Report 6
Intermediate Engineering Analysis
Section 7566
Team 11
Due date: April 11, 2012.
Solved by Daniel Suh
- Find the fundamental period of
and ![{\displaystyle \sin(n\omega x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69230a8e4a3866b9d2fccd6affe4d35bf99f3559)
- Show that these functions also have period
.
- Show that the constants
is also a periodic function with period
.
- Find the fundamental period, and show that these functions have a period of
.
![{\displaystyle L={\frac {\pi }{\omega }}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48045e972150053301c91606407670539092dc71)
![{\displaystyle f(x+np)=f(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e8f795c72269529289997229eb37bfbc24afcf5)
![{\displaystyle f(x)=cos(n\omega x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/33f66b2eda7e42a57b7426af19d6ebd2acc909e7)
![{\displaystyle f(x+n{\frac {2\pi }{n\omega }})=cos(n\omega (x+{\frac {2\pi }{n\omega }}))\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c88b7625e7aa377923bab4e0fabc969364cf811)
![{\displaystyle f(x+{\frac {2\pi }{\omega }})=cos(n\omega x+2\pi )\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/313f15e6c471814c08a4474b072e47c8d8a1a0eb)
![{\displaystyle f(x+2L)=cos(n\omega x+2\pi )\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cad11ec442a2f67dd0c170573b2691d828cd7e5f)
is the fundamental period for
.
![{\displaystyle f(x+np)=f(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e8f795c72269529289997229eb37bfbc24afcf5)
![{\displaystyle f(x)=sin(n\omega x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a779d09434d1a78bfcbb404492b5f87491e02f90)
![{\displaystyle f(x+n{\frac {2\pi }{n\omega }})=sin(n\omega (x+{\frac {2\pi }{n\omega }}))\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d34c8a73ddf24bb4bfff332a6d379755cfd99ea)
![{\displaystyle f(x+{\frac {2\pi }{\omega }})=sin(n\omega x+2\pi )\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ef8089cb22934eaf0cb4173b5ee31d8ad1bbb5a)
![{\displaystyle f(x+2L)=sin(n\omega x+2\pi )\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df554d19c226c39aa51e17e14d3c951254699567)
is the fundamental period for
.
- Show that the constants
is also a periodic function with period
.
From Fourier Series,
![{\displaystyle a_{0}={\frac {1}{2L}}\int _{-L}^{L}f(x)dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e586d5cac06a02ec1e703af35e1ee94b3aa4e98)
![{\displaystyle f(x)=\cos(n\omega x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/260d9a17f93e029fc574c094362400b22b3900ab)
![{\displaystyle a_{0}={\frac {\omega }{2\pi }}\int _{-{\frac {\pi }{\omega }}}^{\frac {\pi }{\omega }}cos(n\omega x)dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77030549c2c0c1f57cb80658fe60c2761991cd1c)
![{\displaystyle a_{0}={\frac {\omega }{2\pi }}[{\frac {sin(n\omega x)}{n\omega }}|_{-{\frac {\pi }{\omega }}}^{\frac {\pi }{\omega }}]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/948cfb833c24a8531890b3ac3410eb6e71569a99)
![{\displaystyle a_{0}={\frac {\omega }{2\pi }}[{\frac {2sin(n\pi )}{n\omega }}]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/91e23e4538e88251593b0a0c69cd95afe7ce8952)
![{\displaystyle a_{0}=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4356f67bf684ab8ea88010454dd931e95a3a20fa)
![{\displaystyle f(x)=\sin(n\omega x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/21934c881ff637c6b4b9500f1c3c13e1058d0e7b)
![{\displaystyle a_{0}={\frac {\omega }{2\pi }}\int _{-{\frac {\pi }{\omega }}}^{\frac {\pi }{\omega }}sin(n\omega x)dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7639e132898705921e8568578a14297b6c32e6ff)
![{\displaystyle a_{0}={\frac {\omega }{2\pi }}[{\frac {cos(n\omega x)}{n\omega }}|_{-{\frac {\pi }{\omega }}}^{\frac {\pi }{\omega }}]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2eba41982a5c9a4390f37c5904e426055cbc52b4)
![{\displaystyle a_{0}={\frac {\omega }{2\pi }}[{\frac {-cos{n\pi }+cos{n\pi }}{n\omega }}]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfa42c9d3dba92ee2ccc1ea8a835a5fe220bcfb2)
![{\displaystyle a_{0}=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4356f67bf684ab8ea88010454dd931e95a3a20fa)
Solve the problems 11 and 12 from Kreyszig p.491
First, for problem 11:
from
where
We know that
so that ![{\displaystyle L=1\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c14f55e9b9f25216c2c4c35b241024920ce54240)
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
:
![{\displaystyle a_{0}={\frac {1}{2L}}\int _{-L}^{L}f(x)dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e586d5cac06a02ec1e703af35e1ee94b3aa4e98)
Plugging in,
![{\displaystyle a_{0}={\frac {1}{2}}\int _{-1}^{1}x^{2}dx={\frac {1}{2}}\left[{\frac {1}{3}}x^{3}\right]_{-1}^{1}={\frac {1}{2}}\left[{\frac {1}{3}}+{\frac {1}{3}}\right]={\frac {1}{3}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9cdcec4b23b99a873211496d4c7284a7e126a84d)
We find ![{\displaystyle a_{n}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bfc054430becea483bc5be46fdb347c6a2413ebb)
![{\displaystyle a_{n}={\frac {1}{L}}\int _{-L}^{L}f(x)cos({\frac {n\pi x}{L}})dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/162595bd2a78e0fce003ccf7bf5df2d771f08891)
![{\displaystyle a_{n}=1\int _{-1}^{1}x^{2}cos(n\pi x)dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/12d607b767b4fd20704b290510efd89bf931e4f9)
We can use the equivalent expression:
![{\displaystyle a_{n}=2\int _{0}^{1}x^{2}cos(n\pi x)dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/33118e75f8d984dadf4130bb24ebc5ea7430c8af)
Now we evaluate the integral by using integration by parts. Below are the steps:
Integration by parts uses ![{\displaystyle \int fdg=fg-\int gdf\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d105268e29a9ea5ef7715a9ad5a36133c4f449a)
![{\displaystyle {\begin{matrix}f=x^{2}&dg=cos(\pi nx)dx\\df=2xdx&g={\frac {sin(\pi nx)}{\pi n}}\end{matrix}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fa70ab53fe66a538e367ee66887f23824ef861fd)
Then, we have:
![{\displaystyle \left[{\frac {x^{2}sin(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {2}{\pi n}}\int _{0}^{1}xsin(\pi nx)dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bad33c22c5b16e79133262bd5ec55d4edda8aad9)
Repeating for the second term:
![{\displaystyle {\begin{matrix}f=x&dg=sin(\pi nx)\\df=dx&g={\frac {-cos(\pi nx)}{\pi n}}\end{matrix}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd4e1835f3ebbe448f822426efe727828cbc0df6)
Then, we have:
![{\displaystyle \left[{\frac {-xcos(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {1}{\pi n}}\int _{0}^{1}cos(\pi nx)dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83b10de0d3425337dcfd8d80e144e393d55c673a)
Bringing all the terms together:
![{\displaystyle 2\left[\left[{\frac {x^{2}sin(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {2}{\pi n}}\left[{\frac {-xcos(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {1}{\pi n}}\int _{0}^{1}cos(\pi nx)dx\right]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a60b2f09abdff1de7c5f4edcb367187a52ca5cb2)
Then,
![{\displaystyle 2\left[\left[{\frac {x^{2}sin(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {2}{\pi n}}\left[{\frac {-xcos(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {1}{\pi n}}\left[{\frac {sin(\pi nx)}{\pi n}}\right]_{0}^{1}\right]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/285797fa8a188f052f64069a7fb8e60d99fabecb)
Evaluating at the boundaries and simplifying:
![{\displaystyle 2\left[\left[0-0\right]+{\frac {2}{\pi ^{2}n^{2}}}\left[cos(\pi n)\right]-{\frac {1}{\pi n}}\left[0-0\right]\right]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40e36f0954f238cf8f595c4bd95b1cef757d0a54)
Further,
![{\displaystyle {\frac {4}{\pi ^{2}n^{2}}}cos(\pi n)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ae10c1fc09db62590d2ca658e68adf471759932)
We can simplify one step further by acknowledging that
.
![{\displaystyle \therefore a_{n}={\frac {4}{\pi ^{2}n^{2}}}(-1)^{n}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9cba6940cf07e00309077b4b2845c04ae2314548)
Finally,
![{\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }a_{n}cos({\frac {\pi n}{L}})x\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83cd61676023aa4aaeefe567b51cacd0e6c2fc83)
The Fourier series is:
![{\displaystyle f(x)={\frac {1}{3}}+\sum _{n=1}^{\infty }{\frac {4}{\pi ^{2}n^{2}}}(-1)^{n}cos(\pi n)x\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/871ea420857996e2f913957ad991e90f8b68dc97)
Now problem 12:
from
where
We know that
so that ![{\displaystyle L=2\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b97d5eec9b31b3c2a7198b9e59c329707513ebd0)
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
:
![{\displaystyle a_{0}={\frac {1}{2L}}\int _{-L}^{L}f(x)dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e586d5cac06a02ec1e703af35e1ee94b3aa4e98)
Plugging in,
![{\displaystyle a_{0}={\frac {1}{4}}\int _{-2}^{2}1-{\frac {x^{2}}{4}}dx={\frac {1}{4}}\left[x-{\frac {1}{12}}x^{3}\right]_{-2}^{2}={\frac {1}{4}}\left[{\frac {24-8}{12}}-{\frac {24+8}{12}}\right]={\frac {2}{3}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/35f2149448f2c8078f965e3d71d3e3d957077c8b)
We find ![{\displaystyle a_{n}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bfc054430becea483bc5be46fdb347c6a2413ebb)
![{\displaystyle a_{n}={\frac {1}{L}}\int _{-L}^{L}f(x)cos({\frac {n\pi x}{L}})dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/162595bd2a78e0fce003ccf7bf5df2d771f08891)
![{\displaystyle a_{n}={\frac {1}{2}}\int _{-2}^{2}(1-{\frac {x^{2}}{4}}cos(n\pi x)dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e84964c43d813cda4279d462d06cd7131629b9a)
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
![{\displaystyle {\frac {2}{\pi ^{2}n^{2}}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/57e67c5b35bae52cefdf240e3ca3bc06582f643d)
Finally,
![{\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }a_{n}cos({\frac {\pi n}{L}})x\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83cd61676023aa4aaeefe567b51cacd0e6c2fc83)
The Fourier series is:
![{\displaystyle f(x)={\frac {2}{3}}+\sum _{n=1}^{\infty }{\frac {-2}{\pi ^{2}n^{2}}}cos({\frac {\pi n}{2}})x\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9dee4ef3fb7beb12e00ad583f0af5277446b3507)
Solved by Francisco Arrieta
Find the Fourier series expansion for
on p. 9.8 as follows:
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
![{\displaystyle p=2L=4\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c533ad3668bb8754f263fc5c4600e8d7e339c73)
![{\displaystyle L=2\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b97d5eec9b31b3c2a7198b9e59c329707513ebd0)
![{\displaystyle \omega ={\frac {2\pi }{p}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0648141d1fbb0fae6e6aa5e455011195098b7ba8)
![{\displaystyle {\bar {a}}_{0}={\frac {1}{4}}\int _{-2}^{2}f({\bar {x}})d{\bar {x}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42a7bdcdf57fefccc6aa967befaee61ccf07aaa8)
![{\displaystyle {\bar {a}}_{0}={\frac {A}{2}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f15fcc5efd24926346d28440de0e786711de585f)
![{\displaystyle {\bar {a}}_{k}={\frac {1}{2}}\int _{-2}^{2}f({\bar {x}})\cos {\frac {k\pi {\bar {x}}}{2}}d{\bar {x}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/184d807ee36c187008c6e98a981366ce2e7bd331)
![{\displaystyle {\bar {a}}_{k}={\frac {2A}{k\pi }}\sin {\frac {k\pi }{2}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c1c5be4d3b115737dffc2ca6b02f00c7d587346)
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
For n=5
After transforming the variable of
Do the same as above, but using
to obtain the Fourier expansion of
; compare to the result obtained above
![{\displaystyle p=2L=4\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c533ad3668bb8754f263fc5c4600e8d7e339c73)
![{\displaystyle L=2\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b97d5eec9b31b3c2a7198b9e59c329707513ebd0)
![{\displaystyle \omega ={\frac {2\pi }{p}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0648141d1fbb0fae6e6aa5e455011195098b7ba8)
![{\displaystyle {\tilde {a}}_{0}={\frac {1}{4}}\int _{0}^{4}f({\tilde {x}})d{\tilde {x}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b27462969d6b70157aaa138adbb2793fbb689fd)
![{\displaystyle {\tilde {a}}_{0}={\frac {A}{2}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f42d2aab36237aee3755be739b822c6c71e78b51)
![{\displaystyle {\tilde {a}}_{k}={\frac {1}{2}}\int _{0}^{4}f({\tilde {x}})\cos k\omega {\tilde {x}}d{\tilde {x}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2aa47cbc8521e3edae5369d0aa56ce652451a022)
![{\displaystyle {\tilde {a}}_{k}={\frac {A}{k\pi }}\sin \pi k\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c0c684ff209c1e2b3bb907abfcf56907be62c864)
Note:
for n=1, 2..
![{\displaystyle {\tilde {b}}_{k}={\frac {1}{2}}\int _{0}^{4}f({\tilde {x}})\sin k{\frac {\pi }{2}}{\tilde {x}}d{\tilde {x}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2f5aad9f47e0a68bb74938717458117a20de4b73)
![{\displaystyle {\tilde {b}}_{k}={\frac {-A}{\pi k}}(\cos \pi k-1)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba1b062f285de6979a51405855b979c61c9e32f9)
Note:
If k is even
If k=1, 3, 5...
Then Fourier series becomes a Fourier sine series:
For n=0:
For n=1:
For n=5:
After transforming the variable of
Page 491, #15,17. Plot the truncated Fourier series for n=2,4,8.
15.
Since
, the function is odd.
![{\displaystyle a_{0}=(1/\pi )*[-1\int _{-\pi }^{-\pi /2}(x+\pi )dx+\int _{-\pi /2}^{\pi /2}(x)dx+\int _{\pi /2}^{\pi }(\pi -x)dx]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3515ec49fdd8ef95aedb6e74ee844ba7b39b6ce)
After simplification, ![{\displaystyle a_{0}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8f3589226b1f07bd27b7c82d8f470a4685fffe2)
.
Using Equation (5) on page 490, the exapansion becomes;
![{\displaystyle f(x)=8k/\pi ^{2}[sin(\pi *x/L)-sin(3\pi *x/L)/9...)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3ee6df1febd26a27d67e60bb464aa6fbf8454a94)
With
this becomes;
for n is "odd".
This means when n is even, ![{\displaystyle f(x)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf85883d74b75fe35ca8d3f2b44802df078e4fa1)
![](//upload.wikimedia.org/wikiversity/en/2/27/R63a.jpg)
17.
![{\displaystyle f(x)=1-|x|,-1\leq x\leq 1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f3bd7723ea233d07072066bba050abdd0eb9726)
![{\displaystyle f(x)=\left\{{\begin{matrix}1+x-\pi ,-1<x<0\\1-x,0<x<1\\\end{matrix}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db1bab6f543a90c4fd1b800fcc0a1391bcd45d31)
![{\displaystyle a_{0}=1/2[\int _{-1}^{0}(1+x)dx+\int _{0}^{1}(1-x)dx]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/663792d020231d208ed32b73f0c4db7045053632)
![{\displaystyle a_{0}=1/2[1-1/2+1-1/2]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/57762d639f689e1bcd7d58eeb81c3590c3473f73)
![{\displaystyle a_{0}=1/2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/82733c82a9a3f4fd483529c221322e82c679eb61)
![{\displaystyle a_{n}=1/1[\int _{-1}^{0}(1+x)cos(n\pi *x/1)dx+\int _{0}^{1}(1-x)cos(n\pi *x/1)dx]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ecc54c05805473116b69ac5965dd79d368827f34)
Simplifying results in;
![{\displaystyle a_{n}=2[1/(n^{2}\pi ^{2})_{(}-1)^{n}/(n^{2}\pi ^{2})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/82a4babc3b94dfce840755f8704669ff666085ce)
So,
when n is odd.
![{\displaystyle b_{n}=1/1[\int _{-1}^{0}(1+x)sin(n\pi *x/1)dx+\int _{0}^{1}(1-x)sin(n\pi *x/1)dx]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b3a3c83592e4975230374d189014a6e9669263f)
Simplifying results in;
![{\displaystyle b_{n}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ecc141d40ca4a13ec7beafd6d264d5a435792a58)
![{\displaystyle f(x)=1/2+\sum _{n=0}^{\infty }4/(n^{2}\pi ^{2})*cos(n\pi *x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/352fe9408d8d2a57b7b7b5a62039e90f176fe464)
Since
for even n,
becomes
for even n.
solved by Luca Imponenti
Consider the L2-ODE-CC (5) p.7b-7 with the window function f(x) p.9-8 as excitation:
![{\displaystyle y''-3y'+2y=f(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67a1e520203f26fbe9ee3cb4dae8a511ae13044a)
and the initial conditions
![{\displaystyle y(0)=1\ ,\ y'(0)=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/583283c322ca54967e07641dbc6f6d3715cebe73)
1. Find
such that:
![{\displaystyle y_{n}''+ay_{n}'+by_{n}=r_{n}(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b35f0b4015cf0503fdd9613b4558d2018b26c0c3)
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:
![{\displaystyle n=0,1\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d956c504c579ff49e5a885c0fe2fe8d2ef849bd)
One period of the window function p9.8 is described as follows
![{\displaystyle f(x)={\begin{cases}0&\ \ -1.75<x<l0.25\\A&\ \ 0.25<x<2.25\\\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/738f0ef1577d40ba1ac25ed0b94d601354e8d880)
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:
![{\displaystyle +{\frac {A}{n\pi }}(cos({\frac {n\pi }{8}})-cos({\frac {9n\pi }{8}}))sin({\frac {n\pi x}{2}})]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8893708df79a090f008049edab53aed382f0f66e)
Considering the homogeneous case of our ODE:
The characteristic equation is
Therefore our homogeneous solution is of the form
Considering the case with f(x) as excitation
![{\displaystyle +(cos({\frac {k\pi }{8}})-cos({\frac {9k\pi }{8}}))sin({\frac {k\pi x}{2}})]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5cd6ccf84a4564bffcd87fc35ab5fb5eccc20fad)
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:
![{\displaystyle {\begin{bmatrix}2&0\\\\0&2\end{bmatrix}}*{\begin{bmatrix}A_{1}\\B_{1}\end{bmatrix}}={\begin{bmatrix}\ {\frac {A}{\pi }}(sin({\frac {9\pi }{8}})-sin({\frac {\pi }{8}}))\\\ {\frac {A}{\pi }}(cos({\frac {\pi }{8}})-cos({\frac {9\pi }{8}}))\end{bmatrix}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4029d03172453f5b329f67078d3b75c65121a36)
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:
![{\displaystyle {\begin{bmatrix}2-{\frac {(\pi k)^{2}}{4}}&{\frac {-3\pi k}{2}}\\{\frac {-3\pi k}{2}}&2-{\frac {(\pi k)^{2}}{4}}\end{bmatrix}}*{\begin{bmatrix}A_{k}\\B_{k}\end{bmatrix}}={\begin{bmatrix}\ {\frac {A}{\pi k}}(sin({\frac {9\pi k}{8}})-sin({\frac {\pi k}{8}}))\\\ {\frac {A}{\pi k}}(cos({\frac {\pi k}{8}})-cos({\frac {9\pi }{8}}))\end{bmatrix}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/898a4982a9e6fce5101ac94f44371201ce154a73)
Results of these calculations are shown below:
The solution to the particular case can be written for all n (assuming A=1):
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:
Using ode45 the following graph was generated for n=0:
and for n=1
Solved by: Gonzalo Perez
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.
The graphs are to be separately graphed as follows.
For n = 3, the code that will generate the graph looks like this:
For n = 6:
For n = 9:
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
For n=4 (from x = 0 to x = 10):
For n=4 (zoom-in plots around x = -0.5):
For n=4 (zoom-in plots around x = 0):
For n=4 (zoom-in plots around x = +0.5):
For n=7 (from x = 0 to x = 10):
For n=11 (from x = 0 to x = 10):
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):
Combined plots for n=7 and n=11 (zoom-in plots around x = 0):
Combined plots for n=7 and n=11 (zoom-in plots around x = 0.5):
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
.
The general code that can be used to graph the plots of n=4,7,11 is:
With the above code, we can add the following code to graph n=4.
For n = 4 (from x = 0.9 to x = 10):
For n = 7 (from x = 0.9 to x = 10):
For n=11 (from x = 0.9 to x = 10):
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:
For n = 4, 7, 11 at x = 1.5:
For n = 4, 7, 11 at x = 2:
For n = 4, 7, 11 at x = 2.5:
For the last part, the first (and unchanged TA's code) is the following:
The second part includes the change that had to be made in order to solve this problem:
Given: For the following differential equation: ![{\displaystyle y_{p}''+4y_{p}'+3y_{p}=2e^{2x}cos(3x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b815a5c8e78c979b1fb1cbc93d7b5509f43d17f)
With a particular solution of the form: ![{\displaystyle y_{p}=xe^{-2x}(Mcos(3x)+Nsin(3x))\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1958b6b5f5d2758b2427af95a2339c7f437e46cb)
Verify that this solution has a final expression of
as follows:
1) Simplify the term ![{\displaystyle y_{p}''\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b2590ee2e704bf3f6439eb76e7cab88105bd5ae)
2) Simplify the 2nd term
and combine with the simplified first term
3) Finally add the 3rd term ![{\displaystyle 13y_{p}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd900c7eed14121691a56d520169c2f5896a0e0d)
4) Find the final expression for ![{\displaystyle y_{p}(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d714a7ed52cff802cf61992f0730e29277806722)
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:
- ↑ www.wolframalpha.com
--Egm4313.s12.team11.sheider (talk) 22:30, 10 April 2012 (UTC)
Solved by Solved by Daniel Suh
Find the separated Ordinary Differential Equations for the heat equation.
![{\displaystyle {\frac {\partial u}{\partial t}}=\kappa {\frac {\partial ^{2}u}{\partial x^{2}}}\cdots (1)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/930a51d42b9c5b00ce667c5f86da16a16eabeb3f)
![{\displaystyle u(x,t)=F(x)\cdot G(t)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b76bbb9e90f006beb4c2e2fccbc99a21f9b1c0ab)
![{\displaystyle {\frac {\partial u}{\partial t}}=F(x)\cdot {\dot {G}}(t)\cdots (2)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/30c021f5f37d0377798914dbf22cdf20f3423d8d)
![{\displaystyle {\frac {\partial ^{2}u}{\partial x^{2}}}=F''(x)\cdot G(t)\cdots (3)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aca0f3a9bb7272f8043dac6d5153a13f2667ea22)
Combine
and
into ![{\displaystyle (1)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6364dbc5bef0269dfc82b7b3acfabdc884b5fd3f)
![{\displaystyle F(x)\cdot {\dot {G}}(t)=\kappa F''(x)\cdot G(t)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5537f55fd2cc821f95acd869bf28b9ab4fcb149e)
(constant)
Separate
![{\displaystyle F''(x)-cF(x)=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4aff50cdcc66f32aeb19f5f8432ca334ab2b3f01)
![{\displaystyle G'(t)-\kappa cG(t)=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0f5a4fa82abce82cb3e5d070000def2154037ff)