Laplace transforms: Simple Electrical Network - Revision history

Victor.shepherd: /* Bode Plots */

2009-12-13T23:35:19Z

Bode Plots

← Older revision		Revision as of 16:35, 13 December 2009
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	==Bode Plots==		==Bode Plots==
	The following are bode plots for the transfer functions		The following are bode plots for the transfer functions
	~~[[Image:I1_.jpg\|400px\|thumb\|left\|Bode I_2 Plot]]~~

	[[Image:I2_.jpg\|400px\|thumb\|left\|~~Bode I_2 Plot~~]]

			[[Image:I1_.jpg\|400px\|thumb\|left\|<math>H(s)_1=\frac{25s+12500}{s(s^2+500s+2500)}</math>]]



			[[Image:I2_.jpg\|400px\|thumb\|left\|<math>H(s)_2=\frac{6250}{s(s^2+500s+2500)}</math>]]

2009-12-13T23:32:06Z

Bode Plots

← Older revision		Revision as of 16:32, 13 December 2009
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	==Bode Plots==		==Bode Plots==
			The following are bode plots for the transfer functions
			[[Image:I1_.jpg\|400px\|thumb\|left\|Bode I_2 Plot]]

			[[Image:I2_.jpg\|400px\|thumb\|left\|Bode I_2 Plot]]

2009-12-13T22:55:38Z

Solution

← Older revision		Revision as of 15:55, 13 December 2009
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	<math>\Rightarrow s^2+500s=-2500</math>		<math>\Rightarrow s^2+500s=-2500</math>

	<math>\Rightarrow s^2+500s+[\frac{500}{2}]^2=-2500+(\frac{500}{2})^2</math>		<math>\Rightarrow s^2+500s+\left(\frac{500}{2}\right)^2=-2500+\left(\frac{500}{2}\right)^2</math>

	<math>\Rightarrow s^2+500s+62500=6000</math>		<math>\Rightarrow s^2+500s+62500=6000</math>

2009-12-12T00:25:10Z

Solution

← Older revision		Revision as of 17:25, 11 December 2009
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	<math>\Rightarrow s^2+500s+62500=6000</math>		<math>\Rightarrow s^2+500s+62500=6000</math>

	<math>\Rightarrow (s^2+250)^2-(20\sqrt{15})^2=0</math)		<math>\Rightarrow (s+250)^2-(100\sqrt{6})^2=0</math>

	~~we do this by divinding <math>(\frac{b}{2})^2</math>~~		Thus

	~~Taking the Inverse Laplace transform yields~~		<math>I_2(s)=\frac{5}{2s}-5\frac{s+250}{(s+250)^2-(100\sqrt{6})^2}-\frac{5\sqrt{6}}{12}\frac{100\sqrt{6}}{(s+250)^2-(100\sqrt{6})^2}</math>

	~~<math>\mathcal{L}^{-1}\left\{I_1(s)\right\}=\frac{5}{8}+\frac{39\sqrt{103}}{824}sin(\frac{25}{2}\sqrt{103}t)</math>~~

			Taking the Inverse Laplace transform gives

	<math>\mathcal{L}^{-1}\left\{I_2(s)\right\}=</math>		<math>\mathcal{L}^{-1}\left\{I_2(s)\right\}= i_2(t) =\frac{5}{2}-5e^{-250t}cosh100\sqrt{6}t-\frac{5\sqrt{6}}{12}5e^{-250t}sinh100\sqrt{6}t</math>

	==Initial Value Theorem==		==Initial Value Theorem==

2009-12-10T09:43:31Z

Initial Value Theorem

← Older revision		Revision as of 02:43, 10 December 2009
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	<math>\lim_{s \to \infty}sI(s)=f(0^+)</math>		<math>\lim_{s \to \infty}sI(s)=f(0^+)</math>

	<math>\lim_{s \to \infty} \frac{25s+12500}{s(s^2+500s+2500)}=i(0)</math>		<math>\lim_{s \to \infty} s\frac{25s+12500}{s(s^2+500s+2500)}=i(0)</math>

	<math> \Rightarrow i(0)=0</math>		<math> \Rightarrow i(0)=0</math>

	<math>\lim_{s \to \infty}\frac{6250}{s(s^2+500s+2500)}=i(0)</math>		<math>\lim_{s \to \infty}s\frac{6250}{s(s^2+500s+2500)}=i(0)</math>

	<math> \Rightarrow i(0)=0</math>		<math> \Rightarrow i(0)=0</math>

2009-12-10T09:43:19Z

Final Value Theorem

← Older revision		Revision as of 02:43, 10 December 2009
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	<math>\lim_{s \to 0}sI(s)=f(\infty)</math>		<math>\lim_{s \to 0}sI(s)=f(\infty)</math>

	<math>\lim_{s \to \infty} \frac{25s+12500}{s(s^2+500s+2500)}=i(\infty)</math>		<math>\lim_{s \to \infty} s\frac{25s+12500}{s(s^2+500s+2500)}=i(\infty)</math>

	<math>\Rightarrow i(\infty)=0</math>		<math>\Rightarrow i(\infty)=0</math>

	<math>\lim_{s \to 0}\frac{6250}{s(s^2+500s+2500)}=i(\infty)</math>		<math>\lim_{s \to 0}s\frac{6250}{s(s^2+500s+2500)}=i(\infty)</math>

	<math>\Rightarrow i(\infty)=0</math>		<math>\Rightarrow i(\infty)=0</math>

	==Bode Plots==		==Bode Plots==

2009-12-10T09:41:26Z

Final Value Theorem

← Older revision		Revision as of 02:41, 10 December 2009
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	<math>\Rightarrow i(\infty)=0</math>		<math>\Rightarrow i(\infty)=0</math>

	<math>~~\Rightarrow~~ \lim_{s \to 0}\frac{6250}{s(s^2+500s+2500)}=i(\infty)</math>		<math>\lim_{s \to 0}\frac{6250}{s(s^2+500s+2500)}=i(\infty)</math>

	<math>\Rightarrow i(\infty)=0</math>		<math>\Rightarrow i(\infty)=0</math>

	==Bode Plots==		==Bode Plots==

2009-12-10T09:41:01Z

Final Value Theorem

← Older revision		Revision as of 02:41, 10 December 2009
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	==Final Value Theorem==		==Final Value Theorem==

	<math>\lim_{s \to 0}sF(s)=f(\infty)</math>		<math>\lim_{s \to 0}sI(s)=f(\infty)</math>

			<math>\lim_{s \to \infty} \frac{25s+12500}{s(s^2+500s+2500)}=i(\infty)</math>

			<math>\Rightarrow i(\infty)=0</math>

	<math>\Rightarrow \lim_{s \to 0}\frac{6250}{s(s^2+500s+2500)}=i(\infty)</math>		<math>\Rightarrow \lim_{s \to 0}\frac{6250}{s(s^2+500s+2500)}=i(\infty)</math>

2009-12-10T09:39:30Z

Initial Value Theorem

← Older revision		Revision as of 02:39, 10 December 2009
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	==Initial Value Theorem==		==Initial Value Theorem==

	<math>\lim_{s \to \infty}sF(s)=f(0^+)</math>		<math>\lim_{s \to \infty}sI(s)=f(0^+)</math>

	<math>\Rightarrow \lim_{s \to \infty}\frac{6250}{s(s^2+500s+2500)}=i(0)</math>		<math>\lim_{s \to \infty} \frac{25s+12500}{s(s^2+500s+2500)}=i(0)</math>

			<math> \Rightarrow i(0)=0</math>

			<math>\lim_{s \to \infty}\frac{6250}{s(s^2+500s+2500)}=i(0)</math>

	<math> \Rightarrow i(0)=0</math>		<math> \Rightarrow i(0)=0</math>

2009-12-10T09:30:33Z

Initial Value Theorem

← Older revision		Revision as of 02:30, 10 December 2009
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	==Initial Value Theorem==		==Initial Value Theorem==
	~~<math></math>~~
	~~\lim_{s \to \infty}sF(s)=f(0^+)~~

	~~\Rightarrow~~ \lim_{s \to \infty}~~\dfrac{s^3}{~~(s~~^2+400~~)(0~~.01s~~^~~2+s~~+~~10000)}=i(0~~)		<math>\lim_{s \to \infty}sF(s)=f(0^+)</math>

	\Rightarrow i(0)=0		<math>\Rightarrow \lim_{s \to \infty}\frac{6250}{s(s^2+500s+2500)}=i(0)</math>

			<math> \Rightarrow i(0)=0</math>

	==Final Value Theorem==		==Final Value Theorem==