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<math>  
 
<math>  
 
\begin{align}
 
\begin{align}
\bar  1) x(t) = sin(6 \pi t), \omega_{o} = 6\pi \\
+
1) x(t) = sin(6 \pi t), \omega_{o} = 6\pi \\
 
& = \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} \\
 
& = \frac{1}{2j} e^{j6\pi t} - \frac{1}{2j} e^{-j6\pi t} \\

Revision as of 02:30, 26 April 2019


Fourier Series Coefficients

A project by Kalyan Mada



Introduction

I am going to compute some fourier series coefficients.


CT signals

$ \begin{align} 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} \\ 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} \end{align} $


DT signals

$ \begin{align} x[n] = 1 + sin(\frac{2\pi}{8}n) + 3cos(\frac{2\pi}{8}n), N=8 --> \omega_{o} = \frac{2\pi}{8} \\ x[n] = -j^n, \omega_o = \frac{\pi}{2} \\ x[n] = \begin{cases} 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} \end{align} $



Questions and comments

If you have any questions, comments, etc. please post them here.


[to 2019 Spring ECE 301 Boutin]


Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett