Question:
The Dirac-delta function \( \delta(t - t_0) \) for \( t, t_0 \in \mathbb{R} \), has the following property
The Laplace transform of the Dirac-delta function \( \delta(t - a) \) for \( a>0 \); \( \mathcal{L}(\delta(t - a)) = F(s) \) is
The Dirac-delta function \( \delta(t - t_0) \) for \( t, t_0 \in \mathbb{R} \), has the following property
The Laplace transform of the Dirac-delta function \( \delta(t - a) \) for \( a>0 \); \( \mathcal{L}(\delta(t - a)) = F(s) \) is
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The Laplace transform of the Dirac delta function \( \delta(t - a) \) is \( e^{-sa} \), which is a standard result from the properties of the delta function in Laplace transforms.
Updated On: Dec 19, 2025
- 0
- \( \infty \)
- \( e^{sa} \)
- \( e^{-sa} \)
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The Correct Option is D
Solution and Explanation
The Dirac delta function \( \delta(t - t_0) \) has the property that for a test function \( \varphi(t) \), the integral involving the delta function evaluates to \( \varphi(t_0) \) if \( t_0 \) lies between the limits of integration, and zero otherwise. This is known as the sifting property of the delta function.
Now, we need to find the Laplace transform of \( \delta(t - a) \), which is defined as: \[ \mathcal{L}(\delta(t - a)) = \int_0^\infty e^{-st} \delta(t - a) \, dt \] Using the sifting property, we get: \[ \mathcal{L}(\delta(t - a)) = e^{-sa} \] Thus, the correct Laplace transform of \( \delta(t - a) \) is \( e^{-sa} \). Final Answer: (D)
Now, we need to find the Laplace transform of \( \delta(t - a) \), which is defined as: \[ \mathcal{L}(\delta(t - a)) = \int_0^\infty e^{-st} \delta(t - a) \, dt \] Using the sifting property, we get: \[ \mathcal{L}(\delta(t - a)) = e^{-sa} \] Thus, the correct Laplace transform of \( \delta(t - a) \) is \( e^{-sa} \). Final Answer: (D)
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