The Laplace transform of exp(at), where a > 0, is defined only for the Laplace parameter, s > a since______________?
Correct answer: B. The Laplace transform of integral of exp(at) has finite values only for s > a
- A. The function is exponential
- B. The Laplace transform of integral of exp(at) has finite values only for s > a
- C. The Laplace transform integral of exp(at) has initial values only for s > a
- D. The function exp(at) is piece-wise continuous only for s > a
Explanation
The transform integral is ∫₀∞e^{-(s-a)t}dt, which converges only when s-a>0. Therefore the Laplace transform exists only for s>a.
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