MTE 484 Lab 1

Part c

Angle = 0 → Potentiometer = 535

• Offset = 535?

Angle = pi/4 → Potentiometer = 482

Angle = -pi/4 → Potentiometer = 595

-0.01482 * reading + 7.93

Part d - stiction

Positive direction: 0.185 Negative direction: 0.182

$$\frac{\frac{C_1{K}_1}{\tau }}{s^2+\frac{1}{\tau }s+\frac{C_1{K}_1}{\tau }}=\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}$$

First use the overshoot measurement to find

$$\zeta =--\frac{\ln (\text{\%OS}/100)}{\sqrt{{\pi }^2+\ln (\text{\%OS}/100){)}^2}}$$

Use $T_s$ and $\zeta$ to find ${\omega }_n$:

$$\begin{array}{r}{T}_s=\frac{4}{\zeta \cdot {\omega }_n}\\ {\omega }_n=\frac{4}{\zeta \cdot T_s}\end{array}$$

Use $\zeta$ and ${\omega }_n$ to find $\tau$:

\begin{align} \frac{1}{\tau} & = 2\zeta \omega_{n} \\ \tau & = \frac{1}{2\zeta \omega_{n}} \end{align}$$ Find $K_{1}$ with $C_{1}, \tau, \omega_{n}$:

\begin{align} \frac{C_{1}K_{1}}{\tau} = \omega_{n}^{2} \[2ex] K_{1} = \frac{\omega_{n}^{2}\tau}{C_{1}} \end{align}

$$Motorinput,currentangle(rad),Error$$