Calculation of the Acceleration of a Stepper Motor I'm working on a stepper motor as part of my studies and while I was doing research I came across this paper I tried to walk my way through the equations and tried to find the proof for the Eq. 13. But I'm having a hard time going from Eq. 2 to Eq. 3. The author calculated the acceleration between c1 and c2. The derivation of the speed gives us the acceleration. But since we don't have the function as a variable of time I cannot calculate the derivative from the equation.
Also I tried to calculate it from the change on the axes but it doesn't have the information on the time axes.
This is the graph for the velocity against time
The paper gives the velocity equation as:
$$w = \frac{αf }{c}$$
It doesn't define the velocity as a function of time. And we need the get to the equation 3 the derivative of w.
$$w' = \frac{2αf^2 (c_1-c_2) }{c_1c_2(c_1+c_2)}$$
I appreciate all your help me making this calculation. Thank you.
 A: The velocity $\omega$ is defined as a piecewise constant function. Therefore you cannot differentiate it to obtain the acceleration (or deceleration). It should be done as illustrated below.
We consider two adjacent timer counts $c_1$ and $c_2$ and express the corresponding delays $\delta t_1$ and $\delta t_2$ as a function of timer count and timer frequency $f$ using the given equation (1).
$$\delta t_1=\dfrac{c_1}{f}\qquad\text{and}\qquad \delta t_2=\dfrac{c_2}{f}$$
At this juncture, we need to assume that the speed specified by the equation (2) occurs at the midpoint of each step interval. The interval of time elapsed between the midpoints of the steps of the considered pair of timer counts is,
$$\Delta t=\dfrac{\delta t_1}{2}+\dfrac{\delta t_2}{2} =\dfrac{c_1}{2f}+\dfrac{c_2}{2f}=\dfrac{1}{2f}\left(c_1+c_2\right).\tag{a}$$
According to the equation (2), the respective motor speeds at these two timer counts are as follows.
$$\omega_1=\dfrac{\alpha f}{c_1}\qquad\text{and}\qquad \omega_2=\dfrac{\alpha f}{c_2}, $$
where $\alpha$ is the motor step angle
This shows that, if $c_1 \gt c_2$, then $\omega_2 \gt \omega_1$, i.e. motor is accelerating. Likewise, if $c_1 \lt c_2$, then the motor is decelerating. The change in motor's speed $\delta\omega$ can be expressed as,
$$\Delta\omega=\omega_2 - \omega_1=\dfrac{\alpha f}{c_2}-\dfrac{\alpha f}{c_1}=\alpha f\left(\dfrac{c_1-c_2}{c_1 c_2}\right).\tag{b}$$
Using (a) and (b), we can determine the sought motor acceleration $\omega ‘$ as,
$$\omega ‘ =\dfrac{\Delta\omega}{\Delta t}=\dfrac{2\alpha f^2\left(c_1 – c_2\right)}{c_1 c_2\left(c_1+c_2\right)}.$$
If we are to use the equation (3) given in the article to calculate both the acceleration and deceleration of the motor speed, then we must pay attention to the sign of $\omega’$ to differentiate between the two phases. If the value of $\omega’$ turns out to be positive, then the motor speed is accelerating and vice versa.
