ASN2 - Something Interesting: Exponential

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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 $cos({\frac {2\pi nt}{T}})\!$ . Where the Fourier series is $x1(t)=\sum _{n=0}^{\infty }a_{n}cos({\frac {2\pi nt}{T}})\!$

However, a more convient way is using an exponential funtion $e^{\frac {j2\pi nt}{T}}\!$.

$x2(t)=\sum _{n=0}^{\infty }a_{n}e^{\frac {j2\pi nt}{T}}\!$

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

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

$x2(t)\!$ . $e^{\frac {-j2\pi mt}{T}}=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}\sum _{n=0}^{\infty }a_{n}e^{\frac {j2\pi nt}{T}}e^{\frac {-j2\pi mt}{T}}dt=\sum _{n=0}^{\infty }a_{n}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{\frac {j2\pi (n-m)t}{T}}dt=\sum _{n=0}^{\infty }a_{n}T\delta _{mn}\!$

Then $a_{m}=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x2(t)e^{\frac {-j2\pi mt}{T}}dt$

$x1(t)\!$ . $cos({\frac {2\pi mt}{T}})=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}\sum _{n=0}^{\infty }a_{n}cos({\frac {2\pi nt}{T}})cos{\frac {2\pi mt}{T}}dt\!$ At this point you should use a trig identity

applying this trig identity gives

${\frac {1}{2}}\sum _{n=0}^{\infty }a_{n}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}cos({\frac {j2\pi (n-m)t}{T}})cos({\frac {j2\pi (n+m)t}{T}})dt={\frac {1}{2}}\sum _{n=0}^{\infty }a_{n}T\delta _{mn}\!$

Then $a_{m}=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x1(t)cos({\frac {2\pi mt}{T}})dt$