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

$$\begin{array}{rl}m& =\frac{y_2-y_1}{x_2-x_1}=\frac{-\frac{\pi }{4}-\frac{\pi }{4}}{588-482}=-0.01482\\ 0& =m(535)+\text{offset}⟹\text{offset}=-m(535)=7.93\\ \therefore \text{rad}& =-0.01482\cdot \text{reading}+7.93\end{array}$$

Part d - stiction

Positive direction: 0.185 Negative direction: 0.182

use $T_p$ somewhere?

$$\begin{array}{rl}{T}_{ry}(s)& =\frac{C(s)G(s)}{1+C(s)G(s)}\\ & =\frac{\frac{K_p{K}_1}{s(\tau s+1)}}{1+\frac{K_p{K}_1}{s(\tau s+1)}}\\ & =\frac{K_p{K}_1}{s(\tau s+1)+K_p{K}_1}\\ & =\frac{K_p{K}_1}{\tau s^2+s+K_p{K}_1}\\ & =\frac{K_p{K}_1/\tau }{s^2+\frac{1}{\tau }s+\frac{K_p{K}_1}{\tau }}\end{array}$$ $$\frac{\frac{K_p{K}_1}{\tau }}{s^2+\frac{1}{\tau }s+\frac{K_p{K}_1}{\tau }}=\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}$$

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 ${\omega }_n$:

$$\begin{array}{rl}{T}_p& =\frac{\pi }{{\omega }_n\sqrt{1-{\zeta }^2}}\\ {\omega }_n& =\frac{\pi }{T_p\sqrt{1-{\zeta }^2}}\end{array}$$

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

$$Motorinput,currentangle(rad),Error=====K13.5_{0}to25.txt=====$$

OS = 4.88% \quad T_p = 110.0 ,\text{ms}

\zeta = -\frac{\ln(%OS/100)}{\sqrt{\pi^2 + \ln^2(%OS/100)}} = -\frac{\ln(0.0488)}{\sqrt{\pi^2 + (\ln(0.0488))^2}} = 0.6929

\omega_n = \frac{\pi}{T_p \sqrt{1-\zeta^2}} = \frac{\pi}{0.1100 \cdot \sqrt{1-(0.6929)^2}} = 39.6107 ,\text{rad/s}

\tau = \frac{1}{2\zeta \omega_n} = \frac{1}{2 \cdot 0.6929 \cdot 39.6107} = 0.0182 ,\text{s}

K_1 = \frac{\omega_n^2 \tau}{C_1} = \frac{(39.6107)^2 \cdot 0.0182}{-13.5000} = -2.1172 ,\frac{\text{rad}}{\text{V·s}}

$$=====K14_{0}to25.txt=====$$

OS = 6.19% \quad T_p = 90.0 ,\text{ms}

\zeta = -\frac{\ln(%OS/100)}{\sqrt{\pi^2 + \ln^2(%OS/100)}} = -\frac{\ln(0.0619)}{\sqrt{\pi^2 + (\ln(0.0619))^2}} = 0.6630

\omega_n = \frac{\pi}{T_p \sqrt{1-\zeta^2}} = \frac{\pi}{0.0900 \cdot \sqrt{1-(0.6630)^2}} = 46.6297 ,\text{rad/s}

\tau = \frac{1}{2\zeta \omega_n} = \frac{1}{2 \cdot 0.6630 \cdot 46.6297} = 0.0162 ,\text{s}

K_1 = \frac{\omega_n^2 \tau}{C_1} = \frac{(46.6297)^2 \cdot 0.0162}{-14.0000} = -2.5117 ,\frac{\text{rad}}{\text{V·s}}

$$=====K15_{0}to20.txt=====$$

OS = 7.28% \quad T_p = 125.0 ,\text{ms}

\zeta = -\frac{\ln(%OS/100)}{\sqrt{\pi^2 + \ln^2(%OS/100)}} = -\frac{\ln(0.0728)}{\sqrt{\pi^2 + (\ln(0.0728))^2}} = 0.6405

\omega_n = \frac{\pi}{T_p \sqrt{1-\zeta^2}} = \frac{\pi}{0.1250 \cdot \sqrt{1-(0.6405)^2}} = 32.7251 ,\text{rad/s}

\tau = \frac{1}{2\zeta \omega_n} = \frac{1}{2 \cdot 0.6405 \cdot 32.7251} = 0.0239 ,\text{s}

K_1 = \frac{\omega_n^2 \tau}{C_1} = \frac{(32.7251)^2 \cdot 0.0239}{-15.0000} = -1.7032 ,\frac{\text{rad}}{\text{V·s}}

$$=====K17_{0}to20.txt=====$$

OS = 9.82% \quad T_p = 90.0 ,\text{ms}

\zeta = -\frac{\ln(%OS/100)}{\sqrt{\pi^2 + \ln^2(%OS/100)}} = -\frac{\ln(0.0982)}{\sqrt{\pi^2 + (\ln(0.0982))^2}} = 0.5942

\omega_n = \frac{\pi}{T_p \sqrt{1-\zeta^2}} = \frac{\pi}{0.0900 \cdot \sqrt{1-(0.5942)^2}} = 43.4001 ,\text{rad/s}

\tau = \frac{1}{2\zeta \omega_n} = \frac{1}{2 \cdot 0.5942 \cdot 43.4001} = 0.0194 ,\text{s}

K_1 = \frac{\omega_n^2 \tau}{C_1} = \frac{(43.4001)^2 \cdot 0.0194}{-17.0000} = -2.1481 ,\frac{\text{rad}}{\text{V·s}}

$$=====K21_{0}to15.txt=====$$

OS = 17.14% \quad T_p = 85.0 ,\text{ms}

\zeta = -\frac{\ln(%OS/100)}{\sqrt{\pi^2 + \ln^2(%OS/100)}} = -\frac{\ln(0.1714)}{\sqrt{\pi^2 + (\ln(0.1714))^2}} = 0.4896

\omega_n = \frac{\pi}{T_p \sqrt{1-\zeta^2}} = \frac{\pi}{0.0850 \cdot \sqrt{1-(0.4896)^2}} = 42.3870 ,\text{rad/s}

\tau = \frac{1}{2\zeta \omega_n} = \frac{1}{2 \cdot 0.4896 \cdot 42.3870} = 0.0241 ,\text{s}

K_1 = \frac{\omega_n^2 \tau}{C_1} = \frac{(42.3870)^2 \cdot 0.0241}{-21.0000} = -2.0614 ,\frac{\text{rad}}{\text{V·s}}

$$=====K23_{0}to15.txt=====$$

OS = 12.05% \quad T_p = 70.0 ,\text{ms}

\zeta = -\frac{\ln(%OS/100)}{\sqrt{\pi^2 + \ln^2(%OS/100)}} = -\frac{\ln(0.1205)}{\sqrt{\pi^2 + (\ln(0.1205))^2}} = 0.5587

\omega_n = \frac{\pi}{T_p \sqrt{1-\zeta^2}} = \frac{\pi}{0.0700 \cdot \sqrt{1-(0.5587)^2}} = 54.1148 ,\text{rad/s}

\tau = \frac{1}{2\zeta \omega_n} = \frac{1}{2 \cdot 0.5587 \cdot 54.1148} = 0.0165 ,\text{s}

K_1 = \frac{\omega_n^2 \tau}{C_1} = \frac{(54.1148)^2 \cdot 0.0165}{-23.0000} = -2.1055 ,\frac{\text{rad}}{\text{V·s}}