Line 19: Line 19:
 
<math>  
 
<math>  
 
\begin{align}
 
\begin{align}
\bar f(x) &= \oint_S g(x) dx \\
+
\bar
&= \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) = 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 \\
Line 32: Line 28:
  
 
==DT signals==
 
==DT signals==
 +
<math>
 +
\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 \\
 +
\end{align}
 +
</math>
 
----
 
----
 
----
 
----

Revision as of 19:40, 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 \\ \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 \\ \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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Prof. Math. Ohio State and Associate Dean
Outstanding Alumnus Purdue Math 2008

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