ASN2 - Something Interesting: Exponential: Difference between revisions

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One way of representing a basis function is with cosine <math>  cos(\frac{ 2 \pi nt}{T}) \!</math>
One way of representing a basis function is with cosine <math>  cos(\frac{ 2 \pi nt}{T}) \!</math>
.
.
Where the Fourier series is <math> x1(t)= \sum_{n=0}^\infty a_n cos(\frac{ 2 \pi nt}{T}) \!</math>
Where the Fourier series is <math> x1(t)= \sum_{n=0}^\infty a_n cos(\frac{ 2 \pi nt}{T}) \!</math>


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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. To find the coefficients perform the dot product ' '''.''' '  as follows.


The procedure to solve for the coefficients is to perform the dot product ' '''.''' '  operation of the basis function with <math> x(t) \!</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>


<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>
Then the 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>


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
<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 the 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>

Latest revision as of 22:52, 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 cos(2πntT) .

Where the Fourier series is x1(t)=n=0ancos(2πntT)

However, a more convient way is using an exponential funtion ej2πntT.

x2(t)=n=0anej2πntT

To solve a Fourier series equation for the coefffients an 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

x2(t) . ej2πmtT=T2T2n=0anej2πntTej2πmtTdt=n=0anT2T2ej2π(nm)tTdt=n=0anδmn

Then the result is

am=T2T2x2(t)ej2πmtTdt

Using cosine basis function

x1(t) . cos(2πmtT)=T2T2n=0ancos(2πntT)cos2πmtTdt At this point you should use a trig identity

applying this trig identity gives

12n=0anT2T2cos(j2π(nm)tT)cos(j2π(n+m)tT)dt=12n=0anTδmn

Then the result is

am=T2T2x1(t)cos(2πmtT)dt