02 - First Order Systems

From mass or energy balances, we get for the output variable \(y(t)\), \[\begin{align*} a_1\frac{dy}{dt} + a_0 y(t) &= b x(t) \\ a_1\frac{dy}{dt} + a_0 y &= b x \\ \text{where $x(t)$ is the input. If $a_0\ne0$, then}\\  \frac{a_1}{a_0}\frac{dy}{dt} + y &= \frac{b}{a_0}x(t) \\ \tau_p \frac{dy}{dt} + y &= K_p x(t) \tag*{(1)} \end{align*}\] where \(\tau_p\) is known as time constant of the process, and \(K_p\) is called the steady-state gain or static gain or simply the gain of the process.

If \(Y(t) = y(t)-y_s\) and \(X(t) = x(t)- x_s\) are in terms of deviation variables around a steady state, then the initial conditions are: \[Y(0) = y(0) - y_s = y_s - y_s = 0 \qquad \text{and} \qquad X(0) = x(0) - x_s = 0\] Using the above conditions, and taking the Laplace transform for Eqn.(1), we get \[G(s) = \frac{Y(s)}{X(s)} = \frac{K_p}{\tau_p s+1}\] Because of the usage of deviation variables, the Laplace transform of the differential equation results in an equation that is free of initial conditions, because the initial values of \(X\) and \(Y\) are zero.