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2-D FIR filter using frequency transformation

`h = ftrans2(b,t)`

`h = ftrans2(b)`

produces the two-dimensional FIR filter `h`

= ftrans2(`b`

,`t`

)`h`

that corresponds to the
one-dimensional FIR filter `b`

using the transform
`t`

. `b`

must be a one-dimensional, Type I
(even symmetric, odd-length) filter such as can be returned by
`fir1`

, `fir2`

, or `firpm`

in
the Signal Processing Toolbox software. The transform matrix `t`

contains coefficients that define the frequency transformation to use.

The transformation below defines the frequency response of the two-dimensional filter
returned by `ftrans2`

.

$${{H}_{({\omega}_{1},{\omega}_{2})}=B(\omega )|}_{\mathrm{cos}\omega =T({\omega}_{1},{\omega}_{2})},$$

where *B(*ω*)* is the Fourier
transform of the one-dimensional filter `b`

:

$$B(\omega )={\displaystyle \sum _{n=-N}^{N}b(n){e}^{-j\omega n}}$$

and
*T(*ω_{1}*,*ω_{2}*)
* is the Fourier transform of the transformation matrix
`t`

:

$$T({\omega}_{1},{\omega}_{2})={\displaystyle \sum _{{n}_{2}}{\displaystyle \sum _{{n}_{1}}t({n}_{1},{n}_{2}){e}^{-j{\omega}_{1}{n}_{1}}{e}^{-j{\omega}_{2}{n}_{2}}}}.$$

The returned filter `h`

is the inverse Fourier transform of
*H(*ω_{1}*,*ω_{2}*)*:

$$h({n}_{1},{n}_{2})=\frac{1}{{\left(2\pi \right)}^{2}}{\displaystyle {\int}_{-\pi}^{\pi}{\displaystyle {\int}_{-\pi}^{\pi}H({\omega}_{1},{\omega}_{2}){e}^{j{\omega}_{1}{n}_{1}}{e}^{j{\omega}_{2}{n}_{2}}d{\omega}_{1}d{\omega}_{2}}}.$$

[1] Lim, Jae S.,
*Two-Dimensional Signal and Image Processing*, Englewood
Cliffs, NJ, Prentice Hall, 1990, pp. 218-237.