| Line 24: | Line 24: | ||
\begin{align} | \begin{align} | ||
x(t)& = \frac{e^{j6\pi t} - e^{-j6\pi t} }{2j} \\ | 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} | + | & = \frac{1}{2j} e^{j6\pi t} - \frac{1}{2j} e^{-j6\pi t} |
| − | + | ||
| − | + | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
| + | |||
| + | |||
| + | By Fourier Series we know that | ||
| + | <math> | ||
| + | x(t) = \sum_{k=-\infty}^\infty a_k e^{jk\omega_o t} = \sum_{k=-\infty}^\infty a_k e^{jk 6 \pi t}. \text{(*)} | ||
| + | </math> | ||
| + | |||
| + | By comparison with (*), we can see that | ||
| + | <math> | ||
| + | a_1 = \frac{1}{2j}, a_{-1} = -\frac{1}{2j}, a_k = 0 \text{ for all other k's}. | ||
| + | </math> | ||
| + | |||
| + | |||
---- | ---- | ||
<math> | <math> | ||
Revision as of 13:53, 26 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} \end{align} $
By Fourier Series we know that
$ x(t) = \sum_{k=-\infty}^\infty a_k e^{jk\omega_o t} = \sum_{k=-\infty}^\infty a_k e^{jk 6 \pi t}. \text{(*)} $
By comparison with (*), we can see that $ a_1 = \frac{1}{2j}, a_{-1} = -\frac{1}{2j}, a_k = 0 \text{ for all other k's}. $
$ \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}) \\ & \text{By Fourier Series we know that} \sum_{k=-\infty}^\infty a_k e^{jk\omega_o t} \\ & \text{Here, } \omega_o = 3 \pi \text{ ,therefore, } \\ & = 2e^{j \omega_o (0)t} + \frac{1}{2}e^{j \omega_o (2)t} + \frac{1}{2}e^{j \omega_o (-2)t} + \frac{1}{4j}e^{j \omega_o (1)t} - \frac{1}{4j}e^{j \omega_o (-1)t} \\\ & \text{So we can say that } a_0 = 2, a_1 = \frac{1}{4j}, a_{-1} = -\frac{1}{4j}, a_2 = a_{-2} = \frac{1}{2}, a_k = 0 \text{ for all other k} \\ \text{3) } x(t) = cos(\frac{2\pi}{10}t), \omega_{o} = \frac{\pi}{10} \\ & = \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{By Fourier Series we know that} \sum_{k=-\infty}^\infty a_k e^{jk\omega_o t} \\ & \text{Here, } \omega_o = \frac{ \pi}{10} \text{ ,therefore, } \\ & = \frac{1}{2}e^{j\omega_o(2) t} + \frac{1}{2}e^{-j\omega_o(-2) t} \\ & \text{So we can say that } a_2 = a_{-2} = \frac{1}{2}, a_k = 0 \text{ for all other k}\\ \text{4) } x(t) = \begin{cases} 3, & \text{if}\ a=1 \\ 0, & \text{otherwise} \end{cases} \end{align} $
DT signals
$ \begin{align} \text{1) } x[n] = sin(12 \pi n) & = \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{By Fourier Series we know that} \sum_{k=-\infty}^\infty a_k e^{jk\omega_o t} \\ & \text{Here, } \omega_o = 6 \pi \text{ ,therefore, } \\ & = \frac{1}{2j} e^{j\omega_o(1) t} - \frac{1}{2j} e^{j\omega_o(-1) t} \\ & \text{So we can say that } a_1 = \frac{1}{2j}, a_{-1} = -\frac{1}{2j}, 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} \\ & = \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{By Fourier Series we know that} \sum_{k=-\infty}^\infty a_k e^{jk\omega_o t} \\ & \text{Here, } \omega_o = 6 \pi \text{ ,therefore, } \\ & = \frac{1}{2j} e^{j\omega_o(1) t} - \frac{1}{2j} e^{j\omega_o(-1) t} \\ & \text{So we can say that } a_1 = \frac{1}{2j}, a_{-1} = -\frac{1}{2j}, a_k = 0 \text{ for all other k} \\ \text{3) } x[n] = -j^n, \omega_o = \frac{\pi}{2} \\ \text{4) } x[n] = \begin{cases} sin(\pi t), & \text{if}\ a=1 \\ 0, & \text{otherwise} \end{cases}\\ \end{align} $
Questions and comments
If you have any questions, comments, etc. please post them here.
[to 2019 Spring ECE 301 Boutin]
