User:Egm4313.s12.team1.essenwein

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Eric Essenwein

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Problem R6.4[edit]

Statement[edit]

Consider the L2-ODE-CC (5) p7b-7 with the window function f(x) p.9-8 as excitation:


(4.0)


and the initial conditions

(4.1)


1. Find such that:

(4.2)


with the same initial conditions (4.1).
Plot for n=2,4,8, for x in [0,10].
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:n=0,1

Part 1 Solution[edit]

First, we shift the excitation f(x) to the left by introducing a new independent variable, t. This allows the period to start at zero.

(4.3)


The piecewise representation of the window function is now (in terms of t) as follows:

for
for

(4.4)


To find the Fourier transform, the period of oscillation is determined:

(4.5)


And the frequency of oscillation:

(4.6)


The general form of a Fourier transform is

(4.7)


The following equations are given values of the constants in (4.7), evaluated for the function in (4.4):

(4.8)


(4.9)


(4.10)


In their simplest forms:

(4.11)


(4.12)


Plugging in r(x)=f(t) in (4.0),

(4.13)


From the general form, a particular solution to (4.13) and its derivatives are as follows:

(4.14)


(4.15)


(4.16)


By plugging in (4.14), (4.15), and (4.16) into (4.13), we find An and Bn in terms of another constant, Cn:

(4.17)


(4.18)


(4.19)


After substituting t with (4.3), the solutions are shown for n=2,4,8.

(4.20)


Y2.jpg


(4.21)



(4.22)


Part 2 Solution[edit]

The excitation for n=0 and n=1 are the same, because .

The homogeneous solution to (4.0) is

(4.23)


For , the particular solution is

(4.24)


Evaluating, we find that



(4.25)


Thus, the complete solution is

(4.26)


Integrating using MATLAB's ode45 command, the following plot is obtained for y:

Author[edit]

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Problem R5.6[edit]

Statement[edit]

Consider the following L2-ODE-CC; see p.6-6:

(6.0)

Homogeneous solution:

(6.1)

Particular solution:

(6.2)

Complete the solution for this problem.

Find the overall solution that corresponds to the initial condition (3b) p.3-7

(6.3)

Solution[edit]

Start by finding and .

(6.4)

(6.5)

Substitute and its derivatives into (6.0) to find and .

(6.6)

Separating terms and setting equal to the excitation from (6.0):

(6.7)

From (6.7), we solve coefficients to get





(6.8)

Solving for and :



(6.9)

Which gives us the particular solution:

(6.10)

For the general solution,

(6.11)

(6.12)

(6.13)

To solve for and , we use initial conditions from (6.3):

(6.14)

Which simplifies to:

(6.15)

For the second initial condition from (6.3):

(6.16)

(6.17)

(6.18)

(6.19)

We can now write with all coefficients known.

(6.20)

Final Equation

(6.21)

5.6.jpg

Author[edit]

Solved and Typed By - Egm4313.s12.team1.essenwein (talk) 23:45, 19 March 2012 (UTC)
Reviewed By -