Process Dynamics: Transfer Function Derivation

The problem asks for the transfer function of a first-order process described by the differential equation: $ \tau_p\frac{dy}{ dt} + y = K_p f(t) $ with initial conditions $y(0) = 0$ and $f(0) = 0$. The transfer function relates the output $Y(s)$ to the input $F(s)$ in the Laplace domain.

Laplace Transform Application

To find the transfer function, we take the Laplace transform of the differential equation. We use the following standard Laplace transform properties:

• $ \mathcal{L}\left\{\frac{dx(t)}{dt}\right\} = sX(s) - x(0) $
• $ \mathcal{L}\{x(t)\} = X(s) $

Applying the Laplace transform to the given equation:

$ \mathcal{L}\left\{\tau_p\frac{dy}{ dt} + y\right\} = \mathcal{L}\{K_p f(t)\} $

$ \tau_p \mathcal{L}\left\{\frac{dy}{ dt}\right\} + \mathcal{L}\{y\} = K_p \mathcal{L}\{f(t)\} $

Applying Initial Conditions

Substitute the Laplace transforms and the initial conditions ($y(0) = 0$, $f(0) = 0$):

$ \tau_p (sY(s) - y(0)) + Y(s) = K_p F(s) $

$ \tau_p (sY(s) - 0) + Y(s) = K_p F(s) $

$ \tau_p s Y(s) + Y(s) = K_p F(s) $

Calculating the Transfer Function

Factor out $Y(s)$ from the left side:

$ Y(s)(\tau_p s + 1) = K_p F(s) $

The transfer function $G(s)$ is the ratio of the output transform $Y(s)$ to the input transform $F(s)$:

$ G(s) = \frac{Y(s)}{F(s)} $

Rearranging the equation to solve for $G(s)$:

$ G(s) = \frac{K_p}{\tau_p s + 1} $

Final Answer

This result matches option 1. The transfer function for the given first-order process is $ \frac{K_p}{\tau_p s + 1} $.