--- title: "Impulse response; ideal LPF formula" subject: "Signals and Systems" chapter: 5 key: "semester_4/signals_and_systems:impulse-response-ideal-lpf" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/impulse-response-ideal-lpf site: BCT Engineering --- # Impulse response; ideal LPF formula Signals and Systems · Chapter 5: Continuous-Time Systems - Asked in: 2078 Baisakh, 6 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/impulse-response-ideal-lpf - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5 ## Question What do you mean by impulse response of a LTI system? Derive formula to calculate impulse response of continuous time ideal low pass filter. ## Short answer Impulse response is system's output to unit impulse δ(t); convolution gives output. For ideal LPF with cutoff ωc, H(jω)=1 for |ω|≤ωc, so inverse FT gives h(t)= sin(ωc t)/(π t). Peak ωc/π, zeros at kπ/ωc, non-causal. ## Answer #### Impulse response of an LTI system The **impulse response** $h(t)$ of an LTI system is its output when the input is the unit impulse $\delta(t)$ and the system is initially at rest: $$ \delta(t) \;\to\; h(t) $$ Because any input can be written as $x(t) = \int x(\tau)\delta(t-\tau)d\tau$, linearity and time invariance give $y(t) = \int x(\tau)h(t-\tau)d\tau = x(t)*h(t)$. So $h(t)$ completely characterises an LTI system: causality ($h(t) = 0$ for $t<0$), stability ($\int|h|\,dt < \infty$) and frequency response ($H(j\omega) = \mathcal{F}\{h(t)\}$) can all be read from it. #### 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 ``` - $h(0) = \omega_c/\pi$; zero crossings at $t = k\pi/\omega_c$ ($k = \pm1, \pm2, \dots$). - Width of the main lobe $2\pi/\omega_c$ shrinks as bandwidth $\omega_c$ grows. - $h(t)$ exists for $t < 0$, so the ideal LPF is **non-causal** and not realizable; it is also of infinite duration. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the impulse response of an LTI system? A. The output when the input is δ(t), with the system initially at rest B. The output when the input is a unit step C. The derivative of the input D. The output when the input is zero Answer: A. h(t) is the response to the unit impulse δ(t). (Key point: Impulse response definition) 2. Why does h(t) completely characterise an LTI system? A. Because h(t) equals the input B. Any input is a sum of shifted impulses, so y = x*h C. Because LTI systems have no memory D. Because h(t) is always causal Answer: B. Writing x as an integral of impulses and applying linearity and time invariance gives the convolution. (Key point: Why h(t) is enough) 3. In deriving the ideal LPF h(t), what is the integral evaluated? A. (1/2π) times the integral of e^(jωt) from 0 to ∞ B. 2π times the integral of e^(-jωt) over all ω C. (1/2π) times the integral of e^(jωt) from -ωc to ωc D. The integral of H(jω) from -ωc to ωc without e^(jωt) Answer: C. The inverse Fourier transform over the passband gives sin(ωc t)/(πt). (Key point: Derivation step) 4. As the cutoff ωc increases, what happens to the main lobe of h(t)? A. It widens B. It stays constant C. It becomes causal D. Its width 2π/ωc shrinks Answer: D. Zeros are at ±π/ωc, so a larger bandwidth gives a narrower pulse. 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