Enlargement and reduction

As in the case of shifting, the way to avoid aliasing during enlargement and reduction is to simulate an operation on a continuous image. First, we will look at what happens to the Fourier transform of a continuous image when it is enlarged or reduced.

   
Fourier zoom property

Consider the zoom operator Z_a, in which a is a magnification factor.

$$Z_a[x](t) = x({t\over a})$$

When a > 1, the resulting image is larger by a factor of a. When 0 < a < 1, it is smaller by a factor of a^-1. A larger image will contain lower frequencies, and vice versa. We can see why this is true by evaluating the Fourier transform of Z_a[x]:

$${\cal F}{[Z_a[x]]} = \int_{-\infty}^{+\infty} Z_a[x](t) e^{-i \omega t}dt$$

$$= \int_{-\infty}^{+\infty} x({t\over a}) e^{-i \omega t}dt$$

With a change of variables $\tau = t/a$, $d\tau = dt/a$,

$$= \int_{-\infty}^{+\infty} x(\tau)e^{-i(a\omega)\tau} d\tau a$$

$$= a Z_{1/a}[{\cal F}{[x]}]$$

Reconstruction, zooming, and resampling

The ideal antialiased zoom operation on a sampled image would involve reconstruction of the continuous signal, continuous zooming, and resampling.

  

Figure 9: Reconstruction, zooming, and resampling

Combination of filtering operations

As in the case of image shifting, the two filtering operations-postfiltering for reconstruction, and prefiltering for resampling-can be combined into one. However, the zooming operation changes the frequency content of the signal, so the cutoff frequencies of the two filters will not match. The operations in Figure 9 may be expressed as:

$$Y(\omega) = aH(\omega) Z_{1/a}[H \hat{X}](\omega) = aH(\omega)H(a\omega)\hat{X}(a\omega)$$

By looking at the product of the two filters we can see why the pair will have a joint cutoff of $\min(\omega_c/a, \omega_c)$:

$$H(a\omega) = \left\{ \begin{array}{ll} 1 & \mbox{if $\vert\... ... \omega_c/a$}\\ 0 & \mbox{otherwise}\\ \end{array} \right.$$

$$H(\omega)H(a\omega) = \left\{ \begin{array}{ll} 1 & \mbox{i... ...eq \omega_c$}\\ 0 & \mbox{otherwise}\\ \end{array} \right.$$

The reason for the min operation is that the zoom process is more constrained by whichever image has lower resolution. When 0 < a < 1, the image is being shrunk, and the target image resolution is the limiting factor. Because the target image has fewer pixels than the source, there are fewer degrees of freedom in which to encode the signal. Therefore unless we do something very tricky, information will be lost. The source image resolution is limiting when a > 1.

$$Y(\omega) = aH(\max(\omega, a\omega))\hat{X}(a\omega)$$

Discrete implementation

The following equations, obtained by applicaton of discrete convolution (Section 2.6), describe how antialiased zooming may be performed directly on discrete images.

$$A = \min(a, 1)$$

$$y[n] = A \sum_{m=-\infty}^{\infty} x[m] h(AT({n\over a} - m))$$

where T is the sampling interval of x[n].