| (13 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
[[Category:2019 Spring ECE 301 Boutin]] | [[Category:2019 Spring ECE 301 Boutin]] | ||
[[Category:ECE301]] | [[Category:ECE301]] | ||
| − | |||
| − | |||
| Line 13: | Line 11: | ||
---- | ---- | ||
==Introduction== | ==Introduction== | ||
| − | I am going to compute some | + | I am going to compute some Fourier series coefficients. I have done 3 in both CT and DT, with explanations as to how I got my answers. Hope you can find this helpful! |
---- | ---- | ||
| Line 43: | Line 41: | ||
---- | ---- | ||
<math> | <math> | ||
| − | \text{2) } x(t) = 2 + cos(6 \pi t) - \frac{1}{2} sin(3 \pi t), \omega_{o} = 3\pi. | + | \text{2) } x(t) = 2 + cos(6 \pi t) - \frac{1}{2} sin(3 \pi t),\text{ the frequency of this signal is } \omega_{o} = 3\pi. |
</math> | </math> | ||
| Line 49: | Line 47: | ||
\begin{align} | \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}) \\ | 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{(*)} | + | & = 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} | \end{align} | ||
</math> | </math> | ||
| Line 65: | Line 63: | ||
---- | ---- | ||
<math> | <math> | ||
| − | \text{3) } x(t) = cos(\frac{2\pi}{10}t), \omega_{o} = \frac{\pi}{10} \\ | + | \text{3) } x(t) = cos(\frac{2\pi}{10}t), \text{ the frequency of this signal is } \omega_{o} = \frac{2\pi}{10} \\ |
</math> | </math> | ||
| Line 77: | Line 75: | ||
By Fourier Series we know that | By Fourier Series we know that | ||
<math> | <math> | ||
| − | x(t) = \int_{-\infty}^\infty a_k e^{jk\omega_o t} = \int_{-\infty}^\infty a_k e^{jk | + | x(t) = \int_{-\infty}^\infty a_k e^{jk\omega_o t} = \int_{-\infty}^\infty a_k e^{jk \frac{2\pi}{10} t} \text{(**)} |
</math> | </math> | ||
| Line 83: | Line 81: | ||
<math> | <math> | ||
a_2 = a_{-2} = \frac{1}{2}, \text{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 99: | Line 88: | ||
---- | ---- | ||
<math> | <math> | ||
| − | \text{1) } x[n] = sin(\frac{2 \pi}{4} n), N = 4 --> \omega_o = | + | \text{1) } x[n] = sin(\frac{2 \pi}{4} n), N = 4 --> \text{ the frequency of this signal is } \omega_o = \frac{2\pi}{4} |
</math> | </math> | ||
| Line 112: | Line 101: | ||
By Fourier Series we know that | By Fourier Series we know that | ||
<math> | <math> | ||
| − | x[n] = \sum_{k=-\infty}^\infty a_k e^{jk\omega_o n} = \sum_{k=-\infty}^\infty a_k e^{ | + | 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{(**)} |
</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 | + | a_1 = \frac{1}{2j}, a_{-1} = -\frac{1}{2j}, \text{and } a_k = 0 \text{ for } a_{k+4}. |
</math> | </math> | ||
However Since this is a discrete signal we must use the period, which in this case is 4. | However Since this is a discrete signal we must use the period, which in this case is 4. | ||
<math> | <math> | ||
</math> | </math> | ||
| − | So in order to change the <math> a_{-1} </math> we must move to the <math> 3^{\text{rd}} </math> value of the previous period since 4 - 1 is 3. So our final answer would | + | So in order to change the <math> a_{-1} </math> we must move to the <math> 3^{\text{rd}} </math> value of the previous period since 4 - 1 is 3. So our final answer would be |
<math> | <math> | ||
| − | a_1 = \frac{1}{2j}, a_{3} = -\frac{1}{2j}, \text{and } | + | a_1 = \frac{1}{2j}, a_{3} = -\frac{1}{2j}, a_k = 0 \text{ for 0,2 }, \text{ and } a_{k+4} = a_k \text{ for all other k}. |
</math> | </math> | ||
---- | ---- | ||
<math> | <math> | ||
| − | \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} \\ | + | \text{2) } x[n] = 1 + sin(\frac{2\pi}{8}n) + 3cos(\frac{2\pi}{8}n), N=8 --> \text{ the frequency of this signal is } \omega_{o} = \frac{2\pi}{8} \\ |
</math> | </math> | ||
| Line 135: | Line 124: | ||
\begin{align} | \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}) \\ | 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} + | + | & = 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} \text{(*)} |
\end{align} | \end{align} | ||
</math> | </math> | ||
| Line 146: | Line 135: | ||
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 | + | 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} . |
</math> | </math> | ||
| + | However Since this is a discrete signal we must use the period, which in this case is 8. | ||
| + | <math> | ||
| + | </math> | ||
| + | So in order to change the <math> a_{-1} </math> we must move to the <math> 7^{\text{th}} </math> value of the previous period since 8 - 1 is 7. So our final answer would be | ||
| + | <math> | ||
| + | a_0 = 1, a_1 = \frac{3}{2} + \frac{1}{2j}, a_{7} = \frac{3}{2} -\frac{1}{2j}, a_k = 0 \text{ for 2,3,4,5,6 }, \text{ and } a_{k+8} = a_k \text{ for all other k}. | ||
| + | </math> | ||
| + | |||
---- | ---- | ||
<math> | <math> | ||
| − | \text{3) } x[n] = -j^n, \omega_o = \frac{\pi}{2} \\ | + | \text{3) } x[n] = -j^n, \text{ the frequency of this signal is } \omega_o = \frac{\pi}{2} \\ |
</math> | </math> | ||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
| − | x[n]& = | + | x[n]& = -e^{j\frac{\pi}{2} n} \text{(*)} |
| − | + | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
| Line 165: | Line 161: | ||
By Fourier Series we know that | By Fourier Series we know that | ||
<math> | <math> | ||
| − | x[n] = \sum_{k=-\infty}^\infty a_k e^{jk\omega_o n} = \sum_{k=-\infty}^\infty a_k e^{jk | + | x[n] = \sum_{k=-\infty}^\infty a_k e^{jk\omega_o n} = \sum_{k=-\infty}^\infty a_k e^{jk \frac{\pi}{2} n} \text{(**)} |
</math> | </math> | ||
By comparing (*) with (**), we can see that | By comparing (*) with (**), we can see that | ||
<math> | <math> | ||
| − | a_1 = | + | a_1 = -1, a_{k} = 0 \text{ for 0,2,3 }, \text{ and } a_{k+4} = a_k \text{ for all other k}. |
</math> | </math> | ||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
---- | ---- | ||
[[https://www.projectrhea.org/rhea/index.php/2019_Spring_ECE_301_Boutin|Back to 2019 Spring ECE 301 Boutin]] | [[https://www.projectrhea.org/rhea/index.php/2019_Spring_ECE_301_Boutin|Back to 2019 Spring ECE 301 Boutin]] | ||
---- | ---- | ||
Latest revision as of 20:55, 30 April 2019
A project by Kalyan Mada
Introduction
I am going to compute some Fourier series coefficients. I have done 3 in both CT and DT, with explanations as to how I got my answers. Hope you can find this helpful!
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),\text{ the frequency of this signal is } \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), \text{ the frequency of this signal is } \omega_{o} = \frac{2\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 \frac{2\pi}{10} 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}\\ $
DT signals
$ \text{1) } x[n] = sin(\frac{2 \pi}{4} n), N = 4 --> \text{ the frequency of this signal is } \omega_o = \frac{2\pi}{4} $
$ \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 be $ a_1 = \frac{1}{2j}, a_{3} = -\frac{1}{2j}, a_k = 0 \text{ for 0,2 }, \text{ and } a_{k+4} = a_k \text{ for all other k}. $
$ \text{2) } x[n] = 1 + sin(\frac{2\pi}{8}n) + 3cos(\frac{2\pi}{8}n), N=8 --> \text{ the frequency of this signal is } \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} \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_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 be $ a_0 = 1, a_1 = \frac{3}{2} + \frac{1}{2j}, a_{7} = \frac{3}{2} -\frac{1}{2j}, a_k = 0 \text{ for 2,3,4,5,6 }, \text{ and } a_{k+8} = a_k \text{ for all other k}. $
$ \text{3) } x[n] = -j^n, \text{ the frequency of this signal is } \omega_o = \frac{\pi}{2} \\ $
$ \begin{align} x[n]& = -e^{j\frac{\pi}{2} 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{\pi}{2} n} \text{(**)} $
By comparing (*) with (**), we can see that $ a_1 = -1, a_{k} = 0 \text{ for 0,2,3 }, \text{ and } a_{k+4} = a_k \text{ for all other k}. $
[to 2019 Spring ECE 301 Boutin]
