Image processing using the frequency domain - duplicate image
1 visualizzazione (ultimi 30 giorni)
Mostra commenti meno recenti
I need to understand the way this picture can be fixed
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/150553/image.jpeg)
need to get only one instance of the dog.
This is what we have learned :
- low pass filter
- high pass filter
- homomorphic filter
is there any way using the above to fix the image? I've included the fft image so maybe you can point out the thing I should be seeing that makes the image blurry as it is.
asking for theoretical help mostly , thanks
2 Commenti
Image Analyst
il 19 Gen 2013
Avoid the problem in the first place. Lay off the caffeine so you can keep the camera still, or train Sean's dog to be more still. ;-)
Risposta accettata
Matt J
il 19 Gen 2013
Modificato: Matt J
il 19 Gen 2013
If F(k) is the spectrum of the dog, then F(k)*exp(-2*pi*j*dot(t,k)) is the spectrum of the dog translated by t and the spectrum of the blurry dog is their superposition
F(k)*(1+exp(-2*pi*j*dot(t,k)))
I suspect you're being asked to filter out the (1+exp(-2*pi*j*dot(t,k))). You need to know t in order to do so, but t dictates the orientation of the sinusoidal pattern of lines you're seeing in the background in the spectrum you've shown. Some sort of edge/line detection should do it.
10 Commenti
Image Analyst
il 20 Gen 2013
Have you ever looked at the spectrum without fftshift()? If so, you'll notice the "origin" is at the 4 corners so your filter will have to go out only until the middle of the image and you'll have to flip things around as you process the other 3 quadrants so that their origins are in the same place, then you'll have to unflip then to put their origins back. It might be easier to fftshift your filter, but if you don't you'll have to extract quadrants, call things like flipud(), fliplr() or transpose to get the origin where it needs to be.
Più risposte (0)
Vedere anche
Categorie
Scopri di più su Frequency Transformations in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!