Question

The output \( y(t) \) of a first-order process is governed by the following differential equation: \[ \tau_p \frac{dy}{dt} + y = K_p f(t) \] where \( \tau_p \) is a non-zero time constant, \( K_p \) is the gain, and \( f(t) \) is the input with \( f(0) = 0 \). Assume \( y(0) = 0 \). The transfer function for this process is (consider \( s \) as the independent variable in the Laplace domain).

• A. \( \frac{K_p}{\tau_p s + 1} \)
• B. \( \frac{\tau_p}{K_p s + 1} \)
• C. \( \frac{\tau_p}{K_p (s + 1)} \)
• D. \( \frac{K_p}{\tau_p (s + 1)} \)

Hint:

The transfer function of a first-order system can be derived by taking the Laplace transform of the system’s governing differential equation.

Answer 1

The Correct Option is A

The given differential equation is:

\[ \tau_p \frac{dy}{dt} + y = K_p f(t) \]
We need to find the transfer function, which is the Laplace transform of the output-to-input ratio.

Step 1: Take the Laplace transform of both sides

Taking the Laplace transform of the differential equation, we get:
\[ \tau_p s Y(s) + Y(s) = K_p F(s) \]
where \( Y(s) \) and \( F(s) \) are the Laplace transforms of \( y(t) \) and \( f(t) \), respectively, and we assume initial conditions \( y(0) = 0 \) and \( f(0) = 0 \).

Step 2: Solve for \( Y(s) / F(s) \)

Rearranging the equation to isolate \( Y(s) \):
\[ Y(s) \left( \tau_p s + 1 \right) = K_p F(s) \]
Thus,
\[ \frac{Y(s)}{F(s)} = \frac{K_p}{\tau_p s + 1} \]

Final Answer:
The transfer function is \( \boxed{\frac{K_p}{\tau_p s + 1}} \).