# FourierGaussFltImg

## DESCRIPTION

Applies a Gaussian frequency domain filter to a frequency domain image. The effect is to convolve a filter with the image in the spatial domain. The frequency domain representation of the filter is as follows:H(u,v) = HF Gain + (DC Gain - HF Gain) *exp {-[(a11 u + a12 v)^{2}+ (a21 u + a22 v)^{2}]}

where

H( ) = transfer function of the filter u,v = 2D frequency coordinates HF Gain = gain of filter at the Nyquist frequency (highest) DC Gain = gain of the filter at zero frequency (DC) Min Half = frequency of half power point along the minor elliptical axis Maj Half = frequency of half power point along the major elliptical axis Theta = angle in degrees of the filter's orientation xSize = x dimension of the source image ySize = y dimension of the source image sigmaL = sqrt (0.693147 / (minHalf*minHalf)) sigmaS = sqrt (0.693147 / (majHalf*majHalf)) phi = 0.017453 * Theta a11 = sigmaS * cos(phi) / xSize a12 = sigmaS * sin(phi) / ySize a21 = -sigmaL * sin(phi) / xSize a22 = sigmaL * cos(phi) / ySize

The input must be a frequency domain image in the packed format produced by the ForwardFFTImg module (see format description in ForwardFFTImg documentation). The Fourier transformation and the cross correlation are discussed in:

Digital Image Processing, Gonzales, R.C., Wintz, P., Addison Wesley, Second Edition, 1987, pp 61--137.

## INPUTS

**Port:** Img In

**Type:** Lattice

**Constraints:** 1..3-D

Source frequency domain image.

## WIDGETS

**Port:** HF Gain

**Type:** Dial

HF Gain constant value.

**Port:** DC Gain

**Type:** Dial

DC Gain constant value.

**Port:** Min Half

**Type:** Dial

Min Half constant value.

**Port:** Maj Half

**Type:** Dial

Maj Half constant value.

**Port:** Theta

**Type:** Dial

Theta constant value.

## OUTPUTS

**Port:** Img Out

**Type:** Lattice

**Constraints:** 1..3-D

Filtered frequency domain image.

## KNOWN PROBLEMS

Selecting MajHalf = 0 or MinHalf = 0 will cause a NaN to be generated in the filter kernel, which will result in NaNs throughout the inverse transformed image downstream. Mathematically, the correct value would be a zero instead of a NaN, but the inputs are not tested for this condition.## SEE ALSO

ForwardFFTImg, InverseFFTImg, FourierExpFltImg.[Documentation Home]

© The Numerical Algorithms Group Ltd, Oxford UK. 2000