Find the Laplace transform of the function u(t - 2π) sin(t - 2π). The given answer is e^{-2π s} * 1/(s^2 + 1).
Answer (2)
The Laplace transform of u ( t − 2 π ) sin ( t − 2 π ) is derived using the shift property of the Laplace transform. The final result is e − 2 π s ⋅ s 2 + 1 1 . This transform accounts for the unit step function that shifts the sine function in time.
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To find the Laplace transform of the function u ( t − 2 π ) sin ( t − 2 π ) , we need to use a property of the Laplace transform related to the unit step function and time shifts.
Understand the Function : The function u ( t − 2 π ) is a unit step function that shifts sin ( t − 2 π ) to begin at t = 2 π . This essentially delays the start of the sine function by 2 π units.
Recall the Time Shift Property : The Laplace transform of a function f ( t ) delayed by a units and multiplied by a unit step function u ( t − a ) is given by: L { u ( t − a ) f ( t − a )} = e − a s F ( s ) where F ( s ) is the Laplace transform of f ( t ) .
Identify f ( t ) and its Laplace Transform : In this problem, the inner function f ( t ) = sin ( t ) has a known Laplace transform: L { sin ( t )} = s 2 + 1 1
Apply the Time Shift Property :
The delay here is a = 2 π . Therefore, applying the property, we have: L { u ( t − 2 π ) sin ( t − 2 π )} = e − 2 π s ⋅ s 2 + 1 1
Write the Final Result :
Combining these steps, the Laplace transform is given by: e − 2 π s × s 2 + 1 1
This matches the provided answer and utilizes the step-by-step application of the time shift property associated with the Laplace transform.
