(New page: <math>x(t) = \sin(4\pi t) + \sin(6\pi t)</math></BR> <math>x(t) = \frac{e^{j4\pi t} - e^{-j4\pi t}}{2}} \frac{e^{j6\pi t} - e^{-j6\pi t}}{2}}\,</math>) |
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| − | <math>x(t) = \sin(4\pi t) | + | [[Category:problem solving]] |
| − | <math>x(t) = \frac{e^{j4\pi t} - e^{-j4\pi t}}{2} | + | [[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"]] | ||
| + | ---- | ||
| + | <math>\ x(t) = \sin(4\pi t) \sin(6\pi t)</math> | ||
| + | |||
| + | |||
| + | <math>\ x(t) = (\frac{e^{j4\pi t} - e^{-j4\pi t}}{2}) (\frac{e^{j6\pi t} - e^{-j6\pi t}}{2j})</math> | ||
| + | |||
| + | |||
| + | <math>\ x(t) = \frac{-1}{4}(e^{j10\pi t} - e{-j2\pi t} - e^{j2\pi t} + e^{-j10\pi t})</math> | ||
| + | |||
| + | |||
| + | <math>\ x(t) = \frac{-1}{4}(e^{5(j2\pi t)} - e^{-1(j2\pi t)} - e^{1(j2\pi t)} + e^{-5(j2\pi t)})</math> | ||
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| + | |||
| + | <math>a_{5} = \frac{-1}{4}, a_{-1} = \frac{1}{4}, a_{1} = \frac{1}{4}, a_{-5} = \frac{-1}{4}</math> | ||
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| + | |||
| + | All other values <math>\ a_{n} = 0</math> | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 11:01, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
$ \ x(t) = \sin(4\pi t) \sin(6\pi t) $
$ \ x(t) = (\frac{e^{j4\pi t} - e^{-j4\pi t}}{2}) (\frac{e^{j6\pi t} - e^{-j6\pi t}}{2j}) $
$ \ x(t) = \frac{-1}{4}(e^{j10\pi t} - e{-j2\pi t} - e^{j2\pi t} + e^{-j10\pi t}) $
$ \ x(t) = \frac{-1}{4}(e^{5(j2\pi t)} - e^{-1(j2\pi t)} - e^{1(j2\pi t)} + e^{-5(j2\pi t)}) $
$ a_{5} = \frac{-1}{4}, a_{-1} = \frac{1}{4}, a_{1} = \frac{1}{4}, a_{-5} = \frac{-1}{4} $
All other values $ \ a_{n} = 0 $
