User talk:Gregory.peterson - Revision history

Gregory.peterson: /* Bode Plot */

2009-10-27T22:51:39Z

Bode Plot

← Older revision		Revision as of 15:51, 27 October 2009
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	<math>\text {This plot is done using the control toolbox in MatLab. }\,</math>		<math>\text {This plot is done using the control toolbox in MatLab. }\,</math>

	[[Image:~~bodeplot~~.~~pdf~~\|700px\|thumb\|left\|Fig (1)]]		[[Image:bodeplotlna.jpeg\|700px\|thumb\|left\|Fig (1)]]

2009-10-27T22:49:25Z

Transfer Function

← Older revision		Revision as of 15:49, 27 October 2009
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	<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math><br /><br />		<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math><br /><br />

			===Bode Plot===

			<math>\text {This plot is done using the control toolbox in MatLab. }\,</math>

			[[Image:bodeplot.pdf\|700px\|thumb\|left\|Fig (1)]]

2009-10-27T22:45:05Z

Transfer Function

← Older revision		Revision as of 15:45, 27 October 2009
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	<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math><br /><br />		<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math><br /><br />

	~~to be continued...~~

	Created by Greg Peterson		Created by Greg Peterson

	Checked by Mark Bernet		Checked by Mark Bernet

2009-10-27T22:44:26Z

Using the Laplace Transform to solve a spring mass system that is critically damped

2009-10-26T17:44:28Z

Transfer Function

2009-10-23T17:03:42Z

← Older revision		Revision as of 10:03, 23 October 2009
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	<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math><br /><br />		<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math><br /><br />

			to be continued...

2009-10-23T08:03:06Z

Applying this to our problem

← Older revision		Revision as of 01:03, 23 October 2009
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	<math>\text {So as you can see the value for the initial position will be 0. }\,</math>		<math>\text {So as you can see the value for the initial position will be 0. Because the infinity in the denominator always makes the function tend toward zero.}\,</math>

	<math>\text {Which makes sense because the system is initially in equilibrium. }\,</math>		<math>\text {Which makes sense because the system is initially in equilibrium. }\,</math>
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	<math>\text {Which appears to mean the system will be right below equilibrium after a long time. }\,</math>		<math>\text {Which appears to mean the system will be right below equilibrium after a long time. }\,</math>


	==Bode Plot of the transfer function==		==Bode Plot of the transfer function==

2009-10-23T08:01:32Z

Given

← Older revision		Revision as of 01:01, 23 October 2009
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	<math>m=\frac{98}{9.81}</math>		<math>m=\frac{98}{9.81}</math>

	<math>\text {k=40}\,</math>		<math>\text {Spring Constant k=40}\,</math>

	<math>\text {Damping ~~constant~~ C=40}\,</math>		<math>\text {Damping Constant C=40}\,</math>

	<math>\text {x(0)=0}\,</math>		<math>\text {x(0)=0}\,</math>

2009-10-23T08:00:53Z

Things we know

← Older revision		Revision as of 01:00, 23 October 2009
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	===~~Things we know~~===		===Given===


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	<math>\text {Standard equation: }\,</math>		<math>\text {Standard equation: }\,</math>

	<math>m\frac{d^2x}{dt^2}+C\frac{dx}{dt}+khx=0</math>		<math>m\frac{d^2x}{dt^2}+C\frac{dx}{dt}+khx=0</math>


	===Solving the problem===		===Solving the problem===

2009-10-23T08:00:12Z

Transfer Function

← Older revision		Revision as of 01:00, 23 October 2009
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	===Transfer Function===		===Transfer Function===

	<math>\mathbf {X}(s)=-\frac{3}{(s+4)^2} </math><br /><br />		<math>\mathbf {X}(s)=-\frac{4}{(s+2)^2} </math><br /><br />