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

To solve this problem, we need to find the transfer function of the given system using the Laplace transform and the given relationship between the input \( x(t) \) and output \( y(t) \).

1. Understanding the Concepts:

- Transfer Function: The transfer function \( H(s) \) of a system relates the Laplace transform of the output to the Laplace transform of the input. It is defined as:

\[ H(s) = \frac{Y(s)}{X(s)} \]

- Laplace Transform: The Laplace transform of a function \( f(t) \) is given by:

\[ F(s) = \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) dt \]

- Step Function \( u(t) \): The step function \( u(t) \) is 0 for \( t < 0 \) and 1 for \( t \geq 0 \). The Laplace transform of \( u(t - a) \) is:

\[ \mathcal{L}\{u(t - a)\} = \frac{e^{-as}}{s} \]

2. Given Values:

The equation relating the input and output is:

\[ \frac{dy(t)}{dt} + y(t) = 3x(t - 3)u(t - 3) \]

Taking the Laplace transform of both sides:

\[ sY(s) + Y(s) = 3X(s)e^{-3s} \]

Rearranging to find the transfer function:

\[ Y(s)\left(s + 1\right) = 3X(s)e^{-3s} \]

\[ H(s) = \frac{Y(s)}{X(s)} = \frac{3e^{-3s}}{s + 1} \]

Final Answer:

The transfer function of the system is \( \frac{3e^{-3s}}{s+1} \).