Fourier Series Coefficients

## Introduction

I am going to compute some fourier series coefficients.

## CT signals

$\text{1) } x(t) = sin(6 \pi t), \text{ the frequency of this signal is } \omega_{o} = 6\pi.$

\begin{align} x(t)& = \frac{e^{j6\pi t} - e^{-j6\pi t} }{2j} \\ & = \frac{1}{2j} e^{j6\pi t} - \frac{1}{2j} e^{-j6\pi t} \text{(*)} \end{align}

By Fourier Series we know that $x(t) = \int_{-\infty}^\infty a_k e^{jk\omega_o t} = \int_{-\infty}^\infty a_k e^{jk 6 \pi t} \text{(**)}$

By comparing (*) with (**), we can see that $a_1 = \frac{1}{2j}, a_{-1} = -\frac{1}{2j}, \text{and } a_k = 0 \text{ for all other k}.$

$\text{2) } x(t) = 2 + cos(6 \pi t) - \frac{1}{2} sin(3 \pi t), \omega_{o} = 3\pi.$

\begin{align} x(t) & = 2 + \frac{1}{2}(e^{j6\pi t} + e^{-j6\pi t}) + \frac{1}{4j} (e^{j3\pi t} -e^{-j3\pi t}) \\ & = 2e^{j\pi t} + \frac{1}{2}e^{j 6\pi t} + \frac{1}{2}e^{-j 6\pi t} + \frac{1}{4j}e^{j 3\pi t} - \frac{1}{4j}e^{j 3\pi t} \text{(*)} \end{align}

By Fourier Series we know that $x(t) = \int_{-\infty}^\infty a_k e^{jk\omega_o t} = \int_{-\infty}^\infty a_k e^{jk 3 \pi t} \text{(**)}$

By comparing (*) with (**), we can see that $a_0 = 2, a_1 = \frac{1}{4j}, a_{-1} = -\frac{1}{4j}, a_2 = a_{-2} = \frac{1}{2}, \text{and } a_k = 0 \text{ for all other k} \\$

$\text{3) } x(t) = cos(\frac{2\pi}{10}t), \omega_{o} = \frac{\pi}{10} \\$

\begin{align} x(t) & = \frac{e^{j\frac{2\pi}{10} t} + e^{-j\frac{2\pi}{10} t} }{2} \\ & = \frac{1}{2}e^{j\frac{2\pi}{10} t} + \frac{1}{2}e^{-j\frac{2\pi}{10} t} \text{(*)} \end{align}

By Fourier Series we know that $x(t) = \int_{-\infty}^\infty a_k e^{jk\omega_o t} = \int_{-\infty}^\infty a_k e^{jk 6 \pi t} \text{(**)}$

By comparing (*) with (**), we can see that $a_2 = a_{-2} = \frac{1}{2}, \text{and } a_k = 0 \text{ for all other k}\\$

$\text{4) } x(t) = \begin{cases} 3, & \text{if}\ a=1 \\ 0, & \text{otherwise} \end{cases}$

## DT signals

$\text{1) } x[n] = sin(\frac{2 \pi}{4} n), N = 4 --> \omega_o = 12 \pi$

\begin{align} x[n]& = \frac{e^{j\frac{2}{4}\pi n} - e^{-j\frac{2}{4}\pi n} }{2j} \\ & = \frac{1}{2j} e^{j\frac{2}{4}\pi n} - \frac{1}{2j} e^{-j\frac{2}{4}\pi n} \text{(*)} \end{align}

By Fourier Series we know that $x[n] = \sum_{k=-\infty}^\infty a_k e^{jk\omega_o n} = \sum_{k=-\infty}^\infty a_k e^{j\frac{2}{4}k \pi n} \text{(**)}$

By comparing (*) with (**), we can see that $a_1 = \frac{1}{2j}, a_{-1} = -\frac{1}{2j}, \text{and } a_k = 0 \text{ for all other k}.$ However Since this is a discrete signal we must use the period, which in this case is 4.

So in order to change the $a_{-1}$ we must move to the $3^{\text{rd}}$ value of the previous period since 4 - 1 is 3. So our final answer would look like $a_1 = \frac{1}{2j}, a_{3} = -\frac{1}{2j}, \text{and } a_k = 0 \text{ for all other k}.$

$\text{2) } x[n] = 1 + sin(\frac{2\pi}{8}n) + 3cos(\frac{2\pi}{8}n), N=8 --> \omega_{o} = \frac{2\pi}{8} \\$

\begin{align} x[n] & = 1 + \frac{1}{2j}(e^{j\frac{2}{8}\pi n} - e^{-j\frac{2}{8}\pi n}) + \frac{3}{2} (e^{j\frac{2}{8}\pi n} + e^{-j\frac{2}{8}\pi n}) \\ & = 1e^{j\pi n} + \frac{1}{2j}e^{j \frac{2}{8}\pi n} + \frac{1}{2j}e^{-j \frac{2}{8}\pi n} + \frac{3}{2}e^{j \frac{2}{8}\pi n} - \frac{3}{2}e^{j \frac{2}{8}\pi n} \text{(*)} \end{align}

By Fourier Series we know that $x[n] = \sum_{k=-\infty}^\infty a_k e^{jk\omega_o n} = \sum_{k=-\infty}^\infty a_k e^{jk \frac{2}{8} \pi n} \text{(**)}$

By comparing (*) with (**), we can see that $a_1 = \frac{1}{2j}, a_{-1} = -\frac{1}{2j}, \text{and } a_k = 0 \text{ for all other k}.$

$\text{3) } x[n] = -j^n, \omega_o = \frac{\pi}{2} \\$

\begin{align} x[n]& = \frac{e^{j12\pi n} - e^{-j12\pi n} }{2j} \\ & = \frac{1}{2j} e^{j12\pi n} - \frac{1}{2j} e^{-j12\pi n} \text{(*)} \end{align}

By Fourier Series we know that $x[n] = \sum_{k=-\infty}^\infty a_k e^{jk\omega_o n} = \sum_{k=-\infty}^\infty a_k e^{jk 12 \pi n} \text{(**)}$

By comparing (*) with (**), we can see that $a_1 = \frac{1}{2j}, a_{-1} = -\frac{1}{2j}, \text{and } a_k = 0 \text{ for all other k}.$

$\text{4) } x[n] = \begin{cases} sin(\pi t), & \text{if}\ a=1 \\ 0, & \text{otherwise} \end{cases}\\$

## Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett