I am aware that an ideal low-pass filter in both continuous time and discrete time has a $\mathrm{sinc}$ impulse response. What would the impulse response of an ideal high-pass or band-pass filter look like (assuming that it is real)? My intuition says that it would be some sort of superposition of shifted $\mathrm{sinc}$ functions but I am not quite sure. Thanks in advance.
EDIT: Here is my attempt at deriving the impulse response $h(t)$ for a continuous-time ideal high pass filter with a zero phase characteristic.
The frequency response in this case can be written as $H(\omega) = u(\omega - \omega_c) + u(-\omega -\omega_c)$ where $u$ is the unit step function and $\omega_c$ is the filter cut-off frequency.
Using the Fourier transform pair:
$$u(t) \leftrightarrow \frac{1}{j\omega} + \pi \delta(\omega)$$
we can exploit the duality property to obtain:
$$\frac{1}{jt} + \pi \delta(t) \leftrightarrow 2\pi u(-\omega)$$
which implies (from linearity):
$$\frac{1}{j2\pi t} + \frac{1}{2} \delta(t) \leftrightarrow u(-\omega)$$
Additionally we obtain from the time reversal property and the scaling property of the delta function:
$$-\frac{1}{j2\pi t} + \frac{1}{2} \delta(t) \leftrightarrow u(\omega)$$
Applying the inverse Fourier transform to:
$$H(\omega) = u(\omega - \omega_c) + u(-\omega -\omega_c) = u(\omega - \omega_c) + u(-(\omega +\omega_c))$$
and using the linearity property, the frequency shift property, and the two Fourier transform pairs I wrote above we obtain:
$$h(t) = e^{j\omega_c t} \{ -\frac{1}{j2\pi t} + \frac{1}{2} \delta(t) \} + e^{-j\omega_c t} \{ \frac{1}{j2\pi t} + \frac{1}{2} \delta(t) \} $$
and I don't know where to go from there (specifically, I don't know how to handle the delta functions that popped up).