Question

Let $\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathit{e}}^{\mathbf{i}\mathbf{\omega }\mathbf{t}}$on$\mathbf{(}\mathbf{-}\mathbf{\pi }\mathbf{,}\mathbf{\pi }\mathbf{)}$. Expand$\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$in a complex exponential Fourier series of period $\mathbf{2}\mathit{\pi }$. (Assume $\mathit{w}\mathbf{\ne }$integer.)

Short answer

The expanded function $f\left(t\right)$in a complex exponential Fourier series of period $2\pi$is$f\left(t\right)=\sum _{n=-\infty }^{\infty }\frac{\sin \left(\pi (\omega -n)\right)}{\pi (\omega -n)}{e}^{\mathrm{int}}$.

Step-by-step solution

Given information

The given function is $f\left(t\right)=e^{iwt}$. The function $f\left(t\right)$ is to be expanded in a complex exponential Fourier series of period$2\pi$.

Write complex series formulas

The following are the formulas for the complex Fourier series,

$\begin{array}{c}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}{\mathbf{c}}_{\mathbf{n}}{\mathbf{e}}^{\mathbf{i}\mathbf{n}\mathbf{\pi }\mathbf{t}\mathbf{/}\mathbf{l}}\\ {\mathbf{c}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{l}}\underset{-l}{\overset{l}{\int }}f\left(t\right)e^{-in\pi t/l}dt\end{array}$

Here the period of $f\left(t\right)$is $2l$and the frequencies of the terms in the series are$n/2l$.

Find the complex exponential Fourier series

First evaluate complex Fourier series coefficients of the series.

$\begin{array}{c}{c}_n=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}{e}^{i\omega t}{e}^{-\mathrm{int}}dt\\ =\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}{e}^{it(\omega -n)}dt\\ ={\frac{1}{2\pi }\frac{1}{i(\omega -n)}{e}^{it(\omega -n)}|}_{-\pi }^{\pi }\\ =\frac{1}{2\pi }\frac{2i\sin \left(\pi (\omega -n)\right)}{i(\omega -n)}\end{array}$

Further solve to get$c_n$.

$c_n=\frac{\sin \left(\pi (\omega -n)\right)}{\pi (\omega -n)}$

Now, evaluate the function$f\left(t\right)$.

$f\left(t\right)=\sum _{n=-\infty }^{\infty }\frac{\sin \left(\pi (\omega -n)\right)}{\pi (\omega -n)}{e}^{\mathrm{int}}$