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| + | [[Category:problem solving]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:Fourier series]] | ||
| + | [[Category:signals and systems]] | ||
| + | == Example of Computation of Fourier series of a CT SIGNAL == | ||
| + | A [[Signals_and_systems_practice_problems_list|practice problem on "Signals and Systems"]] | ||
| + | ---- | ||
| + | |||
| + | |||
CT Periodic Signal : <math>x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4})</math> | CT Periodic Signal : <math>x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4})</math> | ||
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<math>x(t) = 1+\frac {1}{2j} e^{j\omega_0 t}-\frac {1}{2j}e^{-j\omega_0 t}+\frac{1}{2}e^{j(2\omega_0 t+\frac {\pi}{4})}+\frac {1}{2j}e^{-j(2\omega_0 t+\frac {\pi}{4})}</math> | <math>x(t) = 1+\frac {1}{2j} e^{j\omega_0 t}-\frac {1}{2j}e^{-j\omega_0 t}+\frac{1}{2}e^{j(2\omega_0 t+\frac {\pi}{4})}+\frac {1}{2j}e^{-j(2\omega_0 t+\frac {\pi}{4})}</math> | ||
| − | <math>x(t) = | + | <math>x(t) = 1e^{0j\omega_0 t}+\frac {1}{2j} e^{j\omega_0 t}-\frac {1}{2j}e^{-j\omega_0 t}+\frac{1}{2}e^{j\frac {\pi}{4}}e^{2\omega_0 t}+\frac{1}{2}e^{-j\frac {\pi}{4}}e^{2\omega_0 t}</math> |
| + | Hence we get, | ||
| − | + | <math>a_0 = 1</math> | |
| − | + | <math>a_1 = \frac{1}{2j},</math> | |
| − | <math> | + | <math>a_{-1} = -\frac{1}{2j},</math> |
| − | <math> | + | <math>a_2 = \frac{1}{2}e^{j\frac{\pi}{4}}=\frac{\sqrt2}{4}(1+j),</math> |
| − | <math>a_{ | + | <math>a_{-2} = \frac{1}{2}e^{-j\frac{\pi}{4}}=\frac{\sqrt2}{4}(1-j),</math> |
| − | + | ||
| − | + | ||
We can write the function in the following illiterations: | We can write the function in the following illiterations: | ||
| − | <math> | + | <math>a_0 = 1</math> |
| + | |||
| + | <math>a_1 = \frac{1}{2j},</math> | ||
| + | |||
| + | <math>a_{-1} = -\frac{1}{2j},</math> | ||
| − | <math> | + | <math>a_2 = \frac{\sqrt2}{4}(1+j),</math> |
| − | <math>a_{ | + | <math>a_{-2} = \frac{\sqrt2}{4}(1-j),</math> |
| − | <math>a_k = 0 , k \neq | + | <math>a_k = 0 , k \neq 0,1,-1,2,-2\,</math> |
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 10:54, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
CT Periodic Signal : $ x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4}) $
$ x(t) = 1+\frac {1}{2j} (e^{j\omega_0 t}-e^{-j\omega_0 t})+\frac{1}{2}(e^{j(2\omega_0 t+\frac {\pi}{4})}+e^{-j(2\omega_0 t+\frac {\pi}{4})}) $
$ x(t) = 1+\frac {1}{2j} e^{j\omega_0 t}-\frac {1}{2j}e^{-j\omega_0 t}+\frac{1}{2}e^{j(2\omega_0 t+\frac {\pi}{4})}+\frac {1}{2j}e^{-j(2\omega_0 t+\frac {\pi}{4})} $
$ x(t) = 1e^{0j\omega_0 t}+\frac {1}{2j} e^{j\omega_0 t}-\frac {1}{2j}e^{-j\omega_0 t}+\frac{1}{2}e^{j\frac {\pi}{4}}e^{2\omega_0 t}+\frac{1}{2}e^{-j\frac {\pi}{4}}e^{2\omega_0 t} $
Hence we get,
$ a_0 = 1 $
$ a_1 = \frac{1}{2j}, $
$ a_{-1} = -\frac{1}{2j}, $
$ a_2 = \frac{1}{2}e^{j\frac{\pi}{4}}=\frac{\sqrt2}{4}(1+j), $
$ a_{-2} = \frac{1}{2}e^{-j\frac{\pi}{4}}=\frac{\sqrt2}{4}(1-j), $
We can write the function in the following illiterations:
$ a_0 = 1 $
$ a_1 = \frac{1}{2j}, $
$ a_{-1} = -\frac{1}{2j}, $
$ a_2 = \frac{\sqrt2}{4}(1+j), $
$ a_{-2} = \frac{\sqrt2}{4}(1-j), $
$ a_k = 0 , k \neq 0,1,-1,2,-2\, $
