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For the time-shifted impulse signal f(t) = A·δ(t − a), what is its unilateral Laplace transform F(s)?

Difficulty: Easy

Correct Answer: A e^{-a s}

Explanation:


Introduction / Context:
The Laplace transform of shifted impulses is a standard identity used widely in control systems, circuit analysis, and signals. Recognizing how δ(t − a) maps to the s-domain is crucial for modeling inputs that occur at specific times.


Given Data / Assumptions:

  • f(t) = A·δ(t − a), where a ≥ 0 and A is a constant amplitude.
  • Unilateral Laplace transform convention is used.


Concept / Approach:
By definition, the Laplace transform L{δ(t − a)} = e^{-a s}, reflecting a pure delay of a seconds. A constant amplitude A scales the result linearly. Thus, L{A·δ(t − a)} = A e^{-a s}.


Step-by-Step Solution:

F(s) = ∫_0^∞ A·δ(t − a)·e^{-s t} dt.Impulse sampling property: ∫ δ(t − a)·g(t) dt = g(a) for a ≥ 0.Therefore, F(s) = A·e^{-s a}.


Verification / Alternative check:

Time shift in the time domain corresponds to multiplying by e^{-s a} in the s-domain for impulses and for general delayed signals with u(t − a).


Why Other Options Are Wrong:

A e^{a s}: represents a time advance, not a delay.A·a·e^{-a s} or A·a·e^{a s}: incorrectly introduces an extra factor a.A / (s + a): corresponds to L{A·e^{-a t}} for a different function, not a shifted delta.


Common Pitfalls:

Confusing delayed exponentials with delayed impulses; the latter transform to pure exponentials in s.


Final Answer:

A e^{-a s}
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