Question

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

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

Hint:

When dealing with time-delayed inputs like \( x(t - a)u(t - a) \), remember that the Laplace transform introduces a factor of \( e^{-as} \). This is crucial in identifying the correct transfer function in systems involving delays.

Answer 1

The Correct Option is C

Given system: \[ \frac{dy(t)}{dt} + y(t) = 3x(t-3)u(t-3) \] Take Laplace transform of both sides assuming zero initial conditions: \[ sY(s) + Y(s) = 3X(s)e^{-3s} \] Factor \( Y(s) \) on the left-hand side: \[ Y(s)(s+1) = 3X(s)e^{-3s} \] Divide both sides by \( (s+1) \) to get the transfer function: \[ \frac{Y(s)}{X(s)} = \frac{3e^{-3s}}{s+1} \] Hence, the transfer function of the system is: \[ \boxed{\frac{3e^{-3s}}{s+1}} \]