Let's continue to assume our error dynamics can be expressed as a linear ordinary differential equation, such as this second-order differential equation from the previous video. Let's set the mass equal to zero, giving us this first-order differential equation, which says that the force due to the spring and the force due to the damper always sum to zero. We define the time constant b divided by k and rewrite the error dynamics in this standard first-order form, theta_e-dot plus 1 over the time constant times theta_e equals zero. This error differential equation is stable if the time constant is positive. If the time constant is negative, perhaps because of a negative stiffness spring, the differential equation is unstable and initial error grows with time.

The solution to this differential equation is theta_e equals e to the minus t over the time constant times the initial error theta_e at time zero. The unit step error response, where theta_e at time zero is equal to 1, can be plotted as a decaying exponential of time. The decay gets faster as the time constant decreases, either because the spring gets stiffer or the damper gets softer. The steady-state error is zero and the overshoot is zero. The 2 percent settling time, meaning the time for the error to decay to 2 percent of its initial value, is determined by solving for the time t satisfying this equation. Taking the natural log of both sides, we see that the error decays to 2 percent of its initial value after approximately 4 time constants.

In the next video, we will consider the case of a second-order error differential equation.