ASN2 - Something Interesting: Exponential - Revision history

Jodi.Hodge at 05:52, 14 December 2009

2009-12-14T05:52:27Z

← Older revision		Revision as of 22:52, 13 December 2009
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	To solve a Fourier series equation for the coefffients <math> a_n \!</math> using the above expressions result in similar solutions but using the eponetial basis function is simplier to solving. To find the coefficients perform the dot product ' '''.''' ' as follows.		To solve a Fourier series equation for the coefffients <math> a_n \!</math> using the above expressions result in similar solutions but using the eponetial basis function is simplier to solving. To find the coefficients perform the dot product ' '''.''' ' as follows.

	Using exponential basis function		Using exponential basis function

Jodi.Hodge at 05:52, 14 December 2009

2009-12-14T05:52:03Z

← Older revision		Revision as of 22:52, 13 December 2009
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	<math> x2(t)= \sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}} \!</math>		<math> x2(t)= \sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}} \!</math>

	To solve a Fourier series equation for the coefffients <math> a_n \!</math> using the above expressions result in similar solutions but using the eponetial basis function is simplier to solving. To find the coefficients perform the dot product ' '''.''' ' ~~operation of the basis function with <math> x(t) \!</math>~~		To solve a Fourier series equation for the coefffients <math> a_n \!</math> using the above expressions result in similar solutions but using the eponetial basis function is simplier to solving. To find the coefficients perform the dot product ' '''.''' ' as follows.

	Using exponential basis function		Using exponential basis function

Jodi.Hodge at 05:51, 14 December 2009

2009-12-14T05:51:24Z

← Older revision		Revision as of 22:51, 13 December 2009
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	One way of representing a basis function is with cosine <math> cos(\frac{ 2 \pi nt}{T}) \!</math>		One way of representing a basis function is with cosine <math> cos(\frac{ 2 \pi nt}{T}) \!</math>
	.		.

	Where the Fourier series is <math> x1(t)= \sum_{n=0}^\infty a_n cos(\frac{ 2 \pi nt}{T}) \!</math>		Where the Fourier series is <math> x1(t)= \sum_{n=0}^\infty a_n cos(\frac{ 2 \pi nt}{T}) \!</math>

Jodi.Hodge at 04:28, 14 December 2009

2009-12-14T04:28:56Z

← Older revision		Revision as of 21:28, 13 December 2009
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	<math> x2(t) \!</math> '''.''' <math> e^{\frac{ -j2 \pi mt}{T}} = \int_{-\frac{T}{2}}^{\frac{T}{2}}\sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}}e^{\frac{ -j2 \pi mt}{T}} dt =\sum_{n=0}^\infty a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} e^{\frac{ j2 \pi (n-m)t}{T}} dt =\sum_{n=0}^\infty a_n \delta_{mn} \!</math>		<math> x2(t) \!</math> '''.''' <math> e^{\frac{ -j2 \pi mt}{T}} = \int_{-\frac{T}{2}}^{\frac{T}{2}}\sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}}e^{\frac{ -j2 \pi mt}{T}} dt =\sum_{n=0}^\infty a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} e^{\frac{ j2 \pi (n-m)t}{T}} dt =\sum_{n=0}^\infty a_n \delta_{mn} \!</math>

	Then result is		Then the result is

	<math>a_m=\int_{-\frac{T}{2}}^{\frac{T}{2}}x2(t)e^{\frac{ -j2 \pi mt}{T}} dt </math>		<math>a_m=\int_{-\frac{T}{2}}^{\frac{T}{2}}x2(t)e^{\frac{ -j2 \pi mt}{T}} dt </math>
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	<math> \frac{1}{2} \sum_{n=0}^\infty a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} cos({\frac{ j2 \pi (n-m)t}{T}})cos({\frac{ j2 \pi (n+m)t}{T}}) dt =\frac{1}{2}\sum_{n=0}^\infty a_n T \delta_{mn} \!</math>		<math> \frac{1}{2} \sum_{n=0}^\infty a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} cos({\frac{ j2 \pi (n-m)t}{T}})cos({\frac{ j2 \pi (n+m)t}{T}}) dt =\frac{1}{2}\sum_{n=0}^\infty a_n T \delta_{mn} \!</math>

	Then result is		Then the result is

	<math>a_m=\int_{-\frac{T}{2}}^{\frac{T}{2}}x1(t)cos({\frac{ 2 \pi mt}{T}}) dt </math>		<math>a_m=\int_{-\frac{T}{2}}^{\frac{T}{2}}x1(t)cos({\frac{ 2 \pi mt}{T}}) dt </math>

Jodi.Hodge at 04:28, 14 December 2009

2009-12-14T04:28:35Z

← Older revision		Revision as of 21:28, 13 December 2009
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	To solve a Fourier series equation for the coefffients <math> a_n \!</math> using the above expressions result in similar solutions but using the eponetial basis function is simplier to solving. To find the coefficients perform the dot product ' '''.''' ' operation of the basis function with <math> x(t) \!</math>		To solve a Fourier series equation for the coefffients <math> a_n \!</math> using the above expressions result in similar solutions but using the eponetial basis function is simplier to solving. To find the coefficients perform the dot product ' '''.''' ' operation of the basis function with <math> x(t) \!</math>

	~~Preffered method~~		Using exponential basis function

	<math> x2(t) \!</math> '''.''' <math> e^{\frac{ -j2 \pi mt}{T}} = \int_{-\frac{T}{2}}^{\frac{T}{2}}\sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}}e^{\frac{ -j2 \pi mt}{T}} dt =\sum_{n=0}^\infty a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} e^{\frac{ j2 \pi (n-m)t}{T}} dt =\sum_{n=0}^\infty a_n \delta_{mn} \!</math>		<math> x2(t) \!</math> '''.''' <math> e^{\frac{ -j2 \pi mt}{T}} = \int_{-\frac{T}{2}}^{\frac{T}{2}}\sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}}e^{\frac{ -j2 \pi mt}{T}} dt =\sum_{n=0}^\infty a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} e^{\frac{ j2 \pi (n-m)t}{T}} dt =\sum_{n=0}^\infty a_n \delta_{mn} \!</math>
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	<math>a_m=\int_{-\frac{T}{2}}^{\frac{T}{2}}x2(t)e^{\frac{ -j2 \pi mt}{T}} dt </math>		<math>a_m=\int_{-\frac{T}{2}}^{\frac{T}{2}}x2(t)e^{\frac{ -j2 \pi mt}{T}} dt </math>

	~~Secondary method~~		Using cosine basis function

	<math> x1(t) \!</math> '''.''' <math> cos({\frac{ 2 \pi mt}{T}}) = \int_{-\frac{T}{2}}^{\frac{T}{2}}\sum_{n=0}^\infty a_n cos({\frac{ 2 \pi nt}{T}})cos{\frac{ 2 \pi mt}{T}} dt \!</math> At this point you should use a trig identity		<math> x1(t) \!</math> '''.''' <math> cos({\frac{ 2 \pi mt}{T}}) = \int_{-\frac{T}{2}}^{\frac{T}{2}}\sum_{n=0}^\infty a_n cos({\frac{ 2 \pi nt}{T}})cos{\frac{ 2 \pi mt}{T}} dt \!</math> At this point you should use a trig identity

Jodi.Hodge at 04:27, 14 December 2009

2009-12-14T04:27:33Z

← Older revision		Revision as of 21:27, 13 December 2009
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	<math> x2(t)= \sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}} \!</math>		<math> x2(t)= \sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}} \!</math>

	To solve a Fourier series equation for the coefffients <math> a_n \!</math> using the above expressions result in similar solutions but using the eponetial basis function is simplier to solving.		To solve a Fourier series equation for the coefffients <math> a_n \!</math> using the above expressions result in similar solutions but using the eponetial basis function is simplier to solving. To find the coefficients perform the dot product ' '''.''' ' operation of the basis function with <math> x(t) \!</math>

	To find the coefficients perform the dot product ' '''.''' ' operation of the basis function with <math> x(t) \!</math>

	Preffered method		Preffered method

Jodi.Hodge at 04:26, 14 December 2009

2009-12-14T04:26:39Z

← Older revision		Revision as of 21:26, 13 December 2009
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	<math> x2(t)= \sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}} \!</math>		<math> x2(t)= \sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}} \!</math>

	To solve a Fourier series equation for the coefffients <math> a_n \!</math> using the above expressions result in similar solutions~~. The perfered method of solving is to use~~ the eponetial basis function ~~because for is that mathematical it~~ is simplier ~~for~~ solving.		To solve a Fourier series equation for the coefffients <math> a_n \!</math> using the above expressions result in similar solutions but using the eponetial basis function is simplier to solving.

	~~The procedure to solve for~~ the coefficients ~~is to~~ perform the dot product ' '''.''' ' operation of the basis function with <math> x(t) \!</math>		To find the coefficients perform the dot product ' '''.''' ' operation of the basis function with <math> x(t) \!</math>

			Preffered method

	<math> x2(t) \!</math> '''.''' <math> e^{\frac{ -j2 \pi mt}{T}} = \int_{-\frac{T}{2}}^{\frac{T}{2}}\sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}}e^{\frac{ -j2 \pi mt}{T}} dt =\sum_{n=0}^\infty a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} e^{\frac{ j2 \pi (n-m)t}{T}} dt =\sum_{n=0}^\infty a_n T \delta_{mn} \!</math>		<math> x2(t) \!</math> '''.''' <math> e^{\frac{ -j2 \pi mt}{T}} = \int_{-\frac{T}{2}}^{\frac{T}{2}}\sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}}e^{\frac{ -j2 \pi mt}{T}} dt =\sum_{n=0}^\infty a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} e^{\frac{ j2 \pi (n-m)t}{T}} dt =\sum_{n=0}^\infty a_n \delta_{mn} \!</math>

			Then result is

	~~Then~~
	<math>a_m=\int_{-\frac{T}{2}}^{\frac{T}{2}}x2(t)e^{\frac{ -j2 \pi mt}{T}} dt </math>		<math>a_m=\int_{-\frac{T}{2}}^{\frac{T}{2}}x2(t)e^{\frac{ -j2 \pi mt}{T}} dt </math>

			Secondary method

	<math> x1(t) \!</math> '''.''' <math> cos({\frac{ 2 \pi mt}{T}}) = \int_{-\frac{T}{2}}^{\frac{T}{2}}\sum_{n=0}^\infty a_n cos({\frac{ 2 \pi nt}{T}})cos{\frac{ 2 \pi mt}{T}} dt \!</math> At this point you should use a trig identity		<math> x1(t) \!</math> '''.''' <math> cos({\frac{ 2 \pi mt}{T}}) = \int_{-\frac{T}{2}}^{\frac{T}{2}}\sum_{n=0}^\infty a_n cos({\frac{ 2 \pi nt}{T}})cos{\frac{ 2 \pi mt}{T}} dt \!</math> At this point you should use a trig identity
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	<math> \frac{1}{2} \sum_{n=0}^\infty a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} cos({\frac{ j2 \pi (n-m)t}{T}})cos({\frac{ j2 \pi (n+m)t}{T}}) dt =\frac{1}{2}\sum_{n=0}^\infty a_n T \delta_{mn} \!</math>		<math> \frac{1}{2} \sum_{n=0}^\infty a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} cos({\frac{ j2 \pi (n-m)t}{T}})cos({\frac{ j2 \pi (n+m)t}{T}}) dt =\frac{1}{2}\sum_{n=0}^\infty a_n T \delta_{mn} \!</math>

	Then		Then result is

	<math>a_m=\int_{-\frac{T}{2}}^{\frac{T}{2}}x1(t)cos({\frac{ 2 \pi mt}{T}}) dt </math>		<math>a_m=\int_{-\frac{T}{2}}^{\frac{T}{2}}x1(t)cos({\frac{ 2 \pi mt}{T}}) dt </math>

Jodi.Hodge at 04:10, 14 December 2009

2009-12-14T04:10:49Z

← Older revision		Revision as of 21:10, 13 December 2009
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	Here's an demonstration of using the expontential function in an Fourier Series example.		Here's an demonstration of using the expontential function in a Fourier Series example.

	One way of representing a basis function is with cosine.		One way of representing a basis function is with cosine <math> cos(\frac{ 2 \pi nt}{T}) \!</math>
	<math> x1(t)= \sum_{n=0}^\infty a_n cos(\frac{ 2 \pi nt}{T}) \!</math>		.
			Where the Fourier series is <math> x1(t)= \sum_{n=0}^\infty a_n cos(\frac{ 2 \pi nt}{T}) \!</math>

	However, a more convient way is using an exponential funtion.		However, a more convient way is using an exponential funtion <math> e^{\frac{ j2 \pi nt}{T}} \!</math>.
	<math> ~~x1(t)= \sum_{n=0}^\infty a_n~~ e^{\frac{ j2 \pi nt}{T}} \!</math>

	~~To solve for the coefffients~~ <math> a_n \!</math> ~~the solutions for both are almost identical. The benefit of using the eponetialinstead of cosine is that mathematical it is simplier for solving.~~		<math> x2(t)= \sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}} \!</math>

	To solve for the coefficients do the dot product ' '''.''' ' of the basis function ~~and~~ <math> x(t) \!</math>		To solve a Fourier series equation for the coefffients <math> a_n \!</math> using the above expressions result in similar solutions. The perfered method of solving is to use the eponetial basis function because for is that mathematical it is simplier for solving.

			The procedure to solve for the coefficients is to perform the dot product ' '''.''' ' operation of the basis function with <math> x(t) \!</math>

Jodi.Hodge at 03:54, 14 December 2009

2009-12-14T03:54:01Z

← Older revision		Revision as of 20:54, 13 December 2009
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	Fourier Series		Here's an demonstration of using the expontential function in an Fourier Series example.

	~~Using~~ cosine ~~to represent the basis functions~~		One way of representing a basis function is with cosine.
	<math> x1(t)= \sum_{n=0}^\infty a_n cos(\frac{ 2 \pi nt}{T}) \!</math>		<math> x1(t)= \sum_{n=0}^\infty a_n cos(\frac{ 2 \pi nt}{T}) \!</math>

	~~Using~~ an exponential ~~to represent basis functions~~		However, a more convient way is using an exponential funtion.
	<math> x1(t)= \sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}} \!</math>		<math> x1(t)= \sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}} \!</math>

Jodi.Hodge at 16:39, 3 December 2009

2009-12-03T16:39:09Z

← Older revision		Revision as of 09:39, 3 December 2009
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	<math> x2(t) \!</math> '''.''' <math> e^{\frac{ -j2 \pi mt}{T}} = \int_{-\frac{T}{2}}^{\frac{T}{2}}\sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}}e^{\frac{ -j2 \pi mt}{T}} dt =\sum_{n=0}^\infty a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} e^{\frac{ j2 \pi (n-m)t}{T}} dt =\sum_{n=0}^\infty a_n T \delta_{mn} \!</math>		<math> x2(t) \!</math> '''.''' <math> e^{\frac{ -j2 \pi mt}{T}} = \int_{-\frac{T}{2}}^{\frac{T}{2}}\sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}}e^{\frac{ -j2 \pi mt}{T}} dt =\sum_{n=0}^\infty a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} e^{\frac{ j2 \pi (n-m)t}{T}} dt =\sum_{n=0}^\infty a_n T \delta_{mn} \!</math>

			Then
	<math>a_m=\int_{-\frac{T}{2}}^{\frac{T}{2}}x2(t)e^{\frac{ -j2 \pi mt}{T}} dt </math>		<math>a_m=\int_{-\frac{T}{2}}^{\frac{T}{2}}x2(t)e^{\frac{ -j2 \pi mt}{T}} dt </math>

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	applying this trig identity gives		applying this trig identity gives

	<math> \frac{1}{2} \sum_{n=0}^\infty a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} cos({\frac{ j2 \pi (n-m)t}{T}})cos({\frac{ j2 \pi (n+m)t}{T}}) dt =\sum_{n=0}^\infty a_n T \delta_{mn} \!</math>		<math> \frac{1}{2} \sum_{n=0}^\infty a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} cos({\frac{ j2 \pi (n-m)t}{T}})cos({\frac{ j2 \pi (n+m)t}{T}}) dt =\frac{1}{2}\sum_{n=0}^\infty a_n T \delta_{mn} \!</math>

			Then
			<math>a_m=\int_{-\frac{T}{2}}^{\frac{T}{2}}x1(t)cos({\frac{ 2 \pi mt}{T}}) dt </math>