--- title: "Frequency response; ideal LPF impulse response" subject: "Signals and Systems" chapter: 5 key: "semester_4/signals_and_systems:ideal-lpf-cutoff-impulse" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ideal-lpf-cutoff-impulse site: BCT Engineering --- # Frequency response; ideal LPF impulse response Signals and Systems · Chapter 5: Continuous-Time Systems - Asked in: 2082 Chaitra, 2+4 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-cutoff-impulse - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5 ## Question Define frequency response of the discrete LTI system. Derive formula to calculate the impulse response of continuous time ideal low pass filter. ## Short answer For an ideal LPF with cutoff ωc, H(jω)=1 for |ω|≤ωc, 0 otherwise. Inverse Fourier transform yields h(t)= sin(ωc t)/(π t) = (ωc/π) sinc(ωc t/π). It has peak ωc/π at t=0, zeros at kπ/ωc (k≠0), and is non-causal. ## Answer #### Frequency response of a discrete-time LTI system The **frequency response** $H(e^{j\Omega})$ of a discrete-time LTI system is the DTFT of its impulse response $h[n]$: $$ H(e^{j\Omega}) = \sum_{n=-\infty}^{\infty} h[n]\,e^{-j\Omega n} $$ If $x[n] = e^{j\Omega n}$, then $y[n] = H(e^{j\Omega})\,e^{j\Omega n}$, so $H(e^{j\Omega})$ gives the gain $|H|$ and phase shift $\angle H$ at each frequency. Also $Y(e^{j\Omega}) = H(e^{j\Omega})X(e^{j\Omega})$. It is always **periodic in $\Omega$ with period $2\pi$**. Example: $h[n] = a^n u[n]$, $|a| < 1$ gives $H(e^{j\Omega}) = \dfrac{1}{1 - a e^{-j\Omega}}$. #### Impulse response of the CT ideal low-pass filter 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 $H(j\omega)$: $$ \begin{aligned} h(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty} H(j\omega)e^{j\omega t}\,d\omega = \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} e^{j\omega t}\,d\omega \\ &= \frac{1}{2\pi}\cdot\frac{e^{j\omega_c t} - e^{-j\omega_c t}}{jt} = \frac{1}{2\pi}\cdot\frac{2j\sin\omega_c t}{jt} \\ &= \frac{\sin\omega_c t}{\pi t} = \frac{\omega_c}{\pi}\,\text{sinc}\!\left(\frac{\omega_c t}{\pi}\right) \end{aligned} $$ ```text H(jw) h(t) 1 ______ wc/pi | | /\ | | _ _ / \ _ _ ----+------+--> w -' '-' '-/----\-' '-' '-> t -wc 0 wc -pi/wc pi/wc ``` - Peak value $h(0) = \omega_c/\pi$; zeros at $t = k\pi/\omega_c$, $k \ne 0$. - $h(t) \ne 0$ for $t < 0$, so the ideal LPF is non-causal and cannot be built exactly. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the frequency response H(e^jΩ) of a discrete-time LTI system? A. The DTFT of its impulse response h[n] B. The z-transform of x[n] evaluated at z = 0 C. The sum of h[n] over all n only D. The DTFT of the input signal x[n] Answer: A. H(e^jΩ) = sum of h[n] e^(-jΩn), the DTFT of h[n]; it gives gain and phase at each frequency. (Key point: Frequency response definition) 2. How does the discrete-time frequency response H(e^jΩ) behave with Ω? A. It is periodic with period π B. It is periodic with period 2π C. It is non-periodic and decays to zero D. It is periodic with period 1 Answer: B. Since e^(-jΩn) repeats every 2π in Ω, the DTFT is always 2π-periodic. (Key point: Periodicity) 3. Inverse Fourier transform of the ideal LPF (gain 1 for |ω| ≤ ωc) gives which h(t)? A. sin(ωc t)/t B. (ωc/π) cos(ωc t) C. sin(ωc t)/(π t) D. sin(ωc t)/(2π t^2) Answer: C. Integrating e^(jωt)/(2π) from -ωc to ωc gives (1/2π)(2j sin ωc t)/(jt) = sin(ωc t)/(π t). (Key point: Ideal LPF impulse response) 4. For the ideal LPF, what are the peak value h(0) and the zero crossings? A. h(0) = 1; zeros at t = k/ωc B. h(0) = ωc/π; zeros at t = 2kπ/ωc only C. h(0) = 2ωc/π; zeros at t = kπ D. h(0) = ωc/π; zeros at t = kπ/ωc, k ≠ 0 Answer: D. The sinc has peak ωc/π at t=0 and zeros where sin(ωc t)=0, i.e. t = kπ/ωc. 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