Ideal low pass and high pass filters

Frequency response

The frequency response $H(\omega)$ of an LTI system is the Fourier transform of its impulse response. It tells how the system changes the magnitude and phase of a sinusoid of each frequency:

$$H(\omega) = \int_{-\infty}^{\infty} h(t)e^{-j\omega t}dt = \frac{Y(\omega)}{X(\omega)} = |H(\omega)|\,e^{j\theta(\omega)}$$

If the input is $e^{j\omega t}$, the output is $H(\omega)e^{j\omega t}$. $|H(\omega)|$ is the magnitude (gain) response and $\theta(\omega)$ is the phase response.

Impulse response of an ideal low pass filter

Pass band $|\omega| < \omega_c$, unit gain, linear phase (delay $t_0$):

$$H_{LP}(\omega) = \begin{cases} e^{-j\omega t_0}, & |\omega| < \omega_c \\ 0, & \text{otherwise} \end{cases}$$ $$\begin{aligned} h_{LP}(t) &= \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} e^{j\omega(t-t_0)}d\omega \\ &= \frac{1}{2\pi}\cdot\frac{2j\sin\omega_c(t-t_0)}{j(t-t_0)} \\ &= \frac{\sin\omega_c(t-t_0)}{\pi(t-t_0)} = \frac{\omega_c}{\pi}\mathrm{sinc}\left[\omega_c(t-t_0)\right] \end{aligned}$$

This is a sinc pulse centred at $t_0$ with peak $\omega_c/\pi$ (here $\mathrm{sinc}\,x = \sin x / x$).

Impulse response of an ideal high pass filter

The ideal HPF blocks $|\omega| < \omega_c$ and passes everything above. It is the all-pass (delay) system minus the LPF:

$$H_{HP}(\omega) = e^{-j\omega t_0} - H_{LP}(\omega) = \begin{cases} 0, & |\omega| < \omega_c \\ e^{-j\omega t_0}, & |\omega| > \omega_c \end{cases}$$

 |H_LP|              |H_HP|
 1 +----+          1 ---+      +---
   |    |               |      |
 --+----+----> w   -----+--+---+---> w
 -wc  0  wc           -wc  0   wc

Since $e^{-j\omega t_0} \leftrightarrow \delta(t-t_0)$:

$$h_{HP}(t) = \delta(t - t_0) - \frac{\sin\omega_c(t-t_0)}{\pi(t-t_0)}$$

Remarks:

• Both impulse responses are non-zero for $t < 0$ (sinc tails), so both ideal filters are non-causal and cannot be built exactly; practical filters approximate them.
• With $t_0 = 0$: $h_{LP}(t) = \frac{\sin\omega_c t}{\pi t}$ and $h_{HP}(t) = \delta(t) - \frac{\sin\omega_c t}{\pi t}$.