ASN2 - Something Interesting: Exponential

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Here's an demonstration of using the expontential function in an Fourier Series example.

One way of representing a basis function is with cosine. $x1(t)=\sum _{n=0}^{\infty }a_{n}cos({\frac {2\pi nt}{T}})\!$

However, a more convient way is using an exponential funtion. $x1(t)=\sum _{n=0}^{\infty }a_{n}e^{\frac {j2\pi nt}{T}}\!$

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 do the dot product ' . ' of the basis function and $x(t)\!$

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

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