--- title: "Impulse response of DT ideal low pass filter" subject: "Signals and Systems" chapter: 6 key: "semester_4/signals_and_systems:ideal-lpf-discrete-time" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ideal-lpf-discrete-time site: BCT Engineering --- # Impulse response of DT ideal low pass filter Signals and Systems · Chapter 6: Discrete-Time Systems - Asked in: 2076 Baisakh, 5 marks - Video: [Impulse response of ideal discrete time lowpass filter](https://www.youtube.com/watch?v=elnx7_w-Mg8) — ProfKathleenWage - 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-discrete-time - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6 ## Question Derive formula to calculate the impulse response of ideal low pass filter in discrete time. ## Short answer The impulse response of an ideal discrete-time lowpass filter with cutoff ω_c is h[n] = sin(ω_c n)/(π n) = (ω_c/π) sinc(ω_c n/π) for n ≠ 0, and h[0] = ω_c/π. It is non‑causal and not absolutely summable (unstable). ## Answer An **ideal discrete-time low pass filter** passes all frequencies below a cut-off $\omega_c$ with unit gain and blocks the rest. Its frequency response over one period ($-\pi\le\omega\le\pi$) is $$ H(e^{j\omega})=\begin{cases}1, & |\omega|\le\omega_c\\ 0, & \omega_c<|\omega|\le\pi\end{cases} $$ and it repeats every $2\pi$. ```text H(e^jw) 1 +-----+ | | ----+------+-----+------+---> w -pi -wc 0 wc pi ``` #### Derivation The impulse response is the inverse DTFT: $$ \begin{aligned} h[n] &= \frac{1}{2\pi}\int_{-\pi}^{\pi}H(e^{j\omega})e^{j\omega n}\,d\omega = \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c}e^{j\omega n}\,d\omega \\ &= \frac{1}{2\pi}\left[\frac{e^{j\omega n}}{jn}\right]_{-\omega_c}^{\omega_c} = \frac{1}{2\pi}\cdot\frac{e^{j\omega_c n}-e^{-j\omega_c n}}{jn} \\ &= \frac{1}{2\pi}\cdot\frac{2j\sin(\omega_c n)}{jn} = \frac{\sin(\omega_c n)}{\pi n}, \qquad n\ne 0 \end{aligned} $$ At $n=0$: $$ h[0]=\frac{1}{2\pi}\int_{-\omega_c}^{\omega_c}d\omega=\frac{\omega_c}{\pi} $$ which is also the limit of $\frac{\sin\omega_c n}{\pi n}$ as $n\to0$. So $$ h[n]=\frac{\sin(\omega_c n)}{\pi n}=\frac{\omega_c}{\pi}\,\mathrm{sinc}\!\left(\frac{\omega_c n}{\pi}\right) $$ where $\mathrm{sinc}\,x=\frac{\sin \pi x}{\pi x}$ (the normalised sinc). #### Remarks - $h[n]$ is a sampled sinc: maximum $\omega_c/\pi$ at $n=0$, zero crossings where $\omega_c n$ is a multiple of $\pi$. - $h[n]\ne0$ for $n<0$, so the ideal LPF is **non-causal**. - $\sum|h[n]|$ diverges (decays only as $1/n$), so it is **not BIBO stable**. Hence it cannot be built exactly; practical filters truncate/window $h[n]$ and delay it. - Example: for $\omega_c=\pi/2$, $h[n]=\frac{\sin(\pi n/2)}{\pi n}$: $h[0]=0.5$, $h[\pm1]=1/\pi\approx0.318$, $h[\pm2]=0$, $h[\pm3]=-1/(3\pi)\approx-0.106$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The impulse response of the ideal LPF is found by: A. Inverse DTFT: h[n] = (1/2pi) integral of H(e^jw)e^(jwn) dw over one period B. Differentiating H C. Taking the forward DTFT of H D. Integrating over all w from -infinity to infinity Answer: A. h[n] is the inverse DTFT, integrated over -pi to pi where H is 1 only for |w| <= wc. (Key point: Method) 2. The result of the derivation is: A. h[n] = sin(wc n)/(pi n) B. h[n] = cos(wc n)/(pi n) C. h[n] = sin(wc n)/n D. h[n] = wc/pi for all n Answer: A. Integrating gives (e^(jwc n) - e^(-jwc n))/(2 pi j n) = sin(wc n)/(pi n). (Key point: Result) 3. What is h[0]? A. 0 B. wc/pi C. 1 D. infinity Answer: B. h[0] = (1/2pi)(2wc) = wc/pi, also the limit of sin(wc n)/(pi n). (Key point: Value at origin) 4. Why can the ideal LPF not be built exactly? A. It is non-causal (h[n] nonzero for n < 0) and not BIBO stable B. It has a negative gain C. Its h[n] is complex D. It is periodic Answer: A. The sinc h[n] extends to negative n and decays only as 1/n, so it is not absolutely summable. 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