HW 3 - Revision history

Cdxskier at 04:18, 14 October 2010

2010-10-14T04:18:43Z

← Older revision		Revision as of 21:18, 13 October 2010
Line 4:		Line 4:
	* I made a vector "y" equal to "x".		* I made a vector "y" equal to "x".
	* I used the meshgrid function in Octave to make my vectors square matrices xx and yy ([http://www.network-theory.co.uk/docs/octave3/octave_159.html more info here])		* I used the meshgrid function in Octave to make my vectors square matrices xx and yy ([http://www.network-theory.co.uk/docs/octave3/octave_159.html more info here])
	* I then integrated <math>\ e^{(j2\pi xx yy)}</math> and plotted the result.		* I then integrated <math>\ e^{(j2\pixxyy)}</math> and plotted the result.
	As you can see by the graph below, it is indeed very similar to an impulse function. To confirm, I integrated under the plot and found the area to be 0.99621, which is very close to the expected value of 1.		As you can see by the graph below, it is indeed very similar to an impulse function. To confirm, I integrated under the plot and found the area to be 0.99621, which is very close to the expected value of 1.
	[[File:HW3.png]]		[[File:HW3.png]]

Cdxskier at 04:18, 14 October 2010

2010-10-14T04:18:23Z

← Older revision		Revision as of 21:18, 13 October 2010
Line 4:		Line 4:
	* I made a vector "y" equal to "x".		* I made a vector "y" equal to "x".
	* I used the meshgrid function in Octave to make my vectors square matrices xx and yy ([http://www.network-theory.co.uk/docs/octave3/octave_159.html more info here])		* I used the meshgrid function in Octave to make my vectors square matrices xx and yy ([http://www.network-theory.co.uk/docs/octave3/octave_159.html more info here])
	* I then integrated <math>\ e^{(j2\pi xx yy)}\</math> and plotted the result.		* I then integrated <math>\ e^{(j2\pi xx yy)}</math> and plotted the result.
	As you can see by the graph below, it is indeed very similar to an impulse function. To confirm, I integrated under the plot and found the area to be 0.99621, which is very close to the expected value of 1.		As you can see by the graph below, it is indeed very similar to an impulse function. To confirm, I integrated under the plot and found the area to be 0.99621, which is very close to the expected value of 1.
	[[File:HW3.png]]		[[File:HW3.png]]

Cdxskier at 04:18, 14 October 2010

2010-10-14T04:18:02Z

← Older revision		Revision as of 21:18, 13 October 2010
Line 3:		Line 3:
	* I made a vector "x" from -5 to 5 with points spaced every .02 in between. I originally made the vector from -10 to 10 with an even smaller spacing ([http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem the Nyquist theorem] requires that samples must be taken at at least twice the frequency), but the integral took too long.		* I made a vector "x" from -5 to 5 with points spaced every .02 in between. I originally made the vector from -10 to 10 with an even smaller spacing ([http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem the Nyquist theorem] requires that samples must be taken at at least twice the frequency), but the integral took too long.
	* I made a vector "y" equal to "x".		* I made a vector "y" equal to "x".
	* I used the meshgrid function in Octave to make my vectors square matrices ([http://www.network-theory.co.uk/docs/octave3/octave_159.html more info here])		* I used the meshgrid function in Octave to make my vectors square matrices xx and yy ([http://www.network-theory.co.uk/docs/octave3/octave_159.html more info here])
			* I then integrated <math>\ e^{(j2\pi xx yy)}\</math> and plotted the result.
			As you can see by the graph below, it is indeed very similar to an impulse function. To confirm, I integrated under the plot and found the area to be 0.99621, which is very close to the expected value of 1.
	[[File:HW3.png]]		[[File:HW3.png]]


			===Author===
			[[Christopher Garrison Lau I]]

Cdxskier at 04:12, 14 October 2010

2010-10-14T04:12:38Z

← Older revision		Revision as of 21:12, 13 October 2010
Line 1:		Line 1:
	Doing this assignment taught many new lessons, in both sampling and Octave.		Doing this assignment taught many new lessons, in both sampling and Octave.
	An infinite integral in Octave would be impractical. As a result, I did the following:		An infinite integral in Octave would be impractical. As a result, I did the following:
	* I made a vector "x" from -5 to 5 with points spaced every .02 in between. I originally made the vector from -10 to 10 with an even smaller spacing ([http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem] the Nyquist theorem requires that samples must be taken at at least twice the frequency), but the integral took too long.		* I made a vector "x" from -5 to 5 with points spaced every .02 in between. I originally made the vector from -10 to 10 with an even smaller spacing ([http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem the Nyquist theorem] requires that samples must be taken at at least twice the frequency), but the integral took too long.
	* I made a vector "y" equal to "x".		* I made a vector "y" equal to "x".
	* I used the meshgrid function in Octave to make my vectors square matrices ([http://www.network-theory.co.uk/docs/octave3/octave_159.html] more info here)		* I used the meshgrid function in Octave to make my vectors square matrices ([http://www.network-theory.co.uk/docs/octave3/octave_159.html more info here])
	[[File:HW3.png]]		[[File:HW3.png]]

Cdxskier at 04:11, 14 October 2010

2010-10-14T04:11:27Z

← Older revision		Revision as of 21:11, 13 October 2010
Line 1:		Line 1:
	Doing this assignment taught many new lessons, in both sampling and ~~octave~~.		Doing this assignment taught many new lessons, in both sampling and Octave.
			An infinite integral in Octave would be impractical. As a result, I did the following:
			* I made a vector "x" from -5 to 5 with points spaced every .02 in between. I originally made the vector from -10 to 10 with an even smaller spacing ([http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem] the Nyquist theorem requires that samples must be taken at at least twice the frequency), but the integral took too long.
			* I made a vector "y" equal to "x".
			* I used the meshgrid function in Octave to make my vectors square matrices ([http://www.network-theory.co.uk/docs/octave3/octave_159.html] more info here)
	[[File:HW3.png]]		[[File:HW3.png]]

Cdxskier at 04:03, 14 October 2010

2010-10-14T04:03:21Z

← Older revision		Revision as of 21:03, 13 October 2010
Line 1:		Line 1:
	Doing this assignment taught many new lessons, in both sampling and octave.		Doing this assignment taught many new lessons, in both sampling and octave.
			[[File:HW3.png]]

Cdxskier: Created page with 'Doing this assignment taught many new lessons, in both sampling and octave.'

2010-10-14T04:00:15Z

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Doing this assignment taught many new lessons, in both sampling and octave.