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I have some old STM images that appears to be sheared by about 5 degrees; in other words as scanning progressed from top to bottom the scanning area slowly drifted left-to-right, by the end the scanning was shifted to the side by say 5% up to 10% of the length of the scan.

$$x_{new} = x + \alpha y$$

where $|\alpha| < 0.1$ and the original data is within the square: $-1 <= x, y <= 1$.

The straightforward way to fix this would be to re-interpolate the original image to de-shear it, then crop off the triangular bits along the sides so that the FT is performed on a rectangular area of all good data, and I will probably do that.

In fact, since I'm using a 2D Hamming window I might even be able to completely ignore the blank triangular areas that the de-shearing produces (replace unknown data with image median).

Question: But I am curious if there is a transform that I can apply (again by re-interpolation) to the result of the Fourier transform that will produce an equivalent result without de-shearing the original image. I'm only interested in the central are of the FT so I don't care about lost edges.

I have a hunch that the answer is yes and it will turn out to be a similar type of linear transform or shearing of the FT.

In fact it looks like it's very simple and this is what's happening

$$\omega_{y,new} = \omega_{y} + \beta \omega_{x}$$

with $\beta = \alpha$

when I make some plots, but I don't know how to go about showing this to be true.

Related:

uhoh
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1 Answers1

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Your intuition was correct:

Call the original image $I(x,y)$ the sheared image is $I(x+\alpha y, y)$. I will use the Fourier integral and not the sum, but up to discretization, the argument should stand. Define the Fourier integral of $I$: $$\hat I(\omega_x,\omega_y) = \mathcal F \circ I(x,y) = \iint dx dy e^{i(\omega_x x+\omega_y y)}I(x,y) $$ Then $$\iint dx dy e^{i(\omega_x x+\omega_y y)}I(x+\alpha y,y)=\iint dx dye^{i(\omega_x (x +\alpha y)+(\omega_y-\omega_x\alpha) y)}I(x+\alpha y,y)$$ $$=\iint du dv e^{i(\omega_x u+(\omega_y-\omega_x \alpha) v)} I(u,v)=\hat I(\omega_x ,\omega_y-\omega_x\alpha)$$ Thus $$\mathcal F \circ I(x+\alpha y,y) = \hat I(\omega_x, \omega_y -\omega_x\alpha)$$

user619894
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