Exercise: Sawtooth Redone With Exponential Basis Functions - Revision history

Gary.starr at 20:18, 25 January 2010

2010-01-25T20:18:18Z

← Older revision		Revision as of 13:18, 25 January 2010
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	[[Lau, Chris\|Christopher Garrison Lau I]]		[[Lau, Chris\|Christopher Garrison Lau I]]

			Tyler Starr
	==Read By==		==Read By==
	==Comments==		==Comments==

Cdxskier at 22:41, 20 January 2010

2010-01-20T22:41:48Z

← Older revision		Revision as of 15:41, 20 January 2010
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	==Reviewed By==		==Reviewed By==

			[[Lau, Chris\|Christopher Garrison Lau I]]
	==Read By==		==Read By==
	==Comments==		==Comments==

John.hawkins: Finished with primary content.

2010-01-20T04:48:48Z

Finished with primary content.

← Older revision		Revision as of 21:48, 19 January 2010
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	<br />		<br />

	Noting ~~again~~ that our period for this function is <math>T=1</math> and that an obvious choice for <math>c</math> is zero, we proceed:		Noting that our period for this function is <math>T=1</math> and that an obvious choice for <math>c</math> is zero, we proceed:

	<br />		<br />
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	<br />		<br />

	~~Again~~, the case when <math>\ n=0</math> needs to be considered separately. In this case,		For <math>n\neq 0</math>, the above integral is solved easiest using integration by parts. When <math>\ n=0</math>,however, IBP does not work, so the case when <math>\ n=0</math> needs to be considered separately. In this case,

	<br />		<br />
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	<br />		<br />

	For <math>n\neq 0</math>, ~~the above integral is solved easiest using integration by parts. So~~ letting		For <math>n\neq 0</math>, we continue with IBP, letting
	<br />		<br />

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	<br />		<br />
	~~we have~~		This gives
	<br />		<br />

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	<br />		<br />
	==Solution Graphs==		==Solution Graphs==
	I modified the Matlab code used in the [[Exercise: Sawtooth Wave Fourier Transform]] to generate the solution graphs using the equation found above instead of the previously found solution. This code can be found here: [[Sawtooth2 Matlab Code]]. It generates the following analagous three graphs, which as hoped appear exactly identical to those found using the other method. Note that the "terms" mentioned in the titles of the graphs should now be interpreted as the sum of the ~~nth~~ and -~~nth~~ terms.		I modified the Matlab code used in the [[Exercise: Sawtooth Wave Fourier Transform]] to generate the solution graphs using the equation found above instead of the previously found solution. This code can be found here: [[Sawtooth2 Matlab Code]]. It generates the following analagous three graphs, which as hoped appear exactly identical to those found using the other method. Note that the "terms" mentioned in the titles of the graphs should now be interpreted as the sum of the <math>n</math>th and <math>-n</math>th terms.
	<br />		<br />

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	<math>=\frac{1}{-j2\pi n} (\cos 2\pi nt+j\sin 2\pi nt)-\frac{1}{-j2\i n}\left(\cos(-2\pi nt)+j\sin(-2\pi nt)\right)</math>		<math>=\frac{1}{-j2\pi n} (\cos 2\pi nt+j\sin 2\pi nt)-\frac{1}{-j2\pi n}\left(\cos(-2\pi nt)+j\sin(-2\pi nt)\right)</math>

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	</center>		</center>

			<br />

			Therefore,

			<br />
			<center><math>x(t)=\frac{1}{2}-\sum_{n=\pm 1}^{\pm \infty}\frac{1}{j2\pi n}e^{j2\pi nt}=\frac{1}{2}-\sum_{n=1}^\infty\frac{1}{\pi n}\sin 2\pi nt</math></center>

			<br />

			As we had hoped and expected.

			<br />

			==Reviewed By==
			==Read By==
			==Comments==

John.hawkins at 01:01, 20 January 2010

2010-01-20T01:01:03Z

@@ Line 104: / Line 104: @@
 <br />
 ==Solution Graphs==
-I modified the Matlab code used in the [[Exercise: Sawtooth Wave Fourier Transform]] to generate the solution graphs using the equation found above instead of the previously found solution.  This code can be found here: [[Sawtooth2 Matlab Code]].  It generates the following analagous three graphs, which as hoped appear exactly identical to those found using the other method.
+I modified the Matlab code used in the [[Exercise: Sawtooth Wave Fourier Transform]] to generate the solution graphs using the equation found above instead of the previously found solution.  This code can be found here: [[Sawtooth2 Matlab Code]].  It generates the following analagous three graphs, which as hoped appear exactly identical to those found using the other method.  Note that the "terms" mentioned in the titles of the graphs should now be interpreted as the sum of the nth and -nth terms.
 <br />
@@ Line 116: / Line 116: @@
 <br />

John.hawkins: Intermediate progress

2010-01-20T00:45:42Z

Intermediate progress

@@ Line 19: / Line 19: @@
 <br />
-<center><math>x(t)=\sum_{-\infty}^\infty a_n e^{j2\pi nt/T}</math></center>
+<center><math>x(t)=\sum_{n=-\infty}^\infty a_n e^{j2\pi nt/T}</math></center>
 <br />
@@ Line 31: / Line 31: @@
 <br />
-Noting again that our period for this function is <math>\ T=1</math>, we proceed:
+Noting again that our period for this function is <math>T=1</math> and that an obvious choice for <math>c</math> is zero, we proceed:
 <br />
@@ Line 67: / Line 67: @@
 <math>a_n=t\left(\frac{1}{-j2\pi n}\right)e^{-j2\pi nt}\Bigg|_0^1-\int_0^1\left(\frac{1}{-j2\pi n}\right)e^{-j2\pi nt}dt</math>

John.hawkins: First bit of work

2010-01-20T00:17:54Z

First bit of work

← Older revision		Revision as of 17:17, 19 January 2010
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			==Author==

			John Hawkins

	==Problem Statement==		==Problem Statement==
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	<br />		<br />

			Note that this is the same function solved in [[Exercise: Sawtooth Wave Fourier Transform]], but solved differently to compare the two methods.
			==Solution==
			The goal of this method is to find the coefficients <math>a_n</math> such that

			<br />

			<center><math>x(t)=\sum_{-\infty}^\infty a_n e^{j2\pi nt/T}</math></center>

			<br />

			In class we showed not only that this was possible, but also that

			<br />

			<center><math>a_n=\frac{1}{T}\int_c^{c+T}x(t) e^{-j2\pi nt/T}dt</math></center>

			<br />

			Noting again that our period for this function is <math>\ T=1</math>, we proceed:

			<br />
			<center>

			<math>a_n=\frac{1}{1}\int_0^1te^{-j2\pi nt/1}dt</math>

			</center>
			<br />

			Again, the case when <math>\ n=0</math> needs to be considered separately. In this case,

			<br />
			<center><math>a_0=\int_0^1t\ dt=\frac{1}{2}</math></center>
			<br />

			For <math>n\neq 0</math>, the above integral is solved easiest using integration by parts. So letting
			<br />

			<center>

			<math>u=t\quad\Rightarrow\quad du=dt</math>

			<br />

			<math>dv=e^{-j2\pi nt}dt \quad\Rightarrow\quad v=\frac{1}{-j2\pi n}e^{-j2\pi nt}</math>

			</center>

			<br />
			we have
			<br />

			<center>

			<math>a_n=t\left(\frac{1}{-j2\pi n}\right)e^{-j2\pi nt}\Bigg\|_0^1-\int_0^1\left(\frac{1}{-j2\pi n}\right)e^{-j2\pi nt}dt</math>

John.hawkins: New page: ==Problem Statement== Find the Fourier Tranform with exponential basis functions of the sawtooth wave given by the equation
$x(t)=t-\lfloor t \rfloor$
2010-01-19T23:57:19Z

New page: ==Problem Statement== Find the Fourier Tranform with exponential basis functions of the sawtooth wave given by the equation <br /> <center><math>x(t)=t-\lfloor t \rfloor</math></cente...

New page

==Problem Statement==

Find the Fourier Tranform with exponential basis functions of the sawtooth wave given by the equation

<br />

<center><math>x(t)=t-\lfloor t \rfloor</math></center>

<br />