ASN2 - Something Interesting: Exponential: Difference between revisions

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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>

Revision as of 21:26, 13 December 2009

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Here's an demonstration of using the expontential function in a Fourier Series example.

One way of representing a basis function is with cosine . Where the Fourier series is

However, a more convient way is using an exponential funtion .

To solve a Fourier series equation for the coefffients 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

Preffered method

.

Then result is

Secondary method

. At this point you should use a trig identity

applying this trig identity gives

Then result is