---
title: "Impulse response of ideal low pass filter"
subject: "Signals and Systems"
chapter: 5
key: "semester_4/signals_and_systems:ideal-lpf-impulse-response"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ideal-lpf-impulse-response
site: BCT Engineering
---

# Impulse response of ideal low pass filter

Signals and Systems · Chapter 5: Continuous-Time Systems

- Asked in: 2080 Chaitra, 5 marks; 2079 Chaitra, 5 marks; 2071 Bhadra, 5 marks; 2070 Magh, 5 marks (asked 4 times)
- 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-impulse-response
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5

## Question

Derive the expression for impulse response of ideal low pass filter (with cutoff frequency ω_c) and discuss.

## Short answer

The impulse response of an ideal low-pass filter with cutoff ω_c is h(t) = sin(ω_c t)/(π t). It is non-causal and infinite in extent, thus physically unrealizable; its peak at t=0 is ω_c/π and zero crossings occur at t = kπ/ω_c.

## Answer

The ideal low-pass filter passes all frequencies below the cutoff $\omega_c$ with unit gain and blocks all others:
$$
H(j\omega) = \begin{cases} 1, & |\omega| \le \omega_c \\ 0, & |\omega| > \omega_c \end{cases}
$$
(zero phase taken for simplicity).

#### Derivation
Taking the inverse Fourier transform:
$$
\begin{aligned}
h(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty} H(j\omega)e^{j\omega t}\,d\omega = \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} e^{j\omega t}\,d\omega \\
&= \frac{1}{2\pi}\left[\frac{e^{j\omega t}}{jt}\right]_{-\omega_c}^{\omega_c} = \frac{1}{2\pi}\cdot\frac{e^{j\omega_c t} - e^{-j\omega_c t}}{jt} \\
&= \frac{1}{2\pi}\cdot\frac{2j\sin\omega_c t}{jt} = \frac{\sin\omega_c t}{\pi t}
\end{aligned}
$$
In sinc form:
$$
h(t) = \frac{\omega_c}{\pi}\cdot\frac{\sin\omega_c t}{\omega_c t} = \frac{\omega_c}{\pi}\,\text{sinc}\!\left(\frac{\omega_c t}{\pi}\right)
$$

```text
 H(jw)                    h(t)
   1 ______                   wc/pi
    |      |                   /\
    |      |          _  _    /  \    _  _
----+------+--> w   -' '-' '-/----\-' '-' '-> t
  -wc  0   wc               -pi/wc pi/wc
                       zeros at t = k.pi/wc
```

#### Discussion
- **Peak:** at $t = 0$, $h(0) = \omega_c/\pi$ (limit of $\sin\omega_c t/\pi t$).
- **Zero crossings:** at $t = k\pi/\omega_c$, $k = \pm1, \pm2, \dots$. A larger $\omega_c$ gives a narrower main lobe (time–bandwidth inverse relation).
- **Non-causal:** $h(t) \ne 0$ for $t < 0$; the sinc extends to $-\infty$. So the filter would have to respond before the impulse is applied, and it is **not physically realizable**.
- **Infinite duration** and oscillating tails (ringing). Its step response shows overshoot (Gibbs phenomenon, about 9%).
- With linear phase $H(j\omega) = e^{-j\omega t_0}$ in the passband, $h(t) = \dfrac{\sin\omega_c(t-t_0)}{\pi(t-t_0)}$: a delayed sinc that is still non-zero for $t<0$.
- Practical filters (Butterworth, Chebyshev, RC) approximate it with a finite transition band.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is h(t) of the ideal LPF obtained?
   A. Fourier transform of H(jω) over all ω
   B. Differentiating H(jω)
   C. Inverse Fourier transform of H(jω) over |ω| <= ωc
   D. Laplace transform of u(t)
   Answer: C. h(t) = (1/2π)∫_{-ωc}^{ωc} e^{jωt} dω. (Key point: Method)

2. What is h(t)?
   A. sin(ωc t)/(π t)
   B. sin(ωc t)/t
   C. cos(ωc t)/(π t)
   D. e^{-ωc t}u(t)
   Answer: A. Evaluating the integral gives sin(ωc t)/(π t). (Key point: Result)

3. What is h(0)?
   A. 0
   B. ωc/π
   C. 1
   D. π/ωc
   Answer: B. The limit of sin(ωc t)/(πt) as t -> 0 is ωc/π. (Key point: Peak value)

4. Why is the ideal LPF not physically realizable?
   A. h(t) is zero everywhere
   B. It has too small a bandwidth
   C. h(t) is a pure delay
   D. h(t) is non-zero for t < 0 (non-causal)
   Answer: D. The sinc extends to negative time, so output precedes input. (Key point: Realizability)

## Close variants set in other exams

- [Ideal LPF impulse and step response](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aideal-lpf-impulse-step) — Signals and Systems, chapter 5, asked 5×

## Related in Signals and Systems

- [Impulse response of DT ideal low pass filter](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aideal-lpf-discrete-time) — Signals and Systems, chapter 6
- [Frequency response; ideal LPF impulse response](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aideal-lpf-cutoff-impulse) — Signals and Systems, chapter 5, asked 2×
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- [Ideal LPF impulse response; realizable?](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aideal-lpf-impulse-realizable) — Signals and Systems, chapter 5
- [Ideal band pass and low pass filters (DT)](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aideal-bandpass-lowpass-dt) — Signals and Systems, chapter 6

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