---
title: "LTI: causality and stability"
subject: "Signals and Systems"
chapter: 5
key: "semester_4/signals_and_systems:lti-causality-stability"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/lti-causality-stability
site: BCT Engineering
---

# LTI: causality and stability

Signals and Systems · Chapter 5: Continuous-Time Systems

- Asked in: 2083 Bhadra (new course), 5 marks
- Video: [Causal LTI System](https://www.youtube.com/watch?v=xdxHuw7Rs8k) — Neso Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/lti-causality-stability
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5

## Question

Define LTI system. Explain causality and stability property of continuous time LTI system.

## Short answer

An LTI system is linear and time invariant; its output is y(t)=x(t)*h(t). It is causal iff h(t)=0 for t<0, and BIBO stable iff the impulse response is absolutely integrable, integral of |h(t)| dt finite.

## Answer

A **Linear Time-Invariant (LTI) system** is a system that obeys superposition (linear) and whose response to a delayed input is the same response delayed (time invariant). Its output for any input is given by convolution with its impulse response:

$$
y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} h(\tau)\,x(t-\tau)\,d\tau
$$

#### Causality

A system is **causal** if its output at time $t$ depends only on inputs at times $\le t$.

In the convolution, the term $x(t-\tau)$ for $\tau < 0$ is a future input value. For it never to be used, the weight $h(\tau)$ must be zero there:

$$
h(t) = 0 \quad \text{for } t < 0
$$

Then $y(t) = \int_0^{\infty} h(\tau)\,x(t-\tau)\,d\tau$.

- Causal: $h(t) = e^{-2t}u(t)$, $h(t) = u(t) - u(t-1)$.
- Non-causal: $h(t) = e^{-|t|}$, $h(t) = \delta(t+2)$.
- Physically realizable real-time systems must be causal.

#### Stability

A system is **BIBO stable** if every bounded input, $|x(t)| \le M_x$, gives a bounded output.

$$
|y(t)| = \left|\int h(\tau)x(t-\tau)d\tau\right| \le \int|h(\tau)|\,|x(t-\tau)|\,d\tau \le M_x\int_{-\infty}^{\infty}|h(\tau)|\,d\tau
$$

So the output is bounded if

$$
\int_{-\infty}^{\infty} |h(t)|\,dt < \infty
$$

(the impulse response is **absolutely integrable**). This is also necessary: if the integral is infinite, the bounded input $x(t) = \mathrm{sgn}[h(-t)]$ gives $y(0) = \int|h(\tau)|d\tau = \infty$.

- Stable: $h(t) = e^{-2t}u(t)$, since $\int_0^\infty e^{-2t}dt = 0.5$.
- Unstable: $h(t) = u(t)$ (integrator), $h(t) = e^{2t}u(t)$.

## Check yourself

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

1. What is the output of an LTI system?
   A. y(t) = x(t) h(t)
   B. y(t) = integral of h(tau) x(t - tau) d tau
   C. y(t) = x(t) + h(t)
   D. y(t) = dh/dt * x(t)
   Answer: B. LTI output is convolution of the input with the impulse response. (Key point: Convolution form)

2. Why must h(t) = 0 for t < 0 in a causal LTI system?
   A. So that the integral converges
   B. So the system is stable
   C. So the system is linear
   D. Because x(t - tau) for tau < 0 is a future input and must get zero weight
   Answer: D. For negative tau the term x(t - tau) uses future values, so their weights must vanish. (Key point: Causality condition)

3. The BIBO stability condition for LTI is:
   A. h(t) -> 0 as t -> inf only
   B. The integral of |h(t)| over all t is finite
   C. The integral of h(t) is zero
   D. h(t) = 0 for t < 0
   Answer: B. The bound |y| <= Mx times the integral of |h| shows sufficiency. (Key point: Stability condition)

4. Which h(t) represents a stable system?
   A. e^(2t) u(t)
   B. u(t)
   C. e^(-2t) u(t), with integral 0.5
   D. delta(t + 2) and e^(2t)u(t)
   Answer: C. The integral of e^(-2t) from 0 to inf is 0.5; u(t) (integrator) and e^(2t)u(t) are unstable. (Key point: Stable example)

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