1(exp[-k
2((x²+y²)]w(x,y), w(x,y) = 1 for -3 ≤ x ≤ 3 and equals 0 for other x and y.
In other word, h(x,y) is a 7X7 filter. Furthermore, the parameters k
1 and k
2 are real positive values and complete specify h(x,y). We require that Σ[x = -∞ to ∞]Σ[y = -∞ to ∞] h(x,y) = 1.
(a) Sketch the shape of W(f
x, 0) as a function of f
x for a very small value of k
2 (≈ 0).
(b) Sketch the shape of W(f
x, 0) as a function of f
x for a very large value of k
2 (≈ ∞).
(c) Suppose now that the imaging system has been changed: n(x,y) is also filtered by h(x,y) before corrupting the observation. Derive the Wiener filter (should be accurate for all values of k
2). How is it compared to the inverse filter of the original imaging system?
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Here's what have I thought or done.
In part (a) and part (b),
I think k
1 is the scaling factor only since it helps to satisfy Σ[x = -∞ to ∞]Σ[y = -∞ to ∞] h(x,y) = 1. So, if k
2 is very small, the effect of x and y will be larger than the case in which k
2 is very large from the h(x,y) equation in which h(x,y) is directly proportion to exp[-f(x,y)]. So, for small value of k
2, it behaves as a low pass filter and most high frequency components will not be filtered, a good filter. When k
2 is very small, it behaves as a high pass filter in which the Wiener filter cannot function well.
While I have totally no idea in part (c).
Thank you very much for reading such as long question. I am sorry about that. But I hope you can help me, thanks.
Thanks all!!