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 but using the eponetial basis function is simplier to solving. To find the coefficients perform the dot product ' . ' as follows.

Using exponential basis function

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

Then the result is

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

Using cosine basis function

$x1(t)\!$ . $cos({\frac {2\pi mt}{T}})=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}\sum _{n=0}^{\infty }a_{n}cos({\frac {2\pi nt}{T}})cos{\frac {2\pi mt}{T}}dt\!$ 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 the result is

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