Question

If \( F(s) \) denotes the Laplace transform of some function \( f(t) \), then the Laplace transform of \( e^{bt} f(t) \), where \( b \) is a real constant, is:

• A. \( F(s-b) \)
• B. \( F(s+b) \)
• C. \( -F(s) \)
• D. \( F(b-s) \)

Hint:

When multiplying a function by \( e^{bt} \), the Laplace transform shifts by \( b \) in the complex frequency domain. This is a standard property: \( \mathcal{L}\{e^{bt} f(t)\} = F(s-b) \).

Answer 1

The Correct Option is A

The Laplace transform of a function \( f(t) \) is given by the following integral expression:
\[ F(s) = \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt \]
This integral transforms the function \( f(t) \) from the time domain to the frequency domain. Now, we are tasked with finding the Laplace transform of the function \( e^{bt} f(t) \), where \( b \) is a real constant.
To derive this, we can use the property of the Laplace transform that relates the shift in the time domain to the frequency domain. The general property is:
\[ \mathcal{L}\{e^{bt} f(t)\} = F(s - b) \]
This result can be derived by recognizing that multiplying the function by \( e^{bt} \) in the time domain results in a shift in the Laplace transform by \( b \) in the complex frequency domain.
Thus, the Laplace transform of \( e^{bt} f(t) \) is \( F(s-b) \), which corresponds to option (A). Therefore, the correct answer is (A).