Complex Exponential Fourier Series

Inner product is just Ravi's fancy math way of saying dot product for functions, and orthogonal is his fancy math way of saying perpendicular.

V_1 and V_2 are orthonormal if:

If \psi_l and \psi_k are orthonormal on [a,b]:

\int\limits_a^b \psi_k(t) \psi_l^*(t)dt = \begin{cases} 1 & k=l \\ 0 & k \ne l \end{cases}

Consider the interval [0,1]:

\text{Let } \psi_1(t) = 1; t \in [0,1]

\text{Let } \psi_2(t) = \begin{cases} \end{cases}

x(t) = \sum_{k=-\infty}^\infty X_k e^{jk\omega_0 t}

T_0 = 1

\omega_0 = 2\pi

x(t) = 1 + \sum_{k=-\infty, k\ne0}^\infty {2 \sin\left({ \pi k \over 2}\right) \over \pi k}e^{2\pi k t}

x(t) = 1 + \sum_{k=-\infty, k\ne0}^\infty {2 \sin\left({\pi k \over 2}\right) \over \pi k} e^{jk2\pi t}

x(t - 0.1) = 1 + \sum_{k=-\infty, k\ne0}^\infty {2 \sin\left({\pi k \over 2}\right) \over \pi k} e^{jk2\pi (t - 0.1)}

x(t) = c_0 + 2\sum_{k=1}^\infty \left[c_k \cos(k \omega_0 t) + d_k \sin(k \omega_0 t)\right]

A = 1, T = 2, T_0 = 2, \omega_0 = \pi

x(t) = t

x(t) = \sum_k X_k e^{j\pi k t}

X_k = {1\over2} \int\limits_{-1}^1 t e^{-j \pi k t} dt

This gives an imaginary X_k.

X_0 = {1\over2} \int\limits_{-1}^1 t dt = 0

x(t) = c_0 + 2\sum_{k=1}^\infty c_k \cos(\pi k t) + d_k \sin(\pi k t)

Since it's odd:

c_k = 0

d_k = {1\over2} \int\limits_{-1}^1 t \sin(\pi k t) dt

x_t = 2 \sum_{k=1}^\infty {(-1)^{k+1} \over \pi k} \sin(\pi k t)