EGR 103/Fall 2019/Minilab 3

Typographical Errors

• 3.5.4 originally had part of the transfer function wrong; it should be:

$$H=\frac{(j\omega)(j\omega+2000)}{(j\omega+10)(j\omega+1000)}$$

References / Hints

• Problem 3.5.1
  • To load data from an Excel file with headers:

import pandas
 
# %% Load data
edata = pandas.read_excel("file.xlsx")
col_1_stuff = edata.values[:, 0].copy()
col_2_stuff = edata.values[:, 1].copy()

• Python:Plotting
• Python:Interpolation
• Python:Extrema
• scipy.interpolate.CubicSpline¶ at docs.scipy.org

• Problem 3.5.2
  • scipy.integrate.simps at docs.scipy.org
  • Note that the denominator in the line of action calculation is $$f_t$$, which you already calculated.
  • If you want to see a graph of the cross section (i.e. Figure 19.9(b)), assuming you call the height above the bottom $$z$$ and the width at that height $$wz$$, you can add the following code:

fig = plt.figure(num=1, clear=True)
ax = fig.add_subplot(1, 1, 1)
zval = np.block([z[::-1], z[:]])
wval = np.block([-wz[::-1] / 2, wz[:] / 2])
ax.plot(wval, zval, "k-")
ax.axis("equal")

• Problem 3.5.3
  • scipy.integrate.trapz at docs.scipy.org
  • scipy.integrate.simps at docs.scipy.org
  • Since $$n$$ goes from 2 to 30, it is not exactly what you need for an index to an array...think carefully about how to store things.

• Problem 3.5.4
  • matplotlib.pyplot.semilogx at matplotlib.org
  • Python:Extrema
  • Python:Finding roots

Partial Solutions

• Problem 3.5.1
  • Biggest difference happens at about $$x=30$$.
  • Estimates of $$f(50)$$ are 0.471 and 0.521 for natural and not-a-knot, respectively.
• Problem 3.5.2
  • Total force is about 4e9; LOA is in the high 20s.
• Problem 3.5.3
  • Exact answer is 1104.
  • Simps is always better than trapz except for $$n=2$$, where they are the same.
  • After about $$n=1$$ the true percent error goes to almost 0 based on scale of the graph.
• Problem 3.5.4
  • The plot looks like someone started drawing a duck from the back of the neck over the head and out to the end of the bill, then stopped.
  • The maximum value on the graph is about 6.
  • The maximum value of the absolute value of the transfer function is about 2. Note that $$20\log_{10}(2)=6.02$$, which explains why the max of the graph is where it is.
  • The cutoff frequencies are at approximately 10 and 1450.