Chem Differential Eq HW Solutions Fall 2011 24

Chem Differential Eq HW Solutions Fall 2011 24 - 24 Chapter...

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Unformatted text preview: 24 Chapter 2 Fourier Series The modified steady-state is obtained by subtracting y3 from the steady-state. Here is its graph. modifiedsteadystate t_ steadystate t alph 3 Cos 3 t Plot Evaluate modifiedsteadystate t , t, 0, 4 Pi bet 3 Sin 3 t ; 0.1 0.05 2 4 6 8 10 12 -0.05 -0.1 In order to compare, we plot both functions on the same graph. Plot Evaluate steadystate t , modifiedsteadystate t , t, 0, 4 Pi 0.4 0.2 2 4 6 8 10 12 -0.2 -0.4 It seems like we were able to reduce the amplitude of the steady-state solution by a factor of 2 or 3 by removing the third normal mode. Can we do better? Let 2 us analyze the amplitudes of the normal modes. These are equal to α2 + βn . We n have the following numerical values: amplitudes N Table Sqrt alph n ^ 2 bet n ^ 2 , n, 1, 20 0.141312, 0., 0.415652, 0., 0.0169855, 0., 0.00466489, 0., 0.00199279, 0., 0.00104287, 0., 0.000616018, 0., 0.000394819, 0., 0.000268454, 0., 0.000190924, 0. It is clear from these values that y3 has the largest amplitude (which is what we expect) but y1 also has a relatively large amplitude. So, by removing the first component of F , we remove y1 , and this may reduce the oscillations even further. Let’s see the results. We will plot the steady-state solution ys , ys − y3 , and ys − y1 − y3 . ...
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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