Difference between revisions of "2019 Spring ECE301 Boutin Fourier series coefficients" - Rhea

Revision as of 20:30, 30 April 2019

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^{jk\frac{2}{4} \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 } a_{k+4}. $ 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 } a_{k+4}. $

$ \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{3}{2} + \frac{1}{2j})e^{j \frac{2}{8}\pi n} + (\frac{3}{2} - \frac{1}{2j})e^{-j \frac{2}{8}\pi n} \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_0 = 1, a_1 = \frac{3}{2} + \frac{1}{2j}, a_{-1} = \frac{3}{2} -\frac{1}{2j}, \text{and } a_k = 0 \text{ for} a_{k+8} . $ However Since this is a discrete signal we must use the period, which in this case is 8.

So in order to change the $ a_{-1} $ we must move to the $ 7^{\text{th}} $ value of the previous period since 8 - 1 is 7. So our final answer would look like $ a_0 = 1, a_1 = \frac{3}{2} + \frac{1}{2j}, a_{7} = \frac{3}{2} -\frac{1}{2j}, \text{and } a_k = 0 \text{ for } a_{k+8}. $

$ \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}\\ $

[to 2019 Spring ECE 301 Boutin]