Question

Let \( u(t) \) denote the unit step function. The bilateral Laplace transform of the function \( f(t) = e^t u(-t) \) is \(\underline{\hspace{2cm}}\) .

• A. \( \frac{1}{s-1} \) with real part of \( s < 1 \)
• B. \( \frac{1}{s-1} \) with real part of \( s > 1 \)
• C. \( \frac{-1}{s-1} \) with real part of \( s < 1 \)
• D. \( \frac{-1}{s-1} \) with real part of \( s > 1 \)

Hint:

For bilateral Laplace transforms involving unit step functions, pay attention to the region of convergence and the sign changes due to \( u(-t) \).

Answer 1

The Correct Option is C

The given function is \( f(t) = e^t u(-t) \), where \( u(t) \) is the unit step function. The unit step function \( u(-t) \) causes the function to be zero for \( t \geq 0 \), and the Laplace transform of the function will be computed over the interval from \( -\infty \) to \( 0 \). The Laplace transform of \( e^t u(-t) \) is computed using the bilateral Laplace transform, and the result is: \[ \mathcal{L}\{e^t u(-t)\} = \frac{-1}{s-1} \text{for} \text{Re}(s) < 1. \] Thus, the correct answer is (C). Final Answer: \( \frac{-1}{s-1} \) with real part of \( s < 1 \).