University of Florida/Egm4313/s12.team11.R7

Report 7


Intermediate Engineering Analysis
Section 7566
Team 11
Due date: April 25, 2012.

R7.1Edit

Solved by Andrea Vargas

Problem StatementEdit

Verify (4)-(5) pg 19-9:
(4)   for  
(5)   for  

SolutionEdit

From the lecture notes on p.19-9 we know that:
(2)  
(3)  

Then, we have
 

Since the period of   is   and we know that  , we can assume that   for simplicity.

We can compute the result of the previous integral using Wolfram ALpha (Mathematica software). The following is the input given to the software:
int from 0 to pi sin(ax)sin(bx)

where a is   and b is  . The software generates the following answer:

 

Substituting for   and  :
 
We also know that   where c is any integer. We can cancel any terms with   or   as they are equal to zero.
Then, we can verify:

                         (4)    for  

Similarly we can verify (5).
We have
 

Here, we will keep the integration boundaries as   to be consistent with the problem statement.

We can compute the result of the previous integral using Wolfram ALpha (Mathematica software). The following is the input given to the software:
int 0 to L (sin(ax))^2

where a is  . The software generates the following answer:

 

Substituting for  :
 

We know from the previous explanation that  .So,we can apply the same assumption as before that   where c is any integer. This allows us to cancel any terms with   as they are equal to zero.
Then, we can verify:

                         (5)    for  

R7.2Edit

Solved by Francisco Arrieta

Problem StatementEdit

Plot the truncated series   with   and for:

 

 

SolutionEdit

Using:

 


Then:

 

 

  for all even values of j


Plugging back to the truncated series:

 

 


For   :

 

When   :

 

When   :

 

When   :

 

When   :

 

--Egm4313.s12.team11.arrieta (talk) 06:20, 22 April 2012 (UTC)

R7.3Edit

Problem StatementEdit

Find (a) the scalar product, (b) the magnitude of   and   ,(c) the angle between   and   for:

1)  

2)  

Part 1Edit

solved by Kyle Gooding

Scalar ProductEdit

 

 

Using integration by parts;

 

                       
MagnitudeEdit

 

 

 

                        

 

 
 

                        
Angle Between FunctionsEdit

 

 

                       
The two functions are nearly orthogonal.

Part 2Edit

solved by Luca Imponenti

Scalar ProductEdit

 

 

 
 
 

Since all exponents are even, everything in brackets cancels out

                       
MagnitudeEdit

 

 
 
 
 
 
                        

 

 
 
 
 
 
                        
Angle Between FunctionsEdit

 

Since   the two functions are orthogonal

                          

R7.4Edit

Solved by Gonzalo Perez

K 2011 pg.482 pb. 6Edit

Problem StatementEdit

Sketch or graph   which for   is given as follows:

 

SolutionEdit

The MATLAB code shown below was used to developed the graph of  :

 

 

K 2011 pg.482 pb. 9Edit

Problem StatementEdit

Sketch or graph   which for   is given as follows:

 

SolutionEdit

The MATLAB code shown below was used to developed the graph of the piecewise function  :

 

 



Solved by Jonathan Sheider

K 2011 p.482 pb. 12Edit

Problem StatementEdit

Find the Fourier series of the given function which is assumed to have a period of  . Show the details of your work.
Sketch or graph the partial sums up to that including   and  

Given:

 

SolutionEdit

The Fourier series of a function with a period of   is defined:

 

Where:

 
 
 

This particular function given in the problem can be also defined in a piecewise manner over this interval, namely:

 

Calculating the first term  :

 
 
 
 

                                                          


Calculating the coefficient  :

 
 

Using integration by parts with the following substitutions:

  and therefore  
  and therefore  

This yields for the integral:

 
 
 

Note that for all n = 1,2,3... :   as   therefore these terms are evaluated as zero, which yields:

 
 

Note that   therefore:

 
 
 
 

To evaluate the term  :

Note that   for odd n values as  

And that   for even n values as  .

Therefore, it can be concluded that for odd n values:

 

And for even n values:

 

Therefore, for the coefficient  , all even terms will equal zero while all odd terms will have a multiplier of 2, this yields (for all odd n values):

                                             


Calculating the coefficient  :

 
 

Using integration by parts with the following substitutions:

  and therefore  
  and therefore  

This yields for the integral:

 
 
 

Note that for all n = 1,2,3... :   as   therefore these terms are evaluated as zero, which yields:

 
 

Note that   therefore:

 
 

                                                        

Therefore there will be no   terms in the Fourier representation. In conclusion, the Fourier series representation for the given function is as follows:

 

                                     



A graph of the function, and the Fourier series for   is shown below:

 
--Egm4313.s12.team11.sheider (talk) 06:00, 22 April 2012 (UTC)

K 2011 p.482 pb. 13Edit

Problem StatementEdit

Find the Fourier series of the given function which is assumed to have a period of  . Show the details of your work.
Sketch or graph the partial sums up to that including   and  

Given:

 

SolutionEdit

The Fourier series of a function with a period of   is defined:

 

Where:

 
 
 

Calculating the first term  :

 
 
 
 
 

                                                          


Calculating the coefficient  :

 

Using integration by parts with the following substitutions for the integral  :

  and therefore  
  and therefore  

Using integration by parts with the following substitutions for the integral  :

  and therefore  
  and therefore  

This yields for the overall expression:

 
 
 

Note that for all n = 1,2,3... :   as   therefore these terms are evaluated as zero, which yields:

 

 

Note that   therefore:

 

 
 
 

To evaluate the term  :

Note that   for odd n values as  

And that   for even n values as  .

Therefore, it can be concluded that for odd n values:

 

And for even n values:

 

Therefore, for the coefficient  , all even terms will equal zero while all odd terms will have a multiplier of 2, this yields (for all odd n values):

                                             


Calculating the coefficient  :

 

Using integration by parts with the following substitutions for the integral  :

  and therefore  
  and therefore  

Using integration by parts with the following substitutions for the integral  :

  and therefore  
  and therefore  

This yields for the overall expression:

 
 
 
 

Note that for all n = 1,2,3... :   as   therefore these terms are evaluated as zero, which yields:

 
 

Note that   therefore:

 
 
 

To evaluate the term  :

Note that   for odd n values as  

And that   for even n values as  .

Therefore, it can be concluded that for odd n values:

 

And for even n values:

 

Therefore, for the coefficient  , all even terms will equal zero while all odd terms will have a multiplier of 2, this yields (for all odd n values):

                                             


In conclusion, the Fourier series representation for the given function is as follows:

 


                    



A graph of the function, and the Fourier series for   is shown below:

 
--Egm4313.s12.team11.sheider (talk) 06:00, 22 April 2012 (UTC)

R7.5Edit

Solved by Daniel Suh

Problem StatementEdit

Consider the following,
 

 

with  , and  

1. Find the integration with the given data.

2. Confirm the results with Matlab's trapz command for the trapezoidal rule.

SolutionEdit

Part 1Edit

Trigonometric IdentitiesEdit

Angle Sum and Difference Identities

 
 


Rearrange

 
 


Substitute and Combine

 
 
 

Utilize Trig IdentitiesEdit

 

 

 

 

 

 

                       

Part 2Edit

>> X = 0:2*pi/100:2*pi;

>> Y = sin(2*X).*sin(3*X);

>> Z = trapz(X,Y)

Z =

 2.9490e-017