Distortionless transmission; ideal LPF step response

Distortionless transmission

Transmission is distortionless when the output has exactly the same shape as the input, only scaled by a constant $K$ and delayed by a constant $t_0$:

$$y(t) = K\,x(t-t_0) \;\Rightarrow\; H(\omega) = K\,e^{-j\omega t_0}$$

So the system needs (1) constant magnitude $|H(\omega)| = K$ and (2) linear phase $\angle H(\omega) = -\omega t_0$ over the signal bandwidth.

Unit step response of the ideal LPF

Ideal LPF (unit gain, delay $t_0$, cutoff $\omega_c$):

$$H(\omega) = e^{-j\omega t_0}\ \text{for } |\omega| < \omega_c,\quad 0 \text{ otherwise}$$

Its impulse response (inverse Fourier transform) is

$$h(t) = \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} e^{j\omega(t-t_0)}d\omega = \frac{\sin\omega_c(t-t_0)}{\pi(t-t_0)}$$

The step response is the integral of $h(t)$:

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

Put $x = \omega_c(\lambda - t_0)$:

$$\begin{aligned} s(t) &= \frac{1}{\pi}\int_{-\infty}^{\omega_c(t-t_0)}\frac{\sin x}{x}dx \\ &= \frac{1}{\pi}\left[\frac{\pi}{2} + \mathrm{Si}\left(\omega_c(t-t_0)\right)\right] \\ &= \frac{1}{2} + \frac{1}{\pi}\mathrm{Si}\left[\omega_c(t-t_0)\right] \end{aligned}$$

where $\mathrm{Si}(y) = \int_0^{y}\frac{\sin x}{x}dx$ is the sine integral.

• $s(t_0) = 0.5$, $s(\infty) = 1$, $s(-\infty) = 0$.
• Rise time $\approx \pi/\omega_c$, inversely proportional to bandwidth.
• About 9% overshoot with ringing (Gibbs phenomenon), and response before $t = 0$ (non-causal).