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[[Category:2019 Spring ECE 301 Boutin]]
 
[[Category:2019 Spring ECE 301 Boutin]]
 
[[Category:ECE301]]
 
[[Category:ECE301]]
[[Category:ECE]]
 
[[Category:slecture]]
 
  
  
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----
 
----
 
==Introduction==
 
==Introduction==
I am going to compute some fourier series coefficients.  
+
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==
 
==CT signals==
 +
----
 
<math>  
 
<math>  
 +
\text{1) } x(t) = sin(6 \pi t), \text{ the frequency of this signal is } \omega_{o} = 6\pi.
 +
</math>
 +
 +
<math>
 
\begin{align}
 
\begin{align}
\text{1) } x(t) = sin(6 \pi t), \omega_{o} = 6\pi \\
+
x(t)& = \frac{e^{j6\pi t} - e^{-j6\pi t} }{2j} \\
& = \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{(*)}
& = \frac{1}{2j} e^{j6\pi t} - \frac{1}{2j} e^{-j6\pi t} \\
+
\end{align}
& \text{By Fourier Series we know that} \sum_{k=-\infty}^\infty a_k e^{jk\omega_o t} \\
+
</math>
& \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} \\
+
2) x(t) = 2 + cos(6 \pi t) - \frac{1}{2} sin(3 \pi t), \omega_{o} = 3\pi \\
+
3) x(t) = cos(\frac{2\pi}{10}t), \omega_{o} = \frac{\pi}{10} \\
+
4) x(t) =
+
\begin{cases}
+
  3, & \text{if}\ a=1 \\
+
  0, & \text{otherwise}
+
\end{cases}
+
  
 +
 +
By Fourier Series we know that
 +
<math>
 +
x(t) = \int_{-\infty}^\infty a_k e^{jk\omega_o t} =  \int_{-\infty}^\infty a_k e^{jk 6 \pi  t} \text{(**)}
 +
</math>
 +
 +
By comparing (*) with (**), we can see that
 +
<math>
 +
a_1 = \frac{1}{2j}, a_{-1} = -\frac{1}{2j},  \text{and } a_k = 0 \text{ for all other k}.
 +
</math>
 +
 +
 +
----
 +
<math>
 +
\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>
 +
\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}
 
\end{align}
 
</math>
 
</math>
 +
 +
By Fourier Series we know that
 +
<math>
 +
x(t) = \int_{-\infty}^\infty a_k e^{jk\omega_o t} =  \int_{-\infty}^\infty a_k e^{jk 3 \pi  t} \text{(**)}
 +
</math>
 +
 +
By comparing (*) with (**), we can see that
 +
<math>
 +
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>
 +
\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>
 +
\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}
 +
</math>
 +
 +
By Fourier Series we know that
 +
<math>
 +
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>
 +
 +
By comparing (*) with (**), we can see that
 +
<math>
 +
a_2 = a_{-2} = \frac{1}{2},  \text{and } a_k = 0 \text{ for all other k}\\
 +
</math>
 +
 
----
 
----
  
 
==DT signals==
 
==DT signals==
 +
----
 
<math>  
 
<math>  
 +
\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>
 
\begin{align}
 
\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}
 +
</math>
  
  
x[n] = 1 + sin(\frac{2\pi}{8}n) + 3cos(\frac{2\pi}{8}n), N=8 --> \omega_{o} = \frac{2\pi}{8} \\
+
By Fourier Series we know that
x[n] = -j^n, \omega_o = \frac{\pi}{2} \\
+
<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}{4} \pi  n} \text{(**)}
\begin{cases}
+
</math>
  sin(\pi t), & \text{if}\ a=1 \\
+
  0, & \text{otherwise}
+
  \end{cases}\\
+
x[n] =
+
\begin{cases}
+
  4, & \text{if}\ a=1 \\
+
  -4, & \text{otherwise}
+
\end{cases}
+
  
 +
By comparing (*) with (**), we can see that
 +
<math>
 +
a_1 = \frac{1}{2j}, a_{-1} = -\frac{1}{2j},  \text{and } a_k = 0 \text{ for  } a_{k+4}.
 +
</math>
 +
However Since this is a discrete signal we must  use the period, which in this case is 4.
 +
<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 be
 +
<math>
 +
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>
 +
\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>
 +
\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}
 
\end{align}
 
</math>
 
</math>
 +
 +
By Fourier Series we know that
 +
<math>
 +
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>
 +
 +
By comparing (*) with (**), we can see that
 +
<math>
 +
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>
 +
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>
 +
 +
 
----
 
----
----
+
 
==[[2019_Spring_ECE_301_Boutin_21 Savage in MATLAB_Jeff Rodrigues_comments | Questions and comments]]==
+
<math>
If you have any questions, comments, etc. please post them [[2019_Spring_ECE_301_Boutin_21 Savage in MATLAB_Jeff Rodrigues_comments|here]].
+
\text{3) } x[n] = -j^n, \text{ the frequency of this signal is }  \omega_o = \frac{\pi}{2} \\
 +
</math>
 +
 
 +
<math>
 +
\begin{align}
 +
x[n]&  = -e^{j\frac{\pi}{2} n} \text{(*)}
 +
\end{align}
 +
</math>
 +
 
 +
 
 +
By Fourier Series we know that
 +
<math>
 +
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>
 +
 
 +
By comparing (*) with (**), we can see that
 +
<math>
 +
a_1 = -1, a_{k} = 0 \text{ for  0,2,3 }, \text{ and } a_{k+4} = a_k \text{ for all other k}.
 +
</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 21:55, 30 April 2019


Fourier Series Coefficients

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]


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Questions/answers with a recent ECE grad

Ryne Rayburn