Ideal LPF impulse and step response

Impulse response of the ideal LPF

The ideal LPF has constant gain and linear phase in the pass band and zero gain outside:

$$H(\omega) = \begin{cases} e^{-j\omega t_0}, & |\omega| \le \omega_c \\ 0, & |\omega| > \omega_c \end{cases}$$

(unit gain taken; for gain $K$, multiply the results by $K$).

Inverse Fourier transform:

$$\begin{aligned} h(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty} H(\omega)e^{j\omega t}d\omega = \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} e^{j\omega(t-t_0)}d\omega \\ &= \frac{1}{2\pi}\cdot\frac{e^{j\omega_c(t-t_0)} - e^{-j\omega_c(t-t_0)}}{j(t-t_0)} \\ &= \frac{\sin\omega_c(t-t_0)}{\pi(t-t_0)} = \frac{\omega_c}{\pi}\,\mathrm{sinc}\left[\omega_c(t-t_0)\right] \end{aligned}$$

• Peak value $\omega_c/\pi$ at $t = t_0$.
• Zero crossings at $t = t_0 + n\pi/\omega_c$, $n = \pm1, \pm2, \dots$
• Main lobe width $2\pi/\omega_c$: a wider bandwidth gives a narrower, taller pulse.
• $h(t) \ne 0$ for $t < 0$, so the filter is non-causal (not physically realizable).

           h(t)
            /\   wc/pi
  .  .-.   /  \   .-.  .
 -'--' '--/----\--' '--'--> t
             t0

Step response of the ideal LPF

The step response is the running integral of the impulse response:

$$s(t) = \int_{-\infty}^{t} h(\lambda)\,d\lambda = \int_{-\infty}^{t} \frac{\sin\omega_c(\lambda-t_0)}{\pi(\lambda-t_0)}\,d\lambda$$

Let $x = \omega_c(\lambda - t_0)$, so $d\lambda = dx/\omega_c$ and $\lambda - t_0 = x/\omega_c$:

$$\begin{aligned} s(t) &= \frac{1}{\pi}\int_{-\infty}^{\omega_c(t-t_0)} \frac{\sin x}{x}\,dx \\ &= \frac{1}{\pi}\int_{-\infty}^{0} \frac{\sin x}{x}\,dx + \frac{1}{\pi}\int_{0}^{\omega_c(t-t_0)} \frac{\sin x}{x}\,dx \end{aligned}$$

Using $\int_{-\infty}^{0} \frac{\sin x}{x}dx = \frac{\pi}{2}$ and the sine integral $\mathrm{Si}(y) = \int_0^{y}\frac{\sin x}{x}dx$:

$$s(t) = \frac{1}{2} + \frac{1}{\pi}\,\mathrm{Si}\left[\omega_c(t-t_0)\right]$$

Features:

• $s(t_0) = 1/2$; $s(-\infty) = 0$; $s(\infty) = 1$ (since $\mathrm{Si}(\pm\infty) = \pm\pi/2$).
• Rise time: the slope at $t_0$ is $h(t_0) = \omega_c/\pi$, so $t_r \approx \pi/\omega_c = 1/(2B)$ where $B = \omega_c/2\pi$ Hz. Rise time is inversely proportional to bandwidth.
• Overshoot and ringing: the response overshoots by about 9% (peak about 1.09) and oscillates around 1. This is the Gibbs phenomenon; it does not shrink as $\omega_c$ increases.
• The response starts before $t = 0$, again showing the filter is non-causal.

 s(t)
 1.09|          _
 1.0 |   - - - / \_/~~~~~~~
 0.5 |       /
     |~~\_  /
   0 +----------------------> t
            t0