ASN2 - Something Interesting: Exponential

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Fourier Series

Using cosine to represent the basis functions $x1(t)=\sum _{n=0}^{\infty }a_{n}cos({\frac {2\pi nt}{T}})\!$

Using an exponential to represent basis functions $x1(t)=\sum _{n=0}^{\infty }a_{n}e^{\frac {j2\pi nt}{T}}\!$

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