Question

The Laplace transform of \( e^{-2t} \cos(\sqrt{2} t) \) is .......

• A. \( \frac{s - 2}{(s - 2)^2 + 2},\ s > 2 \)
• B. \( \frac{2}{(s + 2)^2 + 2},\ s > -2 \)
• C. \( \frac{\sqrt{2}}{(s + 2)^2 + 2},\ s > -2 \)
• D. \( \frac{s + 2}{(s + 2)^2 + 2},\ s > -2 \)

Hint:

Use the Laplace transform shifting theorem: \( \mathcal{L}\{ e^{-at} f(t) \} = F(s + a) \), where \( F(s) \) is the Laplace of \( f(t) \).

Answer 1

The Correct Option is D

We use the Laplace transform formula: \[ \mathcal{L} \{ e^{at} \cdot f(t) \} = F(s - a) \] Given: \[ f(t) = \cos(\sqrt{2}t), \mathcal{L}\{ \cos(bt) \} = \frac{s}{s^2 + b^2} \] So, \[ \mathcal{L}\{ \cos(\sqrt{2}t) \} = \frac{s}{s^2 + 2} \] Now for: \[ \mathcal{L}\{ e^{-2t} \cos(\sqrt{2}t) \} \] Using the shift property: \[ = \mathcal{L}\{ \cos(\sqrt{2}t) \}(s + 2) = \frac{s + 2}{(s + 2)^2 + 2} \] This is valid for \( s>-2 \).