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&= \frac{\mu_0}{2 \pi a \cdot b}\\ | &= \frac{\mu_0}{2 \pi a \cdot b}\\ | ||
& = \int_a^{-\infty} jzdhfbvzjhvz dt \\ | & = \int_a^{-\infty} jzdhfbvzjhvz dt \\ | ||
− | & = \sum_{k=0}^{-\infty} kzdjfgdzjkfg | + | & = \sum_{k=0}^{-\infty} kzdjfgdzjkfg \ |
x(t) = sin(6 \pi t), \omega_{o} = 6\pi \\ | 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 \\ |
Revision as of 19:39, 25 April 2019
A project by Kalyan Mada
Introduction
I am going to compute some fourier series coefficients.
CT signals
$ \begin{align} \bar 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(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 \\ \end{align} $
DT signals
Questions and comments
If you have any questions, comments, etc. please post them here.
[to 2019 Spring ECE 301 Boutin]