1.13 Classic fourier series

Write the coefficients of the complex Fourier series in Cartesian form as $c_k=A_k+i{B}_k$ and substitute into the expression for the complex Fourier series. $$\sum_{k=()}$$∞ ∞ c k 2 k t T k ∞ ∞ A k B k 2 k t T Simplifying each term in the sum using Euler's formula, $$\begin{array}{lll}(A_k+i{B}_k)e^{i\frac{2\pi kt}{T}}& \text{=}& (A_k+i{B}_k)(\cos \left(\frac{2\pi kt}{T}\right)+i\sin \left(\frac{2\pi kt}{T}\right))\\ & \text{=}& A_k\cos \left(\frac{2\pi kt}{T}\right)-B_k\sin \left(\frac{2\pi kt}{T}\right)+i(A_k\sin \left(\frac{2\pi kt}{T}\right)+B_k\cos \left(\frac{2\pi kt}{T}\right))\end{array}$$ We now combine terms that have the same frequency index in magnitude . Because the signal is real-valued, the coefficients of the complex Fourier series have conjugate symmetry: $c_{-k}=\overline{c_k}$ or $A_{-k}=A_k$ and $B_{-k}=-B_k$ . After we add the positive-indexed and negative-indexed terms, each term in the Fourier series becomes $2A_k\cos \left(\frac{2\pi kt}{T}\right)-2B_k\sin \left(\frac{2\pi kt}{T}\right)$ . To obtain the classic Fourier series , we must have $2A_k=a_k$ and $2B_k=-b_k$ .

The average of a set of numbers is the sum divided by the number of terms. Viewing signal integration as the limit ofa Riemann sum, the integral corresponds to the average.

We found that the complex Fourier series coefficients are given by $c_k=\frac{2}{i\pi k}$ . The coefficients are pure imaginary, which means $a_k=0$ . The coefficients of the sine terms are given by $b_k=-(2\Im (c_k))$ so that $$b_k=\begin{cases}\frac{4}{\pi k} & \text{if $k\text{ odd}$}\\ 0 & \text{if $k\text{ even}$}\end{cases}$$ Thus, the Fourier series for the square wave is