--- title: "Ideal LPF is not realizable" subject: "Signals and Systems" chapter: 5 key: "semester_4/signals_and_systems:ideal-lpf-not-realizable" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ideal-lpf-not-realizable site: BCT Engineering --- # Ideal LPF is not realizable Signals and Systems · Chapter 5: Continuous-Time Systems - Asked in: 2079 Jestha, 7 marks - Video: [why ideal filters are not realizable? | Signal Processing | proved](https://www.youtube.com/watch?v=suMwBbMqZr4) — AIwithAtta - 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-not-realizable - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5 ## Question Show that ideal low pass filter is not practically realizable. ## Short answer Its impulse response h(t)= sin(ωc t)/(π t) is non-zero for t<0, violating causality. Even with a delay, the sinc tail extends to negative infinity. Also the Paley-Wiener criterion fails because stopband gain is zero over a band. ## Answer The ideal low-pass filter passes all frequencies below the cutoff $\omega_c$ with unit gain and rejects all others: $$ H(j\omega) = \begin{cases} 1, & |\omega| \le \omega_c \\ 0, & |\omega| > \omega_c \end{cases} $$ A physically realizable system must be **causal**: for an LTI system this means $h(t) = 0$ for $t < 0$. #### Step 1: Impulse response 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 ``` #### Step 2: Check causality $h(t)$ is an even sinc function. It has its peak $\omega_c/\pi$ at $t = 0$ and oscillating tails on **both** sides, extending to $t = -\infty$. For example, at $t = -\pi/(2\omega_c)$: $$ h\!\left(-\frac{\pi}{2\omega_c}\right) = \frac{\sin(-\pi/2)}{\pi(-\pi/2\omega_c)} = \frac{2\omega_c}{\pi^2} \neq 0 $$ So $h(t) \ne 0$ for $t < 0$: the filter must produce output before the impulse is applied at $t = 0$. #### Step 3: Delay does not help With linear phase, $H(j\omega) = e^{-j\omega t_0}$ for $|\omega| < \omega_c$, giving $$ h(t) = \frac{\sin\omega_c(t - t_0)}{\pi(t - t_0)} $$ This is centred at $t_0$, but its tail still extends to $-\infty$, so $h(t) \ne 0$ for some $t < 0$ for every finite $t_0$. #### Step 4: Paley–Wiener criterion A causal filter with magnitude $|H(j\omega)|$ must satisfy $$ \int_{-\infty}^{\infty}\frac{\big|\ln|H(j\omega)|\big|}{1 + \omega^2}\,d\omega < \infty $$ For the ideal LPF, $|H| = 0$ over the whole stopband, so $|\ln|H|| = \infty$ there and the integral diverges. A causal filter can have zero gain only at isolated frequencies, never over a band. Hence the ideal low-pass filter is **non-causal and not physically realizable**. Practical filters such as Butterworth and Chebyshev only approximate it, accepting a finite transition band, some passband ripple or stopband leakage, and a non-linear phase. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What must the impulse response of a physically realizable LTI system satisfy? A. h(t) = 0 for t < 0 (causal) B. h(t) = 0 for t > 0 C. h(t) must be even D. h(t) must be a sinc Answer: A. A physical system cannot respond before the input is applied, so h(t)=0 for t<0. (Key point: Condition for realizability) 2. Why is the ideal LPF not causal? A. Its impulse response is zero for all t B. Its sinc impulse response is non-zero for t < 0 C. Its gain is greater than 1 D. Its cutoff is infinite Answer: B. h(t)=sin(ωc t)/(πt) is even with tails to -∞, e.g. h(-π/2ωc)=2ωc/π^2 ≠ 0. (Key point: Why ideal LPF fails) 3. Does adding a linear-phase delay t0 make the ideal LPF causal? A. Yes, for t0 larger than π/ωc B. Yes, for any t0 > 0 C. No, the tail still extends to -∞ for any finite t0 D. Only if ωc is made infinite Answer: C. The delayed response sin(ωc(t-t0))/(π(t-t0)) is centred at t0 but its tail never ends at -∞. (Key point: Effect of delay) 4. Why does the ideal LPF violate the Paley-Wiener criterion? A. Its phase is non-linear B. |H| is too large in the passband C. The integral converges to zero D. |H| = 0 over a whole band, so |ln|H|| is infinite and the integral diverges Answer: D. A causal filter can have zero gain only at isolated frequencies, not over a band. 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