ASN2 - Something Interesting: Exponential: Difference between revisions

Latest revision as of 22:52, 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 $c o s (\frac{2 π n t}{T})$ .

Where the Fourier series is $x 1 (t) = \sum_{n = 0}^{\infty} a_{n} c o s (\frac{2 π n t}{T})$

However, a more convient way is using an exponential funtion $e^{\frac{j 2 π n t}{T}}$ .

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

To solve a Fourier series equation for the coefffients $a_{n}$ 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

$x 2 (t)$ . $e^{\frac{- j 2 π m t}{T}} = \int_{- \frac{T}{2}}^{\frac{T}{2}} \sum_{n = 0}^{\infty} a_{n} e^{\frac{j 2 π n t}{T}} e^{\frac{- j 2 π m t}{T}} d t = \sum_{n = 0}^{\infty} a_{n} \int_{- \frac{T}{2}}^{\frac{T}{2}} e^{\frac{j 2 π (n - m) t}{T}} d t = \sum_{n = 0}^{\infty} a_{n} δ_{m n}$

Then the result is

$a_{m} = \int_{- \frac{T}{2}}^{\frac{T}{2}} x 2 (t) e^{\frac{- j 2 π m t}{T}} d t$

Using cosine basis function

$x 1 (t)$ . $c o s (\frac{2 π m t}{T}) = \int_{- \frac{T}{2}}^{\frac{T}{2}} \sum_{n = 0}^{\infty} a_{n} c o s (\frac{2 π n t}{T}) c o s \frac{2 π m t}{T} d t$ At this point you should use a trig identity

applying this trig identity gives

$\frac{1}{2} \sum_{n = 0}^{\infty} a_{n} \int_{- \frac{T}{2}}^{\frac{T}{2}} c o s (\frac{j 2 π (n - m) t}{T}) c o s (\frac{j 2 π (n + m) t}{T}) d t = \frac{1}{2} \sum_{n = 0}^{\infty} a_{n} T δ_{m n}$

Then the result is

$a_{m} = \int_{- \frac{T}{2}}^{\frac{T}{2}} x 1 (t) c o s (\frac{2 π m t}{T}) d t$

@@ Line 2: / Line 2: @@
-Fourier Series
+Here's an demonstration of using the expontential function in a Fourier Series example.
-Using cosine to represent the basis functions
+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>
+.
-Using an exponential to represent basis functions
+Where the Fourier series is <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 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.
+However, a more convient way is using an exponential funtion <math> 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>
+<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 ' '''.''' '  as follows.
-<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>
+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>
+Then the result is
 <math>a_m=\int_{-\frac{T}{2}}^{\frac{T}{2}}x2(t)e^{\frac{ -j2 \pi mt}{T}} dt </math>
+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
@@ Line 24: / Line 29: @@
 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 the result is
+<math>a_m=\int_{-\frac{T}{2}}^{\frac{T}{2}}x1(t)cos({\frac{ 2 \pi mt}{T}}) dt </math>

ASN2 - Something Interesting: Exponential: Difference between revisions

Latest revision as of 22:52, 13 December 2009

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