Question

The input \( x(t) \) to a system is related to its output \( y(t) \) as
\[ \frac{d y(t)}{dt} + y(t) = 3x(t - 3)u(t - 3) \] \text{Here \( u(t) \) represents a unit-step function.}
\text{The transfer function of this system is _________}

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

Hint:

When taking the Laplace transform of a system involving delays, the exponential term \( e^{-as} \) appears, representing the time shift.

Answer 1

The Correct Option is B

Step 1: Taking the Laplace Transform.
The given equation is a differential equation. Taking the Laplace transform of both sides (and using the properties of the Laplace transform), we can transform the time-domain equation into the s-domain. The Laplace transform of \( \frac{d}{dt}y(t) \) is \( sY(s) \), and the Laplace transform of \( x(t-3) \) is \( X(s)e^{-3s} \). The equation becomes: \[ sY(s) + Y(s) = 3X(s)e^{-3s} \] Now, solve for the transfer function \( \frac{Y(s)}{X(s)} \).
Step 2: Simplifying the expression.
The transfer function is: \[ \frac{Y(s)}{X(s)} = \frac{3e^{-3s}}{s + 1} \] Thus, the correct transfer function is \( \frac{3e^{-3s}}{s + 1} \). Step 3: Conclusion.
The correct answer is (B) \( \frac{3e^{-3s}}{s + 1} \).