I have been learning about General Linear Phase Filters and the four standard FIR implementations.
I was wondering in the case $\beta = \frac{\pi}{2}$ or $\beta = \frac{3\pi}{2}$ if there was a way to correct the phase distortion caused by the filter? More generally, what does the the $\beta$ term actually do when looking at the signal in the time domain? It has the potential to multiply in a complex constant containing an imaginary term but I don't understand what that means.
A generalized linear phase filter with $\beta=\pi/2$ or $\beta=3\pi/2$ introduces (apart from a constant delay) a frequency independent phase shift of $\pm 90$ degrees. Two standard filters that do exactly that are a differentiator with frequency response
$$H(\omega)=j\omega\tag{1}$$
and a Hilbert transformer with frequency response
$$H(\omega)=-j\,\text{sign}(\omega)\tag{2}$$
Note that both frequency responses are purely imaginary and conjugate symmetric, i.e. $H(\omega)=H^*(-\omega)$, as required for real-valued systems.
In the time domain, the filters do the following:
$$y(t)=\frac{dx(t)}{dt}\qquad\text{(differentiator)}\\ y(t)=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{x(\tau)}{t-\tau}d\tau\qquad\text{(Hilbert transformer)}$$
So in general these systems introduce severe distortion. It is relatively easy to imagine the effect of differentiation. To get a feeling for what the Hilbert transform does, take a look at this table showing some transform pairs.
These two systems can be inverted (apart from a constant term in the input signal). The differentiator can be inverted by an integrator, and the Hilbert transformer can be inverted by another Hilbert transformer and by sign reversal.
Note that these filters are just two common examples of systems with imaginary frequency responses, but they are representative in the sense that they introduce severe distortions in the time domain waveform, just like all other systems with imaginary frequency responses with conjugate symmetry.
