| Line 21: | Line 21: | ||
\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} | ||
\end{align} | \end{align} | ||
| Line 29: | Line 30: | ||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
| + | |||
f(x) &= \oint_S g(x) dx \\ | f(x) &= \oint_S g(x) dx \\ | ||
&= \int_a^b g(x) dx \\ | &= \int_a^b g(x) dx \\ | ||
| Line 34: | Line 36: | ||
& = \int_a^{-\infty} jzdhfbvzjhvz dt \\ | & = \int_a^{-\infty} jzdhfbvzjhvz dt \\ | ||
& = \sum_{k=0}^{-\infty} kzdjfgdzjkfg \\ | & = \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} | \end{align} | ||
</math> | </math> | ||
Revision as of 20:48, 25 April 2019
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} \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]
