Question

A first-order system has transfer function $H(s) = \frac{1}{(s+2)}$. Its impulse response is:}

• A. $2e^{-t}u(t)$
• B. $2e^{-2t}u(t)$
• C. $e^{-2t}u(t)$
• D. $e^{-t}u(t)$

Hint:

The core Laplace transform pair $\frac{1}{s+a} \longleftrightarrow e^{-at}u(t)$ is a fundamental relationship in signal analysis. Substituting $a=2$ immediately gives the impulse response: $e^{-2t}u(t)$.

Answer 1

The Correct Option is C

Concept: The impulse response, denoted as $h(t)$, of a continuous-time Linear Time-Invariant (LTI) system is the time-domain output produced when an ideal Dirac delta function ($\delta(t)$) is applied to the input under zero initial conditions. Mathematically, the impulse response is the Inverse Laplace Transform of the system's transfer function $H(s)$: \[ h(t) = \mathcal{L}^{-1}\{H(s)\} \] A foundational Laplace transform pair states that a right-sided exponential decay function maps to a simple single-pole fraction in the s-domain: \[ \mathcal{L}\{e^{-at}u(t)\} = \frac{1}{s+a} \quad \text{with } \text{Re}(s) > -a \]

Step 1: Applying the inverse transformation directly.
We are given the system transfer function: \[ H(s) = \frac{1}{s+2} \] To find the impulse response, we apply the inverse Laplace transform operator to both sides of the equation: \[ h(t) = \mathcal{L}^{-1}\left\{ \frac{1}{s+2} \right\} \]

Step 2: Identifying the pole constant.
By comparing our system function $\frac{1}{s+2}$ with the standard template $\frac{1}{s+a}$, we find the pole constant $a$: \[ a = 2 \] Substituting $a = 2$ into the time-domain exponential transform pair yields: \[ h(t) = e^{-2t}u(t) \] Where $u(t)$ represents the standard Heaviside step function, which enforces causality by setting the expression to zero for all negative time values ($t < 0$). This result matches Option (C).