Difference between revisions of "EGR 103/Fall 2019/Minilab 3"

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* Problem 2.5.1
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== Typographical Errors ==
−
** To load data froman Excel file with headers:
+
* 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:
 
::<syntaxhighlight lang=python>
 
::<syntaxhighlight lang=python>
 
import pandas
 
import pandas
 
   
 
   
 
# %% Load data
 
# %% Load data
−
edata = pandas.read_excel('file.xlsx')
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edata = pandas.read_excel("file.xlsx")
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col_1_stuff = edata.values[:,0].copy()
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col_1_stuff = edata.values[:, 0].copy()
−
col_2_stuff = edata.values[:,1].copy()
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col_2_stuff = edata.values[:, 1].copy()
 
</syntaxhighlight>
 
</syntaxhighlight>
 
:* [[Python:Plotting]]
 
:* [[Python:Plotting]]
 
:* [[Python:Interpolation]]
 
:* [[Python:Interpolation]]
 
:* [[Python:Extrema]]
 
:* [[Python:Extrema]]
−
:* [https://docs.scipy.org/doc/scipy-0.18.1/reference/generated/scipy.interpolate.CubicSpline.html scipy.interpolate.CubicSpline¶] at [docs.scipy.org docsscipy.org]
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:* [https://docs.scipy.org/doc/scipy-0.18.1/reference/generated/scipy.interpolate.CubicSpline.html scipy.interpolate.CubicSpline¶] at [http://docs.scipy.org docs.scipy.org]
  
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* Problem 2.5.2
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* Problem 3.5.2
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** [[https://docs.scipy.org/doc/scipy/reference/generated/scipy.integrate.simps.html scipy.integrate.simps]] at [docs.scipy.org docsscipy.org]
+
** [https://docs.scipy.org/doc/scipy/reference/generated/scipy.integrate.simps.html scipy.integrate.simps] at [http://docs.scipy.org docs.scipy.org]
 
** Note that the denominator in the line of action calculation is $$f_t$$, which you already calculated.
 
** 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:
+
** The equation for the line of action, $$d$$, is on the second page of the problem.
−
<syntaxhighlight lang=python>
+
** $$D$$ is the total depth of the water (75)
 +
** 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:
 +
::<syntaxhighlight lang=python>
 
fig = plt.figure(num=1, clear=True)
 
fig = plt.figure(num=1, clear=True)
−
ax = fig.add_subplot(1,1,1)
+
ax = fig.add_subplot(1, 1, 1)
 
zval = np.block([z[::-1], z[:]])
 
zval = np.block([z[::-1], z[:]])
−
wval = np.block([-wz[::-1]/2,wz[:]/2])
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wval = np.block([-wz[::-1] / 2, wz[:] / 2])
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ax.plot(wval, zval, 'k-')
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ax.plot(wval, zval, "k-")
−
ax.axis('equal')
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ax.axis("equal")
 
</syntaxhighlight>
 
</syntaxhighlight>
 +
 +
* Problem 3.5.3
 +
** [https://docs.scipy.org/doc/scipy/reference/generated/scipy.integrate.quad.html scipy.integrate.quad] at [http://docs.scipy.org docs.scipy.org]
 +
*** Be careful!  Look at how many results the quad function gives back - you don't need all of them!
 +
** [https://docs.scipy.org/doc/scipy/reference/generated/scipy.integrate.trapz.html scipy.integrate.trapz] at [http://docs.scipy.org docs.scipy.org]
 +
** [https://docs.scipy.org/doc/scipy/reference/generated/scipy.integrate.simps.html scipy.integrate.simps] at [http://docs.scipy.org 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
 +
** [https://matplotlib.org/3.1.1/api/_as_gen/matplotlib.pyplot.semilogx.html matplotlib.pyplot.semilogx] at [https://matplotlib.org 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 <code>np.logspace(2, 4, 40)</code>
 +
** When you are looking for the maximum value and the cutoff frequencies, note that you are always looking at the absolute value of the transfer function; you can use <code>abs()</code> or <code>np.abs()</code> for that.
 +
** There are four main ways to make plots in terms of the scaling of the axes:
 +
*** <code>ax.plot(x, y)</code> uses linear scaling on both axes
 +
*** <code>ax.semilogx(x, y)</code> uses logarithmic scaling on the $$x$$ axis only
 +
*** <code>ax.semilogy(x, y)</code> uses logarithmic scaling on the $$y$$ axis only
 +
*** <code>ax.loglog(x, y)</code> uses logarithmic scaling on both axes
 +
 +
== 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; the line of action $$d$$ 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.

Latest revision as of 22:14, 3 December 2019

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.
    • The equation for the line of action, $$d$$, is on the second page of the problem.
    • $$D$$ is the total depth of the water (75)
    • 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 3.5.4
    • 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)
    • When you are looking for the maximum value and the cutoff frequencies, note that you are always looking at the absolute value of the transfer function; you can use abs() or np.abs() for that.
    • There are four main ways to make plots in terms of the scaling of the axes:
      • ax.plot(x, y) uses linear scaling on both axes
      • ax.semilogx(x, y) uses logarithmic scaling on the $$x$$ axis only
      • ax.semilogy(x, y) uses logarithmic scaling on the $$y$$ axis only
      • ax.loglog(x, y) uses logarithmic scaling on both axes

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; the line of action $$d$$ 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.
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