User:Egm6322.s12.team2.steele.m2

EGM 6322 Team 2 Report 1

=R8.2 Fourier Series, Euler's Identity, and Fourier Coefficients= Question from 42-11 to 42-12: Verify (4). $$\int_0^T e^{i(n-m) \omega \theta} d\theta = $$ $$ \begin{cases} 0, & \text{if} n \ne m \\ T, & \text{if} n=m \end{cases}$$ Recover from {1} the Fourier coefficients found with the trigonometric identities above. From lecture notes [[media:pea1.f11.mtg42.djvu|Mtg 42]]

Solution
Apply Euler's identity. $$e^{ikx} = \cos kx + i\sin kx$$

$$\int_0^T f(\theta)e^{-im\omega \theta} d\theta = \sum_{n=-\infty}^{+\infty} \tilde a_n \, \int_0^T e^{i(n-m) \omega \theta} d\theta$$ The first part of the solution requires the derivation of (4) from p. 42-11. First, let n=m and substitute into the derivation with the Euler identity. The solution requires expansion of the terms in the integral into terms of sine and cosine. The key is to recognize that the expansion of the Euler identity for the positive and negative exponents as shown below. $$e^{-in\omega \theta} = \cos{n\omega \theta} - i\sin{n\omega \theta}$$ $$e^{in\omega \theta} = \cos{n\omega \theta} + i\sin{n\omega \theta}$$ $$e^{in\omega \theta} + e^{-in\omega \theta} = \cos{n\omega \theta} + \cos{n\omega \theta} - i\sin{n\omega \theta} + i\sin{n\omega \theta}$$ Note that the imaginary terms above go to zero. $$e^{in\omega \theta} + e^{-in\omega \theta} = 2\cos{n\omega \theta}$$

Divide the exponential summation by 2, and we have. $$(e^{in\omega \theta} + e^{-in\omega \theta})/2 = \cos{n\omega \theta}$$ Similarly, the pattern can be repeated for the sine term but leaves the imaginary component.

$$e^{in\omega \theta} - e^{-in\omega \theta} = \cos{n\omega \theta} - \cos{n\omega \theta} + i\sin{n\omega \theta} + i\sin{n\omega \theta}$$ Note that the real terms above go to zero. $$e^{in\omega \theta} + e^{-in\omega \theta} = 2i\sin{n\omega \theta}$$

This final term can be divided by 2i to get the equivalent value below. $$(e^{in\omega \theta} - e^{-in\omega \theta})/2 = \sin{n\omega \theta}$$ The Euler identity applies to the integral below. $$\int_0^T e^{i(n-m) \omega \theta} d\theta = \int_0^T{\cos{(n-m) \omega \theta} + i\sin{(n-m) \omega \theta}}$$ Note that if n=m then the following values occur. $$\cos{(n-m) \omega \theta} = \cos{0} = 1$$ $$\sin{(n-m) \omega \theta} = \sin{0} = 0$$ By substituting the values back into the integral, a simple integration is achieved for m=n. $$ \begin{align} \int_0^T e^{i(n-m) \omega \theta} d\theta & = \int_0^T{\cos{(n-m) \omega \theta} + i\sin{(n-m) \omega \theta}} \\ & = \int_0^T{(1 + 0)} \\ & = T \end{align}$$ Next, we concentrate on the condition where n does not equal m below. In this case, apply the substitution $$\text{T} = 2\pi / \omega$$. $$\begin{align} \int_0^T e^{i(n-m) \omega \theta} d\theta & = \int_0^T{\cos{(n-m) \omega \theta} + i\sin{(n-m) \omega \theta}} \\ & = \frac{1}{(n-m)\omega}(\sin{(n-m)\omega \theta} - i\cos{(n-m)\omega \theta} |_0^T \\ & = \frac{1}{(n-m)\omega}(\sin{(n-m)\omega \theta} - i\cos{(n-m)\omega \theta} |_0^2{\pi / \omega} \\ & = \frac{1}{(n-m)\omega}[(\sin{(n-m)\omega 2{\pi / \omega}} - i\cos{(n-m)\omega 2{\pi / \omega}} - (\sin{0} - i\cos{0})] \\ & = \frac{1}{(n-m)\omega}[(\sin{2\pi (n-m)} - i\cos{2\pi (n-m)}) + i] \\ & = \frac{1}{(n-m)\omega}[0 - i + i] \\ & = 0 \end{align}$$ The 2nd part of problem R8.2 requires the recovery of Fourier coefficients. First, the defintion below must be restated as given for the problem.  $$ \bar a_n := \begin{cases} 2 a_0, & \text{for } n=0 \\ a_n - ib_n, & \text{for } n>0 \\ a_n + ib_n, & \text{for } n<0 \end{cases}$$ Next, we revisit the equation of interest below. $$\int_0^T f(\theta)e^{-im\omega \theta} d\theta = \sum_{n=-\infty}^{+\infty} \tilde a_n \, \int_0^T e^{i(n-m) \omega \theta} d\theta$$ In the equation above, if n=m, then we can re-examine the integral above.  First, in the summation, all terms for n go to 0 except where n=m, so the summation term no longer appears. Subsequently, the integral derivation of (4) on p. 42-11 is applied. So this gives the following equation. $$\int_0^T f(\theta) e^{-in \omega \theta}d\theta = \tilde{a_n} T = \tilde{a_m} T$$ Note that the constant can be expanded into the equation from which we recover the Fourier coefficients. $$\begin{align} \tilde{a_n} & = \frac{1}{2}\bar{a_n} \\ & = \frac{1}{T} \int_0^T f(\theta) e^{-in\omega\theta} d\theta \end{align}$$

The equation can be solved in 3 parts: n=0, n>0, and n<0. Let's start with n=0 for which the Fourier coefficient is defined below. $$\begin{align} \tilde{a_0} & = \frac{1}{2}\bar{a_0} \\ & = frac{1}{2}(2a_0)\int_0^T f(\theta) e^{-in\omega \theta}d\theta \end{align}$$ Next, the Fourier coefficient for n>0 can be derived with the given definitions and the Euler identity. $$\begin{align} \tilde{a_0} & = \frac{1}{2}\bar{a_0} & = \frac{1}{2}\int_0^T f(\theta) e^{-in\omega \theta}d\theta \\ & = \frac{1}{2}\int_0^T f(\theta)(\cos{-in\omega\theta} + i\sin{-n\omega\theta} d\theta \end{align}$$ In the deriviation above for n>0, the Fourier coefficient is the real part of the complex number in the integration. $$a_n = \frac{1}{2} \mathbf{Re}\int_0^T f(\theta)(\cos{-n\omega\theta} + i\sin{-n\omega\theta}) = \frac{1}{2}\int_0^T f(\theta) \cos{-n\omega\theta}$$ Next, we derive the Fourier coefficient for n<0 by simply taking the imaginary component of the previous integral. $$a_n = \frac{1}{2}\mathbf{Im} \int_0^T f(\theta)(\cos{-n\omega\theta} + i\sin{-n\omega\theta}) = \frac{1}{2}\int_0^T f(\theta) \sin{-n\omega\theta}$$

=R8.3 Fourier Series and Real Components= Question from 42-12 Show that, equivalent to (2)-(3) p. 42-10, the Fourier series can be written as $$f(\theta) = \text{Re} \left(\sum_{n={\color{red} 0}}^{+\infty} \tilde a_n e^{in \omega \theta} \right)$$

From lecture notes [[media:pea1.f11.mtg42.djvu|Mtg 42]]

Solution
The first step is to write the original equation. $$ f(\theta) = \frac{1}{2}\sum_{n=-\infty}^{+\infty} \bar{a_n} e^{in\omega\theta} = \frac{1}{2}\sum_{n=-1}^{-\infty}(a_n + ib_n)e^{-in\omega\theta} + 2a_0 e^{in\omega\theta} + \sum_{n=1}^{\infty}(a_n - ib_n)e^{-in\omega\theta}$$ The equation can be simplified by recognizing the conjugate values. $$\bar{a_n}^* = \bar{a_{-n}}$$ The conjugate method can be applied to the summation below. $$\sum_{n=-1}^{-\infty} (a_n + ib_n) e^{in\omega\theta} = \sum_{n=1}^{+\infty} (a_n - ib_n)e^{-in\omega\theta}$$ Substitute the right hand term above back into the main equation for $$f(\theta)$$. $$ \begin{align} f(\theta) & = \frac{1}{2}(\sum_{n=1}^{\infty}(a_n - ib_n)e^{-in\omega\theta} + 2{a_0}e^{in\omega\theta} + \sum_{n=1}^{\infty}(a_n - ib_n)e^{in\omega\theta}) \\ & = \frac{1}{2}(2{a_0}e^{in\omega\theta}(e^{-in\omega\theta} + \sum_{n=1}^{\infty}(a_n - ib_n)e^{in\omega\theta}) \\ & = \frac{1}{2} (2a_0 e^{in\omega\theta} + \sum_{n=0}^{\infty}(a_n - ib_n)\cos{n\omega\theta} \\ & = \frac{1}{2} \sum_{n=0}^{\infty} \bar{a_n} \cos{n\omega\theta} \\ & = \sum_{n=0}^{\infty} \tilde{a_n} \cos{n\omega\theta} \\ & = \text{Re} \sum_{n=0}^{\infty} \tilde{a_n}(cos{n\omega\theta} + i\sin{n\omega\theta}) \\ & = \text{Re} \sum_{n=0}^{\infty} \tilde{a_n} e^{in\omega\theta} \end{align} $$ = References for Report 7 = The key to solve the problem is sum all terms in the "all" column to get the coefficient. $$A_0$$ = sum of numbers in "all" column * $$T_0$$ = 0.745999*$$T_0$$ This value above matches the value calculated in solution 9.1 by 0.7459/0.745999*100, which is almost 100%. This meets the criteria for an approximation with error < 5%. In addition, for $$A_2$$ we have the following table transcribed from the spreadsheet. Again, the key to solve the problem is sum all terms in the "all" column to get the coefficient. $$A_2$$ = sum of numbers in "all" column * $$T_0$$ = -0.7997*$$T_0$$ This value above matches the value calculated in solution 9.1 by 100%, so the error is negligible with a 100% match.

Finally, for $$A_4$$ we have the following table transcribed from the spreadsheet.

Once again we sum all terms in the "all" column to get the coefficient. $$A_4$$ = sum of numbers in "all" column * $$T_0$$ = 0.066715*$$T_0$$ This value above matches the value calculated in solution 9.1 by 0.066715/0.06671*100, which is almost 100%. The close match makes errors negligible and fulfills the criteria of an error < 5%. In summary, we have successfully calculated the coefficients by using the Gauss-Quadrature formulations.

practice

Template

=EGM6321 - Typed Notes from Principles of Engineering Analysis 1, Fall 2011 (Extra Credit)=

Text typed by Manuel Steele over Summer of 2012 with links to images in wiki categories for this class. Extra Credit PEA1