| Line 32: | Line 32: | ||
By Fourier Series we know that | By Fourier Series we know that | ||
<math> | <math> | ||
| − | x(t) = \ | + | x(t) = \int_{-\infty}^\infty a_k e^{jk\omega_o t} = \int_{-\infty}^\infty a_k e^{jk 6 \pi t} \text{(**)} |
</math> | </math> | ||
| Line 55: | Line 55: | ||
By Fourier Series we know that | By Fourier Series we know that | ||
<math> | <math> | ||
| − | x(t) = \ | + | x(t) = \int_{-\infty}^\infty a_k e^{jk\omega_o t} = \int_{-\infty}^\infty a_k e^{jk 3 \pi t} \text{(**)} |
</math> | </math> | ||
By comparing (*) with (**), we can see that | By comparing (*) with (**), we can see that | ||
<math> | <math> | ||
| − | a_0 = 2, a_1 = \frac{1}{4j}, a_{-1} = -\frac{1}{4j}, a_2 = a_{-2} = \frac{1}{2}, and a_k = 0 \text{ for all other k} \\ | + | 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} \\ |
</math> | </math> | ||
| Line 77: | Line 77: | ||
By Fourier Series we know that | By Fourier Series we know that | ||
<math> | <math> | ||
| − | x(t) = \ | + | x(t) = \int_{-\infty}^\infty a_k e^{jk\omega_o t} = \int_{-\infty}^\infty a_k e^{jk 6 \pi t} \text{(**)} |
</math> | </math> | ||
By comparing (*) with (**), we can see that | By comparing (*) with (**), we can see that | ||
<math> | <math> | ||
| − | a_2 = a_{-2} = \frac{1}{2}, and a_k = 0 \text{ for all other k}\\ | + | a_2 = a_{-2} = \frac{1}{2}, \text{and } a_k = 0 \text{ for all other k}\\ |
</math> | </math> | ||
| Line 112: | Line 112: | ||
By Fourier Series we know that | By Fourier Series we know that | ||
<math> | <math> | ||
| − | x[n] = \ | + | 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{(**)} |
</math> | </math> | ||
By comparing (*) with (**), we can see that | By comparing (*) with (**), we can see that | ||
<math> | <math> | ||
| − | a_1 = \frac{1}{2j}, a_{-1} = -\frac{1}{2j}, \text{and } a_k = 0 \text{ for all other k}. | + | 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 apply the principle where we use the period, which in this case is 24. }. |
</math> | </math> | ||
| Line 134: | Line 134: | ||
By Fourier Series we know that | By Fourier Series we know that | ||
<math> | <math> | ||
| − | x[n] = \ | + | 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{(**)} |
</math> | </math> | ||
| Line 158: | Line 158: | ||
By Fourier Series we know that | By Fourier Series we know that | ||
<math> | <math> | ||
| − | x[n] = \ | + | 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{(**)} |
</math> | </math> | ||
Revision as of 20:11, 30 April 2019
A project by Kalyan Mada
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(12 \pi n), \omega_o = 12 \pi $
$ \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. However Since this is a discrete signal we must apply the principle where we use the period, which in this case is 24. }. $
$ \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}\\ $
[to 2019 Spring ECE 301 Boutin]
