Question

\( \mathcal{L}\{ \sin(t-3)u(t-3) \} = \)

• A. \( \frac{1}{s^2 + 1} \)
• B. \( \frac{3}{s^2 + 9} \)
• C. \( \frac{e^{-3s}}{s^2 + 1} \)
• D. \( \frac{e^{-3s}}{s^2 + 9} \)

Hint:

The unit step multiplier \( u(t-a) \) always transforms into an exponential factor \( e^{-as} \) in the \(s\)-domain. Once you factor that out, you only need to compute the standard transform of the unshifted function.

Answer 1

The Correct Option is C

Concept: The problem asks for the Laplace transform of a shifted function multiplied by a unit step function \( u(t-a) \) (also written as \( H(t-a) \)). To solve this, we use the Second Shifting Theorem of Laplace transforms: \[ \mathcal{L}\{ f(t-a) u(t-a) \} = e^{-as} \mathcal{L}\{ f(t) \} = e^{-as} F(s) \] Additionally, we use the standard Laplace transform for a sine wave function: \[ \mathcal{L}\{ \sin(\omega t) \} = \frac{\omega}{s^2 + \omega^2} \]

Step 1: Identifying the parameters from the given function.
The given expression to transform is: \[ \mathcal{L}\{ \sin(t-3)u(t-3) \} \] By comparing this directly with the standard formula \( f(t-a)u(t-a) \), we can identify our parameters:

• The shift constant is \( a = 3 \).

• The shifted function is \( f(t-3) = \sin(t-3) \).
Replacing \( (t-3) \) with an unshifted variable \( t \) gives the base function: \[ f(t) = \sin(t) \]

Step 2: Finding the Laplace transform of the unshifted base function \( f(t) \).
Now, let us calculate the Laplace transform \( F(s) \) of our base function \( f(t) = \sin(t) \). Here, the frequency coefficient is \( \omega = 1 \): \[ F(s) = \mathcal{L}\{ \sin(t) \} = \frac{1}{s^2 + 1^2} = \frac{1}{s^2 + 1} \]

Step 3: Applying the Second Shifting Theorem formula.
Now we combine our results using the Second Shifting Theorem: \[ \mathcal{L}\{ \sin(t-3)u(t-3) \} = e^{-3s} F(s) \] Substituting our calculated value for \( F(s) \): \[ \mathcal{L}\{ \sin(t-3)u(t-3) \} = e^{-3s} \cdot \left( \frac{1}{s^2 + 1} \right) = \frac{e^{-3s}}{s^2 + 1} \] This final form matches Option (C).