# Applying a filter kernel defined by a continuous equation in the frequency domain

I have a filter kernel $K(\omega)$ completely defined in the frequency domain as a continuous function of angular frequency $\omega$. I know $K(\omega)$ defined as a continuous mathematical equation.

Using the discrete FFT and IFFT, I would like to apply this kernel to a discrete time-domain signal using convolution via multiplication in the frequency domain.

What is a good way to deal with this convolution? Should I generate the $U(\omega)$ as a continuous function in the frequency domain, take the IFFT, and then truncate, shift and window the signal as mentioned in the book by Smith?

Alternately, should I take the continuous inverse Fourier transform of the function $K(\omega)$, and generate $k(t)$ completely in the time domain as a discrete signal by evaluating $k(t)$ for discrete values of $t$? Then $k(t)$ can be treated as a FIR filter kernel and taken into the frequency domain.

Although this is a discrete 1D signal, are 2D image filtering kernels generated in the frequency or spatial domains? What is the difference between the application of the kernels in the time and frequency domains?

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Is $K(\omega)$ known analytically? If so, you may be able to calculate its inverse discrete-time Fourier transform to arrive at a discrete time-domain sequence with the desired impulse response. –  Jason R Oct 26 '12 at 17:32
@Jason R: Yes, $K(\omega)$ is rather complicated, and it is analytical. Do I compute the inverse numerically with a IFFT algorithm (i.e. in Matlab, ifft()) or do I compute the inverse symbolically, on paper or using a CAS? –  Nicholas Kinar Oct 26 '12 at 17:37
If you can, a symbolic inverse would be best, because then you'll get the exact discrete time-domain sequence that you want. But that might not be feasible depending upon the form of $K(\omega)$. One other approach is to use arbitrary filter-design methods to design a filter (either FIR or IIR) whose response looks close enough to $K(\omega)$ for what you want. –  Jason R Oct 26 '12 at 18:16
@JasonR: OK, that sounds good. Thanks Jason. –  Nicholas Kinar Oct 26 '12 at 19:23