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> |
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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 |
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. |
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To find the coefficients perform the dot product ' '''.''' ' operation of the basis function with <math> x(t) \!</math> |
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Preffered method |
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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 |
<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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Then result is |
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Then |
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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> |
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Secondary method |
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<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> |
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Then |
Then result is |
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<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 20:26, 13 December 2009
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