MTE 484 Lab 2

$$\begin{array}{r}{K}_1=-2.1079\\ \tau =0.0197\end{array}$$

Motor equation:

$$\begin{array}{rl}\frac{\theta (s)}{V(s)}& =\frac{K_1}{s(\tau s+1)}\\ & =\frac{-2.1079}{s(0.0197s+1)}\\ & =\frac{-2.1079}{s}+\frac{0.04152563}{0.0197s+1}\end{array}$$

Pole at:

$$\begin{array}{rl}0.0197s+1& =0\\ s=-\frac{1}{0.0197}& =-50.76142132\end{array}$$ $$\begin{array}{rl}\text{Plant: }& \frac{K_1/\tau }{s^2+\frac{1}{\tau }s}\\ \frac{K_1}{\tau }& =\frac{-2.1079}{0.0197}=-107\\ \frac{1}{\tau }& =\frac{1}{0.0197}=50.76142132\\ \therefore \text{Plant}& =\frac{-107}{1s^2+50.76142132s}\end{array}$$

% set time step
time_to_plot = 2;
T = 0.005;

%% Continuous Time
num = [-107];
denom = [1 50.7614 0];
cont_TF = tf(num, denom);
G = c2d(cont_TF, T);

pole(G)
zero(G)
zpk(G)

  -0.0012312 (z+0.9189)
  ---------------------
    (z-1) (z-0.7758)

Partial fraction decomposition:

Columns 1 through 5

-0.2137 - 0.0912i -0.2137 + 0.0912i 0.0974 - 0.3139i 0.0974 + 0.3139i 0.3984 - 0.0575i

Columns 6 through 10

0.3984 + 0.0575i 0.3216 + 0.3355i 0.3216 - 0.3355i -0.0377 + 0.5182i -0.0377 - 0.5182i

Columns 11 through 15

-0.4197 + 0.3845i -0.4197 - 0.3845i -0.6142 + 0.0282i -0.6142 - 0.0282i -0.5418 - 0.3721i

Columns 16 through 20

-0.5418 + 0.3721i -0.2461 - 0.6522i -0.2461 + 0.6522i 0.1564 - 0.7180i 0.1564 + 0.7180i

Columns 21 through 25

0.5318 - 0.5578i 0.5318 + 0.5578i 0.7722 - 0.2274i 0.7722 + 0.2274i 0.8186 + 0.1787i

Columns 26 through 30

0.8186 - 0.1787i 0.6664 + 0.5584i 0.6664 - 0.5584i 0.3569 + 0.8262i 0.3569 - 0.8262i

Stiction with Ball and Beam

• Upward (higher than before because we need to fight gravity): -0.48
• Downward (lower than before because gravity is helping us): -0.05

Part E:

When ball is at 0, sensor reading = 310 When ball is at end, sensor reading = 630 Beam length = 41.7

$$m=\frac{y_2-y_1}{x_2-x_1}=0.001303125$$ $$\begin{array}{rl}0& =0.001303125\cdot 310+b\\ b& =-0.40396875\end{array}$$ $$\text{meters}=0.001303125\cdot (\text{sensor reading})-0.40397$$

The motor rotates the motor gear with radius $r$. The gear pulls a lever arm of length $\ell$ attached to the beam. A small rotation of the motor (angle $\theta$) moves the end of the lever by an arc length $\Delta x=r\theta$.

The displacement causes the lever to rotate the beam, which is given by:

$$\sin \varphi \approx \varphi =\frac{\Delta x}{\ell }=\frac{r\theta }{\ell }$$

So the ratio is

$$\frac{\varphi }{\theta }=\frac{r}{\ell }$$

Therefore, $K_2=\frac{r}{\ell }=\frac{2.54}{12}=0.2166667$

$$D(z^{-1})=\frac{-4.1722z^{-1}+11.4610334z^{-2}-15.041011680938z^{-3}+12.301798304500137z^{-4}-5.438853262170353z^{-5}+0.8266580692435093z^{-6}}{1-2.3821z^{-1}+2.28145108z^{-2}-1.15646763878z^{-3}+0.294935656944z^{-4}-0.069482286223z^{-5}+0.047177611809z^{-6}}$$ $$D[z]=\frac{-4.1722z+11.4610334z^2-15.041011680938z^3+12.301798304500137z^4-5.438853262170353z^5+0.8266580692435093z^6}{1-2.3821z+2.28145108z^2-1.15646763878z^3+0.294935656944z^4-0.069482286223z^5+0.047177611809z^6}$$ $$\begin{array}{rl}\frac{Y(s)}{\varphi (s)}& =\frac{K_3}{s^2}\\ \ddot{y}(t)& =K_3\varphi (t)\end{array}$$ $$K_3=\arg \underset{K}{\min }\sum _i({\ddot{y}}_i-K{\varphi }_i{)}^2\Rightarrow K_3=\frac{{\sum }_i{\varphi }_i{\ddot{y}}_i}{{\sum }_i{\varphi }_i^2}.$$