Question

The Laplace transform of \( u(t) \) and its ROC is given as

• A. \( \frac{1}{s}, \, \sigma<0 \)
• B. \( \frac{1}{s}, \, \sigma>0 \)
• C. \( \frac{1}{s - 1}, \, \sigma = 0 \)
• D. \( \frac{1}{1 - s}, \, \sigma \leq 0 \)

Hint:

For the unit step function \( u(t) \), always remember its Laplace transform is \( \frac{1}{s} \) with the region of convergence (ROC) \( \Re(s)>0 \). This ROC is important for ensuring the existence of the transform.

Answer 1

The Correct Option is B

To solve this problem, we need to analyze the Laplace transform of the step function \( u(t) \) and its Region of Convergence (ROC).

1. Understanding the Concepts:

- Laplace Transform of \( u(t) \): The step function \( u(t) \) is defined as:

\[ u(t) = \begin{cases} 0 & t < 0 \\ 1 & t \geq 0 \end{cases} \]

The Laplace transform of \( u(t) \) is given by:

\[ \mathcal{L}\{u(t)\} = \frac{1}{s} \]

- Region of Convergence (ROC): The ROC for the Laplace transform of a function is the range of values for which the Laplace transform converges. For \( u(t) \), the Laplace transform \( \frac{1}{s} \) converges when the real part of \( s \) is greater than 0, i.e., \( \sigma > 0 \), where \( s = \sigma + j\omega \) and \( \sigma \) is the real part.

2. Given Values:

The Laplace transform of \( u(t) \) is:

\[ \mathcal{L}\{u(t)\} = \frac{1}{s} \]

The ROC for this Laplace transform is \( \sigma > 0 \), as the integral converges for positive real parts of \( s \). This is because \( u(t) \) is a causal function that starts at \( t = 0 \) and its Laplace transform converges for \( \sigma > 0 \).

Final Answer:

The Laplace transform of \( u(t) \) and its ROC is \( \frac{1}{s}, \, \sigma > 0 \).