ASN2 - Something Interesting: Exponential: Difference between revisions
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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. |
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. |
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To solve for the coefficients |
To solve for the coefficients do the dot product ' '''.''' ' of the basis function and <math> x(t) \!</math> |
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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> |
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<math>a_m=\int_{-\frac{T}{2}}^{\frac{T}{2}}x2(t)e^{\frac{ -j2 \pi mt}{T}} dt </math> |
Revision as of 08:20, 3 December 2009
Fourier Series
Using cosine to represent the basis functions
Using an exponential to represent basis functions
To solve for the coefffients the solutions for both are almost identical. The benefit of using the eponetialinstead of cosine is that mathematical it is simplier for solving.
To solve for the coefficients do the dot product ' . ' of the basis function and
.