What Is the Fourier Transform of the Unblurred Image Intensity?

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Homework Help Overview

The discussion revolves around finding the Fourier transform of an unblurred image intensity, I_0(x), given a blurred image intensity I(x) that is affected by a point spread function S(x). The Fourier transform of the blurred image intensity is provided, and participants are exploring the relationships between the Fourier transforms of I(x), I_0(x), and S(x).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the convolution theorem to relate the Fourier transforms of I(x) and S(x) to I_0(x). There are attempts to clarify the definitions and relationships between I(k) and F[I(x)]. Some participants express confusion about the calculations and the correct forms of the Fourier transforms.

Discussion Status

The discussion is ongoing, with participants attempting to clarify their understanding of the Fourier transforms involved. Some have provided calculations for F[I] and F[S], while others question the correctness of these calculations and the assumptions made. There is no explicit consensus on the correctness of the approaches taken so far.

Contextual Notes

Participants note that the problem involves specific forms of the Fourier transforms and that assumptions about the functions involved may need to be revisited. There is also mention of potential errors in the calculations of F[S] and the need for clarity regarding the definitions of the functions being transformed.

Hart
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Homework Statement



I(x) is the intensity of an image after passing through a material which
blurs each point according to a point spread function given by:

S\left(x'-x\right)=e^{-a\left|x'-x\right|}

The Fourier transform of I(x) is given by:

I(k) = \frac{A}{\left( a^{2}+k^{2} \right) \left( b^{2}+k^{2} \right)}

Where A is a constant.

(i) Find the Fourier transform I_{0}^{~} of the unblurred image intensity.

(ii) Hence find the original unblurred image intensity I_{0}x


Homework Equations



I(x') = \int_{-\infty}^{\infty}I_{0}(x)S\left(x'-x\right)dx = \left(I_{0}*S \right)(x')

The Attempt at a Solution



F[I_{0}] = \frac{1}{\sqrt{2 \pi}}\left( \frac{F<i>}{F<s>} \right)</s></i>

(Then can inverse Fourier transform this to get the undistorted image intensity [I_{0}k])

Calculation of F:

..after some calculations..

F<s> = \sqrt{\frac{2}{\pi}}\left(\frac{a}{a^{2}+x^{2}} \right) </s>

Calculation of F:

F<i> = A\int_{-\infty}^{\infty}\frac{e^{-ixk}}{\left(a^{2}+k^{2}\right)+\left(b^{2}+k^{2}\right)} dk</i>

F<i> = A\left[\frac{e^{-ixk}}{(-ix)(a^{2}+k^{2})(b^{2}+k^{2})}\right]\right|^{\infty}_{-\infty}</i>

F<i> = A\left[ \frac{2e^{-ix}}{ix(ab)^{2}}\right]</i>

F<i> = \left[ \frac{2Ae^{-ix}}{ix(ab)^{2}}\right]</i>

Therefore can now combine these expressions to get the answer:

F[I_{0}] = \left( \frac{A\left(iax(ab)^{2}\right)e^{-ix}}{a^{2}+x^{2}} \right)

But this looks rather messy, so I assume I've done something wrong somewhere?
 
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Hart said:
Calculation of F:

F<i> = A\int_{-\infty}^{\infty}\frac{e^{-ixk}}{\left(a^{2}+k^{2}\right)+\left(b^{2}+k^{2}\right)} dk</i>

F<i> = A\left[\frac{e^{-ixk}}{(-ix)(a^{2}+k^{2})(b^{2}+k^{2})}\right]\right|^{\infty}_{-\infty}</i>

F<i> = A\left[ \frac{2e^{-ix}}{ix(ab)^{2}}\right]</i>

F<i> = \left[ \frac{2Ae^{-ix}}{ix(ab)^{2}}\right]</i>



I'm not sure what you are doing here. You are given I(k)=F[I(x)], so why are you trying to (inverse?) Fourier transform it and then call the result F?
 
To clarify, for the last part of my attempt I used:

F[I_{0}] = \frac{1}{\sqrt{2 \pi}}\left( \frac{F<i>}{F<s>} \right)</s></i>

.. confused now where to go with this, not really sure how much of my attempt is correct or vaguely right or indeed just completely useless!?
 
Again, you are already given the Fourier transform of I(x)

I(k)\equiv F
 
I(k)\equiv F


.. where does this come from? I don't recall it being stated before within this thread.


anyways, so:

I(k) = F<i> = \left[ \frac{2Ae^{-ix}}{ix(ab)^{2}}\right] </i>

??

..which is the Fourier transform of I(x).
 
Hart said:
.. where does this come from? I don't recall it being stated before within this thread.

Did you not read my first response?

gabbagabbahey said:
... You are given I(k)=F[I(x)]...

It certainly looks to me like I said I(k)=F[I(x)]


anyways, so:

I(k) = F<i> = \left[ \frac{2Ae^{-ix}}{ix(ab)^{2}}\right] </i>

??

..which is the Fourier transform of I(x).


I'm not sure how to make this any clearer. You are told what I(k) is in your problem statement

I(k) = \frac{A}{\left( a^{2}+k^{2} \right) \left( b^{2}+k^{2} \right)}

I(k) and F[I(x)] are two different ways of writing the exact same thing; both represent the Fourier transform of I(x).

F<i>=I(k)=\frac{A}{\left( a^{2}+k^{2} \right) \left( b^{2}+k^{2} \right)}</i>

Does this not make sense to you?
 
...

Hart said:
The Fourier transform of I(x) is given by:

I(k) = \frac{A}{\left( a^{2}+k^{2} \right) \left( b^{2}+k^{2} \right)}

Which is equivalent to stating:

F[I(x)] = I(k) = \frac{A}{\left( a^{2}+k^{2} \right) \left( b^{2}+k^{2} \right)}

Ok.. I do get that now.

So now I need to find Fourier transform I_{0}.. Assume I don't just set k=0 ?
 
Hart said:
So now I need to find Fourier transform I_{0}.. Assume I don't just set k=0 ?

Right, so now you use the convolution theorem:

Hart said:
F[I_{0}] = \frac{1}{\sqrt{2 \pi}}\left( \frac{F<i>}{F<s>} \right)</s></i>


You now know what F is, so if you calculate F, you can use the above convolution theorem. There was nothing wrong with this part of your attempt, however, both your original F and your F were incorrect.

Calculation of F:

..after some calculations..

F<s> = \sqrt{\frac{2}{\pi}}\left(\frac{a}{a^{2}+x^{2}} \right) </s>


You do realize that F is supposed to represent the Fourier transform of S(x), and hence your result should be a function of k, not x, right?...Clearly, you've done something very wrong in your calculation.

Since you are trying to find the Fourier transform of S(x), a good place to start would be to find S(x)...So,, if S(x&#039;-x)=e^{-a|x&#039;-x|}, what is S(x)?.
 
Calculation of F:

f(x) = e^{-\alpha|x|}

\tilda{f}(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx = \sqrt{\frac{2}{\pi}}\left(\frac{\alpha^{2}}{\alpha^{2}+k^{2}}\right)

.. it was actually meant to be with k not x in the final expression, just noticed it when I looked through my calculations again. Hopefully this is actually the correct method and answer.

Then using the Convolution Formula:

F[I_{0}] = \frac{1}{\sqrt{2 \pi}}\left( \frac{F<i>}{F<s>} \right) = \frac{1}{\sqrt{2 \pi}}\left( \frac{\frac{A}{\left( a^{2}+k^{2} \right) \left( b^{2}+k^{2} \right)}}{\sqrt{\frac{2}{\pi}}\left(\frac{\alpha^{2}}{\alpha^{2}+k^{2}}\right)}\right)</s></i>

F[I_{0}] = \frac{1}{\sqrt{2 \pi}}\left({\frac{A\sqrt{\frac{2}{\pi}}\left(\frac{\alpha^{2}}{\alpha^{2}+k^{2}}\right)}{\left( a^{2}+k^{2} \right) \left( b^{2}+k^{2} \right)}\right)

.. which just needs to be simplified.

Any good?!?
 
  • #10
Hart said:
Calculation of F:

f(x) = e^{-\alpha|x|}



Shouldn't this be

S(x) = e^{-a|x|}

...where are \alpha and f(x) coming from? F is the Fruier transform of S(x), not f(x).
 

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