For the linear time invariant systems that are Bounded Input Bounded stable, which one of the following statement is TRUE?
The unit step response will be bounded
Linear Time-Invariant (LTI) systems are fundamental in signal processing and control theory. A key characteristic used to describe their behavior is stability. Specifically, Bounded-Input Bounded-Output (BIBO) stability is a crucial concept. An LTI system is defined as BIBO stable if, for every input signal that remains bounded over time, the corresponding output signal also remains bounded over time.
For an LTI system, BIBO stability is directly related to its impulse response, denoted as $h(t)$ for continuous-time systems or $h[n]$ for discrete-time systems. The condition for BIBO stability is that the impulse response must be absolutely integrable (for continuous-time) or absolutely summable (for discrete-time). Mathematically, this means:
$$ \int_{-\infty}^{\infty} |h(t)| dt < \infty $$
$$ \sum_{n=-\infty}^{\infty} |h[n]| < \infty $$
If this condition is met, the system is guaranteed to be BIBO stable.
Let's analyze each statement in the context of BIBO stable LTI systems:
This statement is incorrect. As explained above, the core condition for BIBO stability is that the impulse response must be absolutely integrable or summable. Simple integrability is not sufficient.
This statement is not always true. Systems whose impulse response has finite support are known as Finite Impulse Response (FIR) systems. All FIR systems are indeed BIBO stable. However, BIBO stable systems are not limited to FIR systems; Infinite Impulse Response (IIR) systems can also be BIBO stable, provided their impulse response meets the absolute integrability/summability condition. Therefore, stating the impulse response *will* have finite support is too restrictive.
This statement is not necessarily true. The unit step response, $y_s(t)$, is the convolution of the impulse response $h(t)$ with the unit step function $u(t)$, i.e., $y_s(t) = h(t) * u(t)$. While BIBO stability ensures the output is bounded for a bounded input like the unit step, it doesn't guarantee that the resulting step response itself is absolutely integrable. The requirement is on the impulse response, not the step response.
This statement is TRUE. The unit step function, $u(t)$ (or $u[n]$), is a classic example of a bounded input signal. By the definition of BIBO stability, if the input is bounded, the output must also be bounded. Therefore, for a BIBO stable LTI system, applying a unit step input will result in a bounded unit step response.
Based on the definition and conditions for BIBO stability in LTI systems, the only statement that is universally true is that the unit step response will be bounded, as the unit step function itself is a bounded input.
The continuous time system described by the equation y(t) = x(t2) comes under the category of -
A continuous time LTI system is described by
\(\dfrac{d^2y(t)}{dt^2} + 4 \dfrac{dy(t)}{dt} + 3y(t) = 2 \dfrac{dx(t)}{dt} + 4x(t)\)
Assuming zero initial conditions, the response y(t) of the above system for the input x(t) = e-2t u(t) is given by
Consider a continuous-time system with input x(t) and output y(t) given by
y(t) = x(t)cos(t)
This system is
Let a causal LTI system be governed by the following differential equation
\(\rm y(t) + \frac{1}{4} \frac{dy}{dt} = 2x(t)\) where x(𝑡) and y(𝑡) are the input and output respectively.
Its impulse response is
Let an input x(t) = 2 sin(10πt) + 5 cos(15πt) + 7 sin(42πt) + 4 cos(45πt) is passed through an LTI system having an impulse response
\(\rm h(t) = 2 \left( \frac{\sin (10 \pi t)}{\pi t} \right) \cos (40 \pi t)\)
The output of the system is