View source for 2019 Spring ECE301 Boutin Fourier series coefficients - Rhea

[[Category:2019 Spring ECE 301 Boutin]]
[[Category:ECE301]]
[[Category:ECE]]
[[Category:slecture]]


<center><font size= 4>Fourier Series Coefficients</font size>

A project by Kalyan Mada

</center>
----
----
==Introduction==
I am going to compute some fourier series coefficients. 

----
==CT signals==
<math> 
\begin{align}
\text{1) } x(t) = sin(6 \pi t), \omega_{o} = 6\pi \\
& = \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(t) = 2 + cos(6 \pi t) - \frac{1}{2} sin(3 \pi t), \omega_{o} = 3\pi \\
& = 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}
</math>
----

==DT signals==
<math> 
\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}
</math>
----
----
==[[2019_Spring_ECE_301_Boutin_21 Savage in MATLAB_Jeff Rodrigues_comments | Questions and comments]]==
If you have any questions, comments, etc. please post them [[2019_Spring_ECE_301_Boutin_21 Savage in MATLAB_Jeff Rodrigues_comments|here]].
----
[[https://www.projectrhea.org/rhea/index.php/2019_Spring_ECE_301_Boutin|Back to 2019 Spring ECE 301 Boutin]]
----