Step response of a first-order system

The first-order model is $\tau\,\dfrac{dq_o}{dt} + q_o = K q_i$. A step input is $q_i = 0$ for $t<0$ and $q_i = q_s$ (constant) for $t\ge 0$, with $q_o(0) = 0$.

Derivation

The complementary solution is $C e^{-t/\tau}$ and the particular solution is $K q_s$. So

$$q_o(t) = K q_s + C e^{-t/\tau}$$

With $q_o(0) = 0$, $C = -K q_s$, giving

$$q_o(t) = K q_s\left(1 - e^{-t/\tau}\right)$$

The normalized response is

$$\frac{q_o}{K q_s} = 1 - e^{-t/\tau}$$

The error from the final value is $e^{-t/\tau}$, which decays exponentially.

Time constant

The time constant $\tau$ is the time taken for the output to reach 63.2% of its final value, $1 - e^{-1}$. Equivalently, it is the time the output would need to reach the final value if it continued at its initial rate. For a thermometer, $\tau = mc/(hA)$.

 q_o/Kqs
 1.0 |            ____---------
     |        _--
 .632|-----.-
     |   ./
     |  /
     | /
     +------+------+-----> t
           tau   2tau

The settling time to 2% is about $4\tau$ and to 5% about $3\tau$. A small $\tau$ means a fast instrument.