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According to the Laplace time-displacement (delay) theorem, a shift of +T in the time domain corresponds to what operation in the s-domain?

Difficulty: Easy

Correct Answer: Multiplication by e^{-sT}

Explanation:


Introduction / Context:
Time-delay modeling is central in control systems, communications, and signal processing. In the Laplace domain, delays appear as exponential factors that modify the phase of the frequency response without changing magnitude for pure transport lags.


Given Data / Assumptions:

  • We analyze the unilateral Laplace transform.
  • A delay of T ≥ 0 is applied to a causal signal.
  • No growth/decay factors other than the delay are present.


Concept / Approach:
The property: £{f(t − T)u(t − T)} = e^{-sT}F(s) says that a time shift of T corresponds to multiplication by e^{-sT} in the s-domain. This is used to incorporate transport lags and piecewise onset times in block-diagram algebra.


Step-by-Step Solution:

Start from definition and substitute t = τ + T.The exponential e^{-sT} factors out of the integral.Remaining integral equals F(s), giving e^{-sT}F(s).


Verification / Alternative check:

For f(t) = u(t), delay T gives transform e^{-sT}/s, which matches standard tables.


Why Other Options Are Wrong:

Division by s or multiplication by s corresponds to integration/differentiation, not delay.Division by e^{-sT} implies time advance.Replacing s by s + T is a different frequency-shift operation.


Common Pitfalls:

Dropping the unit step u(t − T) in the time shift property.


Final Answer:

Multiplication by e^{-sT}
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