University of Florida/Egm4313/s12.team11.R6

From Wikiversity
Jump to navigation Jump to search

Report 6


Intermediate Engineering Analysis
Section 7566
Team 11
Due date: April 11, 2012.

R6.1[edit]

Solved by Daniel Suh

Problem Statement[edit]

  • Find the fundamental period of and
  • Show that these functions also have period .
  • Show that the constants is also a periodic function with period .

Solution[edit]

  • Find the fundamental period, and show that these functions have a period of .







               is the fundamental period for .






               is the fundamental period for .


  • Show that the constants is also a periodic function with period .

From Fourier Series,






          






          

R6.2[edit]

Part A[edit]

Problem Statement[edit]

Solve the problems 11 and 12 from Kreyszig p.491

Solution[edit]

First, for problem 11:
from where We know that so that

We check is the function is even, odd or neither.
Since satisfies the condition we know that the function is even.
Given that the function is even and that for even functions, we exclude the term from the Fourier Series.

We find :

Plugging in,


We find


We can use the equivalent expression:


Now we evaluate the integral by using integration by parts. Below are the steps:
Integration by parts uses

Then, we have:

Repeating for the second term:

Then, we have:

Bringing all the terms together:

Then,

Evaluating at the boundaries and simplifying:

Further,

We can simplify one step further by acknowledging that .


Finally,


The Fourier series is:

                                   



Now problem 12:
from where We know that so that

We check is the function is even, odd or neither.
Since satisfies the condition we know that the function is even.
Given that the function is even and that for even functions, we exclude the term from the Fourier Series.

We find :

Plugging in,


We find


Now we evaluate the integral by using integration by parts. This process is similar to the one shown in the solution for problem 11 above and therefore will be performed in Wolfram Alpha.
The following command was given in Wolfram Alpha: integral (1-(x^2/4))*cos(pi*x) from -2 to 2.
Wolfram Alpha yields the following

Finally,


The Fourier series is:

                              


Part B[edit]

Solved by Francisco Arrieta

Problem Statement[edit]

Find the Fourier series expansion for on p. 9.8 as follows:

Part 1[edit]

Develop the Fourier series expansion of

Plot and the truncated Fourier series

for n=0,1,2,4,8. Observe the values of at the points of discontinuities, and the Gibbs phenomenon. Transform the variable so to obtain the Fourier series expansion of

Solution[edit]



Note:

If k is even

If k=1, 5, 9...

If k=3, 7, 11...


Note:

for n=1, 2...


Then Fourier series becomes a Fourier cosine series:


                                              


For n=0

                                              

For n=1

                                              

N=1.jpg

For n=5

                                              

N=5.jpg

File:Perft.jpg


After transforming the variable of

                                              

Part 2[edit]

Do the same as above, but using to obtain the Fourier expansion of  ; compare to the result obtained above

Solution[edit]



Note:

for n=1, 2..


Note:

If k is even

If k=1, 3, 5...


Then Fourier series becomes a Fourier sine series:

                                              


For n=0:

                                              

For n=1:

                                              

File:2n=1.jpg

For n=5:

                                              

File:2n=5.jpg

File:2perft.jpg


After transforming the variable of

                                              

R6.3[edit]

Problem Statement[edit]

Page 491, #15,17. Plot the truncated Fourier series for n=2,4,8.

Solution[edit]

15.


Since , the function is odd.


After simplification,
. Using Equation (5) on page 490, the exapansion becomes;

With this becomes;
for n is "odd".
This means when n is even,

R63a.jpg

17.




 


Simplifying results in;

So,

 

when n is odd.

Simplifying results in;



Since for even n, becomes for even n.

R63b.jpg


R6.4[edit]

solved by Luca Imponenti

Problem Statement[edit]

Consider the L2-ODE-CC (5) p.7b-7 with the window function f(x) p.9-8 as excitation:

and the initial conditions

1. Find such that:

with the same initial conditions as above.

Plot for for x in

2. Use the matlab command ode45 to integrate the L2-ODE-CC, and plot the numerical solution to compare with the analytical solution.

Level 1:

Fourier Series[edit]

One period of the window function p9.8 is described as follows

From the above intervals one can see that the period, and therefore Applying the Euler formulas from to the Fourier coefficients are computed:

The integral from to can be omitted from this point on since it is always zero.

and

The coefficients give the Fourier series:

Homogeneous Solution[edit]

Considering the homogeneous case of our ODE:

The characteristic equation is

Therefore our homogeneous solution is of the form

Particular Solution[edit]

Considering the case with f(x) as excitation

The solution will be of the form

Taking the derivatives

Plugging these back into the ODE:

Setting the two constants equal

This is valid for all values of n. Since the coefficients of the excitation and are zero for all even n, then the coefficients and will also be zero, so we must only find these coefficients for odd n's. Now carrying out the sum to and comparing like terms yields the following sets of equations. Written in matrix form:

Assuming this matrix can be solved to obtain

For the remaining coefficients to be solved all sums will be used so a more general equation may be written:

Results of these calculations are shown below:

The solution to the particular case can be written for all n (assuming A=1):

      

General Solution[edit]

The general solution is

where

Different coefficients will be calculated for each n. These coefficients are easily solved for by applying the given initial conditions. Below are the calculations for n=2.

Applying the first initial condition

Taking the derivative

Applying the second initial condition

Solving the two equations for two unknowns yields:

So the general solution for n=2 is:

Below is a plot showing the general solutions for n=2,4,8:

Plot64.jpg

Plot64zoom.jpg

Matlab Plots[edit]

Using ode45 the following graph was generated for n=0:

Matlab0.jpg

and for n=1

Matlab1.jpg

R6.5[edit]

Solved by: Gonzalo Perez

Problem Statement[edit]

Part R4.2, p.7c-26[edit]

For each value of n=3,5,9, re-display the expressions for the 3 functions , and plot these 3 functions separately over the interval .

R4.2 p.7-26: Consider the L2-ODE-CC (5) p.7b-7 with as excitation:

(5)p.7-7

(1)p.9-15

and the initial conditions:

(3b)p.3-7

Exact solution: (2)p.9-15

Re-display the expressions for .

Superpose each of the above plot with that of the exact solution.

Solution[edit]

The graphs are to be separately graphed as follows.

For n = 3, the code that will generate the graph looks like this:

File:N=3 code.JPG

File:N=3 picture.jpg

For n = 6:

File:N=6 code.JPG

File:N=6 picture.jpg

For n = 9:

File:N=9 code.JPG

File:N=9 picture.jpg

Part R4.3, p.7c-28[edit]

Understand and run the TA's code to produce a similar plot, but over a larger interval . Do zoom-in plots about the points and comment on the accuracy of different approximations.

R4.3 p.7-28: Consider the L2-ODE-CC (5) p.7b-7 with as excitation:

(5)p.7-7

(1)p.7-28

and the initial conditions:

(2)p.7-28

Solution[edit]

For n=4 (from x = 0 to x = 10):

File:For n-4 (code).JPG

File:For n-4.jpg


For n=4 (zoom-in plots around x = -0.5):

File:For n-4 zoom 1 (code).JPG

File:For n-4 zoom 1.jpg


For n=4 (zoom-in plots around x = 0):

File:For n-4 zoom 2 (code).JPG

File:For n-4 zoom 2.jpg


For n=4 (zoom-in plots around x = +0.5):

File:For n-4 zoom 3 (code).JPG

File:For n-4 zoom 3.jpg


For n=7 (from x = 0 to x = 10):

File:For n-7 (code).JPG

File:For n-7.jpg


For n=11 (from x = 0 to x = 10):

File:For n-11 (code).JPG

File:For n-11.jpg



Where the black line represents n=11 and the blue line represents log(1+x).

The problem asks for n=4, 7, and 11. The TA had up to n=16, but the problem does not ask for this. Manipulating the code and graphs to only account for these n values, n=16 has been disregarded.


Combined plots for n=7 and n=11 (zoom-in plots around x = -0.5):

File:For n-7 and n-11 zoom 1 (code).JPG

File:For n-7 and n-11 zoom 1.jpg


Combined plots for n=7 and n=11 (zoom-in plots around x = 0):

File:For n-7 and n-11 zoom 2 (code).JPG

File:For n-7 and n-11 zoom 2.jpg

File:For n-7 and n-11 zoom 2 ZOOM.JPG



Combined plots for n=7 and n=11 (zoom-in plots around x = 0.5):

File:For n-7 and n-11 zoom 3 (code).JPG

File:For n-7 and n-11 zoom 3.jpg

Part R4.4, p.7c-29[edit]

Understand and run the TA's code to produce a similar plot, but over a larger interval , and for n=4,7. Do zoom-in plots about and comments on the accuracy of the approximations.

R4.4 p.7-29: Extend the accuracy of the solution beyond .

Solution[edit]

The general code that can be used to graph the plots of n=4,7,11 is:

File:4.4 general code.JPG

With the above code, we can add the following code to graph n=4.

For n = 4 (from x = 0.9 to x = 10):

File:N=4 code.JPG

File:N=4.jpg

For n = 7 (from x = 0.9 to x = 10):

File:N=7 code.JPG

File:N=7.jpg

For n=11 (from x = 0.9 to x = 10):

File:N=11 code.JPG

File:N=11.jpg

Repeating the same common part of the code, let's graph the other plots around x = 1, 1.5, 2, 2.5:

For n = 4, 7, 11 at x = 1:

File:N=4,7,11 at x=1 code.JPG

File:N=4,7,11 at x=1.jpg

For n = 4, 7, 11 at x = 1.5:

File:N=4,7,11 at x=1.5 code.JPG

File:This might work for x = 1.5.JPG

For n = 4, 7, 11 at x = 2:

File:N=4,7,11 at x=2 code.JPG

File:N=4,7,11 at x=2.jpg

For n = 4, 7, 11 at x = 2.5:

File:N=4,7,11 at x=2.5 code.JPG

File:X=2.5.jpg


For the last part, the first (and unchanged TA's code) is the following:

File:Might work last part code 1.JPG

The second part includes the change that had to be made in order to solve this problem:

File:Might work last part code 2.JPG

File:Might work last part.JPG


R6.6[edit]

Problem Statement[edit]

Given: For the following differential equation:

With a particular solution of the form:

Verify that this solution has a final expression of as follows:

1) Simplify the term
2) Simplify the 2nd term and combine with the simplified first term
3) Finally add the 3rd term
4) Find the final expression for

Solution[edit]

1) To find the derivative and simplify we will use www.wolframalpha.com[1].

The "Simplify" command will be exploited as well as the notation for taking a derivative in mathematica. The following was entered into www.wolframalpha.com:

Simplify[D[D[x*exp(-2x)*(Mcos(3x) + Nsin(3x)),x],x]]

The answer that was yielded is as follows:






2) In the same fashion, was evaluated and simplified using www.wolframalpha.com

The following was entered into www.wolframalpha.com:

simplify[4*[D[x*exp(-2x)*(Mcos(3x) + Nsin(3x)),x]]]

The answer that was yielded is as follows:



This term can be added to the previous, and then simplified yet again using www.wolframalpha.com

The following was entered into www.wolframalpha.com

Simplify[D[D[x*exp(-2x)*(Mcos(3x) + Nsin(3x)),x],x]+ 4*[D[x*exp(-2x)*(Mcos(3x) + Nsin(3x)),x]]

The answer that was yielded is as follows:




3) The final term is now added to this result, as follows:










4) As verified above, the final term is as follows:



  1. www.wolframalpha.com

--Egm4313.s12.team11.sheider (talk) 22:30, 10 April 2012 (UTC)

R6.7[edit]

Solved by Solved by Daniel Suh

Problem Statement[edit]

Find the separated Ordinary Differential Equations for the heat equation.



Solution[edit]



Combine and into


(constant)

Separate