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 $x1(t)=\sum _{n=0}^{\mathrm{\infty }}{a}_{n}cos(\frac{2\pi nt}{T})$

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

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

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 ' . ' operation of the basis function with $x(t)$

Using exponential basis function

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

Then the result is

$a_m=\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}x2(t)e^{\frac{-j2\pi mt}{T}}dt$

Using cosine basis function

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

applying this trig identity gives

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

Then the result is

$a_m=\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}x1(t)cos(\frac{2\pi mt}{T})dt$