Impulse response of ideal filters

Question

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).

Answer 1

Given that you are ok with the impulse response of the ideal zero phase lowpass real filter, then you can easily derive the impulse response of the ideal high pass zero phase real filter. In continuous-time let $$h_{lpf}(t) = \frac{\sin(\omega_0 t) }{\pi t} $$ be the impulse reponse of an LTI ideal zero phase lowpass filter whose frequency response is by definition: $$ H_{lpf}(\omega) = \begin{cases} 1 ,& \text{ for } |\omega| < \omega_0 \\ 0 ,& \text{ otherwise } \\ \end{cases} $$

Then you can define the frequency response of the ideal zero phase highpass filter as: $$ H_{hpf}(\omega) = \begin{cases} 1 ,& \text{ for } |\omega| > \omega_0 \\ 0 ,& \text{ otherwise } \\ \end{cases} = 1 - H_{lpf}(\omega)$$

By taking the inverse Fourier transform of $H_{hpf}(\omega)$ you can conclude that $$ \boxed{ h_{hpf}(t) = \mathcal{F}^{-1}\{ 1 - H_{lpf}(\omega) \} = \delta(t) - h_{lpf}(t) }$$

Similarly in discrete-time you can conclude that: $$ \boxed{ h_{hpf}[n] = \mathcal{F}^{-1}\{ 1 - H_{lpf}(e^{j\omega}) \} = \delta[n] - h_{lpf}[n] }$$