ASN2 - Something Interesting: Exponential: Difference between revisions

Revision as of 08:54, 3 December 2009

Fourier Series

Using cosine to represent the basis functions $x 1 (t) = \sum_{n = 0}^{\infty} a_{n} c o s (\frac{2 π n t}{T})$

Using an exponential to represent basis functions $x 1 (t) = \sum_{n = 0}^{\infty} a_{n} e^{\frac{j 2 π n t}{T}}$

To solve for the coefffients $a_{n}$ 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 perform the dot product

 $x 2 (t) = \sum_{n = 0}^{\infty} a_{n} e^{\frac{j 2 π n t}{T}}$

@@ Line 5: / Line 5: @@
 Using cosine to represent the basis functions
-<math> x(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
-<math> x(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>
-To obtain the coefffients <math>  a_n \!</math> the solutions are almost identical. The benefit of using the eponetial funtion is that mathematical it is simplier for solving than using the cosine function.
+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 coefficients perform the dot product
+ <math> x2(t)= \sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}} \!</math>

ASN2 - Something Interesting: Exponential: Difference between revisions

Revision as of 08:54, 3 December 2009

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