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& \text{By Fourier Series we know that} \sum_{k=-\infty}^\infty a_k e^{jk\omega_o t} \\ | & \text{By Fourier Series we know that} \sum_{k=-\infty}^\infty a_k e^{jk\omega_o t} \\ | ||
& \text{Here, } \omega_o = 3 \pi \text{ ,therefore, } \\ | & \text{Here, } \omega_o = 3 \pi \text{ ,therefore, } \\ | ||
| − | & = 2e^{j \omega_o (0)t} + \frac{1}{2}e^{j \omega_o (2)t} + \frac{1}{2}e^{j \omega_o (-2)t} + \frac{1}{4j}e^{j \omega_o (1)t} - \frac{1}{4j}e^{j \omega_o (-1)t} \ | + | & = 2e^{j \omega_o (0)t} + \frac{1}{2}e^{j \omega_o (2)t} + \frac{1}{2}e^{j \omega_o (-2)t} + \frac{1}{4j}e^{j \omega_o (1)t} - \frac{1}{4j}e^{j \omega_o (-1)t} \\\ |
& \text{So we can say that } a_0 = 2, a_1 = \frac{1}{4j}, a_{-1} = -\frac{1}{4j}, a_2 = a_{-2} = \frac{1}{2}, a_k = 0 \text{ for all other k} \\ | & \text{So we can say that } a_0 = 2, a_1 = \frac{1}{4j}, a_{-1} = -\frac{1}{4j}, a_2 = a_{-2} = \frac{1}{2}, a_k = 0 \text{ for all other k} \\ | ||
3) x(t) = cos(\frac{2\pi}{10}t), \omega_{o} = \frac{\pi}{10} \\ | 3) x(t) = cos(\frac{2\pi}{10}t), \omega_{o} = \frac{\pi}{10} \\ | ||
Revision as of 12:07, 26 April 2019
A project by Kalyan Mada
Introduction
I am going to compute some fourier series coefficients.
CT signals
$ \begin{align} \text{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}, a_k = 0 \text{ for all other k} \\ 2) x(t) = 2 + cos(6 \pi t) - \frac{1}{2} sin(3 \pi t), \omega_{o} = 3\pi \\ & = 2 + \frac{1}{2}(e^{j6\pi t} + e^{-j6\pi t}) + \frac{1}{4j} (e^{j3\pi t} -e^{-j3\pi t}) \\ & \text{By Fourier Series we know that} \sum_{k=-\infty}^\infty a_k e^{jk\omega_o t} \\ & \text{Here, } \omega_o = 3 \pi \text{ ,therefore, } \\ & = 2e^{j \omega_o (0)t} + \frac{1}{2}e^{j \omega_o (2)t} + \frac{1}{2}e^{j \omega_o (-2)t} + \frac{1}{4j}e^{j \omega_o (1)t} - \frac{1}{4j}e^{j \omega_o (-1)t} \\\ & \text{So we can say that } a_0 = 2, a_1 = \frac{1}{4j}, a_{-1} = -\frac{1}{4j}, a_2 = a_{-2} = \frac{1}{2}, a_k = 0 \text{ for all other k} \\ 3) x(t) = cos(\frac{2\pi}{10}t), \omega_{o} = \frac{\pi}{10} \\ & = \frac{e^{j\frac{2\pi}{10} t} + e^{-j\frac{2\pi}{10} t} }{2} \\ & = \frac{1}{2}e^{j\frac{2\pi}{10} t} + \frac{1}{2}e^{-j\frac{2\pi}{10} t} \\ & \text{By Fourier Series we know that} \sum_{k=-\infty}^\infty a_k e^{jk\omega_o t} \\ & \text{Here, } \omega_o = \frac{ \pi}{10} \text{ ,therefore, } \\ & = \frac{1}{2}e^{j\omega_o(2) t} + \frac{1}{2}e^{-j\omega_o(-2) t} \\ & \text{So we can say that } a_2 = a_{-2} = \frac{1}{2}, a_k = 0 \text{ for all other k}\\ 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
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[to 2019 Spring ECE 301 Boutin]
