Line 19: Line 19:
 
<math>  
 
<math>  
 
\begin{align}
 
\begin{align}
\bar  x(t) = sin(6 \pi t), \omega_{o} = 6\pi \\
+
\bar  x(t) = sin(6 \pi t), \omega_{o} = 6\pi  
 
  x(t) = 2 + cos(6 \pi t) - \frac{1}{2} sin(3 \pi t), \omega_{o} = 3\pi \\
 
  x(t) = 2 + cos(6 \pi t) - \frac{1}{2} sin(3 \pi t), \omega_{o} = 3\pi \\
x(t) = cos(\frac{2\pi}{10}t), \omega_{o} = \frac{\pi}{10}
+
x(t) = cos(\frac{2\pi}{10}t), \omega_{o} = \frac{\pi}{10} \\
 +
x(t) =
 +
\begin{cases}
 +
  0, & \text{if}\ a=1 \\
 +
  1, & \text{otherwise}
 +
\end{cases}
  
 
\end{align}
 
\end{align}

Revision as of 19:57, 25 April 2019


Fourier Series Coefficients

A project by Kalyan Mada



Introduction

I am going to compute some fourier series coefficients.


CT signals

$ \begin{align} \bar x(t) = sin(6 \pi t), \omega_{o} = 6\pi x(t) = 2 + cos(6 \pi t) - \frac{1}{2} sin(3 \pi t), \omega_{o} = 3\pi \\ x(t) = cos(\frac{2\pi}{10}t), \omega_{o} = \frac{\pi}{10} \\ x(t) = \begin{cases} 0, & \text{if}\ a=1 \\ 1, & \text{otherwise} \end{cases} \end{align} $


DT signals

$ \begin{align} f(x) &= \oint_S g(x) dx \\ &= \int_a^b g(x) dx \\ &= \frac{\mu_0}{2 \pi a \cdot b}\\ & = \int_a^{-\infty} jzdhfbvzjhvz dt \\ & = \sum_{k=0}^{-\infty} kzdjfgdzjkfg \\ x[n] = 1 + sin(\frac{2\pi}{8}n) + 3cos(\frac{2\pi}{8}n), N=8 --> \omega_{o} = \frac{2\pi}{8} \end{align} $



Questions and comments

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


[to 2019 Spring ECE 301 Boutin]


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