EGR 103/Fall 2019/Minilab 3
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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()
- 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.quad at docs.scipy.org
- scipy.integrate.trapz at docs.scipy.org
- scipy.integrate.simps at docs.scipy.org
- Since $$n$$ goes from 2 to 100, it is not exactly what you need for an index to an array to save the integral estimates...think carefully about how to store things.
- For the plots with the various integrals, the independent axis will be your $$n$$ values (i.e. range(2, 101))
- Problem 3.5.4
- matplotlib.pyplot.semilogx at matplotlib.org
- Python:Extrema
- Python:Finding roots
- Note that the arguments to the np.logspace() command are the *powers* of 10 and the number of points; that is, to go from 100 to 1000 with 40 points, you would use
np.logspace(2, 4, 40)
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.