--- title: "Ideal LPF realizability; step response" subject: "Signals and Systems" chapter: 5 key: "semester_4/signals_and_systems:ideal-lpf-realizability-step" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ideal-lpf-realizability-step site: BCT Engineering --- # Ideal LPF realizability; step response Signals and Systems · Chapter 5: Continuous-Time Systems - Asked in: 2077 Chaitra, 3+3 marks - Video: [Ideal LPF Impulse Response](https://www.youtube.com/watch?v=fCNLMPL2XI8) — John Buck - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ideal-lpf-realizability-step - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5 ## Question Prove, with necessary derivations, that an ideal low pass filter is not physically realizable. Also derive the expression for step response of ideal low pass filter. ## Short answer Ideal LPF has h(t)=sinc/(pi t), nonzero for t<0, so non-causal and not realizable; delay cannot fix it. Step response is s(t)=1/2+(1/pi)Si(wc t), with about 9% overshoot at t=pi/wc. ## Answer #### Ideal LPF is not physically realizable The ideal low-pass filter passes all frequencies below the cutoff $\omega_c$ with unit gain and rejects all others: $$ H(j\omega) = \begin{cases} 1, & |\omega| \le \omega_c \\ 0, & |\omega| > \omega_c \end{cases} $$ The impulse response is the inverse Fourier transform of $H(j\omega)$: $$ \begin{aligned} h(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty} H(j\omega)e^{j\omega t}\,d\omega = \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} e^{j\omega t}\,d\omega \\ &= \frac{1}{2\pi}\cdot\frac{e^{j\omega_c t} - e^{-j\omega_c t}}{jt} = \frac{1}{2\pi}\cdot\frac{2j\sin\omega_c t}{jt} \\ &= \frac{\sin\omega_c t}{\pi t} = \frac{\omega_c}{\pi}\,\text{sinc}\!\left(\frac{\omega_c t}{\pi}\right) \end{aligned} $$ A realizable (causal) LTI system needs $h(t) = 0$ for $t < 0$. Here $h(t)$ is an even sinc, non-zero on both sides; e.g. $h(-\pi/2\omega_c) = 2\omega_c/\pi^2 \ne 0$. It responds before the impulse arrives, so it is **non-causal**. Adding delay $t_0$ gives $\dfrac{\sin\omega_c(t-t_0)}{\pi(t-t_0)}$, still non-zero for $t < 0$. Also, $|H| = 0$ over a whole band violates the Paley–Wiener condition $\int \frac{|\ln|H(j\omega)||}{1+\omega^2}d\omega < \infty$. Hence it is **not physically realizable**. #### Step response of the ideal LPF The step response is the running integral of $h(t)$: $$ \begin{aligned} s(t) &= \int_{-\infty}^{t} h(\tau)\,d\tau = \int_{-\infty}^{t}\frac{\sin\omega_c\tau}{\pi\tau}\,d\tau \\ &= \int_{-\infty}^{0}\frac{\sin\omega_c\tau}{\pi\tau}\,d\tau + \int_{0}^{t}\frac{\sin\omega_c\tau}{\pi\tau}\,d\tau \end{aligned} $$ Using $\int_0^\infty \frac{\sin\omega_c\tau}{\tau}d\tau = \frac{\pi}{2}$, the first integral is $\frac{1}{\pi}\cdot\frac{\pi}{2} = \frac12$. In the second, put $x = \omega_c\tau$: $$ s(t) = \frac12 + \frac{1}{\pi}\int_0^{\omega_c t}\frac{\sin x}{x}\,dx = \frac12 + \frac{1}{\pi}\,\text{Si}(\omega_c t) $$ where $\text{Si}(z) = \int_0^z \frac{\sin x}{x}dx$ is the sine integral. ```text s(t) overshoot ~9% 1 | _/\_ _ | / ' '----- 0.5 |- - - - / | _ / 0 +-' '--/---------------> t 0 pi/wc ``` - $s(0) = 0.5$; $s(-\infty) = 0$; $s(\infty) = 1$. - Maximum at $t = \pi/\omega_c$: $s = 0.5 + \text{Si}(\pi)/\pi \approx 1.09$ (about 9% overshoot, Gibbs phenomenon). - The response starts before $t = 0$ (non-causal) and rises from 0 to 1 in a time of order $\pi/\omega_c$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the ideal LPF non-causal? A. h(t) is zero for t < 0 B. H(jω) is complex C. Its sinc h(t) is non-zero for t < 0 D. Its cutoff is finite Answer: C. The sinc is even, so it responds before the impulse. (Key point: Non-causal reason) 2. How is the step response s(t) obtained from h(t)? A. s(t) is the derivative of h(t) B. s(t) = h(t)u(t) C. s(t) = h(-t) D. s(t) is the running integral of h(τ) from -∞ to t Answer: D. Step is the integral of the impulse, so the step response integrates h. (Key point: Step response definition) 3. What is the step response of the ideal LPF? A. s(t) = 1/2 + (1/π) Si(ωc t) B. s(t) = 1 - e^(-ωc t) C. s(t) = sin(ωc t)/(πt) D. s(t) = Si(ωc t) Answer: A. Using ∫0^∞ sin/τ = π/2 gives the 1/2, plus the sine integral term. (Key point: Result) 4. What happens at t = π/ωc in the step response? A. s = 0 exactly B. Maximum of about 1.09 (about 9% overshoot, Gibbs) C. s = 0.5 D. s = 1 with no overshoot Answer: B. s = 0.5 + Si(π)/π ≈ 1.09. 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