Convolution - Image Processing

[Hart]

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

[tex]F[I_{0}] = \frac{1}{\sqrt{2 \pi}}\left( \frac{F}{F} \right)[/tex]

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

Calculation of F:

..after some calculations..

[tex] F = \sqrt{\frac{2}{\pi}}\left(\frac{a}{a^{2}+x^{2}} \right) [/tex]

Calculation of F:

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

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

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

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

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?

 

[gabbagabbahey]

Calculation of F:

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

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

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

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

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 [itex]F[/itex]?

 

[Hart]

To clarify, for the last part of my attempt I used:

[tex]F[I_{0}] = \frac{1}{\sqrt{2 \pi}}\left( \frac{F}{F} \right)[/tex]

.. 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!?

 

[gabbagabbahey]

Again, you are already given the Fourier transform of $I(x)$

[tex]I(k)\equiv F[/tex]

 

[Hart]

[tex]I(k)\equiv F[/tex]

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


anyways, so:

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

??

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

 

[gabbagabbahey]

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

Did you not read my first response?

... You are given $I(k)=F[I(x)]$...

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

anyways, so:

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

??

..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 $I(x)$.

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

Does this not make sense to you?

 

[Hart]

...

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 equivelent 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 $$k=0$$ ?

 

[gabbagabbahey]

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:

[tex]F[I_{0}] = \frac{1}{\sqrt{2 \pi}}\left( \frac{F}{F} \right)[/tex]

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

Calculation of F:

..after some calculations..

[tex] F = \sqrt{\frac{2}{\pi}}\left(\frac{a}{a^{2}+x^{2}} \right) [/tex]

You do realize that [itex]F[/itex] 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 $S(x'-x)=e^{-a|x'-x|}$, what is $S(x)$?.

 

[Hart]

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:

[tex]F[I_{0}] = \frac{1}{\sqrt{2 \pi}}\left( \frac{F}{F} \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)[/tex]

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

.. which just needs to be simplified.

Any good?!?

 

[gabbagabbahey]

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? [itex]F[/itex] is the Fruier transform of $S(x)$, not $f(x)$.