Exponential weighting property of Laplace transform: Multiplication by e^{-a t} in time corresponds to which operation in the s-domain?
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ATranslation by +a in s-domain: F(s + a)
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BTranslation by −a in s-domain: F(s − a)
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CMultiplication by e^{-a s} in s-domain
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DNo change in the s-domain
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EConjugation of F(s)
Answer
Correct Answer: Translation by +a in s-domain: F(s + a)
Explanation
Introduction / Context:The frequency-shift (exponential weighting) property of the Laplace transform is a key tool for handling signals multiplied by exponentials, as in modulation, windowing, and system responses with damping or growth.
Given Data / Assumptions:
- Time-domain multiplication: x(t) e^{-a t}.
- Baseline transform: X(s) = L{x(t)}.
- We assume regions of convergence are adjusted accordingly.
Concept / Approach:
The property states L{x(t) e^{-a t}} = X(s + a). Thus multiplying by a decaying exponential shifts the transform to the right by a in the s-plane (i.e., replaces s with s + a).
Step-by-Step Solution:
Start with the definition: L{x(t) e^{-a t}} = ∫{0}^{∞} x(t) e^{-a t} e^{-s t} dt.Combine exponents: = ∫{0}^{∞} x(t) e^{-(s + a) t} dt.Recognize as X(s + a): a shift of +a in the s variable.Verification / Alternative check:
Test with x(t) = 1 (unit step): L{e^{-a t}} = 1/(s + a), which equals X(s + a) for X(s) = 1/s, confirming the property.
Why Other Options Are Wrong:
- Translation by −a corresponds to multiplying by e^{+a t}, not e^{−a t}.
- Multiplication by e^{-a s} in the s-domain is not a standard Laplace property for this operation.
- No change or conjugation are not applicable here.
Common Pitfalls:
- Sign confusion: e^{-a t} → X(s + a); e^{+a t} → X(s − a).
- Ignoring changes in the region of convergence after the shift.
Final Answer:
Translation by +a in s-domain: F(s + a)