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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).

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    $\begingroup$ Why are you not sure? I mean, you sound like you know why it's a sinc for the lpf, so calculate the same transform for the perfect HPF! (by the way, "superposition" in the sense "sum of exactly one summand"). $\endgroup$ – Marcus Müller Feb 25 '18 at 20:02
  • $\begingroup$ I tried to apply it to the case of an ideal high pass filter whose frequency response comprises two shifted unit step functions but the inverse Fourier transform of that involves delta functions that I can't get rid of $\endgroup$ – 0MW Feb 25 '18 at 20:19
  • $\begingroup$ I honestly don't see any delta dirac functions that you need to get rid of here - can you add your calculation to your question (by editing it), please? $\endgroup$ – Marcus Müller Feb 25 '18 at 20:27
  • $\begingroup$ I just added my attempt to my question. $\endgroup$ – 0MW Feb 25 '18 at 20:49
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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: $$ \boxed{ h_{hpf}[n] = \mathcal{F}^{-1}\{ 1 - H_{lpf}(e^{j\omega}) \} = \delta[n] - h_{lpf}[n] }$$

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  • $\begingroup$ Is that Dirac delta impulse problematic at all? $\endgroup$ – 0MW Feb 25 '18 at 23:17
  • $\begingroup$ @0MW what problem do you refer to ? $\endgroup$ – Fat32 Feb 26 '18 at 8:48

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